The original paradox was about conjugate observables, that is, observables that can't be measured simultaneously. The issue was whether quantum mechanics were complete or not. Now we know that is is complete, but non local. Observables that are not conjugate can be correlated too. For example, if the position of the center of mass of a system is well defined, but the positions of the constituent particles are not, then they are correlated.

Let's take a system made of two identical particles, and take a position basis |x>_1 and |x>_2 (Dirac delta.) If the position of the center of mass is 0, consider the entangled state vector:

integral_x { |x>_1 |-x>_2 psi(x)dx }

Then the positions are correlated, since if particle 1 is detected at X, then the state vector is projected onto |X>_1 |-X>_2. When measuring the position of particle 2, the result is -X, we see that it is correlated with the position of particle 1.

@Claude Pierre Massé

My dear Claude Pierre Massé do you mean due to entanglement, if particle 1 is found at X then particle 2 will be found at -X?

Dear Hussainsha,

If you are clear about the momentum measurement on entangled particles, then the same will automatically imply for the position measurement as well, b’cos position and momentum are canonically conjugate pairs.

I really don’t know how much the following explanation will help you to understand the correlation between the position measurement of the entangled particles, but just have a glance.

Consider the EPR case of a pair of well-separated quantum particles which had interacted initially at some position x0 for a very brief time (like the case of elastic collision). When they are separated to a large distance, then it is assumed that there are no more physically known interactions acting between them.

It’s well-known that the accurate position measurement leaves a huge uncertainty in the momentum of a given quantum particle and vice versa which is attributed to the commutation relation [x , p] =i \hbar.

Now, measure the position of particle 1, x1, accurately. Hence its momentum, p1, becomes uncertain. On the particle 2, perform momentum measurement and hence its position gains huge uncertainty. Now, the EPR point is that, since the particles interacted initially, the position of particle 2 can be deduced from the knowledge of the measured position of the particle 1 due to their initial interaction and from the knowledge of measured momentum of particle 2, the momentum of particle 1 can be deduced due to momentum conservation. It means that it becomes possible to accurately know both x1 and p1 simultaneously which is in violation of the commutation relation [x1 , p1] = i \hbar; but, violation should not happen b’cos that commutation relation is the heart of quantum mechanics.

Consider the above situation algebraically;

B’cos the total momentum is conserved, one has

[x1 – x2 , p1 + p2] = [x1 , p1] – [x2 , p2] = 0

where, [x1 , p1] = [x2 , p2] = i \hbar

and

[x1 , p2] = [x2 , p2] = 0 (b'cos particle are well separated and non-interacting)

Since, `x1 – x2’ and `p1 + p2’ commutes, they can be simultaneously measured according to quantum formalism.

Let x0 be some point in space where two particles initially interacted,

Then x1 – x2 = x0 and p1 + p2 = 0 (momentum conservation)

So, from the knowledge of x1 and p2, we can know x2 and p1 as accurately as we wish, which seems to violate the uncertainty relation. Even in this case, the validity of uncertainty relation requires that somehow the measurement of x1 should not allow for the accurate measurement of p2. It means that, measurement of x1 must have an influence on the other particle’s outcome. But look at the algebra; [x1 , p2] = 0. It means that according to quantum formalism, it’s possible to accurately measure both x1 and p2 since they commute; hence implying that x1 (x2) and p1 (p2) can be measured accurately which, in turn, is the violation of the original commutation relations [x1 , p1] = i \hbar and [x2 , p2] = I \hbar which is not at all acceptable as mentioned earlier. Then how to circumvent this situation? It becomes necessary to accept that x1 measurement necessarily influences p2 (and hence x2), even though the particles are very well separated and no more interacting. Einstein called such an influence as spooky b’cos such a action-at-distance is not at all visible at a mere glance at the quantum formalism.

The above algebraic equations were considered by Bohr in the footnote of his reply to the EPR paper. But actually he missed out the following observation:

In the case of entangled particles, x1 & p2 and x2 &, p1 will not commute, respectively; as they were considered by Bohr b’cos he did not make use or impose the momentum conservation to write down the actual set of commutation relations needed to describe the entangled particles considered by EPR.

[x0 , p1] = 0 = [ x1 – x2 , p1] = [x1 , p1] – [ x2 , p1]

Therefore, [x1 , p1] = [x2 , p1] = i \hbar

and similarly, [x2 , p2] = [x1 , p2] = i \hbar

Which still satisfies the original commutation relation, [x1 – x2 , p1 + p2] = 0.

Therefore, in the case of entangled particles, x1 will not commute with p2 and as well x2 with p1, unlike in the case of non-entangled, free particles. Therefore, the actual commutation relations, [x1, p1] = [x2 , p2] = i \hbar and hence the uncertainty principle were unaffected and there is no room to claim that quantum mechanics is incomplete. So, this simple explanation is sufficient to show that quantum mechanics is spooky. The entire Bohr’s reply to EPR paper revolves around the explanation of the commutation relations [x1 , p2] = [x2 , p1] = i \hbar, which he did not explicitly considered, but only with some physical examples. When you are reading Bohr’s reply to EPR, keep these commutation relations, [x1 , p2] = [x2 , p1] = i \hbar, in the back of your mind. Then you will easily understand what he is trying to say.

But, to visualize physically `How entangled particles are able to communicate even though they are well separated? or How actually the spooky-action-at-a-distance is becoming possible?’, will necessarily depend on the physical reality of the Schrodinger wave function.

with best regards, NG.

Ps: Dear Hussainsha,

If you have some difficulty in seeing the following as mentioned above

[x0 , p1] = 0 = [ x1 – x2 , p1] = [x1 , p1] – [ x2 , p1]

Therefore, [x1 , p1] = [x2 , p1] = i \hbar

and similarly, [x2 , p2] = [x1 , p2] = i \hbar

Which still satisfies the original commutation relation, [x1 – x2 , p1 + p2] = 0.

Then, it's much more easy to see the point by the following ( though they are all equivalent and related by some transformation).

Since momentum is conserved,

p1 + p2 = 0

Now just take the commutation with x1 and x2. Then you have the desired result i.e.,

[x1 , p1 + p2] = [x1 , p1] + [x1 , p2] = 0

Hence

[x1 , p1] = [p2 , x1] = i \hbar; and similarly

[x2 , p2] = [p2 , x1] = i \hbar; which are much more straightforward to see.

Best, NG.

@ N. Gurappa...

My dear N. Gurappa thank you for your valuable explanation .....I try to understand your answer...

Yes. x is a dummy variable. If it represents the polarisation, it is exactly the same as the correlation of polarisation. But in this latter case, the polarisation is represented by only two eigenstates |0> and |1> instead of a continuum |x>. That doesn't change the principle.

The important thing is the state of the system, center of mass well defined, or 0 total angular momentum. Then this state should not be factorisable into a tensor product of states of the individual particles.