Quantum and Classical Behaviour
Decoherence is the process by which a quantum-mechanical system that interacts with its environment loses the characteristic properties that distinguish quantum mechanics from classical physics.
The difference between quantum and classical behaviour is exemplified by the famous “double-slit experiment”, in which photons are fired at a barrier containing two slits, and then allowed to fall on a screen opposite the barrier. Classical particles would pass through (at most) one slit at a time, but photons can pass through both simultaneously. The two waves associated with the photon passing through the two different slits fall in and out of phase with each other at different points on the screen — the phase of these waves being related to the total distance the photon travels from source to screen — so they interfere either constructively or destructively, producing a pattern of light and dark bands.
This interference pattern vanishes if any other system interacts with the photon in such a way that, depending on which slit the photon passes through, the second system is affected differently. (For example, devices might be placed at the two slits to discover which one the photon passed through.) The light and dark bands are replaced by a pattern of brightness which is exactly that produced by blocking first the left slit and then the right, and adding the two separate levels of brightness together. In other words, the original pattern, which can only be explained by assuming that each photon passes through both slits simultaneously, is replaced by one that suggests each photon passed through either one slit or the other.
Why does this happen? The short answer is that interference effects can only show up when you have access to an entire quantum system, and not just a part of it.
Pure States
To make these ideas precise, we need to define some concepts, and employ some of the mathematical formalisms of quantum mechanics. (More background to this can be found in the Foundations article on Quantum Mechanics.)
An experiment that can be performed on a quantum system is known as a complete test if the number of different outcomes is as great as it can be for any experiment on this system. To avoid the complications of considering measurements with potentially infinite numbers of possible outcomes, we’ll restrict discussion to systems like the spin of an electron (which is always found to be parallel or antiparallel to any direction in which you choose to measure it), or setups like the double-slit experiment, where a particle can be found travelling along one of a finite number of distinct paths. A complete test is one that doesn’t lump several potentially distinguishable outcomes together; if a photon might be found in any of four places, A, B, C or D, then an experiment that could only distinguish between the two possibilities “it’s in A or it’s in B” and “it’s in C or it’s in D” would not be a complete test.
A quantum system is said to be in a pure state if there is a complete test for which the outcome is 100% predictable. In general, the results of experiments conducted on quantum systems can only be assigned various probabilities, and that will be true for most experiments performed on the system when it’s in a given pure state. But there will always be some particular complete test where the outcome will be a certainty.
For example, suppose we prepare an electron by measuring its spin along some chosen direction, say the z-axis, and discarding it if the measurement shows the spin is not parallel to the direction of measurement. Once we’ve obtained an electron that passes that first test, it’s then certain that a second measurement of its spin along the same direction will yield the outcome “it’s parallel”. In contrast, if the second spin measurement was made along the x-axis, then the outcome would have a 50% probability of being either parallel or antiparallel.
In the mathematical formalism of quantum mechanics, a system that is in a pure state is described by a state vector, which is vector in a complex vector space whose dimension is equal to the number of different outcomes of a complete test on the system. Since any measurement of an electron’s spin can have at most two possible outcomes, parallel or antiparallel, the state vector describing the spin lies in a space with two complex dimensions.
The state space is equipped with an inner product, a rule that assigns a complex number <u, v> to any two vectors u and v, in a manner that gives meaning to the concepts of the size of vectors and the angles between them. For example, if <u, v>=0, then u and v are considered to be orthogonal (perpendicular) to each other. The inner product obeys the rules:
| <u, v> | = | <v, u>* | (1a) |
| <au + bv, w> | = | a* <u, w> + b* <v, w> | (1b) |
| <u, cv + dw> | = | c <u, v> + d <u, w> | (1c) |
| <u, u> | ≥ | 0 | (1d) |
| <u, u> | = | 0 if and only if u=0 | (1e) |
where a, b, c and d are any complex numbers, * means complex conjugate, and u, v and w are any vectors. The norm, or size, of a vector u, written |u|, is defined by:
| |u|2 | = | <u, u> |
The state vector describing a quantum system always has a norm of 1; vectors with this property are known as unit vectors. The inner product of two unit vectors measures how similar in direction they are.
