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?


2 Answers 2


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.

  • $\begingroup$ Hi Jim, thanks for the answer. I hope that you can help me with another doubt; lets suppose that you put a particle under a potential, then its wave function will change according to Schrodinger equation. If this particle is entangled with a second particle then its wave function (that is the same of the first particle) will also change? $\endgroup$ Mar 9, 2017 at 17:19
  • 1
    $\begingroup$ @SiriusFuenmayor if you take an entangled state and do something that affects the wave function, then the wave function changes. If the particles are still entangled afterwards, then they both still have only one (now changed) wave function. However, it is not necessary that these particles remain entangled. If they are not entangled afterward, then the new wave function of the second does not need to be the same as the first, but it has to be different from the original function, because it no longer describes a pair of entangled particles $\endgroup$
    – Jim
    Mar 9, 2017 at 18:53

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."

  • $\begingroup$ "if one spin state is localized "here," then it is also localized "there."" can you clarify it a bit more? $\endgroup$ Mar 9, 2017 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.