What is the concept of nonlocality in quantum mechanics?

[A. Neumaier]

To talk usefully about entanglement , the systems that are considered to be entangled must be distinguishable (usually, by their position or momentum). This is not the case for quarks bound in a proton, or for electrons bound in an atom.

I thought it had to be an INdistinguishable set of states... i.e. HV> + VH> to me is a superposition of states. Whereas if it is either HV> OR VH> AND we simply don't know which... that is a mixed state.

I was talking about systems, not about their state.

For example, two electrons are on the fundamental level indistinguishable; their joint wave function is proportional to |12>-|21>; no other superposition is allowed. Similarly, two photons are on the fundamental level indistinguishable; their joint wave function is proportional to |12>+|21>; no other superposition is allowed. On the other hand, a photon and an electron are intrinsically distinguishable (e.g. by their spin), and arbitrary superposition are possible.

But if, on a more practical level, you know already that (by preparation) you have two electrons or two photons, one moving to the left and one moving to the right, you can use this information to distinguish the electrons or photons; and then you can assign
separate state information to each of them, and also construct arbitrary superpositions.

I agree with you that *general* ignorance of the source may be (usually is) a mixed state. But a special kind of ignorance is that we know "something" about the source but not everything. That can lead to an entangled state, and why I qualified with the word "special".

Can you give an example so that I better understand what you mean?

 

[uzername]

There is always uncertainty with respect to non-commuting observables. That is true even with entangled pairs. There is no "vanishing"!

Doesn't the "wave function" collapse though, once you know? You've reduced the probability to either 1 or 0, right? so how is the uncertainty not vanished?

 

[DrChinese]

Doesn't the "wave function" collapse though, once you know? You've reduced the probability to either 1 or 0, right? so how is the uncertainty not vanished?

Sure it collapses, and the results give you complete certainty for at least one observable - let's say position. That makes momentum, which is non-commuting with respect to position, completely UNcertain.

 

[DrChinese]

Can you give an example so that I better understand what you mean?

Sure. If I have a system of 2 particles with total spin=0, and I don't know which is which (indistinguishable), those particles are ALWAYS spin entangled. And by the term "don't know" I mean: cannot know, in principle.

 

[DrChinese]

Sure. If I have a system of 2 particles with total spin=0, and I don't know which is which (indistinguishable), those particles are ALWAYS spin entangled. And by the term "don't know" I mean: cannot know, in principle.

And just to add to that: those 2 particles can be ANY 2 particles, existing anywhere, with any prior history, at any points in spacetime. I believe that is the rule. Strange as it may seem.

 

[A. Neumaier]

Sure. If I have a system of 2 particles with total spin=0, and I don't know which is which (indistinguishable), those particles are ALWAYS spin entangled.

It is precisely this situation that I had in mind when saying that entanglement is no longer meaningful when particles are indistinguishable. Entanglement requires a tensor product state space, while in this case, you only have the symmetric or antisymmetric part of the tensor product. The state space of indistinguishable particles is very different from that of distinguishable particles. Almost nothing of the usual theory about entanglement survives this change of basic setting.

And by the term "don't know" I mean: cannot know, in principle.

I consider something that cannot be known in principle to be meaningless, not a case for ignorance.

 

[DrChinese]

It is precisely this situation that I had in mind when saying that entanglement is no longer meaningful when particles are indistinguishable.

So I guess you are saying that 2 particles with total spin 0 are NOT entangled if they are indistinguishable. Or ? Perhaps my use of terms is different than yours? Because I have never heard it described otherwise.

 

[A. Neumaier]

So I guess you are saying that 2 particles with total spin 0 are NOT entangled if they are indistinguishable. Or ? Perhaps my use of terms is different than yours? Because I have never heard it described otherwise.

The common assumption in a formal definition of of entangled systems, here quoted from http://en.wikipedia.org/wiki/Entangled_state , is: ''The Hilbert space of the composite system is the tensor product''.

This is violated for indistinguishable bosons or fermions, where the Hilbert space of the composite system is the symmetrized and antisymmetrized tensor product, respectively.

Of course, many people use the term without referring to a precise definition of the context in which it makes sense. And commonly there is a lot of imprecision in working with indistinguishable particles outside of quantum field theory!

But I like to have precise concepts determined by the usefulness of what is defined.

 

[uzername]

Sure it collapses, and the results give you complete certainty for at least one observable - let's say position. That makes momentum, which is non-commuting with respect to position, completely UNcertain.

Yes - I meant only one observable. If we claimed to know them all, is that the same or related concept as hidden variables?

 

[DrChinese]

...Of course, many people use the term without referring to a precise definition of the context in which it makes sense. And commonly there is a lot of imprecision in working with indistinguishable particles outside of quantum field theory!

But I like to have precise concepts determined by the usefulness of what is defined.

Me being one of the guilty!

Thanks for clarifying.