The square of the absolute value of the inner product of two state vectors u and v, |<u, v>|2, gives the probability that when the system is in state u, it will nonetheless pass a test for being in state v. (Here, “a test for being in state v” is any test that the system will pass with certainty whenever it’s in state v.) Unless the inner product <u, v> is 0, i.e. unless the vectors u and v are orthogonal to each other, the probability will be non-zero, and the two states will not be completely distinguishable from each other.
The basis of an n-dimensional space is a set of n vectors, {e1, ..., en}, whose linear combinations (expressions of the form a1e1 + ... + anen, where the ai are any complex numbers) comprise the entire space. In quantum mechanics, it’s common practice to work with a basis that is orthonormal: all the basis vectors have norms of 1, and are mutually orthogonal. This means the basis vectors describe n different states of the system that are completely distinguishable.
For example, in the case of an electron’s spin, we might work with a basis whose two vectors are the states “the spin is parallel to the z-axis” and “the spin is antiparallel to the z-axis”, abbreviated as {eu, ed} (u for “up” and d for “down”). Picking another direction along which to measure the spin amounts to choosing a different basis for the state space. If we choose the x-axis, and call the corresponding basis vectors {er, el} (r for “right” and l for “left”), then the two bases are related by:
| er | = | (1/√2) eu + (1/√2) ed | (2a) |
| el | = | (1/√2) eu – (1/√2) ed | (2b) |
Using the rules for the inner product given by equation (1b), we see that:
| <er, eu> | = | <(1/√2) eu + (1/√2) ed, eu> | |
| = | (1/√2) <eu, eu> + (1/√2) <ed, eu> | ||
| = | 1/√2 | ||
| |<er, eu>|2 | = | 1/2 |
which confirms that when an electron has been prepared in the state eu, if its spin is then measured along the x-axis the probability of finding it to be parallel to that axis is 1/2, or 50%.
If we perform a complete test on a quantum system that lets us distinguish between all the vectors in some basis, {e1, ..., en}, then there will often be a number associated with each basis vector, corresponding to some classical quantity, q, that the test is measuring. For example, for the basis {eu, ed}, the two vectors correspond to values of the electron’s spin along the z-axis of 1/2 and –1/2 (when measured in appropriate units). The manner in which this quantity is measured can then be packaged conveniently in a linear operator on the state space: that is, a function Q that takes one vector and gives you another, and is linear in the sense that:
| Q(au + bv) | = | aQ(u) + bQ(v) |
Specifically, if the basis is {e1, ..., en} and the quantities associated with these basis vectors are {q1, ..., qn}, then Q is defined by:
| Q(u) | = | (q1 <e1, u>) e1 + ... + (qn <en, u>) en | (3) |
Q is known as the observable associated with the quantity q. If we feed Q any one of our basis vectors, say ei, we get:
| Q(ei) | = | qiei | (4) |
because <ei, ei>=1, and all the other inner products in equation (3) are zero. Equation (4) says that ei is an eigenvector of Q, with an eigenvalue of qi; these are general terms for the situation where a vector fed to a linear operator produces some multiple of itself. If we were given nothing but the linear operator Q, we could recover all the state vectors associated with definite values of q, by finding all the eigenvectors of Q.
We can also use Q to give a simple expression for the average value we’d expect to measure for q, if we repeatedly make measurements on a large number of versions of some quantum system that have all been prepared to have the state vector u.
