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?
[Physics] If two particles are entangled and you collapse the wave function of one of the particles. Does the other particle collapse as well
quantum mechanicsquantum-entanglementwavefunction
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As I understand it, the question boils down to, "Why doesn't a mirror collapse the wavefunction of a photon that reflects off the mirror?". The answer is that the photon does not change the state of the mirror. After the photon has been reflected, the mirror is unchanged. There is no way to prove that the photon struck the mirror without also detecting the photon's path downstream. The only state-changing event that occurs to the photon is its detection where it strikes a screen (or camera sensor, or an observer's eye, etc.).
It is not correct that the wavefunction collapses upon interaction. It is "less incorrect" to say that the wavefunction collapses upon detection. "Less incorrect" requires a bit of explanation.
Detection is an interaction that results in an observer "knowing" that the wavefunction has collapsed. Yes, it's a vaguely circular definition. In the many worlds view of quantum mechanics, an observer "detects" the state of a particle and in doing so splits his world into as many different independent alternative worlds as there are possible values for the state of the particle.
In the case of the two-slit interferometer, the observer detects the position ("state") of each photon that strikes a screen. That "detection" (according to the MW view) is not really an observation of what the photon's state (i.e., location) is, but rather a projection of the observer's world onto one of the possible values of the photon's state. In a sense, the observer's world splits into all the possible worlds that would result immediately after the detection event, depending on the possible different state values that the photon might have.
The gedankenexperiment of Schroedinger's cat can be generalized to help explain this. Suppose we put the observer inside a box that is totally isolated from the rest of the universe, and the observer in the box detects a photon on a screen. We cannot know where on that screen the observer detected the photon, until we open the box and look at the observer's records. Moreover, according to Bell's theorem, from our perspective, the observer himself is a quantum object -- so the location of the photon's impact on the screen does not even have an actual value from our perspective until we open the box. The value is not hidden; it is indeterminate. The observer inside the box is sure to think he knows where the photon hit his screen, but from our perspective the observer is in a superposition of states until we "detect" what his state is: until we "open the box". As far as we are concerned, the observer co-exists in all possible states corresponding to different places the photon strikes his screen; and in each state he is sure he knows where the photon struck his screen -- but for each of his different states the photon landed a different place.
All that is background for saying that as long as there is no possible way to know that the photon reflected off a given mirror in an interferometer, the photon wavefunction takes all available paths including those in which it does not reflect of the mirror -- and it will thus form the interference pattern we observe.
If we were to make the mirror so tiny and thin that its recoil could in principle be detected , there would be no interference pattern. I know some folks will ponder what might happen in the gray area between using an extremely tiny & thin mirror and using a normal several-to-many grams mirror. I don't know if anyone has done the experiment, but I'll bet what happens is that the interference pattern's contrast is reduced as the mirror's recoil approaches detectability. It would be an experiment worth doing.
Best Answer
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.