Orthogonal
Throughout these notes, we’ve adopted a system where time and distance are measured in identical units. This is the equivalent of setting the speed of light, c, equal to 1 in our own universe (for example, by measuring distances in metres and using the time it takes for light to travel one metre as the corresponding unit of time). In the Riemannian universe, it amounts to choosing units for time such that Pythagoras’s Theorem holds true, even when one side of the triangle involves an interval of time rather than space. In the novel Orthogonal, it is found empirically that this is the same as setting the speed of blue light equal to 1.
In this section, we will go one step further and choose units for mass and energy such that the “reduced Planck’s constant”, ℏ = h/(2π), is equal to 1. Mass and energy are then measured in units with the dimensions of inverse lengths or spatial frequencies — or, equally, inverse times or time frequencies.
Our particular choice means that the Planck relationship between frequency ν and energy, E = h ν, becomes E = 2 π ν = ω, where ω is the angular frequency of the wave, and the relationship between spatial frequency κ and momentum is p = 2 π κ = k, where k is the angular spatial frequency. The maximum angular frequency ω_{m} that appears in the Riemannian Scalar Wave equation is then simply equal to the rest mass of the associated particle.
In the nonrelativistic quantum mechanics we have discussed so far, we have simply applied the usual Schrödinger equation to the potential energy associated with the force between charged particles, on the basis that nonrelativistic classical dynamics in the Riemannian universe is identical to Newtonian mechanics, so long as we treat kinetic energy as positive and choose the sign for the potential energy to be consistent with that.
For Riemannian relativistic quantum mechanics, we will need to do things slightly differently. The structure of quantum mechanics in its usual formulation is closely linked to the Hamiltonian form of the corresponding classical mechanics, and in the Riemannian case the momentum conjugate to each coordinate in the Hamiltonian sense is the opposite of the relativistic momentum in the same direction.
In nonrelativistic quantum mechanics, we identify energy with the operator i ∂_{t}, but we identify the spatial components of momentum with the operators –i ∂_{x}, –i ∂_{y} and –i ∂_{z}. The same is true in Lorentzian relativistic quantum mechanics, but in the Riemannian case this makes no sense; we need to treat the space and time coordinates identically. In order to retain the usual correspondence with Hamiltonian classical mechanics, we will keep the identification:
E → i ∂_{t}
and we will identify the relativistic momentum with the same kind of operators for the other coordinates, without the usual minus sign:
p_{x} → i ∂_{x}
p_{y} → i ∂_{y}
p_{z} → i ∂_{z}
This will also affect the operators we use for angular momentum. If we define the threedimensional vector operator L = r × p using our Riemannianrelativistic momentum p, it becomes:
L = (x, y, z) × (i ∂_{x}, i ∂_{y}, i ∂_{z})
= i (y ∂_{z} – z ∂_{y}, z ∂_{x} – x ∂_{z}, x ∂_{y} – y ∂_{x})
Like the linear momentum, this differs from the usual operator by a minus sign. As a result, the commutators between the components are different from those in nonrelativistic quantum mechanics:
[L_{i}, L_{j}] = –i ε_{ijk} L_{k}
where the LeviCivita symbol ε_{ijk} is equal to 1 if (i, j, k) is an even permutation of (x, y, z), –1 if it is an odd permutation, and 0 if it contains any coordinate twice. Again, there is a difference of a sign in the righthand side of this equation from the usual case.
Now, for the intrinsic angular momentum of particles to be compatible with this whole scheme, the corresponding operators need to have the same commutator algebra as the orbital angular momentum L. This means that rather than the Pauli spin matrices being the appropriate operators for the components of the angular momentum of a spin½ particle, we will need to use the opposite of the Pauli matrices. The details will be developed as they arise; for now this is just a warning to expect this small twist on the usual conventions.
In our own universe, the Dirac equation is the relativistic wave equation that describes particles such as electrons and positrons: particles with mass, electric charge, and a spin of one half. The wave functions for these particles have four components, but they aren’t the components of a spacetime vector, and they transform under the symmetries of the Lorentz group in a very different fashion to the way spacetime vectors transform.
The corresponding equation for the Riemannian universe is extremely similar, and indeed it can be written in exactly the same form when expressed in terms of gamma matrices: a set of four matrices (each one of dimensions 4 × 4) that act on the wave function components. But the precise properties required of the gamma matrices are different in the two cases, reflecting the different signatures of the underlying spacetime.
Dirac Equation  
m is the mass of the particle ψ is a fourcomponent wave function I_{4} is the 4 × 4 identity matrix We’re using units such that ℏ = 1  
i γ^{μ} ∂_{μ} ψ – m ψ  =  0  (Dirac Equation)  
{γ^{μ}, γ^{ν}}  ≡  γ^{μ} γ^{ν} + γ^{ν} γ^{μ}  
=  2 g^{μν} I_{4}  (Common)  
g^{μν}  =  δ^{μν}  
= 

(Riemannian)  
g^{μν}  =  η^{μν} N.B.: (+ – – –) signature  
= 

(Lorentzian) 
[For most of these notes we’ve used a positive sign for the spacelike dimensions in the Lorentzian case, giving a (– + + +) signature for η. But since virtually all quantum field theory literature uses a (+ – – –) signature, for ease of comparison we’ll switch to that convention.]
In the Dirac equation itself, we’re applying the Einstein Summation Convention to the index μ, which ranges over four spacetime dimensions, combining each of the four gamma matrices with the derivative in the corresponding direction. The wave function ψ at any point in spacetime is a fourcomponent complex vector, with each gamma matrix acting on a derivative of ψ by ordinary matrix multiplication. But the four components of ψ are not indexed by the four spacetime dimensions. We can write out the matrix multiplication explicitly with some extra indices for the relevant components (spinor indices, as opposed to spacetime indices like μ):
i (γ^{μ})^{s}_{r} ∂_{μ} ψ^{r} – m ψ^{s} = 0
where the summation convention now applies to r as well as μ, and we explicitly have four equations, for the four values of the spinor index s.
Where we state the condition we’re imposing on the gamma matrices, the notation {γ^{μ}, γ^{ν}} is known as an anticommutator. This is simply a convenient shorthand, using the definition:
{A, B} ≡ AB + BA
In defining the anticommutator A and B can be anything, but for our purposes they will usually be either matrices or other kinds of linear operators (such as differential operators). We will also make use shortly of the more familiar commutator notation:
[A, B] ≡ AB – BA
If ψ is a solution of the Riemannian Dirac equation, we will have:
i γ^{μ} ∂_{μ} ψ – m ψ = 0
(i γ^{μ} ∂_{μ} – m) ψ = 0
(–i γ^{ν} ∂_{ν} – m) (i γ^{μ} ∂_{μ} – m) ψ = 0
(γ^{ν} γ^{μ} ∂_{ν} ∂_{μ} + m^{2}) ψ = 0
( ½(γ^{μ} γ^{ν} + γ^{ν} γ^{μ}) ∂_{ν} ∂_{μ} + m^{2}) ψ = 0
(δ^{νμ} ∂_{ν} ∂_{μ} + m^{2}) ψ = 0
The step between the thirdlast and secondlast lines follows because, although the individual term γ^{μ} γ^{ν} ≠ γ^{ν} γ^{μ}, the sum over all such terms that we get from the summation convention is symmetrical under an exchange of μ and ν.
What the last line tells us is that each individual component of the fourcomponent field ψ will obey the Riemannian Scalar Wave equation.
There are an infinite number of choices of gamma matrices that satisfy the condition {γ^{μ}, γ^{ν}} = 2 δ^{μν} I_{4}, since given any four matrices with this property we can always change to a different basis of C^{4}, the fourcomplexdimensional space on which these matrices act, and get four new matrices.
But certain choices for the gamma matrices make things simpler. We will give two examples, in what are known as the Weyl basis and the Dirac basis. First, it will be handy to remind ourselves of the Pauli spin matrices that appear in the quantum mechanics of angular momentum and rotations in three dimensions.
Pauli spin matrices  
σ_{x}  = 


σ_{y}  = 


σ_{z}  = 

(Common) 
The commutators of half the Pauli matrices are:
[½σ_{i}, ½σ_{j}] = i ε_{ijk} ½σ_{k}
What ε_{ijk} does here is pick out the single value for k that is distinct from i and j, and select an appropriate sign, giving us:
[½σ_{x}, ½σ_{y}] = i ½σ_{z}
and cyclic permutations of this. As we discussed in the section on the Lie algebra so(3), the generators of rotations around the coordinate axes, J_{i}, have the commutator algebra:
[J_{i}, J_{j}] = ε_{ijk} J_{k}
When describing the isomorphism between the Lie algebras su(2) and so(3), we used a basis {H_{x}, H_{y}, H_{z}} of su(2) comprised of three traceless skewHermitian matrices (matrices whose adjoint is the opposite of the original matrix). When doing quantum mechanics, it is more common to make use of Hermitian matrices, since any realvalued observable quantity in quantum mechanics has a Hermitian operator associated with it. The Pauli matrices are Hermitian matrices, related to our basis of su(2) by:
σ_{k} = i H_{k}
H_{k} = –i σ_{k}
Since the H_{k} are elements of su(2), any linear combination of them with real coefficients can be exponentiated to get an element of SU(2). The Hermitian matrices ½σ_{k} are the quantummechanical operators whose eigenvectors are states of definite spin, for a spin½ particle in nonrelativistic quantum mechanics; the corresponding eigenvalues are ±½. If we want to exponentiate a Hermitian matrix and end up with a unitary operator — an operator that preserves the norm of the quantum wave function — we need to multiply it by ±i first, e.g.:
U_{x}(θ) = exp(–i θ σ_{x}/2)
This is the unitary operator for a rotation by an angle θ around the xaxis, for a spin½ particle in nonrelativistic quantum mechanics.
When we shift from three dimensions to four, the Lie algebra so(4) has six generators J_{μν} for rotations in the six planes spanned by pairs of coordinate directions μ and ν. In order to fix the sign of the matrix J_{μν}, we use the convention that exponentiating a positive multiple of J_{μν} gives an element of SO(4) that rotates the μ axis towards the ν axis. The commutators of these generators are:
[J_{μν}, J_{ρν}] = J_{μρ}
for the case where the two planes have exactly one coordinate axis, ν, in common. Apart from permutations of this with appropriate sign changes, all other commutators are zero. This makes sense, because if the two planes have no axes in common then rotations in them are completely independent, and if they have two axes in common then they are the same plane, and rotations in the same plane commute.
There is also a crucial change that we need to make when we switch from nonrelativistic quantum mechanics to Riemannian relativistic quantum mechanics. Just as we need to use different operators than usual for linear momentum, we need to do the same for angular momentum. As we discussed earlier, for the orbital angular momentum this will happen automatically if we define it in terms of linear momentum, but for intrinsic angular momentum we need to choose the opposite of the usual Pauli matrices to get the correct sign for a particle’s spin. We’ll see a confirmation of this in the section on spin and conservation of angular momentum.
Our first choice for the Riemannian gamma matrices can be written in terms of the Pauli matrices as follows:
Riemannian gamma matrices, Weyl basis  
γ^{t}  = 

=  σ_{x} ⊗ I_{2}  
γ^{x}  = 

=  σ_{y} ⊗ σ_{x}  
γ^{y}  = 

=  σ_{y} ⊗ σ_{y}  
γ^{z}  = 

=  σ_{y} ⊗ σ_{z}  
γ^{μ}  = 

(Riemannian) 
Here, the notation σ_{x} ⊗ I_{2} means we replace each entry in the first matrix, σ_{x}, by the product of that entry and the second matrix, I_{2}, to get a 4 × 4 matrix. In the final line, we give a general form that applies for any index μ, if we use H_{t} = I_{2} as previously defined (but note that unlike the other three H matrices, H_{t} is not in su(2)).
All of these gamma matrices are Hermitian, traceless, and have a determinant of 1. The square of each matrix is the identity matrix, while all nonidentical pairs anticommute: that is, their anticommutator is zero. If we compare these to the choice of gamma matrices in the Weyl basis used in Peskin and Schroeder’s widely used textbook on quantum field theory^{[1]}, our γ^{t} is the same, while the other matrices here are –i times the Lorentzian version. This reflects the fact that Peskin and Schroeder use a (+ – – –) signature for the metric, so their t coordinate already has a spacelike signature by our conventions.
We can use these gamma matrices to derive a representation of the group SO(4), using the formula from ^{[1]}:
S^{μν} = (i/4) [γ^{μ}, γ^{ν}]
Here each S^{μν} is a Hermitian matrix that operates on the fourdimensional complex vector space to which the wave function ψ at each point belongs. It is the image of the so(4) element J_{μν} in some yettobedetermined fourdimensional Lie algebra representation, multiplied by a factor of ±i needed to make it Hermitian. And of course we’re not surprised to see the opposites of the Pauli matrices here, since we know that minus sign is going to be necessary to make the intrinsic spin compatible with everything else.
Riemannian spin matrices, Weyl basis  
S^{xy}  = 

=  –½ I_{2} ⊗ σ_{z}  
S^{yz}  = 

=  –½ I_{2} ⊗ σ_{x}  
S^{zx}  = 

=  –½ I_{2} ⊗ σ_{y}  
S^{xt}  = 

=  ½ σ_{z} ⊗ σ_{x}  
S^{yt}  = 

=  ½ σ_{z} ⊗ σ_{y}  
S^{zt}  = 

=  ½ σ_{z} ⊗ σ_{z}  (Riemannian) 
We can see from the fact that these matrices are blockdiagonal that the fourdimensional representation they define is reducible: it can be split into two subspaces, each of two dimensions, that the action of the representation will never mix up.
What, then, are the two representations of so(4) that are stuck together here? By looking at the spin(½,0) and spin(0,½) representations that we previously described (and which is tabulated in detail at the end of that section), we see that the representation for the first two components of ψ corresponds to spin(½,0), and the representation on the last two components matches spin(0,½); the only difference is that the H matrices of su(2) in that discussion have been replaced by the opposites of the Pauli σ matrices.
The Lie algebra representation is just the derivative of the group representation, so the double cover of SO(4) should act on ψ via the group representation spin(½,0) ⊕ spin(0,½). And as we found when discussing representations that include parity, this can be extended to a representation of the double cover of O(4), in which the parity operation (reversing the signs of x, y and z) swaps the two 2dimensional subspaces of this representation.
If we use ρ as an abbreviation for the spin(½,0) ⊕ spin(0,½) representation, then (up to a choice of sign, coming from the choice between two elements in the double cover of SO(4)) we have ρ:SO(4)→SU(4). That is to say, up to a sign, ρ assigns a 4 × 4 unitary matrix with determinant 1 to every rotation in SO(4). Readers familiar with the Dirac equation in the Lorentzian case will recall that the representation of “boosts” — coordinate changes that involve relative motion — in the analogous representation of SO(3,1) are not unitary, but in the Riemannian case we don’t face that complication.
Explicitly, ρ maps a rotation R in SO(4) to a unitary operator ρ(R) in SU(4) so that:
ρ(exp(J^{μν})) = exp(i S^{μν})
The derivative of this map is the Lie algebra isomorphism from so(4) to su(4):
dρ(J^{μν}) = i S^{μν}
Given this definition, it’s not too hard to verify that, for any rotation R in SO(4):
ρ(R^{–1}) γ^{ν} ρ(R)  =  R^{ν}_{ζ} γ^{ζ}  (1) 
or if we spell out all the matrix multiplication here in index notation:
ρ(R^{–1})^{q}_{p} (γ^{ν})^{p}_{r} ρ(R)^{r}_{s}  =  (R^{ν}_{ζ} γ^{ζ})^{q}_{s}  (2) 
The easiest way to check this is to take the derivative, and use the formula (J^{αβ})^{ν}_{ζ} = δ^{αζ} δ^{βν} – δ^{αν} δ^{βζ}:
[γ^{ν}, S^{αβ}]  =  –i (J^{αβ})^{ν}_{ζ} γ^{ζ}  
[γ^{ν}, (i/4) [γ^{α}, γ^{β}]]  =  i (δ^{αν} γ^{β} – δ^{βν} γ^{α})  
¼ [γ^{ν}, [γ^{α}, γ^{β}]]  =  δ^{αν} γ^{β} – δ^{βν} γ^{α}  
¼ [γ^{ν}, 2 δ^{αβ} I_{4} – 2 γ^{β} γ^{α}]  =  ¼ ((γ^{α} γ^{ν} + γ^{ν} γ^{α}) γ^{β} + γ^{β} (γ^{α} γ^{ν} + γ^{ν} γ^{α}) – (γ^{β} γ^{ν} + γ^{ν} γ^{β}) γ^{α} – γ^{α} (γ^{β} γ^{ν} + γ^{ν} γ^{β}))  
–½ (γ^{ν} γ^{β} γ^{α} – γ^{β} γ^{α} γ^{ν})  =  ¼ (γ^{ν} γ^{α} γ^{β} + γ^{β} γ^{α} γ^{ν} – γ^{ν} γ^{β} γ^{α} – γ^{α} γ^{β} γ^{ν})  
0  =  γ^{ν} γ^{α} γ^{β} + γ^{ν} γ^{β} γ^{α} – γ^{α} γ^{β} γ^{ν} – γ^{β} γ^{α} γ^{ν}  
0  =  γ^{ν} (2 δ^{αβ} I_{4}) – (2 δ^{αβ} I_{4}) γ^{ν}  
0  =  0 
Working backwards through these steps gives a derivation of the original claim.
What equations (1) and (2) tell us is that if we transform the spinor coordinates of the γ matrices using ρ(R), the result is the same as rotating the four γ matrices among themselves as if they were the components of a spacetime vector acted on by the SO(4) element R.
The point of this is to understand what happens to the Dirac equation when we rotate our coordinates. If we make a change of coordinates from x^{μ} to x'^{ν} = R^{ν}_{μ} x^{μ}, then the coordinate derivatives ∂_{μ} are replaced by ∂'_{ν} = (R^{–1})^{μ}_{ν} ∂_{μ}, while our spinor wave function ψ^{s} is transformed into ψ'^{p} = ρ(R)^{p}_{s} ψ^{s}. Writing the Dirac equation for everything in the new coordinates, we have:
i (γ^{ν})^{p}_{r} ∂'_{ν} ψ'^{r} – m ψ'^{p}  =  0  (3) 
Substituting for the transformed derivatives and wave function in terms of the originals:
i (γ^{ν})^{p}_{r} (R^{–1})^{μ}_{ν} ∂_{μ} ρ(R)^{r}_{s} ψ^{s} – m ρ(R)^{p}_{s} ψ^{s}  =  0  (4) 
Multiplying on the left by ρ(R^{–1}) gives:
ρ(R^{–1})^{q}_{p} [i (γ^{ν})^{p}_{r} (R^{–1})^{μ}_{ν} ∂_{μ} ρ(R)^{r}_{s} ψ^{s} – m ρ(R)^{p}_{s} ψ^{s}]  =  0  
i ρ(R^{–1})^{q}_{p} (γ^{ν})^{p}_{r} (R^{–1})^{μ}_{ν} ∂_{μ} ρ(R)^{r}_{s} ψ^{s} – m ρ(R^{–1})^{q}_{p} ρ(R)^{p}_{s} ψ^{s}  =  0  
i ρ(R^{–1})^{q}_{p} (γ^{ν})^{p}_{r} (R^{–1})^{μ}_{ν} ∂_{μ} ρ(R)^{r}_{s} ψ^{s} – m ψ^{q}  =  0 
Making use of (2), this becomes:
i (R^{ν}_{ζ} γ^{ζ})^{q}_{s} (R^{–1})^{μ}_{ν} ∂_{μ} ψ^{s} – m ψ^{q}  =  0  
i (γ^{μ})^{q}_{s} ∂_{μ} ψ^{s} – m ψ^{q}  =  0  (5) 
Equation (5) is the Dirac equation for the original ψ. So we’ve shown that the Dirac equation is SO(4)invariant: the solutions we get if we rotate our coordinates with R are solutions of the original equation, transformed by ρ(R).
It’s worth stressing that when we transform ψ, we don’t treat the γ matrices as some kind of field that should itself be transformed. Rather, the gamma matrices bring together the spacetime coordinate transformations of the derivatives and the corresponding spinor transformation of ψ in such a way that all the changes cancel out. The best way to think about the gamma matrices is as an intertwining operator, or intertwiner: a linear map from one vector space to another that commutes with representations of the same group on the two spaces. The terms ∂_{μ} ψ^{s} transform under the tensor product of two representations of SO(4): the dual of the fundamental vector representation due to the spacetime coordinate derivatives, and the spin(½,0) ⊕ spin(0,½) Dirac spinor representation. By multiplying both sides of equation (2) with (R^{–1})^{μ}_{ν} ρ(R)^{w}_{q}, rearranging the order of some terms (which we can do freely here, because the fully indexed terms are just scalars) and then applying both sides of the result to ∂_{μ} ψ^{s}, we get:
(γ^{ν})^{w}_{r} ρ(R)^{r}_{s} (R^{–1})^{μ}_{ν} ∂_{μ} ψ^{s}  =  ρ(R)^{w}_{q} (γ^{μ})^{q}_{s} ∂_{μ} ψ^{s}  (6) 
On the righthand side, γ maps a term in the tensor product, ∂_{μ} ψ^{s}, to a Dirac spinor with index q, then ρ(R) acts on that to give another Dirac spinor, with index w. On the lefthand side, R^{–1} acts on the dual coordinate index μ while ρ(R) acts on the spinor coordinate s, and then γ takes the result (which at this point is still in the tensor product space) and maps it to a Dirac spinor with index w. That we can apply the representations either before or after using the map γ and get exactly the same result is what we mean by saying that γ is an intertwiner.
We chose the Weyl basis for the gamma matrices in order to make it as easy as possible to see which representation of SO(4) applied to ψ. Another basis, which will be useful in the next section, is the Dirac basis. In this basis, only γ^{t} changes, with all the other matrices exactly the same as in the Weyl basis.
Riemannian gamma matrices, Dirac basis  
γ^{t}  = 