| <u, Q(u)> | = | <u, (q1 <e1, u>) e1 + ... + (qn <en, u>) en> | |
| = | q1 <e1, u><u, e1> + ... + qn <en, u><u, en> | ||
| = | q1 <e1, u><e1, u>* + ... + qn <en, u><en, u>* | ||
| = | q1 |<e1, u>|2 + ... + qn |<en, u>|2 | (5) |
When we measure q for the quantum system, we’re essentially just subjecting it to the complete test whose outcomes correspond to the basis vectors ei and their associated values of q, qi. Since the probability of the state vector u passing a test for being ei is |<ei, u>|2, the RHS of equation (5) is just the sum of all possible measured values for q, weighted by their respective probabilities, which is the expectation value for q given that the system is in state u. We’ll write this as:
| E(q|u) | = | <u, Q(u)> | (6) |
To give a specific example, let the quantum system in question be an electron’s spin, and let’s apply what we’ve done above to the basis {eu, ed}. Associated with these two basis vectors are spins sz=1/2 and sz=–1/2. Using equation (3), the observable Sz is then defined by:
| Sz(u) | = | ((1/2) <eu, u>) eu + ((–1/2) <ed, u>) ed |
and the expectation value for the z-axis spin, sz, is:
| E(sz|u) | = | <u, Sz(u)> | |
| = | (1/2) |<eu, u>|2 + (–1/2) |<ed, u>|2 |
For an electron prepared in the state er, with its spin parallel to the x-axis:
| E(sz|er) | = | <er, Sz(er)> | |
| = | (1/2) |<eu, er>|2 + (–1/2) |<ed, er>|2 | ||
| = | (1/2)(1/2) + (–1/2)(1/2) | ||
| = | 0 |
Because sz is equally likely to be 1/2 or –1/2, the expectation value averages out to zero.
Superpositions
Given two mutually orthogonal state vectors for a quantum system, u and v, and two complex numbers a and b which satisfy the equation:
| |a|2 + |b|2 | = | 1 |
the vector w = au + bv will be a unit vector, and hence another valid state vector for the system. This is known as the principle of superposition.
The vector w defined in this way has a probability of |a|2 for passing a test for the state u, and a probability of |b|2 for passing a test for the state v, which is why these two quantities must add up to one.
More generally, given an orthonormal basis {e1, ..., en}, and n complex numbers ai that satisfy the equation:
| |a1|2 + ... + |an|2 | = | 1 |
the vector
| w | = | a1e1 + ... + anen |
will also describe a possible state of the system, with probability |ai|2 for passing a test for the state ei.
Colloquially, the term “superposition” suggests a quantum system in a superposition of two or more states of the kind that we’d normally expect it to be in, such as an atom in a superposition of two states in which it’s in different locations. Mathematically, though, every state vector can be written as a superposition of other state vectors. For example, in the case of an electron’s spin, the spin-right state er is a superposition of basis vectors from the z-axis spin basis {eu, ed}, since:
| er | = | (1/√2) eu + (1/√2) ed |
but equally, eu and ed are both superpositions of basis vectors from the x-axis spin basis {er, el}:
| eu | = | (1/√2) er + (1/√2) el | |
| ed | = | (1/√2) er – (1/√2) el |
Mixtures and Density Matrices
If a quantum system is not in a pure state, it is said to be in a mixed state, or a mixture. For a system in a mixed state, there is no complete test whose outcome can be predicted with certainty.
To prepare a system in a mixed state, you could toss a coin and then take an electron from either of two sources of pure states, depending on whether the coin showed heads or tails. Whatever the state vector of an electron, there is a corresponding direction in which you can measure the spin so that electrons in that state will always give a result of “parallel”. But if you toss a coin, choose between eu and ed, and then hand the electron to someone else who didn’t see the way the coin fell, there is no direction they can choose for a spin measurement that will guarantee the result with certainty.
Mixed states introduce a form of probability that’s quite different from the kind that quantum measurements display: they reflect ignorance, not just about the results of a particular measurement, but about the state vector that actually describes the system in question. A mixture is thus very different from a superposition. The superposition
| er | = | (1/√2) eu + (1/√2) ed |
has a 50% chance of passing a test for eu, just like a mixed state where you’ve tossed a coin and selected either eu or ed, but the probabilities for other tests will not be the same. For example, a test for the system being in the state er will obviously be passed by this superposition with certainty, whereas the mixed state will only have a 50% chance, and a test for the system being in the state el will never be passed by the superposition, but the mixed state will again have a 50% chance.
