If two particles are entangled and you collapse the wave function of one of the particles. Does the other particle collapse as well?

Question

Let's suppose you entangled two photons, you separate the photons, and then you measure the polarization of one the photons collapsing its wave function. The wave function of the other photon collapses also?

Answer 1

Each photon does not have its own wave function. They are entangled. By definition, there is only one wave function between them. One function describes both particles simultaneously. If you do something to one particle that alters the wave function, then that's it; the wave function is altered.

Here's an analogy: I have a bag with two apples in it. Then I pose this question. If I were to tie a knot in the top of the first apple's bag, would the second apple's bag remain unchanged? The answer is obvious: both apples are in the same bag so if you make changes to the bag of the first apple, the bag of the second can't remain unchanged.

It's the same with entangled particles. The wave function is like the bag; there's only one that describes both particles.

Answer 2

If you have two spins in an entangled state they define a wave function $$ |\psi\rangle~=~\frac{1}{\sqrt{2}}\left(|+\rangle|-\rangle~+~e^{-i\phi}|-\rangle|+\rangle\right) $$ in a singlet state of entanglement. What exists is the entangled state. In effect the individual spin states do not exist. A measurement of one spin state does mean that the total qubit information of the entangled state is now in spins, which means the other spin appears to the. So if Alice measure a spin, then Bob necessarily has the opposite spin.

We can think of this as a mutual collapse. the so called collapse of a wave just means the observables of some system becomes localized in a way that does not obey Schroedinger or any quantum dynamics. This is how we identify states with particles. In the case of entanglement it is the case that if one spin state is localized "here," then it is also localized "there."