=  σ_{z} ⊗ I_{2}  
γ^{x}  = 

=  σ_{y} ⊗ σ_{x}  
γ^{y}  = 

=  σ_{y} ⊗ σ_{y}  
γ^{z}  = 

=  σ_{y} ⊗ σ_{z}  (Riemannian) 
The resulting spin matrices are unchanged when only spatial coordinates are involved, so again they are blockdiagonal, but the matrices that generate rotations involving the t coordinate mix the two halves of the spinor. This doesn’t mean that the representation of SO(4) has changed. The representation in the Dirac basis is equivalent to the representation in the Weyl basis.
Riemannian spin matrices, Dirac basis  
S^{xy}  = 

=  –½ I_{2} ⊗ σ_{z}  
S^{yz}  = 

=  –½ I_{2} ⊗ σ_{x}  
S^{zx}  = 

=  –½ I_{2} ⊗ σ_{y}  
S^{xt}  = 

=  –½ σ_{x} ⊗ σ_{x}  
S^{yt}  = 

=  –½ σ_{x} ⊗ σ_{y}  
S^{zt}  = 

=  –½ σ_{x} ⊗ σ_{z}  (Riemannian) 
Suppose we look for planewave solutions of the Dirac equation, of the form:
ψ(x) = u exp(–i k · x)
Here x is the coordinate fourvector. With our choice of units the propagation vector k is exactly the same as the energymomentum vector associated with the plane wave, and of course k = m. The spinor u is a constant element of C^{4}.
The Dirac equation then becomes a linear equation for u:
(k_{μ} γ^{μ} – m I_{4}) u = 0
So long as k = m the matrix here always has a determinant of 0, so there will always be nontrivial solutions. The easiest way to find a solution is to set k = m e_{t}, the energymomentum vector of a stationary wave. In the Dirac basis this gives us:

=  0 
This is solved by any spinor with the last two components equal to zero, say:
u_{0} = (ξ, 0)
where we’re writing ξ for an arbitrary element of C^{2}, and the 0 here is also to be taken as being in C^{2}.
If we transform our coordinates in the zt plane with the SO(4) matrix exp(–θ J_{zt}), in the new coordinates the original vector m e_{t} will become the energymomentum vector of a wave moving in the positive z direction with a velocity of tan θ. To find the spinor that accompanies this, we multiply u_{0} by exp(–i θ S^{zt}).
k = exp(–θ J_{zt}) m e_{t} = m (cos θ e_{t} + sin θ e_{z})
u(k) = exp(–i θ S^{zt}) u_{0} = (ξ cos(θ/2), i σ_{z} ξ sin(θ/2))
As the plane wave moves more rapidly, the first half of the Dirac spinor shrinks and the second half grows. At infinite velocity, θ=π/2, the two halves have equal magnitude.
Now, it’s tempting to assume that we can simply rotate beyond θ=π/2 to obtain a description of a Riemannian electron being transformed into a Riemannian positron by a change of coordinates, but the Dirac equation can’t quite deal with that. The problem is that a plane wave with a propagation vector pointing backwards in time would give us a negative energy, whereas we ought to measure a positive energy for a positron, just as we do for an electron. To deal correctly with the relationship between particles and antiparticles, we really need the more sophisticated formalism of quantum field theory.
What happens if we switch to the Weyl basis? We’ll look for solutions of the form:
ψ(x) = (ξ_{L}, ξ_{R}) exp(–i k · x)
When we substitute this into the Dirac equation, the resulting linear equation for ξ_{L} and ξ_{R} is:
(k_{μ} γ^{μ} – m I_{4}) (ξ_{L}, ξ_{R}) = 0
which, using the Weyl basis form of the gamma matrices expressed in terms of H matrices, we can write as:
k^{μ} H_{μ} ξ_{R} – m ξ_{L} = 0
k^{μ} H_{μ}* ξ_{L} – m ξ_{R} = 0
As we’d expect from what we saw in the Dirac basis, these equations are not independent. Since k^{μ} H_{μ} is k times a unitary matrix (i.e. a matrix whose inverse is its Hermitian adjoint), we have:
(k^{μ} H_{μ})(k^{μ} H_{μ}*) = k^{2} I_{2} = m^{2} I_{2}
So if we multiply the second of our matrix Dirac equations on the left with k^{μ} H_{μ}, then divide through by m, we obtain the first equation again.
For a stationary particle, with k = m e_{t}, solutions take the form:
ψ(x) = (ξ, ξ) exp(–i m t)
If we boost this in the zt plane — using the Weyl basis spin matrices, of course — we have:
k = exp(–θ J_{zt}) m e_{t} = m (cos θ e_{t} + sin θ e_{z})
(ξ_{L}, ξ_{R}) = exp(–i θ S^{zt}) (ξ, ξ) = (exp(–½i θ σ_{z}) ξ, exp(½i θ σ_{z}) ξ)
The matrices exp(±½i θ σ_{z}) are diagonal, with entries equal to the phases exp(±½i θ), so unlike the case with the Dirac basis the norms of each half of the spinor are unchanged by the boost.
In the Weyl basis, we can easily construct a plane wave solution for any value of k:
(ξ_{L}(k), ξ_{R}(k)) = (√[(k^{μ} H_{μ})/m] ξ, √[(k^{μ} H_{μ}*)/m] ξ)
Substituting this into the first of the matrix Dirac equations, k^{μ} H_{μ} ξ_{R} – m ξ_{L} = 0, and multiplying through by √m gives:
k^{μ} H_{μ} √[k^{μ} H_{μ}*] ξ – m √[k^{μ} H_{μ}] ξ = 0
√[k^{μ} H_{μ}] √[(k^{μ} H_{μ})(k^{μ} H_{μ}*)] ξ – m √[k^{μ} H_{μ}] ξ = 0
m √[k^{μ} H_{μ}] ξ – m √[k^{μ} H_{μ}] ξ = 0
which verifies that it is a solution. In case you’re groaning at the idea of having to take the square root of a matrix, we can actually describe this kind of square root quite easily:
√[k^{μ} H_{μ}] = (k^{μ} H_{μ} + m H_{t}) / √[2(m+k^{t})]
Suppose we pick an orthonormal basis {ξ_{s}} of C^{2}. A useful example would be the normalised eigenvectors of one of the Pauli spin matrices – and by convention, the usual choice is σ_{z}, in which case we’d use the labels s=±½. We can then write a pair of normalised plane wave solutions of the Riemannian Dirac equation for any 4momentum as:
Plane Wave Solutions of the Riemannian Dirac Equation in the Weyl Basis  
k is a 4momentum, k=m {ξ_{s}} is an orthonormal basis of C^{2} H_{μ} are the unit quaternions as matrices (defined here) Sum k^{j}H_{j} ranges over spatial indices x, y, z.  
ψ_{k, s}(x)  =  u(k, s) exp(–i k · x)  (Plane wave) 
u(k, s)  =  ( [(m+k^{t})H_{t} + k^{j}H_{j}] ξ_{s}, [(m+k^{t})H_{t} – k^{j}H_{j}] ξ_{s}) / (2 √[m(m+k^{t})])  (4spinor part) 
u(k, s_{1})† u(k, s_{2})  =  ξ_{s1}† ξ_{s2}  
=  δ_{s1, s2}  (Orthonormality) 
Now, if we pursue the results of the previous section describing plane waves in the Weyl basis, and identify both the space H of 2×2 complex matrices spanned by the H_{μ} and the twocomplexdimensional space C^{2} with the quaternions, as we previously described, we can take the first equation between the spinors ξ_{L} and ξ_{R} and the energymomentum vector k as an equation between three quaternions:
k^{μ} H_{μ} ξ_{R} – m ξ_{L} = 0
→
Q_{V} q_{R} – m q_{L} = 0
where Q_{V} = k^{μ} H_{μ} is a quaternion that transforms as a vector, q_{L} is a quaternion that transforms as a lefthanded spinor, and q_{R} is a quaternion that transforms as a righthanded spinor. In other words, the action of an element (g, h) of SU(2)×SU(2) on these three quaternions is given by:
ρ_{(½, ½)}(g, h) Q_{V} = g Q_{V} h^{–1}
ρ_{(½, 0)}(g, h) q_{L} = g q_{L}
ρ_{(0, ½)}(g, h) q_{R} = h q_{R}
The product Q_{V} q_{R} transforms in the same way as q_{L}, making the equation invariant. Following our previous discussion, if we choose a unit vector v_{0} in C^{2} we can explicitly indentify ξ_{L} = q_{L} v_{0} and ξ_{R} = q_{R} v_{0}. In that case, multiplying our quaternion equation on the right with v_{0} gives the corresponding spinor equation.
So we see that the Dirac equation for a plane wave can be interpreted as giving an extremely simple linear relationship between three quaternions that transform under different representations of SU(2)×SU(2), with a vector for the energymomentum of the particle, and a pair of left and righthanded spinors for the wave itself.
Riemannian Dirac Equation for Quaternions  
(q_{L}, q_{R}) is Dirac spinor as a pair of quaternions Q_{V} is the energymomentum vector of a plane wave solution: ψ(x) = (q_{L}, q_{R}) exp(–i Q_{V} · x)  
Q_{V} q_{R} – m q_{L}  =  0  (Riemannian) 
To understand this quaternionic description better, let’s look at how the usual twocomponent spinor version of the angular momentum of a spin½ particle can be translated into quaternionic terms. We first have to choose a complex structure: a linear operator J on H that corresponds to multiplication by i on C^{2}. Squaring J must amount to multiplication by –1, and J should commute with any linear operators on H that correspond to complexlinear operators on C^{2}. We can achieve these requirements by making J equal to rightmultiplication by some quaternion that is a square root of –1, and implementing other linear operators as leftmultiplication. To be concrete, let’s choose:
J(q) = q H_{z}
Given this choice, we set v_{0}=(0,1), an eigenvector of H_{z} with eigenvalue i. We then identify H and C^{2} via T:H→C^{2}, where
T(q) = q v_{0}
T(J(q)) = q H_{z} (0, 1) = i q (0, 1) = i T(q)
The choice of J identifies two planes in H that we can think of as the complex planes of the two components of the usual spinor: the zt plane and the xy plane. Multiplication of a quaternion q by any complex number a + b i (where a and b are real) is implemented as rightmultiplication by a + b H_{z}, and this operation maps both of these planes into themselves.
We give H the spinor inner product < >_{S}:
< p, q >_{S} = Re_{H}(p* q) – Re_{H}(p* J(q)) i = Re_{H}(p* q) – Re_{H}(p* q H_{z}) i
where “Re_{H}” means the real part of a quaternion, as distinct from the real part of a complex number, which we’ll write “Re_{C}”. We’ll use “*” to denote the conjugate of a quaternion, and “conj” to denote ordinary complex conjugation. If we represent the quaternions as 2×2 complex matrices, quaternionic conjugation is just taking the Hermitian adjoint or complexconjugatetranspose of those matrices, so q* = conj(q^{T}).
Our spinor inner product is really just the standard inner product on C^{2} transferred to H, so that:
< p, q >_{S} = < T(p), T(q) >
To see this, start with the real part:
Re_{C} < T(p), T(q) > = ½(< T(p), T(q) > + < T(q), T(p) >)
= ½(< p v_{0}, q v_{0} > + < q v_{0}, p v_{0} >)
= ½(< v_{0}, p* q v_{0} > + < v_{0}, q* p v_{0} >)
= < v_{0}, ½(p* q + q* p) v_{0} >
= Re_{H}(p* q) < v_{0}, v_{0} >
= Re_{H}(p* q)
We get from the thirdlast line to the secondlast because ½(p* q + q* p) = Re_{H}(p* q) is just a pure real number — or, in terms of matrices, the same real number times the 2×2 identity matrix. For the imaginary part:
Im_{C} < T(p), T(q) > = Re_{C}(–i < T(p), T(q) >) i
= – Re_{C}(< T(p), i T(q) >) i
= – Re_{C}(< T(p), T(q H_{z}) >) i
= – Re_{H}(p* q H_{z}) i
Now, the usual spin eigenstates ξ_{k±} in the twocomponent spinor formalism obey the equations (for k=x, y, z):
–σ_{k} ξ_{k±} = ±ξ_{k±}
Note that we’ve used the opposite of the Pauli matrices here, for reasons we discussed previously. In the quaternion formalism, we need to replace the Pauli matrices σ_{k} (which aren’t elements of H) with the basis quaternions H_{k} = –i σ_{k} and replace the C^{2} spinors ξ_{k±} with the quaternion spinors q_{k±} that satisfy ξ_{k±} = q_{k±} v_{0}. This leads us to the quaternion equations:
H_{k} q_{k±} = ±q_{k±} H_{z}
If we multiply this equation by –i, multiply v_{0} by each side, and use the fact that v_{0} is an eigenvector of H_{z} with eigenvalue i, we recover the previous equation for spinors in C^{2}.
The solutions to these quaternion equations are entire planes in H for each choice of sign, though of course they are planes that are mapped into themselves by multiplication with any complex number, corresponding to onecomplexdimensional eigenspaces in C^{2}.
For H_{x}, the positive eigenspace is spanned by H_{t}+H_{y} and H_{x}+H_{z}; the negative eigenspace is spanned by H_{t}–H_{y} and H_{x}–H_{z}.
For H_{y}, the positive eigenspace is spanned by H_{t}–H_{x} and H_{y}+H_{z}; the negative eigenspace is spanned by H_{t}+H_{x} and H_{y}–H_{z}.
For H_{z}, the positive eigenspace is spanned by H_{t} and H_{z}; the negative eigenspace is spanned by H_{x} and H_{y}.
(All of this extends to the left and righthanded spinor representations of the double cover of SO(4), where we measure spins in various coordinate planes in R^{4}. The various spin operators are all (half) H matrices or their opposites, as listed at the end of this section, so we can get everything we need from the eigenspaces described above. For example, spin in the zt coordinate plane for the spin(½,0) lefthanded spinor representation is measured by –½ H_{z}, so its positive and negative eigenspaces are just the negative and positive eigenspaces of H_{z} described above.)
Any spinor in C^{2} will be the positive spin eigenstate of the spin measured along some direction in space. In the quaternion formalism, this direction for a quaternion q is very easy to describe. For our choice of complex structure, it’s simply:
s(q) = q H_{z} q^{–1}
Obviously this satisfies:
s(q) q = q H_{z}
The spin axis s(q) will always be a purely spatial vector, since H_{z} is spatial to begin with, and all rotations in SO(4) produced by ρ_{(½, ½)}(g, g) Q_{V} = g Q_{V} g^{–1} leave the time axis unchanged, and so are actually rotations in SO(3). We can also very easily find a spinor quaternion whose spin is precisely the opposite of that of q, which we’ll call rev_{S}(q):
rev_{S}(q) = q H_{y}
s(rev_{S}(q)) = q H_{y} H_{z} H_{y}^{–1} q^{–1} = q H_{y} H_{z} (–H_{y}) q^{–1} = –q H_{x} H_{y} q^{–1} = –s(q)
The spinreversal function rev_{S} is conjugatelinear according to our complex structure of rightmultiplication by H_{z}:
rev_{S}(i q) = rev_{S}(q H_{z}) = q H_{z} H_{y} = –q H_{y} H_{z} = –i rev_{S}(q)
It’s easy to see that rev_{S} commutes with the action of SU(2)×SU(2), since that action multiplies quaternions on the left while rev_{S} multiplies them on the right.
Any rotation in SO(3) that leaves the vector s(q) fixed can be produced by ρ_{(½, ½)}(g, g) with g = cos(θ) H_{t} + sin(θ) s(q). Such rotations will act on the spinor q (whether it’s left or righthanded) by leftmultiplication with g, giving:
g q = (cos(θ) H_{t} + sin(θ) s(q)) q
= (cos(θ) H_{t} + sin(θ) q H_{z} q^{–1}) q
= cos(θ) q + sin(θ) q H_{z}
= (cos(θ) + i sin(θ)) q
= exp(i θ) q
In other words, the rotation merely changes the overall phase of the spinor q.
In a relativistic context where the particle can be in motion with some arbitrary velocity, we can no longer think of rotations as having an axis. Instead, we identify a rotation with a pair of unit quaternions (s_{L}, s_{R}), acting via v → s_{L} v s_{R}^{–1}. If we make an arbitrary change of coordinates corresponding to a rotation (g, h), then our original rotation is transformed to (g s_{L} g^{–1}, h s_{R} h^{–1}). This comes from composing three operations: (1) inverting the coordinate change to take things back to the coordinates of the original rotation, (2) performing the original rotation, and (3) applying the coordinate change to return to the new coordinates.
If we have a particle whose state is the Dirac spinor (q_{L}, q_{R}) and q_{L}≠0, q_{R}≠0, we can define a pair of unit quaternions to use for a rotation:
s_{L}(q_{L}) = q_{L} H_{z} q_{L}^{–1}
s_{R}(q_{R}) = q_{R} H_{z} q_{R}^{–1}
Note that if we apply a change of coordinates corresponding to (g, h), and use the spinor transformation laws q_{L} → g q_{L} and q_{R} → h q_{R}, the rotation (s_{L}(q_{L}), s_{R}(q_{R})) transforms exactly as a rotation should under coordinate changes. It also follows from these definitions that s_{L}(q_{L})^{2} = s_{R}(q_{R})^{2} = –H_{t}, so applying the rotation twice, v → s_{L}(q_{L})^{2} v s_{R}(q_{R})^{–2} just gives the identity transformation.
Clearly s_{L}(q_{L}) and s_{R}(q_{R}) will satisfy:
s_{L}(q_{L}) q_{L} = q_{L} H_{z}
s_{R}(q_{R}) q_{R} = q_{R} H_{z}
The physical meaning of this is that a rotation of vectors produced by the pair (s_{L}(q_{L}), s_{R}(q_{R})) will be accompanied by a transformation of the Dirac spinor (q_{L}, q_{R}) → (s_{L}(q_{L}) q_{L}, s_{R}(q_{R}) q_{R}) = (q_{L} H_{z}, q_{R} H_{z}), which amounts merely to multiplying the whole Dirac spinor by i according to the complex structure we’ve put on H.
In a frame where the particle is at rest the energymomentum vector Q_{V} will equal m H_{t}, so the equation:
Q_{V} q_{R} – m q_{L} = 0
will give us q_{R} = q_{L}, which in turn means s_{L}(q_{L}) = s_{R}(q_{R}), and the associated rotation will be threedimensional. Because we know that the rotation squared is the identity, it must be a halfturn in a particular plane. That fact won’t change when we switch to a frame where the particle is in motion, so in general we can treat (s_{L}(q_{L}), s_{R}(q_{R})) as identifying a plane of rotation associated with the particle’s spin, in place of the spin axis s(q) used in the threedimensional context.
What’s more, we can identify two vectors orthogonal to the plane (and to each other) that will be unchanged by the rotation: the energymomentum vector Q_{V} = m q_{L} q_{R}^{–1}, and the vector:
s_{LR}(q_{L}, q_{R}) = q_{L} H_{z} q_{R}^{–1}
The latter coincides with the spin axis s(q) when the particle is stationary and q_{R} = q_{L}.
If we apply the spinreversal function rev_{S} to both quaternions in a Dirac spinor, all four quaternions related to the spin geometry are reversed: s_{L}(q_{L}), s_{R}(q_{R}), Q_{V} and s_{LR}(q_{L}, q_{R}).
It’s easy to construct a Lagrangian that yields the Riemannian Dirac equation:
Lagrangian for Dirac Equation  
L_{RD}  =  ψ† (i γ^{μ} ∂_{μ} ψ – m ψ)  
=  i ψ† γ^{μ} ∂_{μ} ψ – m ψ†ψ  (Riemannian)  
L_{LD}  =  ψ† γ^{t} (i γ^{μ} ∂_{μ} ψ – m ψ)  
=  i ψ† γ^{t} γ^{μ} ∂_{μ} ψ – m ψ† γ^{t} ψ  (Lorentzian) 
Here ψ† is the Hermitian adjoint, or conjugate transpose, of the 4 × 1 column vector ψ, so it is a 1 × 4 row vector that we multiply into the objects on its right by matrix multiplication.
In the Lorentzian case, the matrices that represent Lorentz transformations on ψ are generally not unitary, and as a consequence ψ†ψ is not a scalar: it is not invariant under Lorentz transformations. So it’s necessary to replace ψ† by ψ† γ^{t}; this gives a Lorentzinvariant inner product, ψ† γ^{t} ψ, and allows a Lorentzinvariant Lagrangian to be constructed. (In the QFT literature, ψ† γ^{t} is abbreviated as ψ with a bar over the top.) But those minor headaches belong solely to the Lorentzian case. In the Riemannian case, all the spin matrices are Hermitian, the representation of SO(4) is unitary, and ψ†ψ is SO(4)invariant:
ψ → U ψ
ψ† → (U ψ)† = ψ† U† = ψ† U^{–1}
ψ†ψ → ψ† U^{–1} U ψ = ψ†ψ
The first term of L_{RD} is also SO(4)invariant, as can easily be shown with the help of equation (2) from the section on gamma matrices.
We obtain the Dirac equation from the Lagrangian by means of the EulerLagrange equations. We treat ψ† as an independent field, and write:
∂_{μ} [ ∂_{∂μ ψ†}L_{RD} ] = ∂_{ψ†}L_{RD}
0 = i γ^{μ} ∂_{μ} ψ – m ψ
where the expression in square brackets is zero because the Lagrangian doesn’t depend on any derivative of ψ†. The corresponding calculation using ψ itself as the field gives the same answer, merely with more work:
∂_{μ} [ ∂_{∂μ ψ}L_{RD} ] = ∂_{ψ}L_{RD}
∂_{μ} [ i ψ† γ^{μ} ] = –m ψ†
i ∂_{μ} ψ† γ^{μ} + m ψ† = 0
–(i ∂_{μ} ψ† γ^{μ} + m ψ†)† = 0
i γ^{μ} ∂_{μ} ψ – m ψ = 0
Now that we have the Lagrangian for the Dirac equation, we can use Noether’s Theorem to identify a conserved current. Suppose we multiply our Dirac spinor ψ by any unitmagnitude complex number, also known as a “phase”: exp(i α) for a realvalued α. Since ψ† ends up with the opposite phase, the Lagrangian is unchanged.
Noether’s Theorem associates a conserved current with every such symmetry of the Lagrangian:
j^{μ} = (∂_{∂μ ψ}L_{RD}) (∂_{α} exp(i α) ψ)_{α=0}
= –ψ† γ^{μ} ψ
We can multiply this by any constant and the current will still be conserved. Specifically, we’ll define:
Current for Dirac Equation  
e is charge on particle  
j^{μ}  =  e ψ† γ^{μ} ψ  (Riemannian) 
j^{μ}  =  e ψ† γ^{t} γ^{μ} ψ  (Lorentzian) 
What we mean by calling this a “conserved current” is that its divergence is zero, so it doesn’t appear or disappear out of thin air.
∂_{μ} j^{μ} = e ∂_{μ} (ψ† γ^{μ} ψ)
= e ((∂_{μ} ψ†) γ^{μ} ψ + ψ† γ^{μ} (∂_{μ} ψ))
= e ((i m ψ†) ψ + ψ† (–i m ψ))
= 0
Consider the stationary plane wave solution, using the Dirac basis:
ψ(x) = (ξ, 0) exp(–i m e_{t} · x)
where ξ is any twocomponent spinor. For this solution, we have simply:
j = e (ξ†ξ) e_{t}
because the structure of the gamma matrices apart from γ^{t} makes all the other components zero. Of course with an infinite plane wave we face the usual problems of normalisation, but we’ll ignore that for now and just focus on the direction of j, which is aligned with the particle’s propagation vector in this particular case — and will continue to be aligned if we switch to a reference frame in which the wave is moving. So j behaves just as we’d expect for an electric current density.
We can combine the Lagrangian for the Dirac equation with that we previously derived for the Riemannian Proca equation, the equation governing electromagnetism. We can then make use of our expression for the current density in terms of the Dirac field, to give a total Lagrangian expressed entirely in terms of the elementary fields. [Note that for the Lorentzian versions we’re using a (+ – – –) signature on this page, for ease of comparison with QFT textbooks. This means that some of these Lorentzian equations are not quite what you’d expect from the versions in the notes on electromagnetism.]
Lagrangian for Quantum Electrodynamics  
m_{ph} is the mass of the photon m is the mass of the Dirac particle and e is its charge Electromagnetic field F_{μν} = ∂_{μ} A_{ν} – ∂_{ν} A_{μ}  
L_{RQED}  =  ψ† (i γ^{μ} ∂_{μ} ψ – m ψ) + ¼ F_{μν}F^{μν} – ½ m_{ph}^{2} A_{μ} A^{μ} – A_{μ} j^{μ}  
=  ψ† (i γ^{μ} ∂_{μ} ψ – m ψ) + ¼ F_{μν}F^{μν} – ½ m_{ph}^{2} A_{μ} A^{μ} – e A_{μ} ψ† γ^{μ} ψ  (Riemannian)  
L_{LQED}  =  ψ† γ^{t} (i γ^{μ} ∂_{μ} ψ – m ψ) – ¼ F_{μν}F^{μν} – A_{μ} j^{μ}  
=  ψ† γ^{t} (i γ^{μ} ∂_{μ} ψ – m ψ) – ¼ F_{μν}F^{μν} – e A_{μ} ψ† γ^{t} γ^{μ} ψ  (Lorentzian) 
The EulerLagrange equations for these Lagrangians give us both the equation for the electromagnetic field coupled to the Dirac source, and the Dirac equation in the presence of an electromagnetic field.
Field Equations for Quantum Electrodynamics  
m_{ph} is the mass of the photon m is the mass of the Dirac particle and e is its charge Electromagnetic field F_{μν} = ∂_{μ} A_{ν} – ∂_{ν} A_{μ}  
∂_{μ}∂^{μ} A^{ν} + m_{ph}^{2} A^{ν} + e ψ† γ^{ν} ψ  =  0  
i γ^{μ} ∂_{μ} ψ – m ψ – e A_{μ} γ^{μ} ψ  =  0  (Riemannian) 
∂_{μ} F^{μν} – e ψ† γ^{t} γ^{ν} ψ  =  0  
i γ^{μ} ∂_{μ} ψ – m ψ – e A_{μ} γ^{μ} ψ  =  0  (Lorentzian) 
The Riemannian Dirac equation in an electromagnetic field is:
i γ^{μ} ∂_{μ} ψ – m ψ – e A_{μ} γ^{μ} ψ = 0
Let’s rewrite this in an explicitly Hamiltonian form, by multiplying through on the left by γ^{t} (whose square, like that of all the Riemannian gamma matrices, is the identity matrix) and separating out the time derivative:
i ∂_{t} ψ = H ψ
H = –i γ^{t} γ^{j} ∂_{j} + m γ^{t} + e A_{μ} γ^{t} γ^{μ}
Here the sum over the repeated index j covers spatial coordinates only, while the sum over μ covers all four dimensions.
We will call the gamma matrices multiplied on the left by γ^{t} the alpha matrices, and like the gamma matrices they take a very simple form in the Weyl basis:
Riemannian alpha matrices, Weyl basis  
α^{μ}  =  γ^{t} γ^{μ}  
= 

