# Why do Bell tests give perfect correlations?

Suppose some decay process emits 2 electrons in opposite directions, and their spin is measured by a Stern-Gerlach type device in a particular direction, say Sz. The books say that if 2 detectors have the same orientation, then the spin measurements are 100% correlated (or anti-correlated). If one has Sz=+1/2, then the other has Sz=-1/2.

But what is the quantum state of the electrons before the measurements? It seems to me that if it is anything but an eigenstate of Sz, then the perfect correlations are impossible.

Is there any quantum mechanical explanation of the perfect correlations, other than with an action-at-a-distance collapse of the wave function?

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I'd guess your confusion is coming from the term Sz. There are 3 different Sz's to consider here: The Sz of electron 1 (Sz1), the Sz of electron 2 (Sz2), and the Sz of electron 1 + electron 2 (Sztot). In the system you're talking about, the system is prepared in an eigenstate of Sztot=0, ensuring that a measurement of (Sz1 + Sz2) will always be 0. However, the system is not in an eigenstate of Sz1 (thus its outcome is uncertain) nor in an eigenstate of Sz2 (thus its outcome is uncertain). – Anonymous Coward Aug 15 '11 at 6:17
There are two part in your question: 1) "But what is the quantum state of the electrons before the measurements?" and 2) "Is there any quantum mechanical explanation of the perfect correlations...?". The first cannot be answered based on information you provide. The answer to 2) depends on the answer to 1). – Slaviks Aug 17 '11 at 4:52

Quantum mechanics has one peculiar and highly counterintuitive property:

If you know the complete state of a compound system $S$, this does not necessarily imply you know the complete state of either of the subsystems!

The quantum state you are after looks like this: $$\frac{1}{\sqrt{2}} \left( \langle \uparrow \downarrow | - \langle \downarrow \uparrow| \right)$$

Since this is an entangled state, this means that you cannot decompose this into some product of the form $$\langle \psi_A | \otimes \langle \psi_B |$$

In a way, it is therefore nonsensical to ask what the state of one of the electrons is! If you want to talk about one of the subsystems only, you must make use of the density-matrix formalism. An application of this would show that for the state I've depicted above, the individual spins are as undetermined as possible, being up or down with exactly equal probability.

Now to your second question. Sure, there are other explanations apart from the collapse of the wave function. But from a practical point of view, those are all equivalent and you can basically choose whichever you like best. There are the various flavors of the many-world interpretation, for example.

Note that the correlation does not rely on the condition that the two measurements occur in such a way that light would be able to travel from one lab to the other in the time between the measurements! Indeed, if we call the two observers $A$ and $B$, it does not matter how far apart the labs are or when exactly the measurement takes place. How such experiments are carried out is that for both labs, a list of the spin is written down, which will read like $\uparrow \uparrow \downarrow \uparrow \downarrow \downarrow \dots$. After the experiment, $A$ and $B$ will compare their lists and find a $100\%$ correlation (well, barring some measurement errors that destroy coherence).

I don't think it is possible to give a satisfying explanation relying on classical terms, because nature isn't classical.

I'm not sure I understand what you mean with "there's no eigenstate of $S_z$". The total state as I have written down is a linear combination of product states, which are formed from the basis states of systems $A$ and $B$. So of course, when $A$ or $B$ perform their measurements of the electron spin, they will find only eigenvalues of $S_z$. The important part is that the overall state of the system cannot be written as a pure product state, no matter what basis you choose. This is the hallmark of entanglement, because if you had a product state $\langle \psi_A | \otimes \langle \psi_B$, a measurement on system $A$ would have no effect on the state of system $B$.

You can view the process of measuring the electron spin in systems $A$ and $B$ and then talking about the results as a measurement of the total spin's $z$-component. In terms of total angular momentum, the state I have written is equivalent to $\langle 0, 0|$, i.e. a state of total spin zero. Since this state is an eigenstate of the total spin with eigenvalue $0$, every measurement of the total spin's $z$-component must yield result $0$.

