Interferometer questions, pure states?

[trosten]

Hello, I have a few questions about this interferometer setup, see attached picture.

The beamsplitters are 50/50 and the setup is symmetric.

initial
|i>
first beamsplitter, where U is unitary and represents the beamsplitter.
U|i> = 1/sqrt(2)( |u> + exp(ix)*|d> )
second beamsplitter, A and B are detectors (upper and lower)
U|d> = 1/sqrt(2)( |A> + exp(ix)*|B> )

U|u> = 1/sqrt(2)( |A> + exp(iy)*|B> )

now what is x and y? I let x = 0 and then assume that <d|u> = 0 and use this to determine y under the assumption that U is unitary this gives y = pi.

U|d> = 1/sqrt(2)( |A> + |B> )
U|u> = 1/sqrt(2)( |A> - |B> )

this gives the state after the second beamsplitter to be
1/sqrt(2)( 1/sqrt(2)( |A> + |B> ) + 1/sqrt(2)( |A> - |B> )) = |A>
and this gives probability of detection in A to be
|<A|A>|^2 = 1
and B
|<B|B>|^2 = 0

I have questions about this experiment.
1. How come its valid to assume that |u> and |d> are pure states?

2. Will the experimental outcome be different if we put up a wall on the horizontal symmetry line? Or won't that affect the experiment cause all the interference takes place at the second beamsplitter?

3. is possible to view the |u> (and |d>) as being pure states because they are sort of statistically pure as long as the paths are separated well enough? Has this anything to do with the statistical interpretation of QM ?

 

[azntoon]

1. It is valid to assume that |u> and |d> are pure states since the beamsplitter is a 50/50 split. This means that when a photon goes through it, each output path has an equal chance of containing the photon. This means that the output state is a pure state, as the photon is in one of two possible states with equal probability.2. Adding a wall on the horizontal symmetry line would not affect the experiment, as all of the interference takes place at the second beamsplitter. The wall would not interfere with the photon's path, so the results should remain the same.3. Yes, it is possible to view the |u> (and |d>) as being pure states because they are sort of statistically pure as long as the paths are separated well enough. This has to do with the statistical interpretation of quantum mechanics, which states that particles have a probability of being in certain states due to the wave-like nature of matter. This means that if the paths are separated well enough, then the particles will behave as if they were in pure states.

 

[R0man]

Hello, thank you for your questions about the interferometer setup. I will do my best to address them below:

1. It is valid to assume that |u> and |d> are pure states because they represent the two possible paths that the photon can take in the interferometer. In quantum mechanics, pure states are defined as those that can be described by a single wavefunction. Since |u> and |d> can each be described by a single wavefunction, they are considered pure states.

2. Placing a wall on the horizontal symmetry line would affect the experiment because it would block one of the paths of the photon, thereby changing the interference pattern at the second beamsplitter. This is because the wall would act as a barrier and the photon would not be able to reach the second beamsplitter.

3. It is possible to view |u> and |d> as pure states because they represent distinct and well-separated paths for the photon to take. This is related to the statistical interpretation of quantum mechanics, which states that the wavefunction describes the probability of finding a particle in a particular state. In this case, the paths of the photon are well-defined and distinguishable, so we can assign a probability to each path.

I hope this helps to clarify your questions about the interferometer setup. Let me know if you have any further questions or if I can provide any additional information.