(Riemannian) 
Now, an operator for an observable will describe a conserved quantity if it commutes with the Hamiltonian. If we have a Riemannian electron in a radially symmetric electrostatic field we would expect its angular momentum to be conserved, so let’s compute the commutator between the Hamiltonian for that situation and the orbital angular momentum operator L = r × p. In an electrostatic field with potential V(r) we have A_{t} = –V(r), while all other A_{j} = 0, giving us a Hamiltonian:
H = –i α^{j} ∂_{j} + m γ^{t} – e V(r)
= – α^{j} p_{j} + m γ^{t} – e V(r)
where p_{j} = i ∂_{j} is the jth component of the linear momentum; don’t forget that we have to use a different convention for the operator for Riemannian relativistic momentum. The orbital angular momentum operator has components:
L_{i} = ε_{ijk} x^{j} p_{k}
To find its commutator with the Hamiltonian, we first compute:
[x^{j} p_{k}, p_{s}] = –[x^{j} ∂_{k}, ∂_{s}]
= –( x^{j} ∂_{k} ∂_{s} – ∂_{s}(x^{j} ∂_{k}) )
= –( x^{j} ∂_{k} ∂_{s} – x^{j} ∂_{s} ∂_{k} – δ_{s}^{j} ∂_{k} )
= –i δ_{s}^{j} p_{k}
This gives us:
[L_{i}, p_{s}] = –i ε_{isk} p_{k}
L_{i} will clearly commute with the second term in the Hamiltonian, which involves only multiplication by a fixed matrix. What about the third term? For any function f that depends solely on the squared radial distance r^{2} = x^{2} + y^{2} + z^{2} = x^{l} x^{l}, and any function g we have:
[x^{j} p_{k}, f(x^{l} x^{l})] g = i [x^{j} ∂_{k}, f(x^{l} x^{l})] g
= i ( x^{j} ∂_{k} ( f(x^{l} x^{l}) g ) – f(x^{l} x^{l}) x^{j} ∂_{k} g )
= i ( 2 g x^{j} x^{k} f '(x^{l} x^{l}) + f(x^{l} x^{l}) x^{j} ∂_{k} g – f(x^{l} x^{l}) x^{j} ∂_{k} g )
= 2 i x^{j} x^{k} f '(x^{l} x^{l}) g
Writing the result as a linear operator without the function g:
[x^{j} p_{k}, f(x^{l} x^{l})] = 2 i x^{j} x^{k} f '(x^{l} x^{l})
Since this expression is symmetric in the indices j and k, the antisymmetric ε in L_{i} = ε_{ijk} x^{j} p_{k} will give a total of zero for [L_{i}, f(x^{l} x^{l})], and since the third term in the Hamiltonian can be expressed as a function with the form of f, that term will commute with L_{i}.
That leaves only the first term in the Hamiltonian. So we have:
[L_{i}, H] = [L_{i}, – α^{j} p_{j}]
= – α^{j} [L_{i}, p_{j}]
= i ε_{ijk} α^{j} p_{k}
Since this is nonzero, orbital angular momentum will not be conserved. However, if we attribute spin angular momentum to the electron itself in the right manner, total angular momentum can still be conserved.
We previously described the Hermitian spin matrices S^{μν} that represent the generators of rotations. If we define the spatial spin matrices S = (S^{yz}, S^{zx}, S^{xy}), in the Weyl basis we have S_{i} = –½ I_{2} ⊗ σ_{i} = –½ i I_{2} ⊗ H_{i}.
Spatial spin matrices, Weyl basis  
S_{i}  = 