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what if it is a relativistic field theory, where there is no causality outside the light cone? Or an interpretation with no objective collapse of the wave function? How can there be a 100% correlation if there is no eigenstate of Sz? – Roger Aug 15 '11 at 0:17
I find it very odd to see the states written as $\langle\psi|$ (instead of $|\psi\rangle$) in this context. – David Z Aug 15 '11 at 0:38
ooops. that's me not having access to my beloved \ket latex macro :D – Lagerbaer Aug 15 '11 at 1:02
Yeah, that annoys me too. Hopefully someday they'll let us add in custom LaTeX macros on a per-site basis. – David Z Aug 15 '11 at 1:17
Why are you saying with certainty that the initial state is the singlet state? Entanglement is not assumed in the question. – Slaviks Aug 17 '11 at 6:04

But what is the quantum state of the electrons before the measurements? It seems to me that if it is anything but an eigenstate of Sz, then the perfect correlations are impossible.

Electrons have "wave particle duality" which sounds like philosophical garbage but is detectable in experiments such as this one. Rather than emitting the electrons in an unknown state, let's assume a device that specifically emits a pair of electrons that are 100% anti-correlated in Sx rather than Sz.

We will work in the usual Sz basis, and since Sx can be taken to be real in this basis, I'll ignore the difference between bra and ket vectors (and for convenience, write them all as bras). So one of the electrons is Sx+ = $(1,1)/\sqrt{2}$, the other is Sx- = $(1,-1)/\sqrt{2}$. The combined wave function is defined mathematically as a x product as in:
$$|x+-\rangle = (1,1)\times(1,-1)/2 = (1,1,-1,-1)/2.$$
But since electrons are indistinguishable, we could have the two electrons reversed. This would be the state:
$$|x-+\rangle = (1,-1)\times(1,1)/2 = (1,-1,1,-1)/2.$$
Note that the above two vectors are orthonormal.

Since electrons are fermions, we take the anti-symmetric combination. This means we take the difference between the above two joint wave functions:
$$|x+/-\rangle = (|x+-\rangle - |x-+\rangle)/\sqrt{2} = (0,1,-1,0)/\sqrt{2}.$$

Now let's redo the calculation with two electrons that have spin Sz anti-correlated so that Sz+ =$(1,0)$ and Sz- = $(0,1)$. The two possible cases are:
$$|z+-\rangle = (1,0)\times (0,1) = (0,0,1,0)$$
$$|z-+\rangle = (0,1)\times (1,0) = (0,1,0,0)$$
and the anti-symmetric combination is:
$$|z+/-\rangle = (|z+-\rangle - |z-+\rangle)/\sqrt{2} = (0,-1,1,0)/\sqrt{2}.$$ This is the same as we got for the x case $|x+/-\rangle$ (other than an overall factor of -1 which is just the usual arbitrary complex phase).

Thus we see that according to the rules for quantum mechanics, it doesn't matter which two electrons we begin with. They could be spin in the x direction or z direction, or any direction; you get the same joint wave function.

Now the above explanation may be unsatisfying in that it is mathematical. If you want a more physical explanation for what is going on, maybe the following will help.

You can think of the electrons as being excitations, that is, as waves. When you measure a wave, you alter the wave and change it. In the case of the anti-correlated electrons, the combined wave has just enough stuff to give you two electrons with opposite correlated spin. It doesn't matter which way the spins are arranged, there's just enough to give you a pair of them.

One must be careful when one talks about the passage of time in these experiments. The measurement typically takes place long after the electron wave function has been divided into two portions, say one going to one detector, the other to the other detector. It might be several nanoseconds before an apparatus is able to permanently record which electron had which spin. During THAT (i.e. measurement) time, the electron's wave function has to be changed into another form, one which is suitable for measurement. This cannot happen instantaneously and it does not happen AFTER the electron wave function is split. Until the measurement is made, the electron, along with the measuring apparatus, is all in a Schroedinger Cat sort of combined wave function. Either possible result is possible, but the electron wave function only has enough juice to do one of the two possibilities.

The same "explanation" applies to things like Wheeler's "delayed choice" experiment. There actually is no delay in the choice in that the combination of experiment and particle are in a combined wave state until the measurement becomes permanent.