A mixed state can’t be described by a state vector, but there is a mathematical formalism that encompasses both pure and mixed states. The objects used to represent either kind of state are known as density matrices. Like an observable, a density matrix is a linear operator. (Why is it called a matrix? Any finite-dimensional linear operator can be written as a matrix of complex numbers, once you choose a particular basis in which to do so.)
The density matrix associated with a given pure state, u, is the operator:
| ρu(v) | = | (<u, v>) u | (7) |
This particular density matrix is also called “the projector onto u”, because whatever vector you feed it, you get a kind of shadow cast by the original vector onto the part of the state space that consists only of multiples of u.
Just as we can find the expectation value for some observable Q by means of a state vector, we can do the same with the corresponding density matrix. The expectation value given by equation (6) can be rewritten:
| E(q|u) | = | q1 |<e1, u>|2 + ... + qn |<en, u>|2 | |
| = | <e1, Q(ρu(e1))> + ... + <en, Q(ρu(en))> | ||
| = | tr(Q ρu) | (8) |
In the final line here, we’ve used the notation “tr”, which stands for trace. The trace of a linear operator T is the sum, over any basis, of the ei component of T(ei), with i ranging from 1 to n. Equivalently, if you write the linear operator T as a matrix Tij, the trace is just the sum of all the diagonal elements of the matrix, T11 + ... + Tnn. The notation Q ρu means the linear operator you get by composing Q and ρu, i.e. forming a single operator by applying ρu first and then Q.
Now, suppose that instead of a pure state, we wish to describe a mixed state. Specifically, imagine that there are probabilities p1, ..., pm that a system is in one of the pure states u1, ..., um. The pi, being probabilities, must all be real numbers between 0 and 1, and the sum of them all must come to 1. We define the density matrix for this mixed state by simply adding up the density matrices for all the possible pure states, weighted by their respective probabilities:
| ρ(v) | = | p1 ρu1(v) + ... + pm ρum(v) | (9) |
The expectation value of q for a mixed state will be the sum of the expectation values for all possible pure states, weighted by their respective probabilities, so we have:
| E(q|ρ) | = | tr(Q ρ) | (10) |
If someone hands you a density matrix, how can you tell whether it describes a mixed state or a pure state? The density matrix for a pure state is a projector onto a state vector, which means that if you apply the operator twice, it has exactly the same effect as applying it once. (This is much the same as saying that if a shadow could itself cast a shadow, there’d be no difference between that second shadow and the original.)
| ρu(ρu(v)) | = | (<u, ρu(v)>) u | |
| = | (<u, (<u, v>) u>) u | ||
| = | (<u, v>)(<u, u>) u | ||
| = | (<u, v>) u | ||
| = | ρu(v) | (11) |
If the density matrix for a pure state is written out as a matrix of numbers, then squaring the matrix will give you back the original matrix again.
Subsystems
Our ultimate goal is to understand what a quantum system looks like when we only have access to part of it. To do this, we first need to work in the other direction, and understand how a quantum system behaves when it’s made up of more than one part.
Suppose we have two quantum systems, A and B, whose state spaces have bases {e1, ..., en} and {f1, ..., fp} respectively. This means that a complete test on system A will have n different outcomes, and one on system B will have p different outcomes. For the combined system C, then, a complete test will have np different outcomes, and we can represent them by writing vectors of the form ei ⊗ fj. The symbol ⊗ is called a tensor product; there is a formal mathematical definition of this, but for our purposes we can just take it to mean that ei ⊗ fj is the state of the combined system C in which system A is in the state ei and system B is in the state fj. A basis for the state space of C is then given by the np vectors {e1 ⊗ f1, e1 ⊗ f2, ..., e1 ⊗ fp, ..., en ⊗ f1, en ⊗ f2, ..., en ⊗ fp}. The state space for C is the tensor product of those for A and B, and it inherits an inner product from them in a natural way:
| <u ⊗ v, w ⊗ x> | = | <u, w><v, x> |
Using this inner product, the basis we’ve given for the combined system is orthonormal if those for the two subsystems are orthonormal.