(Riemannian) 
The commutators of the H matrices with spatial indices are:
[H_{j}, H_{k}] = 2 ε_{jkl} H_{l}
Now in the alpha matrices with spatial indices, the adjoint matrix H_{j}* = –H_{j}. From this, we can easily compute the commutators of the spatial spin matrices with spatial alpha matrices:
[S_{i}, α^{j}] = –i ε_{ijl} α^{l}
The matrix S_{i} will commute with all but the first term of the Hamiltonian, and so:
[S_{i}, H] = – [S_{i}, α^{j}] p_{j}
= i ε_{ijl} α^{l} p_{j}
= i ε_{ikj} α^{j} p_{k} [Relabelling indices]
= –i ε_{ijk} α^{j} p_{k}
This is the opposite of the commutator with the orbital angular momentum component L_{i}, so we have, for the total angular momentum:
[L+S, H] = 0
The fact that we have ended up with a conserved quantity for the total angular momentum confirms that we have made the correct choice for the spin matrices!
We previously discussed how pairs of quaternions (q_{L}, q_{R}) could be interpreted as Dirac spinors, and how the angular momentum of a spin½ particle could be described in that formalism. We treat each copy of the quaternions as a twocomplexdimensional Hilbert space, in place of the usual spinor space C^{2}, by giving it a complex structure — and our particular choice was to use multiplication on the right by the quaternion H_{z} as multiplication by i, the square root of minus 1. With these conventions, the Dirac equation for a free plane wave in the Weyl basis becomes extraordinarily simple:
q_{V} q_{R} – m q_{L} = 0
Here (q_{L}, q_{R}) is the Dirac spinor, with the first component transforming as a lefthanded spinor and the second as a righthanded spinor, while q_{V} is a vector quaternion that gives the 4momentum of the plane wave. The quaternionvalued plane wave is:
ψ(x) = (q_{L}, q_{R}) exp(–i q_{V} · x)
We can also develop a description of the angular momentum of a spin1 particle, such as a Riemannian photon, that uses the quaternions. The vector representation of SU(2)×SU(2), the double cover of SO(4), is really just a representation on quaternions, in which the group SU(2) corresponds to the unit quaternions, and a pair of unit quaternions (g, h) in SU(2)×SU(2) acts on the quaternions themselves via:
(g, h) q → gqh^{–1} = gqh*
If we identify the quaternions, H, with fourdimensional Euclidean space, R^{4}, then this representation corresponds to multiplying vectors in R^{4} by elements of SO(4): real, orthogonal 4×4 matrices with a determinant 1. This is fine when we’re doing classical physics and simply want to talk about rotating real fourvectors, but the angular momentum of a particle with spin 1 is a vector in a complex Hilbert space. How can we connect the two?
The solution is to “complexify” the quaternions. Just as we write a general complex number as a + b i where a and b are real numbers, we write a general complexified quaternion as q + s i, where q and s are quaternions. The i here should not be taken to be any of the square roots of minus 1 within the quaternions themselves, or to be the same as the rightmultiplication by H_{z} that we use on the spinor quaternions. (Only when we multiply a complexified vector quaternion by a spinor to get another spinor do we convert factors of i to rightmultiplication by H_{z}.) Within the complexified quaternions themselves, i is just a symbol that commutes with everything and satisfies i^{2}=–1. So if we multiply two complexified quaternions, we have:
(q + s i)(p + r i) = (qp – sr) + (sp + qr) i
In working with the complexified quaternions, we’ll use the following conventions: “Re_{H}” means the real part of a quaternion, as distinct from the real part of a complex number, which we’ll write “Re_{C}”. We’ll use “*” to denote the conjugate of a quaternion, and “conj” to denote ordinary complex conjugation. So for example, in the equation below we take the opposite of all the quaternionic components orthogonal to H_{t}; we don’t reverse the sign of the coefficient of i.
((H_{t} + H_{x}) + (H_{t} + H_{y}) i)* = (H_{t} – H_{x}) + (H_{t} – H_{y}) i
There’s a real inner product, or dot product, defined on the quaternions:
q · s = Re_{H}(q* s) = ½(q* s + s* q)
This is invariant under the action of SU(2)×SU(2); if we act with the pair of unit quaternions (g, h) we get:
q · s → (g q h^{–1}) · (g s h^{–1})
= ½((g q h^{–1})* (g s h^{–1}) + (g s h^{–1})* (g q h^{–1}))
= ½(h q* g^{–1} g s h^{–1} + h s* g^{–1} g q h^{–1})
= ½(h q* s h^{–1} + h s* q h^{–1})
= ½(h (q* s + s* q) h^{–1})
= h Re_{H}(q* s) h^{–1}
= Re_{H}(q* s)
= q · s
We extend this dot product to the complexified quaternions simply by requiring it to be complexlinear in each vector:
(q + s i) · (p + r i) = (q · p – s · r) + (s · p + q · r) i
We also need an inner product on the Hilbert space of complexified quaternions that is linear in the second argument and conjugatelinear (in the complex structure sense!) in the first argument:
< q + s i, p + r i >_{V} = conj(q + s i) · (p + r i)
= (q – s i) · (p + r i)
= (q · p + s · r) + (q · r – s · p) i
We use a subscript of “V” on this inner product, standing for vector, to distinguish it from the inner product on the spinor quaternions:
< p, q >_{S} = Re_{H}(p* q) – Re_{H}(p* q H_{z}) i
We’ve shown that the dot product is invariant under the action of SU(2)×SU(2), so our complex inner product < >_{V} will be invariant too — or in other words, our representation of SU(2)×SU(2) is unitary, as it must be.
So far, we’ve taken a largely purist approach where we treat the quaternions as being fundamental mathematical objects, and this gives us formulas that are useful for many kinds of calculations. But sometimes it’s worth recalling that we can treat the ordinary quaternions as a subspace of the 2×2 complex matrices, spanned by real linear combinations of the four matrices H_{μ} — and in that context we can treat the complexified quaternions as arbitrary 2×2 complex matrices, which can be obtained as complex linear combinations of the H_{μ}.
Thinking of complexified vector quaternions as 2×2 complexvalued matrices, and spinor quaternions as lying in the smaller subspace spanned by the real linear combinations of the H_{μ}, the vector and spinor inner products can be written as:
< p, q >_{V} = ½ tr(conj(p^{T}) q)
< p, q >_{S} = < p, q – i q H_{z} >_{V} = ½ tr(conj(p^{T}) (q – i q H_{z}))
When we treat the complexified quaternions as the Hilbert space for a Riemannian photon’s spin, we take the “pure” quaternions with no factor of i as states of linear polarisation, corresponding to the direction of the fourpotential vector A in a classical Riemannian vector plane wave. In a reference frame in which the photon is at rest, the quaternions H_{x}, H_{y} and H_{z} are state vectors for the three linear polarisations of a plane wave, all of them orthogonal to its propagation vector (which in this case is H_{t}, since we’re in the rest frame).
The three vectors H_{x}, H_{y} and H_{z} can also be thought of as the states of a spin1 particle whose component of spin in the x, y and z directions respectively is zero. This ties in precisely with the nonrelativistic quantum mechanics of a spin1 particle (one with mass, unlike photons in our universe), where choosing a basis of C^{3} consisting of states with a spin component of zero in each of three orthogonal directions allows the spin1 representation of SU(2) on C^{3} to be identified with the vector representation of SO(3) on R^{3}. In this basis, the spin1 unitary matrices that are applied to C^{3} to represent rotations become realvalued — and in fact become exactly the same as the SO(3) matrices for the rotations.
What about the states where the particle has a spin component of plus or minus 1 along some axis? We can find these states by mimicking the relationship between circular polarisation and linear polarisation in classical waves. In a classical plane wave, circular polarisation around the zaxis, say, occurs when a wave with linear polarisation along the xaxis is combined with another with linear polarisation along the yaxis that is 90° out of phase with the first. For a complex exponential the 90° phase shift can be created by multiplication by i. So we expect the normalised photon states with one unit of spin along the zaxis to be given by (H_{x} ± i H_{y})/√2. However, because of the same twist that leads us to use the opposite of the Pauli spin matrices for fermions, we need to choose the signs here carefully: the sum is the negative spin eigenstate and the difference is the positive spin eigenstate.
Complexified vector quaternion states for a spin1 particle  
Complexified vector quaternion  Angular momentum component 
H_{x}  m_{x} = 0 
H_{y}  m_{y} = 0 
H_{z}  m_{z} = 0 
(H_{y} ± i H_{z})/√2  m_{x} = –1, 1 
(H_{z} ± i H_{x})/√2  m_{y} = –1, 1 
(H_{x} ± i H_{y})/√2  m_{z} = –1, 1 
H_{t}  m_{x} = m_{y} = m_{z} = 0 
From these states we can construct operators for the spin along each axis, by projecting onto the spin eigenstates and multiplying by the spin. For example:
J_{z}(v) = – [(H_{x} + i H_{y})/√2] < (H_{x} + i H_{y})/√2, v >_{V} + [(H_{x} – i H_{y})/√2] < (H_{x} – i H_{y})/√2, v >_{V}
= i (H_{x} < H_{y}, v >_{V} – H_{y} < H_{x}, v >_{V})
= i (H_{x} (H_{y} · v) – H_{y} (H_{x} · v))
In going from the secondlast line to the last, we’ve used the fact that the complex inner product becomes identical to the dot product when the first argument has no imaginary part.
We will write this as:
J_{z} = i (H_{x} ⊗ H_{y} – H_{y} ⊗ H_{x})
with the convention that this acts on any complexified quaternion via the dot product with the righthand term in each tensor product. Similarly, we have:
J_{x} = i (H_{y} ⊗ H_{z} – H_{z} ⊗ H_{y})
J_{y} = i (H_{z} ⊗ H_{x} – H_{x} ⊗ H_{z})
Apart from a factor of –i, if we treated these operators as matrices acting on R^{3} they would just be the so(3) elements generating rotations around their respective axes.
Of course the full Hilbert space we’re working in has four complex dimensions, not three. Whatever we choose as a basis for the threedimensional subspace of spins available to a photon at rest, we need a fourth vector such as H_{t} to allow us to deal with the general case of photons in motion.
In order to describe the interactions between Riemannian photons and electrons, we need to be able to construct intertwiners between the representations of SU(2)×SU(2) that apply to these particles, both in the case of individual particles and when we have, say, a system comprised of a photon and an electron. When we’re dealing with such a composite system, the representation that applies to the particles’ joint quantum state is the tensor product of the representations that apply individually:
ρ_{photon and electron}(g, h) φ_{photon} ⊗ ψ_{electron} = (ρ_{photon}(g, h) φ_{photon}) ⊗ (ρ_{electron}(g, h) ψ_{electron})
The representations that apply to individual particles are irreducible, but these tensor product representations are not. Rather, they are equivalent to a direct sum of two or more irreducible representations. What this means is that there will be subspaces within the tensor product vector space that are invariant under the representation — no vector within these subspaces is taken out of them by the representation — and each subspace will be equivalent to some irreducible representation. The simplest example of this is the tensor product of two spin½ representations of SU(2), which contains one subspace equivalent to the spin0 representation of SU(2) and another equivalent to the spin1 representation of SU(2). We can write this in a kind of shorthand as ½ ⊗ ½ = 0 ⊕ 1. We actually defined the spinj representation as sitting inside a tensor product of 2j copies of the spin½ representation, so that discussion is a good place to see this idea worked out in more detail.
More generally, if we tensor the spinj representation of SU(2) with the spink representation, the resulting space splits up into subspaces with spins ranging from j–k to j+k in steps of 1:
j ⊗ k = j–k ⊕ j–k+1 ⊕ ... ⊕ j+k
When it comes to representations of SU(2)×SU(2), which are described by a pair of spins, we have:
(j_{1}, j_{2}) ⊗ (k_{1}, k_{2}) = (j_{1}–k_{1}, j_{2}–k_{2}) ⊕ (j_{1}–k_{1}+1, j_{2}–k_{2}) ⊕ ... ⊕ (j_{1}+k_{1}, j_{2}+k_{2})
with the lefthand spins taking on every value, in integer steps, between j_{1}–k_{1} and j_{1}+k_{1}, and the righthand spins independently taking on every value, in integer steps, between j_{2}–k_{2} and j_{2}+k_{2}.
Now, our goal is to describe processes such as an electron absorbing a photon, in which we start with a composite system consisting of the electron and the photon, and end up with only the electron. We can build up a description of this from more elementary pieces, in which rather than starting with a photon and a complete Dirac spinor, we start with a photon and either a left or righthanded spinor. In these cases, quaternion multiplication gives us an extremely simple intertwiner between the representation for the initial composite system and that which applies to a spinor of the opposite chirality. In other words, we can describe a lefthanded spinor absorbing a photon and becoming a righthanded spinor, or a righthanded spinor absorbing a photon and becoming a lefthanded spinor.
Before proceeding, we’ll tabulate the interpretation of the basis quaternions as spinors.
Quaternion states for a spin½ particle  
Complex structure is rightmultiplication by H_{z}  
Basis quaternion  Interpreted as a spinor state  Angular momentum component 
H_{t}  UP  m_{z} = ½ 
H_{z}  i UP  m_{z} = ½ 
H_{y}  –DOWN  m_{z} = –½ 
H_{x}  –i DOWN  m_{z} = –½ 
The overall choice of phase is arbitrary, but this assignment gives us a consistent scheme for interpreting all the quaternions as spin states of a spin½ particle. Recall that the whole tworealdimensional plane spanned by H_{t} and H_{z} is to be thought of as a single complex plane, and the same is true for the plane spanned by H_{x} and H_{y}.
Let’s start by supposing that a particle described by a righthanded spinor q_{R} absorbs a photon q_{V} and turns into a lefthanded spinor, q_{L}. For now, what we’re exploring is just the spin, so we’re not concerned at all with the energy and momentum of these particles. Under a rotation described by the pair of unit quaternions (g, h), each of these quaternions transforms differently:
q_{R} → h q_{R}
q_{V} → g q_{R} h^{–1}
q_{L} → g q_{L}
However, we can see that the product, q_{V} q_{R}, transforms just like a lefthanded spinor:
(q_{V} q_{R}) → (g q_{V} h^{–1}) (h q_{R})
= g (q_{V} q_{R})
If this business of a righthanded spinor becoming lefthanded seems a bit mysterious when it’s done with quaternions, we can see why this is required by looking at the representations of SU(2)×SU(2):
(½, ½) ⊗ (0,½) = (½, 0) ⊕ (½, 1)
The (½, ½) representation is the photon, the (0,½) representation is the righthanded spinor, and the (½, 0) representation is a subspace of the tensor product that transforms just like a lefthanded spinor. The remaining subspace in the tensor product, which transforms according to a (½, 1) representation, involves a total spin of 3/2; we’re assuming that no elementary particle with that spin exists, so this representation is only realised by the righthanded spinor and the photon remaining as separate entities.
As a candidate for an intertwiner from the space describing systems of photons and righthanded spinors to the space describing lefthanded spinors, we will propose:
T_{1}(q_{V} ⊗ q_{R}) = q_{V} q_{R} / 2
The reason for the particular normalisation factor of ½ will be explained shortly. Let’s see what this intertwiner gives us for various initial composite systems.
T_{1}(Photon with m_{z} = 0, UP spinor)
= T_{1}(H_{z} ⊗ H_{t})
= H_{z} H_{t} / 2
= H_{z} / 2
= [i/2] UP
A spinUP righthanded spinor can absorb a photon with no spin along the zaxis to produce a spinUP lefthanded spinor, with an amplitude for this process of i/2.
T_{1}(Photon with m_{z} = 0, DOWN spinor)
= T_{1}(H_{z} ⊗ (–H_{y}))
= –H_{z} H_{y} / 2
= H_{x} / 2
= [–i/2] DOWN
A spinDOWN righthanded spinor can absorb a photon with no spin along the zaxis to produce a spinDOWN lefthanded spinor, with an amplitude of –i/2.
T_{1}(Photon with m_{z} = –1, UP spinor)
= T_{1}((H_{x} + i H_{y})/√2 ⊗ H_{t})
= (H_{x} + i H_{y}) H_{t} / (2 √2)
= (H_{x} + i H_{y}) / (2 √2)
= (H_{x} + H_{y} H_{z}) / (2 √2)
= (2 H_{x}) / (2 √2)
= [–i/√2] DOWN
In going from the fourth line to the fifth, we’ve made use of the complex structure of the spinor quaternions, converting the factor of i that started out as an independent number in the complexified vector quaternions into rightmultiplication by H_{z}. The conclusion: a spinUP righthanded spinor can absorb a photon with a spin of –1 along the zaxis to produce a spinDOWN lefthanded spinor, with an amplitude of –i/√2.
T_{1}(Photon with m_{z} = –1, DOWN spinor)
= T_{1}((H_{x} + i H_{y})/√2 ⊗ (–H_{y}))
= –(H_{x} + i H_{y}) H_{y} / (2 √2)
= –(H_{z} – i H_{t}) / (2 √2)
= –(H_{z} – H_{z}) / (2 √2)
= 0
A spinDOWN righthanded spinor and a photon with a spin of –1 along the zaxis have ZERO amplitude to produce any lefthanded spinor.
T_{1}(Photon with m_{z} = 1, UP spinor)
= T_{1}((H_{x} – i H_{y})/√2 ⊗ H_{t})
= (H_{x} – i H_{y}) H_{t} / (2 √2)
= (H_{x} – i H_{y}) / (2 √2)
= (H_{x} – H_{y} H_{z}) / (2 √2)
= (H_{x} – H_{x}) / (2 √2)
= 0
A spinUP righthanded spinor and a photon with a spin of 1 along the zaxis have ZERO amplitude to produce any lefthanded spinor.
T_{1}(Photon with m_{z} = 1, DOWN spinor)
= T_{1}((H_{x} – i H_{y})/√2 ⊗ (–H_{y})
= –(H_{x} – i H_{y}) H_{y} / (2 √2)
= –(H_{z} + i H_{t}) / (2 √2)
= –(H_{z} + H_{t} H_{z}) / (2 √2)
= –(2 H_{z}) / (2 √2)
= [–i/√2] UP
A spinDOWN righthanded spinor can absorb a photon with a spin of 1 along the zaxis to produce a spinUP lefthanded spinor, with an amplitude of –i/√2.
Finally, we note that:
T_{1}(H_{t} ⊗ UP) = ½ UP
T_{1}(H_{t} ⊗ DOWN) = ½ DOWN
H_{t} is not a possible state for the spin of a photon at rest, but it certainly has a zcomponent of spin of 0, i.e. J_{z}(H_{t}) = 0, so it makes sense that it can leave the direction of spin of our spinor unchanged.
As we would hope, we get an amplitude of zero whenever there is no lefthanded spinor we can produce while conserving the zcomponent of spin. For the other cases, though, the probabilities — the squared norms of the amplitudes — aren’t 1. For example, we don’t have a probability of 1 for a spinUP spinor and a photon with zero spin along the zaxis to yield another spinUP spinor. Why not? Because this state is not an eigenstate for total spin. It is an eigenstate for the zcomponent of the spin, m_{z}=½, but there’s a nonzero amplitude for the total spin j of the composite system to be 3/2 rather than ½, as we can see by recalling our decomposition of the tensor product space:
(½, ½) ⊗ (0,½) = (½, 0) ⊕ (½, 1).
The normalisation we’ve chosen for the intertwiner makes the squared norms of the amplitudes for all the different ways of creating a spinUP spinor add up to 1:
T_{1}(Photon with m_{z} = 0, UP spinor)^{2} + T_{1}(Photon with m_{z} = 1, DOWN spinor)^{2} + T_{1}(H_{t} ⊗ UP)^{2}
= i/2^{2} + –i/√2^{2} + ½^{2}
= ¼ + ½ + ¼
= 1
The same holds for the different ways of creating spin DOWN. This lets us think of each of the amplitudes we get from the intertwiner as the inner product between one of our initial states and one of two normalised vectors sitting in the composite system’s Hilbert space, in the subspace that transforms under the (½, 0) representation. For both these vectors the total spin j is ½, and the zcomponent m_{z} is ½ or –½ respectively. So the intertwiner, normalised this way, is giving us amplitudes for measuring j and m_{z} and finding them to have these particular values.
We can do something similar for a lefthanded spinor absorbing a photon and turning into a righthanded spinor. We define:
T_{2}(q_{V} ⊗ q_{L}) = –q_{V}* q_{L} / 2
The minus sign will turn out to be a convenient choice later. Remember that the * operation here conjugates quaternions, but has no effect on the number i in our complexified quaternions. The result transforms as a righthanded spinor under the rotation (g, h):
(q_{V}* q_{L}) → (g q_{V} h^{–1})* (g q_{L})
= (h q_{V} g^{–1}) (g q_{L})
= h (q_{V} q_{L})
where g^{–1}=g* and h^{–1}=h* because these are unit quaternions.
We can combine the two intertwiners for left and righthanded spinors to give an intertwiner for a Riemannian Dirac particle absorbing a photon:
T_{3}(q_{V} ⊗ (q_{L}, q_{R})) = (q_{V} q_{R}, –q_{V}* q_{L}) / 2
Since the Dirac particle has both left and righthanded components, the same representation applies before and after it absorbs the photon. Keep in mind, though, that this still has nothing to do with energy and momentum conservation; we’re only looking at spin.
Before we go on to look at other interactions involving photons and electrons, we’ll describe some intertwiners that connect the representations for photons and electrons to the trivial representation of SO(4). The trivial representation describes the way a particle of spin 0 transforms: it’s completely unchanged by a change of coordinates. The Hilbert space associated with its spin is onecomplexdimensional, i.e. it’s just the complex numbers C. So an intertwiner that describes two photons combining to yield a particle of spin 0 will be a function from the tensor product of two copies of the complexified quaternions to the complex numbers, such that the value of the function is unchanged when the same rotation is applied to the two photons.
Such a function is very easy to find: it’s just the dot product! With a suitable normalisation factor, we define:
T_{4}(q_{V1} ⊗ q_{V2}) = q_{V1} · q_{V2} / 2
It’s more or less a definition of what a rotation is that the value of the dot product will be unchanged when the same rotation is applied to q_{V1} and q_{V2}. And it’s easy to check that a state where two photons have a spin component of zero along the same axis, such as H_{z} ⊗ H_{z}, yields a nonzero result, as does a state where one photon has the opposite spin to the other, while a state where two photons have the same nonzero spin will yield zero.