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Add to that the Delayed Choice Quantum Eraser for yet another cool experiment demonstrating both entanglement and particle-wave duality. – Lagerbaer Aug 15 '11 at 1:47
Thanks. I did not realize that the anti-symmetrized Sx eigenstate is the same as the anti-symmetrized Sz eigenstate. I was also thinking of the electrons as 2 distinct electrons. Maybe it is better to think of them as one wave, and they only become electrons after observations. – Roger Aug 16 '11 at 0:28
"Since electrons are fermions, we take the anti-symmetric combination." The fact that the particles are fermions and that the total $S_z=0$ does not imply that they must be in the singlet state that you wrote. This is only so if this spin part is combined with a symmetric orbital wave-function. If the orbital wave-function is anti-symmetric, then the spin part can be symmetric, and this means a different entangled state (triplet Sz=0 component). And if the experiment does not distinguish the two, no quantum mechanical property can be inferred from the fact of perfect correlation. – Slaviks Aug 17 '11 at 6:10
@Slaviks; You're right, I simplified in order to avoid complicating the lesson. The simplification is the assumption that the complete description of the electron is given by what I wrote down, that is, the spin of the electron. In every other respect, we assume that the electron states are identical. This is compatible with the original poser's description of the problem, which makes no mention of dependency on position. – Carl Brannen Aug 17 '11 at 15:14
@Carl: "the original poser's description of the problem, which makes no mention of dependency on position." except that they are detected by two space-time separated events (you also discuss "two portions of wavefucntion"). But this is hair splitting on my part, the essential point is that the described experiment alone does not prove that quantum correlations are any stronger than classical ones. Only assuming non-zero initial entanglement (very quantum notions) requires us to invoke quantum explanation down the road – Slaviks Aug 17 '11 at 16:22

The correlations you describe seem to be trivial and completely explained by a conservation law (conservation of total Sz in this case). Classical example of 100% correlation: if a rocket with knowen total mass and velocity splits into to two parts then measuring the momentum of one of the parts gives the perfect knowledge of the momentum of the second part.

The fact that the particle you mention are in a pure entangled quantum mechanical state is not relevant to this 100% correlation. Correlations entering Bell inequalities are more subtle and necessarily involve measuring non-commuting observables.

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Contrary to Slaviks' opinion, I believe Roger was right on the money when he identified the 100% correlation as the deepest part of the quantum mystery of entanglement. The detailed Bell arguments about polarizers at 22.5 degrees are terribly clever but completely miss the essential point: the perfect correlations with polarizers aligned were already hugely problematic from a classical wave perspective. Carl and Lagerbaer both correctly identify key aspects of this, especially Carl's explicit demonstration that the singlet state maps to itself when rotated 90 degrees. – Marty Green Aug 17 '11 at 2:12
Sorry, cannot agree with you. The answers of Carl and Lagerbaer are perfectly fine, but they assume that the initial state of the systems is a pure singlet. But this is not stated in the question! The initial state may as well be a symmetric mixture of singlet and zero component of the triplet, and non-locality will be lost. There will be just classical uncertainty. I'm a newbie here, should I edit my answer to make the point more explicit? – Slaviks Aug 17 '11 at 4:50
I wouldn't ask you to edit your answer; you've made your case and ideally we should be able to argue it out. Unfortunately, for reasons I don't really understand, the structure of this forum does not really lend itself to good discussions. Any takers our there? – Marty Green Aug 17 '11 at 8:33
The 100% correlation is easy to understand if you apply the conservation principle and forget about quantum mechanics. The 22.5 degree Bell correlations are not so hard either, as it is plausible that probabilistic models could explain them. The difficulty I had was going back to those 100% correlations once I accepted a probabilistic model. They seem to require entanglement. – Roger Aug 18 '11 at 4:42
@Roger: Then indeed I misunderstood the main point of your question. A mixed state with equal weights to singlet and triplet components allow a basis change to unentangled products states and 100% correlation is recovered without entaglement. – Slaviks Aug 18 '11 at 5:54