Suppose we perform a test on system A which can distinguish between all the ei, but which measures nothing about system B. In terms of the whole system, C, what we’re really performing then is an incomplete test, which can distinguish ei ⊗ fj with different i, but lumps together all those states with the same i but different j. If there are quantities qi associated with the basis states ei, then the observable Q for measuring q on system A can be extended to an observable Q ⊗ 1 on the combined system C; formally, this is the tensor product of the observable on A with an “identity” observable 1 on B, which measures nothing about B. Q ⊗ 1 is written in terms of all np basis vectors, but it assigns the same value, qi, to all p vectors with the same i.
| Q ⊗ 1(y) | = | q1 (<e1 ⊗ f1, y> e1 ⊗ f1 + ... + <e1 ⊗ fp, y> e1 ⊗ fp) + | |
| q2 (<e2 ⊗ f1, y> e2 ⊗ f1 + ... + <e2 ⊗ fp, y> e2 ⊗ fp) + | |||
| ... + | |||
| qn (<en ⊗ f1, y> en ⊗ f1 + ... + <en ⊗ fp, y> en ⊗ fp) | |||
| = | Σi qi Σj <ei ⊗ fj, y> ei ⊗ fj |
Now, suppose the whole system C is in some arbitrary pure state, z:
| z | = | Σi,j ci,j ei ⊗ fj |
where the ci,j are np complex numbers, and the sum of the squares of their absolute values comes to 1. We can compute Q ⊗ 1(z):
| Q ⊗ 1(z) | = | Σi qi Σj ci,j ei ⊗ fj |
The expectation value for q is given by:
| E(q|z) | = | <z, Q ⊗ 1(z)> | |
| = | Σi qi Σj ci,j* ci,j | ||
| = | Σi qi Σj |ci,j|2 | (12) |
This is just what you’d expect: all the possible values for q, weighted by their respective probabilities. But it turns out that we can obtain this expression by another route.
If we define ρA,z by:
| ρA,z(v) | = | Σi,j,k ci,k cj,k* <ej, v> ei | (13) |
then, applied to a basis vector em, we have:
| ρA,z(em) | = | Σi,j,k ci,k cj,k* <ej, em> ei | |
| = | Σi,k ci,k cm,k* ei |
Since ei is an eigenvector of Q with eigenvalue qi, we have:
| Q(ρA,z(em)) | = | Σi,k ci,k cm,k* qi ei |
which lets us compute the trace:
| tr(Q ρA,z) | = | Σm <em, Q(ρA,z(em))> | |
| = | Σm,k cm,k cm,k* qm | ||
| = | Σm,k |cm,k|2 qm | ||
| = | Σm qm Σk |cm,k|2 | ||
| = | E(q|z) | (14) |
The operator ρA,z is known as the reduced density matrix for subsystem A. Equation (14) shows that the expectation value of any observable that relates only to A can be obtained from this matrix. The question, then, is whether ρA,z behaves like the density matrix of a pure state, or that of a mixed state.
In the preceding section, we saw that the density matrix of a pure state, when squared, is equal to itself again. What happens when we square ρA,z?
| ρA,z(ρA,z(v)) | = | ρA,z(Σm,r,s cm,s cr,s* <er, v> em) | |
| = | Σi,j,k ci,k cj,k* <ej, Σm,r,s cm,s cr,s* <er, v> em> ei | ||
| = | Σi,j,k,r,s ci,k cj,k* cj,s cr,s* <er, v> ei | (15) |
In some special cases, equation (15) can be simplified. Suppose that z can be written as a single tensor product between two states:
| z | = | (Σi ai ei) ⊗ (Σj bj fj) | (16) |
where Σi |ai|2=1, and Σj |bj|2=1. This means that:
| ci,j | = | ai bj |
Equations (13) and (15) become:
| ρA,z(v) | = | Σi,j,k ai bk aj* bk* <ej, v> ei | |
| = | (Σk |bk|2) Σi,j ai aj* <ej, v> ei | ||
| = | Σi,j ai aj* <ej, v> ei | (17a) | |
| ρA,z(ρA,z(v)) | = | Σi,j,k,r,s ai bk aj* bk* aj bs ar* bs* <er, v> ei | |
| = | (Σj |aj|2) (Σk |bk|2) (Σs |bs|2) Σi,r ai ar* <er, v> ei | ||
| = | Σi,r ai ar* <er, v> ei | (17b) |
The right-hand sides of equations (17a) and (17b) are equal, so in this case the reduced density matrix does behave like that of a pure state. States like z as defined in equation (16) are known as product states, and they describe situations where the individual states of the two subsystems are completely independent of each other. If two systems have been prepared separately (in this case, both in pure states), and haven’t interacted with each other, then the combined system’s state will be a product state, and measurements on either subsystem alone will still yield results consistent with it being in a pure state.