It’s not much harder to describe an intertwiner that goes the other way, taking a particle with spin 0 and giving a state in the Hilbert space of twophoton systems that, like the spin0 particle, is invariant under rotations.
T_{5}(z) = (z/2) (H_{x} ⊗ H_{x} + H_{y} ⊗ H_{y} + H_{z} ⊗ H_{z} + H_{t} ⊗ H_{t})
We choose the normalisation so that when z is a unit complex number, T_{5}(z) is a unit vector. The easiest way to see why T_{5}(z) is invariant under rotations is to think of it as being like the metric tensor in flat Euclidean space: a matrix with 1 for every diagonal entry. Just as the metric is unchanged by rotations, so is this state. (Strictly speaking, if we’re think of the quaternions as vectors this tensor state is like the metric on the dual space of linear functions of vectors, but that’s unchanged by rotations just like the metric on the vector space.)
Our previous interwiner, T_{4}, can be thought of as giving the inner product between the normalised invariant vector T_{5}(1) and its argument:
T_{4}(q_{V1} ⊗ q_{V2}) = < T_{5}(1), q_{V1} ⊗ q_{V2} >_{V⊗V}
We get an inner product on the tensor product space simply by multiplying inner products between the individual terms in each tensor:
< a ⊗ b, c ⊗ d >_{V⊗V} = < a, c >_{V} < b, d >_{V}
If we compose these two intertwiners, we get:
T_{4}(T_{5}(z)) = ¼ z (H_{x} · H_{x} + H_{y} · H_{y} + H_{z} · H_{z} + H_{t} · H_{t}) = z
What about an invariant state for two spinors? We define:
T_{6}(z) = (z/√2) (UP ⊗ DOWN – DOWN ⊗ UP)
= (z/√2) (H_{y} ⊗ H_{t} – H_{t} ⊗ H_{y})
This state can be interpreted either as two lefthanded spinors or two righthanded spinors; in either case it will be invariant under SO(4). The effect of a rotation is to multiply every quaternion here on the left by the same unit quaternion; we’ll leave a general proof of invariance as an exercise, and only work through one specific example, where the quaternion from the rotation is H_{x}.
H_{y} ⊗ H_{t} – H_{t} ⊗ H_{y} → (H_{x} H_{y}) ⊗ (H_{x} H_{t}) – (H_{x} H_{t}) ⊗ (H_{x} H_{y})
= (H_{z}) ⊗ (H_{x}) – (H_{x}) ⊗ (H_{z})
= – (H_{z} H_{z}) ⊗ (H_{x} H_{z}) + (H_{x} H_{z}) ⊗ (H_{z} H_{z}) [Multiplying first tensor factor by –i and second by i]
= – H_{t} ⊗ H_{y} + H_{y} ⊗ H_{t}
Finally, we need an intertwiner that takes a pair of spinors of the same chirality (both lefthanded or both righthanded) and gives a complex number that’s invariant when we rotate the spinors. An obvious candidate is the inner product of the twospinor state with our rotationallyinvariant state, T_{6}(1).
T_{7}(q_{L1} ⊗ q_{L2}) = < T_{6}(1), q_{L1} ⊗ q_{L2} >_{S⊗S}
= (1/√2) [ < H_{y}, q_{L1} >_{S} < H_{t}, q_{L2} >_{S} – < H_{t}, q_{L1} >_{S} < H_{y}, q_{L2} >_{S} ]
Though the simplification to the form below is laborious when performed with stepbystep transformations, it’s easy to verify. Both expressions are complexlinear in q_{L1} and q_{L2}, and both give identical results when evaluated on four basis vectors: H_{t} ⊗ H_{t}, H_{y} ⊗ H_{y}, H_{y} ⊗ H_{t} and H_{t} ⊗ H_{y}.
T_{7}(q_{L1} ⊗ q_{L2}) = (1/√2) < q_{L2} H_{y}, q_{L1} >_{S}
= (1/√2) [Re_{H}(q_{L1}* q_{L2} H_{y}) + i Re_{H}(q_{L1}* q_{L2} H_{x})]
If we compose these two intertwiners between spin 0 and spin ½ ⊗ spin ½, we get:
T_{7}(T_{6}(z)) = ½ z [ Re_{H}(H_{y}* H_{t} H_{y}) + i Re_{H}(H_{y}* H_{t} H_{x}) – Re_{H}(H_{t}* H_{y} H_{y}) – i Re_{H}(H_{t}* H_{y} H_{x}) ]
= ½ z [ 1 + 0 – (–1) – 0 ]
= z
If we want to describe a process where a spinor emits a photon, we can make use of the following trick: we pretend that our spinor is accompanied by a spin0 particle, which decays into two photons. The spinor absorbs one photon, while the other goes on its way. Since we’re only concerned with spin representations, not energy and momentum, the initial state of a spinor and a spin0 particle transforms just like a spinor alone, and we end up with a photon and a spinor of the opposite chirality, just as if it were the spinor that emitted the photon. We don’t have to take any of this literally; the point is that it gives us a map between the spaces we need with exactly the right transformation properties.
So, we define:
T_{8}(q_{R}) = T '_{1}(T_{5}(1) ⊗ q_{R})
T_{5}(1) gives us a pair of photons that we tensor with our righthanded spinor, then we use a new intertwiner T '_{1} to describe the absorption of either of those photons, leaving the other untouched. We obtain T '_{1} by extending the definition of T_{1} to allow for an extra photon:
T '_{1}(q_{V1} ⊗ q_{V2} ⊗ q_{R}) = ½ (q_{V1} ⊗ (q_{V2} q_{R}) + q_{V2} ⊗ (q_{V1} q_{R}))
Spelling this out, we get:
T_{8}(q_{R}) = ½ T '_{1}((H_{x} ⊗ H_{x} + H_{y} ⊗ H_{y} + H_{z} ⊗ H_{z} + H_{t} ⊗ H_{t}) ⊗ q_{R})
= ½ [H_{x} ⊗ (H_{x} q_{R}) + H_{y} ⊗ (H_{y} q_{R}) + H_{z} ⊗ (H_{z} q_{R}) + H_{t} ⊗ (H_{t} q_{R})]
= ½ Σ_{μ} [H_{μ} ⊗ (H_{μ} q_{R})]
The first space in the final tensor product here is the photon that’s emitted, while the second space is the nowlefthanded spinor.
For example, suppose our original spinor has spin UP, q_{R} = H_{t}. Then the result is:
T_{8}(UP) = ½ [H_{x} ⊗ H_{x} + H_{y} ⊗ H_{y} + H_{z} ⊗ H_{z} + H_{t} ⊗ H_{t}]
= ½ [–i H_{x} ⊗ DOWN – H_{y} ⊗ DOWN + i H_{z} ⊗ UP + H_{t} ⊗ UP]
= ½ [–i (H_{x} – i H_{y}) ⊗ DOWN + i (H_{z} – i H_{t}) ⊗ UP]
The first term here, (H_{x} – i H_{y}) ⊗ DOWN, describes a photon with m_{z}=1 and a spinor with spin DOWN (i.e. m_{z}=–½), so their total zcomponent of spin is ½, the same as the original spinor. In the second term, (H_{z} – i H_{t}) ⊗ UP, the photon part is an eigenstate of J_{z} with an eigenvalue of zero, so again we have conservation of the zcomponent of spin, although this photon polarisation is only possible for a particle in motion.
To describe a Dirac spinor emitting a photon, we combine versions of the construction that use T '_{1} and a similarly defined T '_{2}:
T_{9}(q_{L}, q_{R}) = ½ (T '_{1}((H_{x} ⊗ H_{x} + H_{y} ⊗ H_{y} + H_{z} ⊗ H_{z} + H_{t} ⊗ H_{t}) ⊗ q_{R}), T '_{2}((H_{x} ⊗ H_{x} + H_{y} ⊗ H_{y} + H_{z} ⊗ H_{z} + H_{t} ⊗ H_{t}) ⊗ q_{L}))
= ½ Σ_{μ} [H_{μ} ⊗ (H_{μ} q_{R}, –H_{μ}* q_{L})]
What if we have a Dirac spinor emit a photon with T_{9}, and then feed the result into T_{3}, so that the spinor reabsorbs the photon it emitted?
T_{3}(T_{9}(q_{L}, q_{R})) = ¼ Σ_{μ} (–H_{μ} H_{μ}* q_{L}, –H_{μ}* H_{μ} q_{R}) = –(q_{L}, q_{R})
So we get back the original Dirac spinor, multiplied by minus one.
When an electron and a positron annihilate, the only way that energy and momentum can be conserved is for two photons to be created. In fact, the same criterion applies to a free electron absorbing or emitting a photon: the electron has to scatter an incoming photon, it can’t absorb or emit a single photon.
But in this section we’re only concerned with spin, not energy and momentum, so we’ll go ahead and construct an intertwiner for an electron and a positron coming together and creating a single photon. This is actually a useful thing to have, because this kind of interaction, where just three elementary particles are involved, can be seen as one step within a larger process where energy and momentum are conserved: one vertex in a larger Feynman diagram.
To combine two spinors to make a vector, one spinor must be lefthanded and the other righthanded. We can use two of the intertwiners we’ve constructed previously to meet our needs here: T_{8} to turn a righthanded spinor into a lefthanded spinor and a photon, and then T_{7} to combine the new lefthanded spinor with the original lefthanded spinor to get a spin0 particle — a complex number we can simply multiply into our result.
Recall that:
T_{8}(q_{R}) = ½ Σ_{μ} [H_{μ} ⊗ (H_{μ} q_{R})]
T_{7}(q_{L1} ⊗ q_{L2}) = < (H_{y} ⊗ H_{t} – H_{t} ⊗ H_{y})/√2, q_{L1} ⊗ q_{L2} >_{S⊗S}
Instead of simply combining these intertwiners, though, we’ll create a version that treats the left and righthanded spinors equally, having the initially lefthanded spinor play a role in emitting the photon as well.
T_{10}(q_{L} ⊗ q_{R}) = 1/(2√2) Σ_{μ} < H_{y} ⊗ H_{t} – H_{t} ⊗ H_{y}, q_{L} ⊗ (H_{μ} q_{R}) – q_{R} ⊗ (H_{μ}* q_{L}) >_{S⊗S} H_{μ}
= 1/(2√2) Σ_{μ} [< H_{y}, q_{L} >_{S} < H_{t}, H_{μ} q_{R} >_{S} – < H_{t}, q_{L} >_{S} < H_{y}, H_{μ} q_{R} >_{S}
– < H_{y}, q_{R} >_{S} < H_{t}, H_{μ}* q_{L} >_{S} + < H_{t}, q_{R} >_{S} < H_{y}, H_{μ}* q_{L} >_{S}] H_{μ}
Simplifying the expression above is a lot of work, but we can actually go a long way to anticipating the answer from some basic principles. The only combination of left and righthanded spinors that will transform as a complexified vector quaternion would be:
q_{L} (a + i b) q_{R}*
In order for this expression to be complexlinear, if we rightmultiply either q_{L} or q_{R} with H_{z} the result must be the same as multiplying through by i, and so:
H_{z} b = –b H_{z} = a
H_{z} a = –a H_{z} = –b
Setting b = H_{x} gives a = H_{y} in the first of these equations, and these values satisfy the second equation as well. Any real multiple of the same choices will work, and it’s not hard to verify that we can match our original definition precisely if we set:
T_{10}(q_{L} ⊗ q_{R}) = –(i/√2) q_{L} (H_{x} – i H_{y}) q_{R}*
For example:
T_{10}(UP ⊗ UP) = T_{10}(H_{t} ⊗ H_{t}) = –(i/√2) (H_{x} – i H_{y})
The result is –i times our normalised photon state with m_{z}=1. We also have:
T_{10}(DOWN ⊗ DOWN) = T_{10}((–H_{y}) ⊗ (–H_{y}))
= (i/√2) H_{y} (H_{x} – i H_{y}) H_{y}
= (i/√2) (H_{x} + i H_{y})
So the result is just i times our normalised photon state with m_{z}=–1. For other combinations of spinors we get:
T_{10}(UP ⊗ DOWN) = T_{10}(H_{t} ⊗ (–H_{y}))
= –(i/√2) (H_{x} – i H_{y}) H_{y}
= –(i/√2) (H_{z} + i H_{t})
T_{10}(DOWN ⊗ UP) = T_{10}((–H_{y}) ⊗ H_{t})
= (i/√2) H_{y} (H_{x} – i H_{y})
= –(i/√2) (H_{z} – i H_{t})
In both cases the result is a normalised eigenstate with m_{z}=0. At first glance it might seem like a mistake or a problem that these results give us:
T_{10}((UP ⊗ DOWN – DOWN ⊗ UP)/√2) = H_{t}
since that initial state looks like it ought to be an eigenstate with total spin 0, not 1! But the fact that one spinor is lefthanded and the other righthanded makes all the difference here; the tensor product of the chiral representations is (½,0)⊗(0,½)=(½,½), i.e. the entire space has a total spin of 1.
T_{10} is an intertwiner for two single spinors. For two Dirac particles, we have:
T_{11}((q_{L1}, q_{R1}) ⊗ (q_{L2}, q_{R2})) = –(i/√2) (q_{L1} (H_{x} – i H_{y}) q_{R2}* + q_{L2} (H_{x} – i H_{y}) q_{R1}*)
The reverse of the process we discussed in the last section involves a photon turning into two spinors, one lefthanded and one righthanded. As with our other examples, this is not something a free photon can do while conserving energy and momentum, but we’re still interested in understanding the relationship between the spins.
The trick we’ll use to get this intertwiner is to imagine a spin0 particle decaying into a pair of righthanded spinors, one of which simply goes on its way, while the other one absorbs our photon and turns into a lefthanded spinor. We combine this with a version where we get a pair of lefthanded spinors instead, one of which goes on its way, while the other absorbs the photon and becomes righthanded. As with our other constructions, this isn’t meant to be taken literally; we’re simply exploiting the fact that tensoring our initial state with a spin0 representation has no effect on the way it transforms.
The pairs of spinors (whether both lefthanded or both righthanded) that come from our fictitious spin0 particle are given by the normalised state:
T_{6}(1) = (1/√2) (UP ⊗ DOWN – DOWN ⊗ UP)
= (1/√2) (H_{y} ⊗ H_{t} – H_{t} ⊗ H_{y})
We absorb the photon in processes modelled on our intertwiners T_{1} and T_{2}, leaving a state with one lefthanded spinor and one righthanded spinor:
T_{12}(q_{V}) = (1/(2√2)) ((q_{V} H_{y}) ⊗ H_{t} – (q_{V} H_{t}) ⊗ H_{y} + H_{y} ⊗ (q_{V}* H_{t}) – H_{t} ⊗ (q_{V}* H_{y}) )
Some examples:
T_{12}(Photon with m_{z}=1) = T_{12}((H_{x} – i H_{y})/√2)
= ¼ [((H_{x} – i H_{y}) H_{y}) ⊗ H_{t} – ((H_{x} – i H_{y}) H_{t}) ⊗ H_{y} – H_{y} ⊗ ((H_{x} – i H_{y}) H_{t}) + H_{t} ⊗ ((H_{x} – i H_{y}) H_{y}) ]
= ¼ [(H_{z} + H_{z}) ⊗ H_{t} – (H_{x} – H_{x}) ⊗ H_{y} – H_{y} ⊗ (H_{x} – H_{x}) + H_{t} ⊗ (H_{z} + H_{z})]
= ½ [H_{z} ⊗ H_{t} + H_{t} ⊗ H_{z}]
= i UP ⊗ UP
T_{12}(Photon with m_{z}=–1) = T_{12}((H_{x} + i H_{y})/√2)
= ¼ [((H_{x} + i H_{y}) H_{y}) ⊗ H_{t} – ((H_{x} + i H_{y}) H_{t}) ⊗ H_{y} – H_{y} ⊗ ((H_{x} + i H_{y}) H_{t}) + H_{t} ⊗ ((H_{x} + i H_{y}) H_{y}) ]
= ¼ [(H_{z} – H_{z}) ⊗ H_{t} – (H_{x} + H_{x}) ⊗ H_{y} – H_{y} ⊗ (H_{x} + H_{x}) + H_{t} ⊗ (H_{z} – H_{z})]
= –½ [H_{x} ⊗ H_{y} + H_{y} ⊗ H_{x}]
= –i DOWN ⊗ DOWN
T_{12}(Photon with m_{z}=0) = T_{12}(H_{z})
= (1/(2√2)) [(H_{z} H_{y}) ⊗ H_{t} – (H_{z} H_{t}) ⊗ H_{y} – H_{y} ⊗ (H_{z} H_{t}) + H_{t} ⊗ (H_{z} H_{y}) ]
= –(1/(2√2)) [H_{x} ⊗ H_{t} + H_{z} ⊗ H_{y} + H_{y} ⊗ H_{z} + H_{t} ⊗ H_{x}]
= –(1/(2√2)) [(–i DOWN) ⊗ UP + (i UP) ⊗ (–DOWN) + (–DOWN) ⊗ (i UP) + UP ⊗ (–i DOWN)]
= (i/√2) (DOWN ⊗ UP + UP ⊗ DOWN)
T_{12}(H_{t})
= (1/√2) (H_{y} ⊗ H_{t} – H_{t} ⊗ H_{y})
= (1/√2) (UP ⊗ DOWN – DOWN ⊗ UP)
If we apply the intertwiner T_{10} to any of these specific results, we get back the original photon state. Since they constitute a set of basis vectors, this must be true for any state at all:
T_{10}(T_{12}(q_{V})) = q_{V}
We have constructed intertwiners that we can use to describe:
We should stress, once again, that we’re only looking at the spins of these particles; conserving energy and momentum is a separate issue.
Now, from a completely objective point of view in which we refuse to favour any particular observer, all of these processes are examples of the same thing: an interaction between two spinors and a photon. By rotating our frame of reference, we can turn any one of these processes into any other.
In the case where two observers have the same account of which particles are present before and after the interaction, the rotational invariance of the result can be seen very easily. For example, suppose one observer, Carla, says an electron absorbed a photon, and attributed the state q_{V} to the photon and the Dirac spinor (q_{L1}, q_{R1}) to the electron initially. If she is interested in the amplitude for the electron to end up with the Dirac spinor (q_{L2}, q_{R2}), she will use the intertwiner T_{3} to compute this as:
A = < (q_{L2}, q_{R2}), T_{3}(q_{V} ⊗ (q_{L1}, q_{R1})) >_{D}
= ½ < (q_{L2}, q_{R2}), (q_{V} q_{R1}, q_{V}* q_{L1}) >_{D}
= ½ [ < q_{L2}, q_{V} q_{R1} >_{S} + < q_{R2}, q_{V}* q_{L1} >_{S} ]
Here we have defined an inner product on Dirac spinors, < >_{D}, as the sum of the spinor inner products < >_{S} between the left and right hand spinors.
If another observer, Patrizia, agrees that a photon absorbed an electron, but is using a set of coordinates rotated with respect to Carla’s by a pair of unit quaternions (g, h), she will attribute different spin states to the particles:
q'_{V} = g q_{V} h^{–1}
(q'_{L1}, q'_{R1}) = (g q_{L1}, h q_{R1})
(q'_{L2}, q'_{R2}) = (g q_{L2}, h q_{R2})
Using the same intertwiner, but applying it to her own versions of the spin states, Patrizia will predict an amplitude that’s exactly the same as Carla’s:
A' = ½ [ < q'_{L2}, q'_{V} q'_{R1} >_{S} + < q'_{R2}, q'_{V}* q'_{L1} >_{S} ]
= ½ [ < g q_{L2}, g q_{V} h^{–1} h q_{R1} >_{S} + < h q_{R2}, h q_{V}* g^{–1} g q_{L1} >_{S} ]
= ½ [ < g q_{L2}, g q_{V} q_{R1} >_{S} + < h q_{R2}, h q_{V}* q_{L1} >_{S} ]
= ½ [ < q_{L2}, q_{V} q_{R1} >_{S} + < q_{R2}, q_{V}* q_{L1} >_{S} ]
= A
To go from the thirdlast line to the secondlast, we use the fact that the spinor inner product is invariant when both arguments are multiplied on the left by the same unit quaternion, which we established previously.
The situation becomes trickier, though, when different observers disagree over which particles were present before the interaction and which are present after. To compare the measurements and predictions of such observers, not only do we need to switch from one of our four intertwiners to another, we need to do more than transform the particles’ spin states with the usual representations.
Why? Classically, spin depends on the direction that an object is rotating as you watch it move forwards in time. Just as there is no absolute energymomentum vector, there is no absolute spin vector (or tensor); an object spinning counterclockwise around the zaxis as seen from above, by one observer, will be seen to be spinning clockwise by an observer with the opposite arrow of time.
However, we can talk objectively about the direction in which an object is spinning as it approaches a particular interaction. This will agree with the direction of spin for an observer who thinks of the object as being present before the interaction, but will be the opposite to the direction of spin for an observer who thinks of it as being present after the interaction. If we agree to express all spins in these terms, then (rather like the observerindependent approach to conservation of energymomentum), we can see that angular momentum is conserved at every interaction.
Quantum mechanically, we can reverse the spins of our quaternionic spinors with the rev_{S} function we described previously:
rev_{S}(q) = q H_{y}
For the complexified vector quaternions we use for photon states, to reverse the spin we simply conjugate them in the complexstructure sense:
rev_{V}(p + q i) = con(p + q i) = p – q i
Once we’ve spinreversed any of the states that we originally measured leaving the interaction rather than approaching it, we should find the tensor product of all three states to have a total spin of 0. We can quantify this with an intertwiner from the tensor product to spin 0, which we can construct by imagining the two spinors annihilating, and then taking the dot product between the resulting photon and the original:
T_{vert}(q_{V} ⊗ q_{L} ⊗ q_{R}) = T_{4}(q_{V} ⊗ T_{10}(q_{L} ⊗ q_{R})) = –i/(2√2) q_{V} · (q_{L} (H_{x} – i H_{y}) q_{R}*)
The intertwiner T_{vert} as we’ve constructed it is normalised in the sense that the sum, over any orthonormal basis of the tensor product space, of the squared magnitudes of the results is 1. These results can be thought of as inner products between a single normalised vector sitting in the tensor product space, an eigenstate of total spin with an eigenvalue of j=0, and whatever tripleparticle state we feed to the intertwiner. However, we get some extra factors when we compare T_{vert} with our other intertwiners:
< q_{L}, T_{1}(q_{V} ⊗ q_{R}) >_{S} = (√2) T_{vert}(q_{V} ⊗ rev_{S}(q_{L}) ⊗ q_{R}) [Righthanded spinor absorbs photon to become lefthanded]
< q_{R}, T_{2}(q_{V} ⊗ q_{L}) >_{S} = (√2) T_{vert}(q_{V} ⊗ q_{L} ⊗ rev_{S}(q_{R})) [Lefthanded spinor absorbs photon to become righthanded]
< q_{V} ⊗ q_{L}, T_{8}(q_{R}) >_{V⊗S} = (√2) T_{vert}(rev_{V}(q_{V}) ⊗ rev_{S}(q_{L}) ⊗ q_{R}) [Righthanded spinor emits photon and becomes lefthanded]
< q_{V} ⊗ q_{R}, T_{8L}(q_{L}) >_{V⊗S} = (√2) T_{vert}(rev_{V}(q_{V}) ⊗ q_{L} ⊗ rev_{S}(q_{R})) [Lefthanded spinor emits photon and becomes righthanded]
< q_{V}, T_{10}(q_{L} ⊗ q_{R}) >_{V} = 2 T_{vert}(rev_{V}(q_{V}) ⊗ q_{L} ⊗ q_{R}) [Left and righthanded spinors annihilate to create photon]
< q_{L} ⊗ q_{R}, T_{12}(q_{V}) >_{S⊗S} = 2 T_{vert}(q_{V} ⊗ rev_{S}(q_{L}) ⊗ rev_{S}(q_{R})) [Photon turns into pair of left and righthanded spinors]
[We didn’t bother to explicitly construct what we’ve called T_{8L} here, a version of T_{8} that starts with a lefthanded spinor, but the precise definition can be seen in the righthanded component of T_{9} for a full Dirac spinor, at the end of the section on electrons emitting photons.]
The factors of √2 and 2 appear because we normalised these various intertwiners according to which particle(s) are present initially. For example, we normalised T_{1} and T_{2} to give us the amplitude for the initial state to have a total spin of ½ and for the final state to be a particular spinor. Since the dimension of the subspace of the initial state space with a total spin of ½ is 2, if we sum over orthonormal bases for all the initial and final states we get a total for all the squared magnitudes of 2. But since T_{vert} gives amplitudes for the total spin to be 0, and only a 1dimensional subspace of the initial state space meets that condition, the same kind of sum yields just 1. The factor of √2 is the square root of the ratio between the two.
To extract something completely observerindependent from this, suppose one observer has seen an interaction that definitely involved a lefthanded spinor, a righthanded spinor, and a photon. They measured the states of all three particles, and then applied the spinreversal function to any states that they saw as outgoing. Given the bases used for the three measurements, and the actual results for any two of the particles, we ought to be able to compute the relative probabilities for all the basis elements for the third particle, regardless of whether the original observer saw it as an incoming particle or an outgoing one.
We compute the relative probabilities as T_{vert}(q_{V} ⊗ q_{L} ⊗ q_{R})^{2}, holding two of the states fixed at their known results while setting the third state to each element of the measurement basis in turn. We can then divide through by the sum of these values to get true probabilities. Because any overall factors cancel out in the normalisation, carrying out the same procedure with any of the six intertwiners listed above will give exactly the same probabilities as we get from T_{vert}. All the differences between the intertwiners apart from the overall normalisation are dealt with by the spin reversals.
Now that we’ve seen how to treat both vectors and spinors as quaternions, it’s not hard to convert other definitions and equations in Riemannian quantum electrodynamics into quaternionic form. As one simple example, we’ll look at the interaction term in the Riemannian QED Lagrangian:
L_{inter} = –e A_{μ} ψ† γ^{μ} ψ
In quaternionic terms, and working in the Weyl basis, we have:
ψ  =  (q_{L} v_{0}, q_{R} v_{0})  
ψ†  =  (v_{0}† q_{L}*, v_{0}† q_{R}*)  
q_{V}  =  A_{μ} H_{μ}  
γ^{μ}  = 

