I Two particle system, products and entanglement

[PeterDonis]

if we are talking about indistinguishable particles, like photons, are you saying that the above four term superposition expression described in #28 above would not work because it is not symmetric as it should instead be?

It will work as long as $c_2 = c_3$, since the second and third terms are the exchanges of each other. The first and fourth terms are their own exchanges, so their coefficients are not constrained (except by normalization).

 

[fog37]

Thank you so much.

So if the particles are bosons, the coefficients $c_2$ and $c_3$ must be the same. If the particles are fermions, then $c_2 = - c_3$.

Once a measurement is carried out, the two-photon system collapses into any of the four states $|\Psi>_1 |\Psi>_2$, $|\beta>_1 |\beta>_2$, etc. in the superposition. The two particles, being photons, are still bosons and require a symmetric state. But is each one of the tensor products $|\Psi>_1 |\Psi>_2$, $|\beta>_1 |\beta>_2$, etc. a symmetric state?

 

[PeterDonis]

So if the particles are bosons, the coefficients $c_2$ and $c_3$ must be the same. If the particles are fermions, then $c_2 = - c_3$.

Yes.

Once a measurement is carried out, the two-photon system collapses into any of the four states

Yes. But note that these are all separable, so now you don't have a two-photon system any more; you have two one-photon systems. In order to have a two-photon system again, you would have to make the photons interact somehow, and the interaction would change the state.

The two particles, being photons, are still bosons and require a symmetric state.

No, because you now have two one-photon systems, so there is no way to exchange the particles any more; you know which one is which because you measured them. As above, to make it a two-photon system again, the photons would have to interact, and the interaction would change the state, so you wouldn't be able to keep track of which photon was which any more; then the new state of the two-photon system would have to be symmetric.

 

[fog37]

Yes.
Yes. But note that these are all separable, so now you don't have a two-photon system any more; you have two one-photon systems. In order to have a two-photon system again, you would have to make the photons interact somehow, and the interaction would change the state.

No, because you now have two one-photon systems, so there is no way to exchange the particles any more; you know which one is which because you measured them. As above, to make it a two-photon system again, the photons would have to interact, and the interaction would change the state, so you wouldn't be able to keep track of which photon was which any more; then the new state of the two-photon system would have to be symmetric.

So, after the measurement on the composite system in an entangled state, we end up with two one-photon systems and that is what physically happens. So the composite system went from being an entangled two-photon system to a system composed to two independent, non-interacting one-photon subsystems. If I make the photons interact, I would get a system of interacting photons but that system will not necessarily be in an entangled state...

 

[fog37]

Hi PeterDonis,

I am digesting the last comments you made in this thread. Once we make a measurement on the entangled state of the composite system, we end up with two one-photon systems described by a single product state which means the two resulting sub-systems are independent. You mention that

"...now have two one-photon systems, so there is no way to exchange the particles any more; you know which one is which because you measured them. As above, to make it a two-photon system again, the photons would have to interact, and the interaction would change the state, so you wouldn't be able to keep track of which photon was which any more; then the new state of the two-photon system would have to be symmetric."

So, when the total wavefunction is separable (i.e. it can be factored), it describes two systems that are two things:
a) Indistinguishable
b) Independent from each other

When we learn about symmetric and antisymmetric wavefunctions, we talk about multiple electrons (which are always indistinguishable, like photons are) forming a system, being independent and requiring an antisymmetric wavefunction. But I guess it is possible to have, after the measurement on an entangled system formed by one-two electrons system, two subsystems (the electrons) that are independent and now also distinguishable, hence the use of a product state is justified and we don't need to worry about the state being antisymmetric...

 

[PeterDonis]

So, when the total wavefunction is separable (i.e. it can be factored), it describes two systems that are two things:
a) Indistinguishable
b) Independent from each other

b), yes. a), no. There is no requirement that the two systems described by a total wave function that is separable be indistinguishable.

I guess it is possible to have, after the measurement on an entangled system formed by one-two electrons system, two subsystems (the electrons) that are independent and now also distinguishable

They're distinguishable because we measured them, which really means that the "distinguishable" systems are not just the electrons by themselves, but the electrons plus the measuring devices that recorded the results.

Also, as I said before, you can always take the electrons you just measured and have them interact again (which could be as simple as just bringing them back into the same small region of space), which will make them indistinguishable again, so now you have to treat them as a two-particle system with a wave function that has to be antisymmetric under exchange. In other words, the "distinguishability" only lasts as long as you don't change the state of the electrons from the one you measured.