In contrast, suppose the two subsystems interact, and as a consequence they end up in a state:
| z | = | Σi ai ei ⊗ fi | (18) |
where Σi |ai|2=1, and we’re assuming either that n=p (i.e. the two subsystems have state spaces of identical dimension), or that the sum here is over the lesser of the two. This means that:
| ci,j | = | ai, if i=j | |
| = | 0, otherwise |
Equations (13) and (15) become:
| ρA,z(v) | = | Σk ak ak* <ek, v> ek | |
| = | Σk |ak|2 <ek, v> ek | (19a) | |
| ρA,z(ρA,z(v)) | = | Σk ak ak* ak ak* <ek, v> ek | |
| = | Σk |ak|4 <ek, v> ek | (19b) |
Since the |ak| must be real numbers between 0 and 1, and the sum of their squares must come to 1, clearly (19a) and (19b) can only agree if just a single one of the |ak| is equal to 1, and the rest are 0. In fact, equation (19a) is precisely the same as the definition given in equation (9) of a density matrix for a mixture which has the probabilities |ak|2 for being in the various pure states ek.
States like z as defined in equation (18) are known as entangled states or correlated states, and they occur when two subsystems have interacted in such a way that one system’s state will depend on the other’s. For example, subsystem A might have started out in the state:
| u | = | Σi ai ei |
and subsystem B in the state v, giving a combined state which was a product state, u ⊗ v. But if the total system was then subject to an interaction U that effectively tested which ei basis state A was in, and set the state of B accordingly:
| U(ei ⊗ v) | = | ei ⊗ fi |
then since all dynamics in quantum mechanics is described by linear operators, U's effect on u ⊗ v would be:
| U(u ⊗ v) | = | Σi ai ei ⊗ fi |
A deliberate measurement of the quantity q (the quantity associated with the ei basis) would obviously be one way to cause such an interaction, but merely letting a quantum system interact freely with its environment could easily produce the same result.
In this way, decoherence explains why there is no need whatsoever to invoke the notion of a “collapse of the wave” that occurs to quantum systems when they’re “measured” or “observed” — as if these actions were somehow special. Rather, any interaction that entangles a quantum system with a second system will make the original system (when observed in isolation) appear to change from a pure state into a mixture.
How Entanglement Hides Quantum Interference
Finally, we’ll give an explicit example of how the interference effects that characterise quantum systems appear to vanish when the system becomes entangled with something else.
Suppose we have a source of electrons, which supplies them in a pure state:
| u(θ) | = | (1/√2) eu + (1/√2) exp(i θ) ed |
Here θ, the phase difference between the spin-up and spin-down components, is a parameter that we can freely vary by twiddling some control on our electron source. (Doing something like this in practice would probably be easier with photons, but we won’t worry about the details of the hardware.)