Here v_{0} is a unit vector in C^{2} that we use to map C^{2} to the quaternions; our usual choice is v_{0}=(0,1) to be compatible with rightmultiplication by H_{z} as the complex structure on the quaternions. We’re using † for the conjugatetranspose of complex vectors and matrices, and the * we use for quaternionic conjugation is really the same if we think of quaternions as 2×2 complex matrices.
So we have:
L_{inter} = –e A_{μ} [ v_{0}† q_{L}* H_{μ} q_{R} v_{0} + v_{0}† q_{R}* H_{μ}* q_{L} v_{0} ]
= –e v_{0}† [ q_{L}* q_{V} q_{R} + q_{R}* q_{V}* q_{L} ] v_{0}
The quantity in brackets in the last line above is a pure real number as a quaternion, since it’s equal to its own quaternionic conjugate, so as a 2×2 matrix it’s just a multiple of the identity matrix. So we have:
L_{inter} = –e (v_{0}† v_{0}) [ q_{L}* q_{V} q_{R} + q_{R}* q_{V}* q_{L} ]
= –e [ q_{L}* q_{V} q_{R} + q_{R}* q_{V}* q_{L} ]
It’s not hard to see that this is invariant under the usual transformations, given a pair of unit quaternions (g, h) in SU(2)×SU(2):
q_{V} → g q_{V} h^{–1}
q_{L} → g q_{L}
q_{R} → h q_{R}
You might be wondering why we can’t use this function, L_{inter}, as an intertwiner from photonandtwospinor states to spin 0. But the Lagrangian is a real number, and in this context q_{V} is just an ordinary quaternion, not a complexified quaternion. The intertwiner T_{vert} that we already constructed is complexlinear in all three spin states. It is unique (up to an overall constant factor), and we can’t use the Lagrangian in its place.
As a check on our result, suppose that the left and righthanded spinor quaternions actually satisfy the quaternionic Dirac equation for a plane wave with energymomentum vector Q_{V} = k^{μ} H_{μ}, which is:
Q_{V} q_{R} – m q_{L} = 0
We can write the interaction energy as the opposite of the Lagrangian, and then make the substitution q_{L} = Q_{V} q_{R} / m:
H_{inter} = –L_{inter}
= e [ q_{L}* q_{V} q_{R} + q_{R}* q_{V}* q_{L} ]
= (e/m) [ q_{R}* Q_{V}* q_{V} q_{R} + q_{R}* q_{V}* Q_{V} q_{R} ]
= (e/m) q_{R}* [ Q_{V}* q_{V} + q_{V}* Q_{V} ] q_{R}
= (q_{R}* q_{R}) (e/m) [ Q_{V}* q_{V} + q_{V}* Q_{V} ]
= ½ (e/m) [ k^{μ} H_{μ}* A_{ν} H_{ν} + A_{ν} H_{ν}* k^{μ} H_{μ} ]
= ½ (e/m) k^{μ} A_{ν} [ H_{μ}* H_{ν} + H_{ν}* H_{μ} ]
= ½ (e/m) k^{μ} A_{ν} (2 δ^{ν}_{μ})
= (e/m) k^{μ} A_{μ}
where in going from the fourth line to the fifth we have used the fact that the expression in brackets is real (being the sum of Q_{V}* q_{V} and its conjugate) and hence commutes with q_{R}, and we then use q_{R}* q_{R} = q_{R}^{2} = ½, which is the appropriate normalisation if we want ψ†ψ = 1.
This result is obviously invariant, and in a frame where the electron is at rest and k = m e_{t} it becomes:
H_{inter} = e A_{t} = –e V
where V is the potential energy. We expect a minus sign, since true, relativistic energy in the Riemannian universe is the opposite of potential energy.
We’ll begin our treatment of Riemannian quantum field theory by summarising some results from the nonrelativistic quantum mechanics of a simple harmonic oscillator.
In fact, we’ll start with the classical mechanics of a simple harmonic oscillator. In classical physics, suppose we have a point particle of mass M, free to move in one dimension described by the coordinate x, and subject to a force F = –K x. This force, linear in the distance of the particle from the origin, corresponds to a potential energy of V = ½ K x^{2}, an “energy valley” shaped like a parabola. The particle will oscillate back and forth in this valley with an angular frequency ω=√(K/M).
We can write the Hamiltonian for this system, the energy expressed in terms of the coordinate x and the associated momentum p, as:
H_{SHO} = p^{2} / (2M) + ½ M ω^{2} x^{2}
where we’ve written M ω^{2} in place of the “spring constant” K since the two quantities are identical, and ω will be of more interest to us than K.
If we put a dot above a quantity to represent its time derivative, Hamilton’s equations for this system are:
p^{•} = –∂_{x}H_{SHO}
∂_{t} p = –M ω^{2} x
x^{•} = ∂_{p}H_{SHO}
∂_{t} x = p / M
Combined, these give us:
∂_{t}^{2}x = –ω^{2} x
which has the general solution:
x(t) = A cos(ωt) + B sin(ωt)
If we describe the system instead with a Lagrangian, the kinetic energy minus the potential energy, expressed in terms of the coordinate x and its rate of change with time x^{•}:
L_{SHO} = ½ M (x^{•}^{2} – ω^{2} x^{2})
then the EulerLagrange equation of motion gives us:
∂_{t} (∂_{x•}L_{SHO}) = ∂_{x}L_{SHO}
∂_{t} (M x^{•}) = –M ω^{2} x
∂_{t}^{2}x = –ω^{2} x
which is of course the same equation as we obtained from the Hamiltonian.
When we analyse the same system in quantum mechanics, the numbers x and p that describe the particle’s position and momentum are replaced by “observables”: selfadjoint operators on a Hilbert space. One way to think of that Hilbert space is as a space of complexvalued wave functions ψ(x, t). The number x is replaced by the operator that consists of multiplying a wave function by x, while the number p is replaced by the differential operator –i ∂_{x}. [We’re doing nonrelativistic quantum mechanics here, so we don’t change the sign of the momentum operator the way we do with Riemannian relativistic QM.] Requiring the system to be described by a particular Hamiltonian then amounts to requiring that the wave function satisfies the Schrödinger equation:
i ∂_{t} ψ(x, t) = H ψ(x, t)
= (p^{2} / (2M) + ½ M ω^{2} x^{2}) ψ(x, t)
= –1 / (2M) ∂_{x}^{2}ψ(x, t) + ½ M ω^{2} x^{2} ψ(x, t)
But rather than going ahead and finding solutions to this differential equation, it’s more useful for our present purposes if we simply think of the Hilbert space for our quantised oscillator as being spanned by vectors with definite energies, or eigenvectors of the Hamiltonian, without worrying about the specifics of the corresponding wave functions. A more detailed treatment can be found in most introductory quantum mechanics textbooks^{[2]}.
First, we note that the commutator between the momentum and position operators is given by:
[x, p] = x p – p x = i
This is easily checked in the wave function representation of the operators, simply by noting that for any function f of x:
[x, p] f = [x, –i ∂_{x}] f
= x (–i ∂_{x} f ) – (–i ∂_{x} (x f ))
= –i x ∂_{x} f + i (f + x ∂_{x} f )
= i f
From x and p we can construct two new operators, known as ladder operators:
a = i (p – i M ω x) / √(2 M ω)
a† = –i (p + i M ω x) / √(2 M ω)
Here a† is the Hermitian adjoint of a, and p and x are selfadjoint, so finding a† from a just amounts to changing the sign of every i in the first definition. From the commutator between x and p it’s not hard to show that:
[a, a†] = a a† – a† a = 1
From this, it’s easy to derive a number of results that we’ll make use of repeatedly:
a a† = a† a + 1
a† a a† = a† (a† a + 1)
a† a = a a† – 1
a† a a = (a a† – 1) a = a (a† a – 1)
We can express the Hamiltonian operator very simply in terms of the ladder operators:
H_{SHO} = p^{2} / (2M) + ½ M ω^{2} x^{2}
= ω (a†a + ½)
Now, suppose ψ_{E} is any eigenvector of the Hamiltonian, with:
H_{SHO} ψ_{E} = E ψ_{E}
It then follows that:
H_{SHO} (a† ψ_{E})
= ω (a†a + ½) (a† ψ_{E})
= a† ω (a†a + ½ + 1) ψ_{E}
= a† H_{SHO} ψ_{E} + ω (a† ψ_{E})
= (E + ω) (a† ψ_{E})
So a† turns any eigenvector of the Hamiltonian into another eigenvector with an energy that is greater by ω. Similarly:
H_{SHO} (a ψ_{E})
= ω (a†a + ½) (a ψ_{E})
= a ω (a†a + ½ – 1) ψ_{E}
= a H_{SHO} ψ_{E} – ω (a ψ_{E})
= (E – ω) (a ψ_{E})
The operator a turns any eigenvector of the Hamiltonian into another eigenvector with an energy that is less by ω. But that process can’t go on forever, because the energy eigenvalues must be nonnegative. So there must be some state ψ_{0} such that:
a ψ_{0} = 0
This is the “ground state” of the harmonic oscillator. And despite the subscript we’ve used, the energy of this ground state won’t be zero. Rather:
H_{SHO} ψ_{0} = ω (a†a + ½) ψ_{0} = (ω/2) ψ_{0}
So the energy of ψ_{0} is ω/2, and there’s an infinite succession of higherenergy eigenstates, with their energies equal to integerplusahalf multiples of ω:
ψ_{n} = (a†)^{n} ψ_{0}
H_{SHO} ψ_{n} = (n + ½) ω ψ_{n}
When the oscillator is in an energy eigenstate — that is, when it possesses a definite energy — that energy can be thought of as the ground state energy, ω/2, plus a certain number n of additional “quanta”, each with energy ω. (If you’re used to thinking of quanta as having energy ℏ ω, remember that we’re using units where ℏ=1.) The operator a can be thought of as “annihilating” one quantum, because it turns an energy eigenstate into another with one less quantum. Similarly, the operator a† can be thought of as “creating” one quantum. So we will call a an annihilation operator and a† a creation operator. But although these terms are nice and evocative, they can’t be taken literally: these operators do not describe a physical process that takes place over time, they’re just an easy way to map states with definite numbers of quanta to other states with one less or one more.
Now, while everything we’ve done here describes a nonrelativistic point particle undergoing simple harmonic motion, exactly the same analysis can be applied whenever the degrees of freedom of a system are governed by the same kind of Hamiltonian: a sum of terms proportional to the square of a coordinate and the square of the corresponding momentum. Since we’ll be making extensive use of these results, we summarise them in the following table.
Simple Harmonic Oscillator  
M is the mass of the particle ω is the frequency of oscillation We’re using units such that ℏ = 1  
H_{SHO}  =  p^{2} / (2M) + ½ M ω^{2} x^{2}  (Hamiltonian) 
L_{SHO}  =  ½ M (x^{•}^{2} – ω^{2} x^{2})  (Lagrangian) 
∂_{t}^{2}x  =  –ω^{2} x  (Equation of Motion) 
a  =  i (p – i M ω x) / √(2 M ω)  (Annihilation Operator) 
a†  =  –i (p + i M ω x) / √(2 M ω)  (Creation Operator) 
Observables expressed in terms of creation and annihilation operators  
H_{SHO}  =  ω (a†a + ½)  (Hamiltonian) 
x  =  (a + a†) / √(2 M ω)  (Coordinate) 
p  =  –i √(M ω / 2) (a – a†)  (Momentum) 
Commutators  
[x, p]  =  i  
[a, a†]  =  1 
The first classical field we’ll try to “quantise” will consist of complexvalued solutions of the Riemannian Scalar Wave equation. In fact, the “Riemannian Scalar Wave equation” is the Riemannian version of what’s usually known as the KleinGordon equation, and virtually every textbook on quantum field theory includes a discussion of the quantisation of the Lorentzian version, either with realvalued solutions^{[3]} or complexvalued solutions^{[4]}.
As we’ve discussed previously, the only way to tame the Riemannian wave equations and prevent exponential solutions is to assume that the Riemannian universe is finite in all four dimensions: for example a 4torus, T^{4}, or a 4sphere, S^{4}. For simplicity, we will assume a flat 4torus, but our results will not be tied too much to any specific shape or topology. We described some aspects of electromagnetism in such a universe in this section.
If there are to be any solutions at all to the free (that is, sourceless) RSW equation, the dimensions of the 4torus L_{x}, L_{y}, L_{z} and L_{t} must have a special relationship with the maximum frequency, ν_{max}, such that there are integers n_{x}, n_{y}, n_{z} and n_{t} satisfying:
(n_{x} / L_{x})^{2} + (n_{y} / L_{y})^{2} + (n_{z} / L_{z})^{2} + (n_{t} / L_{t})^{2} = ν_{max}^{2}
This allows an integral number of cycles of the wave to occur along each of the dimensions, with the sum of the squares of all the frequencies equal to the required value. For generic values of the dimensions of the torus and ν_{max} there will be no such solutions, but if the dimensions are chosen so that there are solutions, and the universe is extremely large compared to the minimum wavelength, there will be a large but finite number of different integer solutions.
We won’t go into the number theory involved in counting such solutions; we’ll simply assume that there are so many that for most purposes their discrete nature is not experimentally detectable, and also that the directions of the plane waves with the corresponding frequencies are uniformly distributed in all directions. In other words, the possible propagation fourvectors k for the free wave are a finite, discrete set scattered uniformly on a 3sphere in R^{4}, with radius ω_{m}. We’ll exploit the fact that the set is discrete to make some formulas and calculations simpler, but we’ll also feel free to approximate sums over these vectors with integrals when that’s useful.
We’ll call the set of possible propagation fourvectors for the free wave K_{4}, and write N_{modes} for the number of vectors in this set. We’ll choose a particular coordinate across the torus to be our time coordinate, and give the name K_{3} to the set of projections of the vectors in K_{4} into the three dimensions excluding t. For each k in K_{3} there will be two vectors in K_{4} that project to it, one with a positive time component and one with a negative time component. We’ll call the subset of K_{4} with positive time components K_{4}^{+} and the subset with negative time components K_{4}^{–}, and for the sake of simplicitly we’ll assume that there are no modes with a time component of zero. We will refer to the absolute value of the time component as E_{k}:
E_{k} = k^{t}, for k in K_{4}
E_{k} = √(ω_{m}^{2} – k^{2}) = √(m^{2} – k^{2}), for k in K_{3}
Here m is the mass of the particle associated with the scalar field, and in our units m = ℏ ω_{m} = ω_{m} = 2π ν_{max}.
To start with, we will write the Riemannian Scalar Wave equation and the corresponding Lagrangian and Hamiltonian in two different forms: the first in terms of a function φ(x, t) of the space and time coordinates, and the second in terms of the (timedependent) coefficients φ_{k}(t) of a mode expansion over certain periodic functions of space.
Classical Riemannian Scalar Wave Equation  
Fourspace coordinate form, complexvalued wave  
0  =  ∂_{μ} ∂^{μ} φ(x, t) + m^{2} φ(x, t)  (Equation of Motion) 
L_{RSW}  =  ∫ (∂_{μ} φ(x, t)) (∂^{μ} φ^{*}(x, t)) – m^{2} φ(x, t) φ^{*}(x, t) d^{3}x  (Lagrangian) 
Π(x, t)  =  ∂_{t} φ^{*}(x, t)  
Π^{*}(x, t)  =  ∂_{t} φ(x, t)  (Momenta) 
H_{RSW}  =  ∫ Π(x, t) Π^{*}(x, t) – (∂_{i} φ(x, t)) (∂^{i} φ^{*}(x, t)) + m^{2} φ(x, t) φ^{*}(x, t) d^{3}x  (Hamiltonian) 
Mode expansion form k is in K_{3}  
f_{k}(x)  =  √(1/V) exp(–i k · x)  
φ(x, t)  =  Σ_{k in K3} φ_{k}(t) f_{k}(x)  (Mode Expansion) 
∂_{t}^{2} φ_{k}(t)  =  –E_{k}^{2} φ_{k}(t)  (Equation of Motion) 
L_{RSW}  =  Σ_{k in K3}( ∂_{t} φ_{k}(t) ∂_{t} φ^{*}_{k}(t) – E_{k}^{2} φ_{k}(t) φ^{*}_{k}(t) )  (Lagrangian) 
Π_{k}(t)  =  ∂_{t} φ^{*}_{k}(t)  
Π^{*}_{k}(t)  =  ∂_{t} φ_{k}(t)  (Momenta) 
H_{RSW}  =  Σ_{k in K3}( Π_{k}(t) Π^{*}_{k}(t) + E_{k}^{2} φ_{k}(t) φ^{*}_{k}(t) )  (Hamiltonian) 
In the first set of equations, we are treating φ(x, t) and its complex conjugate φ^{*}(x, t) as two independent fields, and the Lagrangian density L_{RSW}(x) (the quantity we integrate over space to get the total Lagrangian) yields the equation of motion directly via the EulerLagrange equation for the complex conjugate field:
∂^{μ} (∂_{∂μ φ*(x, t)} L_{RSW}(x)) = ∂_{φ*(x, t)}L_{RSW}(x)
∂^{μ} ∂_{μ} φ(x, t) = – m^{2} φ(x, t)
The momenta associated with the field and its complex conjugate are defined as the derivative of the Lagrangian density with respect to the time derivative of those two fields:
Π(x, t) = ∂_{∂t φ(x, t)} L_{RSW}(x) = ∂_{t} φ^{*}(x, t)
Π^{*}(x, t) = ∂_{∂t φ*(x, t)} L_{RSW}(x) = ∂_{t} φ(x, t)
The Hamiltonian density is found by taking the sum over the two fields, φ(x, t) and its complex conjugate, of the product of the momentum and the time derivative of the field, and then subtracting the Lagrangian density:
H_{RSW}(x) = Π(x, t) ∂_{t} φ(x, t) + Π(x, t)^{*} ∂_{t} φ(x, t)^{*} – L_{RSW}(x)
= 2 Π(x, t) Π(x, t)^{*} – L_{RSW}(x)
In the formula for the Hamiltonian shown in the table, the repeated index i is summed over space coordinates only, with the time derivatives accounted for by the momenta.
Next, we write φ(x, t) as a finite sum of periodic functions of space, f_{k}(x), proportional to exp(–i k · x) for each k in K_{3}. We call the coefficients in this expansion φ_{k}(t). We normalise the f_{k} with a factor √(1/V), where V is the volume of space across the fourtorus, V = L_{x} L_{y} L_{z}, so that:
∫ f_{k}(x) f_{q}^{*}(x) d^{3}x = δ_{k,q}
The new Lagrangian in terms of the mode coefficients φ_{k}(t) and their complex conjugates is found by substituting the mode expansion into the original integral for L_{RSW}, then making use of the orthogonality of the mode functions. The sum of the squares of the spatial derivatives for each mode is just k^{2}, which combines with the m^{2} term to give E_{k}^{2}. The formulas for the momenta and the Hamiltonian can then be found from the Lagrangian.
Each term in the mode coefficient version of the Hamiltonian almost matches the form of the Hamiltonian of a simple harmonic oscillator, but these degrees of freedom φ_{k}(t) and φ^{*}_{k}(t) and their associated momenta are complex numbers, whereas the position and momentum of a simple harmonic oscillator are real numbers. So we need to translate the Hamiltonian into a function of realvalued degrees of freedom. This isn’t hard: each term is a product of a complex number and its conjugate, which equals the absolute value of that complex number squared, which equals the sum of the squares of the real part and the imaginary part (where we construe the imaginary part as real, i.e. we divide out the factor of i). So we can rephrase everything in terms of:
Re_{k}(t) = Re φ_{k}(t)
Im_{k}(t) = [ Im φ_{k}(t) ] / i
After doing this first in the Lagrangian we can compute the associated momenta, then we end up with a Hamiltonian that takes precisely the harmonic oscillator form, with two real degrees of freedom for every mode.
Classical Riemannian Scalar Wave Equation  
Realvalued degrees of freedom: Re_{k}(t) and Im_{k}(t)
k is in K_{3}  
φ_{k}(t)  =  Re_{k}(t) + i Im_{k}(t)  (Real Variables) 
∂_{t}^{2} Re_{k}(t)  =  –E_{k}^{2} Re_{k}(t)  
∂_{t}^{2} Im_{k}(t)  =  –E_{k}^{2} Im_{k}(t)  (Equations of Motion) 
L_{RSW}  =  Σ_{k in K3}( (∂_{t} Re_{k}(t))^{2} – E_{k}^{2} Re_{k}(t)^{2} + (∂_{t} Im_{k}(t))^{2} – E_{k}^{2} Im_{k}(t)^{2} )  (Lagrangian) 
Π_{Re k}(t)  =  2 ∂_{t} Re_{k}(t)  
Π_{Im k}(t)  =  2 ∂_{t} Im_{k}(t)  (Momenta) 
H_{RSW}  =  Σ_{k in K3}( ¼ Π_{Re k}(t)^{2} + E_{k}^{2} Re_{k}(t)^{2} + ¼ Π_{Im k}(t)^{2} + E_{k}^{2} Im_{k}(t)^{2} )  (Hamiltonian) 
If we compare each term in the Hamiltonian in the last line of this table with the Hamiltonian for the simple harmonic oscillator, we see that we can make the two match up precisely by putting the harmonic oscillator particle’s mass M equal to 2 (this M has nothing to do with the mass of the particle we’re describing with our scalar field), putting the harmonic oscillator frequency ω equal to E_{k} for each of our modes, and identifying the harmonic oscillator’s coordinate x with either Re_{k} or Im_{k} and the harmonic oscillator’s momentum p with either Π_{Re k} or Π_{Im k}. We then define creation and annihilation operators for every k for both the real and imaginary degrees of freedom.
Quantised Riemannian Scalar Field  
Realvalued degrees of freedom: Re_{k} and Im_{k}
k is in K_{3}  
a_{Re k}  =  i / (2 √(E_{k})) (Π_{Re k} – 2 i E_{k} Re_{k})  
a_{Im k}  =  i / (2 √(E_{k})) (Π_{Im k} – 2 i E_{k} Im_{k})  (Annihilation Operators) 
a_{Re k}†  =  –i / (2 √(E_{k})) (Π_{Re k} + 2 i E_{k} Re_{k})  
a_{Im k}†  =  –i / (2 √(E_{k})) (Π_{Im k} + 2 i E_{k} Im_{k})  (Creation Operators) 
Observables expressed in terms of creation and annihilation operators  
H_{RSW}  =  Σ_{k in K3}( E_{k} (a_{Re k}†a_{Re k} + a_{Im k}†a_{Im k} + 1) )  (Hamiltonian) 
Re_{k}  =  (a_{Re k} + a_{Re k}†) / (2 √(E_{k}))  
Im_{k}  =  (a_{Im k} + a_{Im k}†) / (2 √(E_{k}))  (Coordinates) 
Π_{Re k}  =  –i √(E_{k}) (a_{Re k} – a_{Re k}†)  
Π_{Im k}  =  –i √(E_{k}) (a_{Im k} – a_{Im k}†)  (Momenta) 
Commutators  
[Re_{k}, Π_{Re q}]  =  i δ_{k,q}  
[a_{Re k}, a_{Re q}†]  =  δ_{k,q}  
[Im_{k}, Π_{Im q}]  =  i δ_{k,q}  
[a_{Im k}, a_{Im q}†]  =  δ_{k,q} 
The creation and annihilation operators here work in exactly the same way as those for the simple harmonic oscillator: the creation operators a_{Re k}† and a_{Im k}† turn eigenstates of the Hamiltonian with energy E into new eigenstates with energy E+E_{k}, while the annihilation operators a_{Re k} and a_{Im k} turn eigenstates with energy E into new eigenstates with energy E–E_{k}. We define the ground state ψ_{0} of the system to be the vector such that:
a_{Re k} ψ_{0} = 0
a_{Im k} ψ_{0} = 0
for every k in K_{3}. Combining these equations with the Hamiltonian, we see that the energy of ψ_{0} is:
E_{0} = Σ_{k in K3} E_{k}
≈ [½N_{modes} / π^{2}] ∫_{γ=0}^{π/2} m cos(γ) 4 π sin(γ)^{2} dγ
= 2 N_{modes} m / (3 π)
where the approximation comes from treating the ½N_{modes} vectors in K_{3} as the projection of a continuum of vectors uniformly distributed on the t > 0 half of a threesphere of radius m. In the Lorentzian case, E_{0} is infinite and needs to be subtracted out by defining it as the zero of a new energy scale; this sounds questionable, but since differences in energy, rather than actual values, are what matters in most of quantum mechanics it works well enough. In our case we don’t really need to do that, and whether or not we include E_{0} when describing the energies of quantum states, the specific value it takes will show up in some formulas in its own right.
The separate modes of the field — for the various vectors k in K_{3} and for the real and imaginary parts — act just like completely independent harmonic oscillators. In quantum mechanics, independent systems are described by a Hilbert space that is the tensor product of the Hilbert spaces for the individual systems, so we can think of the Hilbert space for our field as a tensor product of N_{modes} copies of the harmonic oscillator Hilbert space. All the operators that are indexed with Re k or Im k act only on their particular factor of the whole Hilbert space, and act as the identity on all the other factors. Nothing in any mode messes with what’s going on in any other mode.
That’s very unrealistic, of course, because the electromagnetic and Diractype fields we ultimately want to deal with are coupled to each other. But treating a sourceless field is a useful warmup exercise before tackling harder problems.
Having found degrees of freedom that we could quantise by analogy with the harmonic oscillator, we’re now going to translate everything back to our original, complexvalued degrees of freedom, which are easier to interpret physically. The relationship between the real and complex degrees of freedom is already fixed; that doesn’t quite tell us how to define creation and annihilation operators for the complexvalued modes, but we can check that the choices here give the correct commutators and put the Hamiltonian into the right form. (Though it might have been more straightforward to put a_{k} = (1/√2) (a_{Re k} + i a_{Im k}) and b_{k} = (1/√2) (a_{Re k} – i a_{Im k}) rather than the versions below in which the righthand sides have been multiplied by an extra factor of –i, these phases have been chosen to give field variables that agree with Weinberg^{[4]} for ease of comparison.)
Quantised Riemannian Scalar Field  
Complexvalued degrees of freedom: φ_{k} and φ_{k}†
k is in K_{3}  
a_{k}  =  (1/√2) (a_{Im k} – i a_{Re k})  
=  1 / √(2E_{k}) (Π_{k}† – i E_{k} φ_{k})  
b_{k}  =  –(1/√2) (a_{Im k} + i a_{Re k})  
=  1 / √(2E_{k}) (Π_{k} – i E_{k} φ_{k}†)  (Annihilation Operators)  
a_{k}†  =  (1/√2) (a_{Im k}† + i a_{Re k}†)  
=  1 / √(2E_{k}) (Π_{k} + i E_{k} φ_{k}†)  
b_{k}†  =  –(1/√2) (a_{Im k}† – i a_{Re k}†)  
=  1 / √(2E_{k}) (Π_{k}† + i E_{k} φ_{k})  (Creation Operators)  
Observables expressed in terms of creation and annihilation operators  
H_{RSW}  =  Σ_{k in K3}( E_{k} (a_{k}†a_{k} + b_{k}†b_{k} + 1) )  (Hamiltonian) 
φ_{k}  =  i (a_{k} – b_{k}†) / √(2E_{k})  
φ_{k}†  =  –i (a_{k}† – b_{k}) / √(2E_{k})  (Coordinates) 
Π_{k}  =  √(E_{k}/2) (a_{k}† + b_{k})  
Π_{k}†  =  √(E_{k}/2) (a_{k} + b_{k}†)  (Momenta) 
Commutators  
[φ_{k}, Π_{q}]  =  i δ_{k,q}  
[a_{k}, a_{q}†]  =  δ_{k,q}  
[φ_{k}†, Π†_{q}]  =  i δ_{k,q}  
[b_{k}, b_{q}†]  =  δ_{k,q} 
Now φ_{k} and Π_{k} are nonHermitian operators on our Hilbert space, rather than complexvalued functions of time, and the operators associated with the complex conjugates of those functions are the Hermitian adjoints φ_{k}† and Π_{k}†. We normally think of observables as being Hermitian operators, but that’s only true if they’re measuring realvalued quantities, and these nonHermitian operators are perfectly good observables for the complexvalued mode coefficients we introduced for the classical field.
In place of the operators for the two kinds of realvalued degrees of freedom, we now have annihilation operators we’ve named a_{k} and b_{k}, and their Hermitian adjoints as the creation operators. The Hamiltonian takes exactly the same form in terms of these new operators as it did for the old ones, so again we will have a ground state ψ_{0} with the energy E_{0}. And as before, operating on an energy eigenstate with either of the creation operators a_{k}† or b_{k}† will give a new eigenstate with its energy increased by E_{k}.
What is the physical meaning of the states we get when we add a quantum of energy to the ground state with a_{k}† or b_{k}†? To answer that, we need to look back to the Lagrangian for the classical field, expressed in terms of the complex mode coefficients:
L_{RSW} = Σ_{k in K3}( ∂_{t} φ_{k}(t) ∂_{t} φ^{*}_{k}(t) – E_{k}^{2} φ_{k}(t) φ^{*}_{k}(t) )
If we multiply every φ_{k} by a phase, exp(–i α), each φ^{*}_{k} is multiplied by exp(i α) and the Lagrangian itself is unchanged. So uniformly rotating the phase of the entire field is a symmetry of the Lagrangian. Any symmetry of this kind gives us a conserved quantity, and the version of Noether’s Theorem that applies to discrete variables tells us that the quantity is:
C = Σ_{k in K3}( (∂_{∂t φk} L_{RSW}) (∂_{α} exp(–i α) φ_{k}(t))_{α=0} + (∂_{∂t φ*k} L_{RSW}) (∂_{α} exp(i α) φ^{*}_{k}(t))_{α=0} )
= –i Σ_{k in K3}( (∂_{t} φ^{*}_{k}(t)) φ_{k}(t) – (∂_{t} φ_{k}(t)) φ^{*}_{k}(t) )
= –i Σ_{k in K3}( Π_{k}(t) φ_{k}(t) – Π^{*}_{k}(t) φ^{*}_{k}(t) )
If we translate C into an operator expressed in terms of the creation and annihilation operators, we get:
C = Σ_{k in K3}( a_{k}† a_{k} – b_{k}†b_{k} )
Using an argument much like that we used to find the energy eigenstates, it’s easy to show that a_{k}† a_{k} counts the number of quanta that have been added to the ground state with a_{k}†, i.e.
(a_{k}† a_{k}) ψ_{0} = 0
(a_{k}† a_{k}) (a_{k}†)^{n} ψ_{0} = n (a_{k}†)^{n} ψ_{0}
Similarly, the operator b_{k}†b_{k} counts the number of quanta that have been added to the ground state with b_{k}†:
(b_{k}† b_{k}) (b_{k}†)^{n} ψ_{0} = n (b_{k}†)^{n} ψ_{0}
The difference of the two, summed over all k, is thus the total number of quanta created with a† operators minus the total number created with b† operators. This number is conserved; we can check that this holds true in the quantum version by checking that C commutes with the Hamiltonian. (In fact, the two number operators are individually conserved, but that’s because we’re dealing with a free field theory where nothing much happens. The difference of the number operators, arising as it does from a uniform change in the phase of φ, will be conserved even in an interacting field theory, so long as the interaction term has factors of both φ and φ^{*}.)
What this is telling us is that the a_{k}† operators are creating particles and the b_{k}† operators the corresponding antiparticles, in the sense that these two kinds of quanta have a “charge” associated with them that is opposite in the two cases, and whose net value remains unchanged over time. If φ was coupled to the electromagnetic field this would be an electric charge, but at this point it’s just a conserved quantum number with no wider meaning. For the most general energy eigenstate ψ, we have:
ψ = (a_{k1}†)^{q1} (a_{k2}†)^{q2} (a_{k3}†)^{q3} ... (b_{k1}†)^{n1} (b_{k2}†)^{n2} (b_{k3}†)^{n3} ... ψ_{0}
H ψ = (E_{0} + (q_{1} + n_{1}) E_{k1} + (q_{2} + n_{2}) E_{k2} + (q_{3} + n_{3}) E_{k3} + ... ) ψ
C ψ = ((q_{1} + q_{2} + q_{3} + ...) – (n_{1} + n_{2} + n_{3} + ... )) ψ
In retrospect, we can now see what kind of quanta the operators a_{Re k}† and a_{Im k}† were creating: they were equal combinations of the particle and antiparticle states, with no net charge!
In the Lorentzian universe, observers in relative motion will agree about the classification of states into particles and antiparticles, but that’s not true in the Riemannian universe. Suppose k is a threevector such that E_{k} is quite small; this means the fourvector k = (E_{k}, k) will be pointing in an almost spatial direction, in the frame in which these quantities are being measured, and it won’t take much relative motion for another observer to give it a negative time component.
We can write things in a less framebound way by defining an annihilation operator indexed by the fourvectors in K_{4}:
c_{(E, k)}  = 