Now, suppose we measure the electrons’ spin along the x-axis. The observable for this is:
| Sx(v) | = | (1/2) <er, v> er + (–1/2) <el, v> el |
which we convert into an expression that uses the z-axis basis:
| Sx(v) | = | (1/2) <(1/√2) eu + (1/√2) ed, v> ((1/√2) eu + (1/√2) ed) + | |
| (–1/2) <(1/√2) eu – (1/√2) ed, v> ((1/√2) eu – (1/√2) ed) | |||
| = | (1/2) <eu, v> ed + (1/2) <ed, v> eu |
This allows us to compute the expectation value for the spin along the x-axis, sx:
| E(sx|u(θ)) | = | <u(θ), Sx(u(θ))> | |
| = | (1/4) (exp(i θ) + exp(–i θ)) | ||
| = | (1/2) cos(θ) | (20) |
So, the expectation value for sx varies cyclically with θ, reflecting the phase difference between the two terms in the superposition u(θ). (We can easily check this for the two cases when u(θ) is an eigenvector of Sx. When θ=0, u(θ)=er, so E(sx|u(θ))=1/2, and when θ=π, u(θ)=el, so E(sx|u(θ))=–1/2).
Now, suppose that some second system becomes entangled with the electrons’ spin along the z-axis. This could be due to a deliberate measurement, or an uncontrolled interaction with the environment. In any case, the electrons available to us are now part of a larger system, and the states for that larger system are:
| z(θ) | = | (1/√2) eu ⊗ fu + (1/√2) exp(i θ) ed ⊗ fd |
where fu and fd are two orthogonal states of the second system; it makes no real difference exactly what the second system is, or what these states describe.
The state z(θ) is an entangled state, in the same general form as equation (18), and the corresponding reduced density matrix is, using equation (19a):
| ρA,z(θ)(v) | = | (1/2) <eu, v> eu + (1/2) <ed, v> ed |
Here, the phase θ has vanished completely, because the complex coefficients in the entangled state only show up in the density matrix (19a) as absolute values.
From this, we can compute the expectation value of sx:
| E(sx|ρA,z(θ)) | = | tr(Sx ρA,z(θ)) | |
| = | <eu, Sx ρA,z(θ)(eu)> + <ed, Sx ρA,z(θ)(ed)> | ||
| = | (1/2) <eu, Sx(eu)> + (1/2) <ed, Sx(ed)> | ||
| = | (1/4) <eu, ed> + (1/4) <ed, eu> | ||
| = | 0 | (21) |
So, with access only to the electrons, the interference we saw in equation (20) is now completely hidden, and the electrons behave as if they were in a mixed state with an equal chance of being either spin-up or spin-down.
However, whether the phase θ really is irretrievably lost will depend on the nature of the second system. If the second system remains accessible, and is itself amenable to simple quantum measurements, then a suitable measurement on both systems will recover the phase dependence.
To be specific, suppose we subject the second system to a test for the state analogous to er:
| fr | = | (1/√2) fu + (1/√2) fd |
(The second system doesn’t have to be a spin for this to make sense, but the details of how the test is carried out will depend on exactly what the second system is.) If we assign the quantity tx=1/2 to states that pass this test, and tx=–1/2 to states that fail it, tx will be a quantity similar to sx, and the associated observable will be:
| Tx(v) | = | (1/2) <fu, v> fd + (1/2) <fd, v> fu |
If we measure both sx and tx and multiply them together, the observable associated with the product of these quantities will be:
| Sx ⊗ Tx(u ⊗ v) | = | (1/4) <eu, u> ( <fu, v> ed ⊗ fd + <fd, v> ed ⊗ fu) + | |
| (1/4) <ed, u> ( <fu, v> eu ⊗ fd + <fd, v> eu ⊗ fu) |
The expectation value of sx tx can then be calculated:
| E(sx tx|z(θ)) | = | <z(θ), Sx ⊗ Tx(z(θ))> | |
| = | (1/8) (exp(i θ) + exp(–i θ)) | ||
| = | (1/4) cos(θ) | (22) |
demonstrating that the phase information hasn’t been lost.
Despite how things look when our observations are confined to the electron alone, the electron certainly hasn’t been “collapsed” into either the spin-up or the spin-down state; the original superposition remains, if you know how to look for it.
References: Quantum Theory: Concepts and Methods by Asher Peres (Kluwer, Dordrecht, 1993). Decoherence and the Appearance of a Classical World in Quantum Theory by D. Giulini, E. Joos, C. Kiefer, J. Kupsch, I.-O. Stamatescu and H.D. Zeh (Springer, Berlin, 1996).