with the corresponding creation operator equal to its Hermitian adjoint:
c_{(E, k)}†  = 

Whether the state c_{(E, k)}† ψ_{0} is seen as a particle or an antiparticle in the frame in which we’re working, if we look at another physical system related to the original one by an arbitrary fourspace rotation R it will be described by the state c_{R (E, k)}† ψ_{0}.
In terms of these operators, by indexing everything with fourvectors in K_{4} we can encompass both particle and antiparticle states in one sweep. The modes and momenta where we previously specified Hermitian adjoints separately can now be found by negating the fourvector index, with φ_{(–Ek, –k)} in our new definition taking the place of φ_{k}†.
Quantised Riemannian Scalar Field  
Complexvalued degrees of freedom: φ_{k} k is now a fourvector from K_{4}  
c_{k}  =  1 / √(2E_{k}) (Π_{–k} – i E_{k} φ_{k})  (Annihilation Operators) 
c_{k}†  =  1 / √(2E_{k}) (Π_{k} + i E_{k} φ_{–k})  (Creation Operators) 
Observables expressed in terms of creation and annihilation operators  
φ_{k}  =  i (c_{k} – c_{–k}†) / √(2E_{k})  (Coordinates) 
Π_{k}  =  √(E_{k}/2) (c_{k}† + c_{–k})  (Momenta) 
H_{RSW}  =  Σ_{k in K3}( Π_{k} Π†_{k} + E_{k}^{2} φ_{k} φ†_{k} )  
=  Σ_{k in K4}( E_{k} (c_{k}†c_{k} + ½) )  (Hamiltonian)  
Commutators  
[φ_{k}, Π_{q}]  =  i δ_{k,q}  
[c_{k}, c_{q}†]  =  δ_{k,q} 
Now that we have operators for the variables associated with the modes, let’s see what their expectation values are for some states of the quantised field. In order to calculate expectation values we’ll need to know the inner products between various states, and we’ll start by assuming that the ground state ψ_{0} is normalised:
< ψ_{0}, ψ_{0} > = 1
What about the state ψ_{q} = c_{q}† ψ_{0}, where we’ve put one quantum into the field for some mode q in K_{4}?
< ψ_{q}, ψ_{q} >
= < c_{q}† ψ_{0}, c_{q}† ψ_{0} >
= < ψ_{0}, c_{q} c_{q}† ψ_{0} >
= < ψ_{0}, (c_{q}† c_{q} + 1) ψ_{0} >
= < ψ_{0}, ψ_{0} >
= 1
Here we’ve made use of a general property of adjoints and inner products, < A x, y > = < x, A† y >, the commutator between the annihilation and creation operators, and the fact that any annihilation operator applied to the ground state gives zero.
For the moment we’re going to gloss over the fact that either these states (in the Schrödinger picture), or the operators (in the Heisenberg picture) should be seen as varying over time; in the cases we’re looking at here the time variation amounts to an oscillating phase that has no effect on the expectation values.
The expectation value of φ_{k} for the state ψ_{q} is:
< ψ_{q}, φ_{k} ψ_{q} >
= < c_{q}† ψ_{0}, φ_{k} c_{q}† ψ_{0} >
= < ψ_{0}, c_{q} φ_{k} c_{q}† ψ_{0} >
= [i / √(2E_{k})] < ψ_{0}, c_{q} (c_{k} – c_{–k}†) c_{q}† ψ_{0} >
= 0
This must be zero, because no odd number of creation and annihilation operators can bring the ground state back to the ground state. What’s more, it shouldn’t surprise us that this expectation value is zero, because it’s a bit like asking for the mean position of the particle in a simple harmonic oscillator, which will of course be x = 0. But the position squared should have a nonzero average, or in the case of this complex field amplitude, the expectation value of φ_{k}† φ_{k} (which we can also write as φ_{–k} φ_{k}) should be nonzero.
We’ll compute this first for the ground state. Note that φ_{k} and φ_{k}† commute, so the order in which we write these operators makes no difference.
< ψ_{0}, φ_{–k} φ_{k} ψ_{0} >
= < ψ_{0}, (i (c_{–k} – c_{k}†) / √(2E_{k})) (i (c_{k} – c_{–k}†) / √(2E_{k})) ψ_{0} >
= –< ψ_{0}, (c_{–k} – c_{k}†) (c_{k} – c_{–k}†) ψ_{0} > / (2E_{k})
= 1 / (2E_{k})
Now for the state ψ_{q}:
< ψ_{q}, φ_{–k} φ_{k} ψ_{q} >
= < ψ_{0}, c_{q} φ_{–k} φ_{k} c_{q}† ψ_{0} >
= –< ψ_{0}, c_{q} (c_{–k} – c_{k}†) (c_{k} – c_{–k}†) c_{q}† ψ_{0} > / (2E_{k})
= (1 + δ_{q, k} + δ_{q, –k}) / (2E_{k})
The squared momenta associated with the modes have expectation values that differ from these only by a factor of E_{k}^{2}.
< ψ_{0}, Π_{–k} Π_{k} ψ_{0} > = E_{k} / 2
< ψ_{q}, Π_{–k} Π_{k} ψ_{q} > = [E_{k} / 2] (1 + δ_{q, k} + δ_{q, –k})
What we’re seeing here is that these squaredamplitude and squaredmomentum observables are sensitive to groundstate energy: they will give us nonzero expectation values even when the state of the field has no quanta in the mode the observable is tuned to. And energy eigenstates such as ψ_{q} are highly nonclassical: the energy is sharply defined at the expense of a completely indeterminate phase for the mode.
Though we won’t go into the details, it’s worth briefly sketching how a semiclassical state for the field could be constructed. In the simple harmonic oscillator, you can take the ground state wave function — a Gaussian wave packet — and displace its centre any distance you like from the centre of the energy well. The translated wave now contains components from an infinite number of energy eigenfunctions, but as they evolve over time the wave packet simply moves back and forth in the well, without changing its shape, very much like a classical particle ^{[2]}.
We can do something similar with one of the modes of our complex scalar field, and if we work with the real and imaginary parts it’s easy to mimic the harmonic oscillator construction. By setting up Gaussian wave packets in the real and imaginary parts that are oscillating 90 degrees out of phase, the result is a complex mode amplitude that rotates in the complex plane: a twodimensional Gaussian circling the origin at a constant distance. For a large enough squared amplitude, this state will resemble a classical wave with a welldefined phase.
In Lorentzian quantum field theory, the normal development of the subject involves writing the field φ, its Hermitian adjoint, and their conjugate momenta as operatorvalued functions of the space coordinates, using the original spatial modes and the quantised mode coefficient operators. That can be done in the Riemannian case too, but the result has to be treated with caution – because φ(x) defined that way will involve a sum over K_{3}, making it dependent on the observer’s choice of a t coordinate. The Lorentzian equivalent, which is an integral over all propagation vectors on the positive mass hyperboloid (the set of timelike vectors such that k · k = –m^{2} and k^{t} > 0), is only tied to a choice of an overall sign for the time coordinate, and transforms sensibly between observers in relative motion. But a Riemannian field φ(x) is locked to a specific choice of time axis, because observers with even slightly different time axes will disagree as to what is a particle and what is an antiparticle.
There is another problem that complicates the Riemannian case, and we can see this even in the classical version. Ideally, we’d like the change of variables between the field at each point in space and the mode coefficients to be a canonical transformation. In Hamiltonian mechanics, we define the Poisson bracket as a differential operator on pairs of functions:
{f, g}_{PB} = (∂_{qj} f ) (∂_{pj} g) – (∂_{pj} f ) (∂_{qj} g)
where we’re summing over the repeated index j. The coordinates q_{i} and momenta p_{i} have the Poisson brackets:
{q_{i}, q_{k}}_{PB} = 0
{p_{i}, p_{k}}_{PB} = 0
{q_{i}, p_{k}}_{PB} = δ_{ik}
For continuous degrees of freedom such as field values, the discrete indices are replaced by spatial coordinates, the sums by integrals, and the Kronecker delta is replaced by a Dirac delta function. In a canonical transformation between variables, these equations should hold for the new set of coordinates and momenta when the Poisson brackets are defined in terms of the original variables^{[5]}. Explicitly, if the new coordinates are Q_{i} and the new momenta P_{i}, we should have:
{Q_{i}, Q_{k}}_{PB} = (∂_{qj} Q_{i}) (∂_{pj} Q_{k}) – (∂_{pj} Q_{i}) (∂_{qj} Q_{k}) = 0
{P_{i}, P_{k}}_{PB} = (∂_{qj} P_{i}) (∂_{pj} P_{k}) – (∂_{pj} P_{i}) (∂_{qj} P_{k}) = 0
{Q_{i}, P_{k}}_{PB} = (∂_{qj} Q_{i}) (∂_{pj} P_{k}) – (∂_{pj} Q_{i}) (∂_{qj} P_{k}) = δ_{ik}
Things get a little trickier when one set of variables are continuous and the other discrete, but in some cases everything still works perfectly. For example, if we were expanding over the countably infinite set of Fourier modes of the field that met periodic boundary conditions on space alone, with suitable normalisation we would get a Kronecker delta in these equations when we changed from spatial coordinates to mode coefficients (with the sum over the index j replaced by an integral over the spatial coordinates), and a Dirac delta function when we performed the reverse transformation.
In the present case, though, what do we get? The new coordinates and momenta in terms of the old are:
φ(x, t) = Σ_{k in K3} φ_{k}(t) f_{k}(x)
Π(x, t) = ∂_{t} φ^{*}(x, t) = Σ_{k in K3} (∂_{t} φ^{*}_{k}(t)) f_{k}^{*}(x) = Σ_{k in K3} Π_{k}(t) f_{k}^{*}(x)
so we have the partial derivatives between old and new variables (where for brevity we’ll now suppress the explicit timedependence):
∂_{φk} φ(x) = f_{k}(x)
∂_{Πk} Π(x) = f_{k}^{*}(x)
and the Poisson bracket between coordinates and momenta:
{φ(x), Π(y)}_{PB}
= Σ_{k in K3}( (∂_{φk} φ(x)) (∂_{Πk} Π(y)) – (∂_{Πk} φ(x)) (∂_{φk} Π(y)) )
= Σ_{k in K3} f_{k}(x) f_{k}^{*}(y)
= (1/V) Σ_{k in K3} exp(i k · (y – x))
= (1/V) Σ_{k in K4+} exp(i k · (y – x))
≈ [N_{modes}/(2 π^{2} V)] ∫_{u over S3+} exp(i m u · (y – x))
where in the secondlast line we’ve used the fact that x and y are purely spatial, so it makes no difference whether we take their dot product with some k in K_{3} or the k in K_{4}^{+} that projects to it, and in the last line we approximate the sum over the vectors in K_{4}^{+} as an integral over the positive half of the unit threesphere in R^{4}. We can exploit the fact that this integral will be invariant under rotations of threespace to choose coordinates such that y – x = x – y e_{z}. In polar coordinates (γ, θ, φ) on S^{3+}, we have 0 < γ < π/2, 0 < θ < π, and 0 < φ < 2π, the measure is sin(γ)^{2} sin(θ), and the unit fourvector u is:
u = (cos(γ), sin(γ) sin(θ) cos(φ), sin(γ) sin(θ) sin(φ), sin(γ) cos(θ))
So the dot product we need is:
u · (x – y e_{z}) = x – y sin(γ) cos(θ)
Since the integrand is independent of the azimuthal coordinate φ (not to be mistaken for the field amplitude), we can integrate that out immediately, leaving:
{φ(x), Π(y)}_{PB}
≈ [N_{modes}/(π V)] ∫_{θ=0}^{π} ∫_{γ=0}^{π/2} exp(i m x – y sin(γ) cos(θ)) sin(γ)^{2} sin(θ) dγ dθ
= [2 N_{modes}/(π V)] [1 / (m x – y)] ∫_{γ=0}^{π/2} sin(m x – y sin(γ)) sin(γ) dγ
= [N_{modes}/V] J_{1}(m x – y) / (m x – y)
Here J_{1} is a Bessel function of the first kind. J_{1} is roughly cyclic, and is zero at the origin, but when we divide through by its argument we get a function that peaks at the origin and then dies away, while oscillating. The first zero occurs when m x – y ≈ 4, and when we recall that m = ω_{m} we see that the width of the peak is of the same order as the minimum wavelength associated with the RSW equation. But although this function has a peak, it certainly isn’t a Dirac delta function, so we do not have a canonical transformation of variables.
The problem is that the modes available to us can’t be combined to produce a perfectly localised field. This has nothing to do with quantum mechanics, and it isn’t because the set of modes is finite; even when we pretend we have a continuum of modes with propagation vectors spread across half a threesphere, there is still a minimum wavelength to those modes, and we can’t expect to build a function from them with an arbitrarily narrow peak.
So instead of trying to deal with the field’s value at every single point, it makes more sense to find a new basis for the space spanned by the mode functions f_{k}(x), consisting of new functions that are as localised as possible, given that they belong to that space.
Let’s define a set of functions g_{χ}(x) as:
g_{χ}(x) = Σ_{k in K3} √(E_{k}/E_{0}) f_{k}(x–χ)
= Σ_{k in K3} √(E_{k}/E_{0}) exp(i k · χ) f_{k}(x)
where χ for the moment is any threevector. Any g_{χ} is a linear combination of the mode functions f_{k}, so unlike a Dirac delta function it lies in the subspace of functions we can actually construct. We can approximate g_{0} using an integral:
g_{0}(x) = Σ_{k in K3} √(E_{k}/E_{0}) f_{k}(x)
≈ [N_{modes}/(2 π^{2} √(V E_{0}))] ∫_{u over S3+} exp(–i m u · x) √(m u^{t})
= √[3 N_{modes}/(2 π V)] ∫_{θ=0}^{π} ∫_{γ=0}^{π/2} exp(–i m x sin(γ) cos(θ)) sin(γ)^{2} (√cos(γ)) sin(θ) dγ dθ
= √[6 N_{modes}/(π V)] / (mx) ∫_{γ=0}^{π/2} sin(m x sin(γ)) (sin(γ) √(cos(γ)) dγ
= √[3 N_{modes}/V][Γ(3/4)/2^{1/4}] J_{5/4}(m x) / (m x)^{5/4}
This Bessel function divided by a power of its argument gives a very similar shape to the function we found for the Poisson bracket, with a peak whose width is of the same order as the minimum wavelength. So it’s a reasonably localised function, and we wouldn’t expect to be able to do much better with any linear combination of the modes.
Now, the function g_{χ}(x) = g_{0}(x – χ) just translates this peak from the origin to another location, so all functions of this kind will be equally localised in space. We will now assume that it’s possible to choose a set of ½N_{modes} different points χ such that all the g_{χ} are mutually orthogonal; we’ll call that choice of points Χ. If we normalise these functions, they will then comprise an alternative orthonormal basis for the function space spanned by the f_{k}.
What if that assumption isn’t true? Well, we do know that if we chose any distinct ½N_{modes} points χ, the g_{χ} would still give us a basis for the function space. We could then turn that basis into an orthonormal basis with the GramSchmidt orthogonalisation process. The resulting functions would no longer all take the form g_{χ}(x), but if we’d scattered our points uniformly in space the result should still be a set of localised functions qualitatively very much like the g_{χ}.
Since the f_{k} certainly are orthogonal, the assumption of orthogonality for the g_{χ} is equivalent to:
< g_{χ1}, g_{χ2}> = Σ_{k in K3} (E_{k}/E_{0}) exp(i k · (χ_{2} – χ_{1})) = δ_{χ1, χ2} [Σ_{k in K3} (E_{k}/E_{0})] = δ_{χ1, χ2}
In place of φ(x), the field at an arbitrary point, we will write φ_{χ} for the coefficient of g_{χ}(x) in the field:
Σ_{k in K3} φ_{k} f_{k}(x) = Σ_{χ in Χ} φ_{χ} g_{χ}(x)
Using the orthonormality of the g_{χ}, we can extract one coefficient by taking the inner product with the corresponding g_{χ}, giving us:
φ_{χ} = Σ_{k in K3} √(E_{k}/E_{0}) exp(–i k · χ) φ_{k}
The inverse formula is:
φ_{k} = Σ_{χ in Χ} √(E_{k}/E_{0}) exp(i k · χ) φ_{χ}
From these formulas and the orthogonality assumption, it’s not hard to show that the transformation of the classical variables from the mode coefficients φ_{k} to the spatial coefficients φ_{χ} is a canonical transformation:
{φ_{χ1}, Π_{χ2}}_{PB} = δ_{χ1, χ2}
We can now write the operators corresponding to these new variables, simply by using the operator form of the φ_{k}.
Quantised Riemannian Scalar Field Schrödinger picture  
The vector χ is a spatial threevector from the set Χ. The operators are unchanging over time, and apply to a timedependent state vector.  
φ_{χ}  =  Σ_{k in K3} √(E_{k}/E_{0}) exp(–i k · χ) φ_{k}  
=  i/√(2E_{0}) Σ_{k in K4+} (c_{k} – c_{–k}†) exp(–i k · χ)  
φ_{χ}†  =  Σ_{k in K3} √(E_{k}/E_{0}) exp(i k · χ) φ_{k}†  
=  i/√(2E_{0}) Σ_{k in K4–} (c_{k} – c_{–k}†) exp(–i k · χ)  
Inverse formula for planewave mode coefficients.  
φ_{k}  =  Σ_{χ in Χ} √(E_{k}/E_{0}) exp(i k · χ) φ_{χ}  
Orthogonality assumption for χ in Χ.  
Σ_{k in K3} E_{k} exp(i k · (χ_{2} – χ_{1}))  =  δ_{χ1, χ2} E_{0} 
In writing the sums indexed by fourvectors in K_{4}^{+} and K_{4}^{–}, we’ve made use of the fact that the dot product of the purely spatial vector χ with these fourvectors is the same as the dot product with their projection in K_{3}, and in writing the second form for the Hermitian adjoint φ_{χ}†, we’ve changed the indexing vector from k in K_{4}^{+} to its opposite in K_{4}^{–}.
You’ll note that the factor of 1/√E_{k} that’s present in the operators for the mode coefficients cancels out in these spatially localised operators. The reason for inserting a factor of √E_{k} into the definition of the functions g_{χ} was precisely to achieve this cancellation! The equivalent formulas in the Lorentzian theory are integrals over all spatial threevectors, and they need the factor of 1/√E_{k} in order to ensure that certain products turn out to be Lorentzinvariant — a factor of 1/E_{k} being exactly what’s needed to make an integral over spatial vectors into a Lorentzinvariant integral over the mass hyperboloid. But our sum is over fourvectors that are already uniformly distributed over half the threesphere, so we don’t want any functions of the framedependent energy hanging around making it harder to formulate SO(4)invariant quantities.
This field of operators is defined in the Schrödinger picture, where some operators are considered to be fixed for all time (like the operators x and p in our original harmonic oscillator), and the state vector changes over time. But in quantum field theory it’s more common to use the Heisenberg picture, where the state vector is taken to be fixed for all time, and all operators become functions of time.
To make the switch to the Heisenberg picture, if the state ψ is to be fixed at ψ(0), operators need to change with time according to:
A(t) = exp(i H t) A(0) exp(–i H t)
Now, let’s consider the specific case of an annihilation operator a_{k} and our Hamiltonian H_{RSW}. Apart from the term E_{k} a_{k}†a_{k}, the operator a_{k} commutes with everything else in H_{RSW}, so we have:
[a_{k}, H_{RSW}] = [a_{k}, E_{k} a_{k}†a_{k}] = E_{k} [a_{k}, a_{k}†] a_{k}
a_{k} H_{RSW} – H_{RSW} a_{k} = E_{k} a_{k}
which gives us:
H_{RSW} a_{k} = a_{k} (H_{RSW} – E_{k})
H_{RSW}^{n} a_{k} = a_{k} (H_{RSW} – E_{k})^{n}
Treating the exponential of the Hamiltonian as a power series, this tells us that:
exp(i H_{RSW} t) a_{k} = a_{k} exp(i (H_{RSW} – E_{k}) t)
exp(i H_{RSW} t) a_{k} exp(–i H_{RSW} t) = a_{k} exp(–i E_{k} t)
Taking the Hermitian adjoint of both sides of this equation yields the transformation property for the corresponding creation operator:
exp(i H_{RSW} t) a_{k}† exp(–i H_{RSW} t) = a_{k}† exp(i E_{k} t)
Obviously the b_{k} and b_{k}† transform in the same way. All our c operators are just a or b renamed, and so long as we are clear that E_{k} is the absolute value of the time component associated with a fourvector k in K_{4}, the same rules apply to c_{k} and c_{k}† as well. Using these transformation rules, we can write our field operators and the associated momenta in the Heisenberg picture, as functions of both space and time.
Quantised Riemannian Scalar Field Heisenberg picture  
The vector χ is a spatial threevector from the set Χ. The operators are timedependent, and apply to an unchanging state vector.  
φ_{χ}(t)  =  i/√(2E_{0}) Σ_{k in K4+} (c_{k} exp(–i E_{k} t) – c_{–k}† exp(i E_{k} t)) exp(–i k · χ)  
φ_{χ}†(t)  =  i/√(2E_{0}) Σ_{k in K4–} (c_{k} exp(–i E_{k} t) – c_{–k}† exp(i E_{k} t)) exp(–i k · χ)  (Coordinates) 
Π_{χ}(t)  =  ∂_{t} φ_{χ}†(t)  
=  1/√(2E_{0}) Σ_{k in K4–} E_{k} (c_{k} exp(–i E_{k} t) + c_{–k}† exp(i E_{k} t)) exp(–i k · χ)  
=  1/√(2E_{0}) Σ_{k in K4+} E_{k} (c_{–k} exp(–i E_{k} t) + c_{k}† exp(i E_{k} t)) exp(i k · χ)  
Π_{χ}†(t)  =  ∂_{t} φ_{χ}(t)  
=  1/√(2E_{0}) Σ_{k in K4+} E_{k} (c_{k} exp(–i E_{k} t) + c_{–k}† exp(i E_{k} t)) exp(–i k · χ)  
=  1/√(2E_{0}) Σ_{k in K4–} E_{k} (c_{–k} exp(–i E_{k} t) + c_{k}† exp(i E_{k} t)) exp(i k · χ)  (Momenta) 
Let’s look at some expectation values for the operators φ_{χ}(t). Since they are linear combinations of the mode amplitudes they will have some properties in common with them: the expectation value of each operator itself on states such as ψ_{0} and ψ_{q} will be zero. So once again we’ll look at the squared amplitude, which we get from φ_{χ}†(t) φ_{χ}(t).
< ψ_{0}, φ_{χ}†(t) φ_{χ}(t) ψ_{0} >
= N_{modes} / (4E_{0})
≈ 3π / (8m)
So as with the mode amplitudes, these spatial coefficient operators are sensitive to the groundstate energy. For a state with one quantum in mode q, we have:
< ψ_{q}, φ_{χ}†(t) φ_{χ}(t) ψ_{q} >
= (N_{modes} + 2) / (4E_{0})
≈ 3π (1 + 2/N_{modes}) / (8m)
So the spatial coefficient is completely blind to which singlequantum planewave mode is present, and can barely distinguish these states from the ground state.
None of that’s too surprising; what we expect these spatial operators to be sensitive to is the difference between spatially localised states. We could construct an orthonormal set of spatially localised states by mimicking the way we constructed the spatially localised modes g_{χ} from the planewave modes f_{k}, defining the states ψ_{χ} by:
Maybe ψ_{χ} = Σ_{k in K3} √(E_{k}/E_{0}) exp(i k · χ) ψ_{k}?
However, these states don’t give simple expectation values for the squared spatial coefficient operators, and there’s no easy way to construct SO(4)invariant quantities from them. But it turns out that we can get some useful localised states by applying the spatial coefficient operators themselves to the ground state:
ψ^{(+)}_{χ, t} = 2 √(E_{0}/N_{modes}) φ_{χ}†(t) ψ_{0} = –i √(2/N_{modes}) Σ_{k in K4+} exp(i (E_{k} t + k · χ)) ψ_{k}
ψ^{(–)}_{χ, t} = 2 √(E_{0}/N_{modes}) φ_{χ}(t) ψ_{0} = –i √(2/N_{modes}) Σ_{k in K4–} exp(i (E_{k} t + k · χ)) ψ_{k}
The intention here is that the state is localised around the threevector χ at the specific time t, and can be expected to spread out spatially at other times; the superscript (±) specifies a particle or antiparticle.
Let’s compute the inner products between the particle states:
< ψ^{(+)}_{χ1, t1}, ψ^{(+)}_{χ2, t2} >
= (4E_{0}/N_{modes}) < ψ_{0}, φ_{χ1}(t_{1}) φ_{χ2}†(t_{2}) ψ_{0} >
= (2/N_{modes}) Σ_{k in K4+} exp(i (E_{k} (t_{2}–t_{1}) + k · (χ_{2} – χ_{1})))
= (2/N_{modes}) Σ_{k in K4+} exp(i (k^{t} (t_{2}–t_{1}) + k_{(3)} · (χ_{2} – χ_{1})))
= (2/N_{modes}) Σ_{k in K4+} exp(i k · δ)
where we have defined the fourvector:
δ = (t_{2}–t_{1}, χ_{2} – χ_{1})
The states for different χ and/or t are not orthogonal, but the inner product will peak when the location and time are identical and undergo a J_{1}(m χ_{2} – χ_{1}) / (m χ_{2} – χ_{1}) falloff as the separation in space increases.
Between the antiparticle states we have:
< ψ^{(–)}_{χ1, t1}, ψ^{(–)}_{χ2, t2} >
= (4E_{0}/N_{modes}) < ψ_{0}, φ_{χ1}†(t_{1}) φ_{χ2}(t_{2}) ψ_{0} >
= (2/N_{modes}) Σ_{k in K4–} exp(i (E_{k} (t_{2}–t_{1}) + k · (χ_{2} – χ_{1})))
= (2/N_{modes}) Σ_{k in K4–} exp(i ((–k^{t}) (t_{2}–t_{1}) + k_{(3)} · (χ_{2} – χ_{1})))
= (2/N_{modes}) Σ_{k in K4–} exp(i ((–k^{t}) (t_{2}–t_{1}) – k_{(3)} · (χ_{2} – χ_{1})))
= (2/N_{modes}) Σ_{k in K4–} exp(–i k · δ)
In going between the thirdlast and secondlast lines here, we’re using the fact that for every fourvector in K_{4}^{–}, the vector with the opposite spatial part and the same time component is also in K_{4}^{–}. This result is identical to the inner product between particle states.
All the particle and antiparticle states are orthogonal to each other:
< ψ^{(+)}_{χ2, t2}, ψ^{(–)}_{χ1, t1} > = 0
One of the most important things to know in quantum field theory is the propagator for the field, which gives the amplitude for finding a particle at a certain time and place, say t_{2} and x_{2}, given that it is definitely present at x_{1} at time t_{1}. In Lorentzian QFT, this is done in a manner that clearly distinguishes between particles and antiparticles, and can be phrased in a way that respects the usual arrow of time. For example, the Feynman propagator for the events (t_{1}, x_{1}) and (t_{2}, x_{2}) is defined so that if t_{2} > t_{1} it is the amplitude for a particle to go from (t_{1}, x_{1}) to (t_{2}, x_{2}), but if t_{2} < t_{1} it is the amplitude for an antiparticle to go from (t_{2}, x_{2}) to (t_{1}, x_{1}). In both cases, the amplitude is found by integrating propagation vectors over the positive mass hyperboloid to cover all the different plane waves that could carry the particle or antiparticle from the earlier time to the later. This amplitude is Lorentzinvariant, if you restrict yourself to orthochronous Lorentz transformations: those in which two observers can agree which mass hyperboloid is the positive one, because they agree on the overall positive direction for time.
In Riemannian QFT, the distinction between particles and antiparticles depends on the observer’s state of motion. What’s more, in a T^{4} universe, in which time is cyclic, it’s meaningless to say that one time comes before another. However, if we drop the part in the description of the Feynman propagator where we switch between two options depending on which event comes first and instead simply add the amplitude for a particle at Event 1 to be detected at Event 2 to the amplitude for an antiparticle at Event 2 to be detected at Event 1, we obtain an SO(4)invariant quantity:
½ (< ψ^{(+)}_{χ1, t1}, ψ^{(+)}_{χ2, t2} > + < ψ^{(–)}_{χ2, t2}, ψ^{(–)}_{χ1, t1} >)
= (2E_{0}/N_{modes}) < ψ_{0}, (φ_{χ1}(t_{1}) φ_{χ2}†(t_{2}) + φ_{χ2}†(t_{2}) φ_{χ1}(t_{1})) ψ_{0} >
= (1/N_{modes}) Σ_{k in K4} exp(i k · δ)
≈ 2 J_{1}(m δ) / (m δ), where δ = (t_{2}–t_{1}, χ_{2} – χ_{1})
Clearly this is SO(4)invariant: it is a function solely of the length of the fourvector δ, a quantity that all observers will agree on. And if we view χ_{1} and t_{1} as fixed and treat χ_{2} and t_{2} as our coordinates, this is a solution of the sourceless RSW equation with fourrotational symmetry around (t_{1}, χ_{1}).
Of course the Bessel function approximation is just that: an approximation. It has the advantage of perfect SO(4)invariance and ease of computation, but it doesn’t satisfy the T^{4} boundary conditions, so if you go around the torus in different directions you’ll end up with different values. In practice, using the shortest fourdimensional distance between any two events will give the appropriate answer.
If we swap our definitions of “particle” and “antiparticle” in the propagator, this amounts to swapping the order of the terms in both inner products, and hence taking the complex conjugate of the whole expression. But since the final value is a real number, this makes no difference. For the same reason, swapping the labels on the two events also makes no difference.
The diagram on the left illustrates what the propagator is measuring. The two events are connected by plane waves with propagation vectors pointing in every possible direction, and there is no invariant meaning to the classification of these waves into particle or antiparticle states.
However, for a given observer we can describe those waves whose propagation vectors point backwards in time as antiparticles travelling forwards in time, and those waves whose propagation vectors point forwards in time as particles, again travelling forwards in time.
Crucially, though, we always include both possibilities. If we followed the Lorentzian prescription for the Feynman propagator and did not include the antiparticle states in this example because some observer had classified Event 1 as coming before Event 2, then even another observer who agreed on that ordering would select a different halfthreesphere of particle states (represented by the red arrows here), and would end up with a different amplitude.
Suppose we repeat the initial steps of the construction we performed when we prepared to quantise a complex scalar field, but in place of the Riemannian scalar wave equation we use the Riemannian Dirac equation. In our earlier discussion of the Dirac equation, we were treating it as describing a quantum wave function, but for our present purposes we will initially pretend that we’re studying a “classical” Dirac field, where the field consists of four complex components and obeys the Dirac equation, but has nothing to do with quantum mechanics. The fields that satisfy this equation form a representation of the Euclidean group; we actually derived the Dirac equation on those terms, in one of the sections on group representations. So initially we’re just mimicking, for this new representation, what we did for the scalar representation previously.
To keep the notation simple, we’ll again refer to the field’s maximum angular frequency (soon to become the particle mass) as m, and the set of possible propagation fourvectors compatible with that mass and the dimensions of the T^{4} universe as K_{4}, with K_{3} as the projections of the vectors in K_{4} to three dimensions and E_{k} as the unique positive energy associated with any k in K_{3}:
E_{k} = √(m^{2} – k^{2})
If we were describing a situation with more than one kind of field, we’d need to distinguish between the different versions of all these things pertaining to each field, since of course there’s no reason why they’d be the same in the various cases.
To expand our Dirac field ψ in Fourier modes, we will make use of the following solutions of the Dirac equation, which generalise the plane wave solutions we discussed earlier:
Plane Wave Solutions of the Riemannian Dirac Equation in the Weyl Basis  
k is in K_{3} {ξ_{s}} is an orthonormal basis of C^{2} H_{μ} are the unit quaternions as matrices (defined here) Sum k^{j}H_{j} ranges over spatial indices x, y, z  
Positive frequency solutions  
ψ_{+, k, s}(x, t)  =  u(k, s) exp(–i (k · x + E_{k}t))  (Plane wave) 
u(k, s)  =  ( [(m+E_{k})H_{t} + k^{j}H_{j}] ξ_{s}, [(m+E_{k})H_{t} – k^{j}H_{j}] ξ_{s}) / (2 √[m(m+E_{k})])  
u(k, s_{1})† u(k, s_{2})  =  δ_{s1, s2}  
u(k, s_{1})† γ^{t} u(k, s_{2})  =  δ_{s1, s2} E_{k} / m  
Negative frequency solutions  
ψ_{–, k, s}(x, t)  =  v(k, s) exp(+i (k · x + E_{k}t))  (Plane wave) 
v(k, s)  =  ( [(m+E_{k})H_{t} + k^{j}H_{j}] ξ_{s}, –[(m+E_{k})H_{t} – k^{j}H_{j}] ξ_{s}) / (2 √[m(m+E_{k})])  
v(k, s_{1})† v(k, s_{2})  =  δ_{s1, s2}  
v(k, s_{1})† γ^{t} v(k, s_{2})  =  –δ_{s1, s2} E_{k} / m  
Products of u and v spinors  
u(k, s_{1})† v(k, s_{2})  =  0  
v(k, s_{1})† u(k, s_{2})  =  0  
u(k, s_{1})† γ^{t} v(–k, s_{2})  =  0  
v(–k, s_{1})† γ^{t} u(k, s_{2})  =  0 
If you’ve studied the normal, Lorentzian versions of all this, the orthogonality relations will look a bit strange! In Lorentzian QFT, you need to insert γ^{t} between any pair of Dirac spinors to get a Lorentzinvariant product, and you have:
u(k, s_{1})† γ^{t} u(k, s_{2}) = δ_{s1, s2} [Lorentzian]
But here, because the rotation group SO(4) has a unitary representation on the Dirac spinors, we can simply take an ordinary inner product between u spinors, or between v spinors, and it will be SO(4)invariant.
In the mode expansion of the field given below, we extract the purely spatial parts of these two planewave solutions ψ_{±, k, s}(x, t) into two sets of functions f_{k, s}(x) and g_{k, s}(x), and then expand a general solution ψ(x, t) as a sum over these functions multiplied by timedependent coefficients. For ψ(x, t) to be a solution of the Dirac equation, the timedependent coefficients will satisfy the same equations as the timedependent parts of ψ_{±, k, s}(x, t), i.e. exp(–i (± E_{k}t)).
“Classical” Riemannian Dirac Equation  
Fourspace coordinate form Sum γ^{j} ∂_{j} ranges over spatial indices x, y, z  
0  =  i γ^{μ} ∂_{μ} ψ(x, t) – m ψ(x, t)  (Equation of Motion) 
L_{RD}  =  ∫ ψ†(x, t) (i γ^{μ} ∂_{μ} ψ(x, t) – m ψ(x, t)) d^{3}x  (Lagrangian) 
Π(x, t)  =  i ψ†(x, t) γ^{t}  
Π^{†}(x, t)  =  0  (Momenta) 
H_{RD}  =  ∫ ψ†(x, t) (–i γ^{j} ∂_{j} ψ(x, t) + m ψ(x, t)) d^{3}x  (Hamiltonian) 
Mode expansion form k is in K_{3}, s is spin ±½  
f_{k, s}(x)  =  √(1/V) u(k, s) exp(–i k · x)  
g_{k, s}(x)  =  √(1/V) v(k, s) exp(+i k · x)  (Spatial modes) 
ψ(x, t)  =  Σ_{k in K3, s=±½} (A_{k, s}(t) f_{k, s}(x) + B_{k, s}(t) g_{k, s}(x))  (Mode Expansion) 
∂_{t} A_{k, s}(t)  =  –i E_{k} A_{k, s}(t)  
∂_{t} B_{k, s}(t)  =  +i E_{k} B_{k, s}(t)  (Equations of Motion) 
L_{RD}  =  Σ_{k in K3, s=±½}
(E_{k}/m) [
i (A_{k, s}(t)^{*}∂_{t} A_{k, s}(t)
– B_{k, s}(t)^{*}∂_{t} B_{k, s}(t)) – E_{k} (A_{k, s}(t)^{*}A_{k, s}(t) + B_{k, s}(t)^{*}B_{k, s}(t))] 
(Lagrangian) 
Π_{A, k, s}(t)  =  i (E_{k}/m) A_{k, s}(t)^{*}  
Π_{B, k, s}(t)  =  –i (E_{k}/m) B_{k, s}(t)^{*}  
Π_{A*, k, s}(t)  =  0  
Π_{B*, k, s}(t)  =  0  (Momenta) 
H_{RD}  =  Σ_{k in K3, s=±½} (E_{k}^{2}/m) (A_{k, s}(t)^{*}A_{k, s}(t) + B_{k, s}(t)^{*}B_{k, s}(t))  
=  –i Σ_{k in K3, s=±½} E_{k} [Π_{A, k, s}(t) A_{k, s}(t) – Π_{B, k, s}(t) B_{k, s}(t)]  (Hamiltonian) 
Although it takes a fair bit of work to derive the Lagrangian and Hamiltonian in terms of the mode coefficients A_{k, s}(t) and B_{k, s}(t), it’s not hard to verify the results, by checking that they give the required equations of motion from the EulerLagrange and Hamilton’s equations respectively.
To turn the classical Hamiltonian into an operator on a Hilbert space of quantum states of the field, we will need to modify the procedure we used when quantising the harmonic oscillator and the scalar field modes. There, we found that the commutators between the creation and annihilation operators were given by:
[a_{α}, a_{β}†] = δ_{α, β}
[a_{α}, a_{β}] = 0
[a_{α}†, a_{β}†] = 0
where α and β label the mode. But because the Dirac field describes fermions, we want the states we get by applying creation operators to the vacuum state ψ_{0} to be antisymmetric in the mode labels:
a_{α}† a_{β}† ψ_{0} = – a_{β}† a_{α}† ψ_{0}
This can be achieved if we replace one of the commutators with an anticommutator:
{a_{α}†, a_{β}†} = a_{α}† a_{β}† + a_{β}† a_{α}† = 0
Taking the Hermitian conjugate of that gives us:
{a_{α}, a_{β}} = 0
Following Zee^{[8]}, we can also justify a third anticommutator:
{a_{α}, a_{β}†} = δ_{α, β}
by requiring that the number operator, N = Σ_{α} a_{α}† a_{α}, satisfies:
N a_{β}† = a_{β}† N + a_{β}†
This is a reasonable requirement, because applied to a state ψ_{n} with n quanta – none of which are yet in the mode β, so that a_{β}† ψ_{n} = ψ_{n+1} – it becomes:
N a_{β}† ψ_{n} = a_{β}† N ψ_{n} + a_{β}† ψ_{n}
N ψ_{n+1} = (n+1) ψ_{n+1}
Given the proposed anticommutator, along with the earlier claim that all creation operators should anticommute, we can derive the required identity:
N a_{β}† = Σ_{α} a_{α}† a_{α} a_{β}†
= Σ_{α} a_{α}† (δ_{α, β} – a_{β}† a_{α})
= a_{β}† – Σ_{α} a_{α}† a_{β}† a_{α}
= a_{β}† + a_{β}† Σ_{α} a_{α}† a_{α}
= a_{β}† N + a_{β}†
Now, the classical Hamiltonian we’ve found:
H_{RD} = –i Σ_{k in K3, s=±½} E_{k} [Π_{A, k, s}(t) A_{k, s}(t) – Π_{B, k, s}(t) B_{k, s}(t)]
is in a form that would make perfect sense if we identified these coordinates and momenta more or less directly with annihilation and creation operators for each mode:
A_{k, s}(t) → √[m/E_{k}] a_{k, s}
B_{k, s}(t) → √[m/E_{k}] b_{k, s}†
Π_{A, k, s} = i (E_{k}/m) A_{k, s}(t)^{*} → i √[E_{k}/m] a_{k, s}†
Π_{B, k, s} = –i (E_{k}/m) B_{k, s}(t)^{*} → –i √[E_{k}/m] b_{k, s}
One question that arises, though, is whether we’re entitled simply to choose the order of products of the coordinates and momenta any way we like, when we convert them to operators. It would be slightly less ad hoc to do something symmetrical that includes both ways of ordering these noncommuting operators – or, since the field is fermionic, something antisymmetrical. So we will apply the rules:
Π_{A, k, s}(t) A_{k} → [i/2] (a_{k, s}† a_{k, s} – a_{k, s} a_{k, s}†) = i (a_{k, s}† a_{k, s} – ½)
Π_{B, k, s}(t) B_{k} → [–i/2] (b_{k, s}† b_{k, s} – b_{k, s} b_{k, s}†) = –i (b_{k, s}† b_{k, s} – ½)
Putting everything together, we end up with:
Quantised Riemannian Dirac Field  
H_{RD}  =  Σ_{k in K3, s=±½} E_{k} [a_{k, s}† a_{k, s} + b_{k, s}† b_{k, s} – 1]  (Hamiltonian) 
Admittedly, what we’ve done here is very far from rigorous! Nevertheless, the Hamiltonian operator we’ve arrived at certainly attributes the correct energy, E_{k}, to each quantum added to the field.
If we take the result seriously, the energy of the vacuum state (for this field alone) will be negative:
H_{RD} ψ_{0} = E_{0} ψ_{0}
E_{0} = – Σ_{k in K3, s=±½} E_{k}
≈ – 4 N_{modes} m / (3 π)
Here N_{modes} is the number of scalar modes consistent with the particle mass and the geometry of the universe, and we’ve included a doubling in the other factors to account for the two spins.
A negative vacuum energy for fermions isn’t peculiar to Riemannian physics; the same thing happens in Lorentzian QFT. As with the scalar field, the difference in the Riemannian case is that we have a finite number of modes, so the vacuum energy is finite.
[1] An Introduction to Quantum Field Theory by Michael E. Peskin and Daniel V. Schroeder, Westview Press, 1995. Section 3.2.
[2] For example, Quantum Mechanics by Leonard I. Schiff, McGrawHill, 1968. Section 13 derives the harmonic oscillator wave functions, while section 25 uses the ladder operator approach.
[3] Peskin and Schroeder, op. cit., chapter 2.
[4] The Quantum Theory of Fields, Volume I: Foundations by Steven Weinberg, Cambridge University Press, 1995. Sections 1.2, 5.2.
[5] Mechanics by L.D. Landau and E.M. Lifshitz, ButterworthHeinemann, 1976. Section 45. N.B.: the Poisson bracket defined by Landau and Lifshitz in their eqn (42.5) is the opposite of the convention we’re using.
[6] Quantum Mechanics and Path Integrals by Richard P. Feynman, Albert R. Hibbs and Daniel F. Styer, Dover, 2010.
[7] Spinors and Trialities by John Baez.
[8] Quantum Field Theory in a Nutshell by A. Zee, Princeton University Press, 2010. Chapter II.2