PBS Interferometer: Testing Polarization of Photons

[StevieTNZ]

Hi there

In "Dance of the Photons" by Anton Zeilinger (pgs. 82-84), Zeilinger has a polarising beam splitter interferometer as such-
http://img.photobucket.com/albums/v608/steviet/IMG_20140905_0001_zpsc8add996.jpg
(assume the length of paths are the same)

The input photon is polarised 45 degrees. The output is that a 45 degree polarised photon is found in the lower output.

Would the output be the same if a photon of no definite polarisation (say, as part of an entangled pair) is inputted into the interferometer?

The experiment is designed to test whether the photon takes on |H> or |V> polarisation at the PBS.

Many thanks
Stevie

 

[atyy]

Measurements on a subsystem will produce the same results as long as the reduced density matrix of the subsystem is the same. Only a measurement that looks at the full density matrix will be able to tell the difference.

But I don't think I can get a photon whose reduced density matrix is the same as a pure state at 45°, and also have it be entangled to have no definite polarization. The reason is that entanglement usually causes the state of the subsystem to appear mixed, not pure. I know I can get an entangled pair, so that when you look at one photon, half of them will appear to be polarized at 45° and the other half will be polarized at 135°.

So I think your question cannot be set-up.

 

[StevieTNZ]

But I don't think I can get a photon whose reduced density matrix is the same as a pure state at 45°, and also have it be entangled to have no definite polarization.

In the diagram I draw a photon polarised at 45 degrees just as an example (as per the book).

I was wondering whether we replaced that photon with another photon in no definite polarisation would we get the same output?

 

[atyy]

When looking at one photon in an entangled pair (let's say something like a Bell state), the photon will behave half the time as if it had a definite polarization in one direction, and the other half the time as if it had a definite polarization in the orthogonal direction.

So you can treat it as if each photon had a definite polarization on any particular run of the experiment, but different definite polarizations on different runs.

So let's say it's entangled so that if you look only at one photon, half the time it will have a definite polarization of 45°, and the other half the time it has a definite polarization of 135°. You already know how the system will treat a 45° photon, so you just need to figure out how the system will treat a 135° photon. (I'm going to guess that on symmetry, a 135° photon will always come out of the side that you marked "0 photons" for the 45° case. If that is correct, then you should get equal numbers of photons coming out in both final directions.)

 

[StevieTNZ]

Wouldn't the photon be in a superposition of being reflected and transmitted at the first PBS, and likewise at the 2nd PBS?

 

[atyy]

Wouldn't the photon be in a superposition of being reflected and transmitted at the first PBS, and likewise at the 2nd PBS?

I don't think so, because if you only look at one photon, it should behave as if it's a mixture of photons each with a definite state, not a superposition. The state of both photons is a superposition, but not the state (reduced density matrix) of one photon in the entangled pair.

 

[StevieTNZ]

I don't think so, because if you only look at one photon, it should behave as if it's a mixture of photons each with a definite state, not a superposition. The state of both photons is a superposition, but not the state (reduced density matrix) of one photon.

To get the correct result, obviously we would need to take into account the state of the other photon, wouldn't we?

 

[atyy]

To get the correct result, obviously we would need to take into account the state of the other photon, wouldn't we?

Yes, and (if I understand correctly, which I'm not sure I do), I have already taken the other photon into account when I say that if you look only at one photon, it behaves like a mixture of photons with one of two definite polarizations.

One way to see this is to assume that someone made a measurement on the other photon and broke the entanglement. Then we know the state will collapse into an unentangled state, and looking at just one photon will be a mixture.

However, because the other side cannot send signals faster than light, and also because simultaneity is relative, we should be able to choose a frame in which we do the measurement before the other side. And if relativity and quantum mechanics go together, we should get the same result as in a different frame where the other side measured first. So even if the other side has not measured and not broken the entanglement, we will get the same results as if the other side had measured and broken the entanglement.

Here I have assumed the results are "real" and the same in every frame, while unreal things like wave function, entanglement, collapse are different for each frame, and that's ok as long as combining these unreal things in different ways always produces the same reality.

 

[naima]

I like this pdf
You find detailled calculus.

 

[StevieTNZ]

If the input photon was polarized as |H> (Horizontal), will the photon be reflected at the 1st PBS, and also at the 2nd PBS? Or at the 2nd PBS would the photon be transmitted as it is coming towards that PBS at a different direction?

 

[naima]

did you read the link i gave you?
You wil see what happens for polarized photons in a setup with 4 PBS, 2 rotators, 2 mirrors and 2 detectors.
Is there a problem along one of the 4 paths?

 

[StevieTNZ]

did you read the link i gave you?
You wil see what happens for polarized photons in a setup with 4 PBS, 2 rotators, 2 mirrors and 2 detectors.
Is there a problem along one of the 4 paths?

I did. However by adding more than what I have in my set-up complicates following what would happen from what is written in the link you provided.

I'm looking in Zeilinger's book, where at the first PBS a |V> polarized photon would be transmitted through, and at the 2nd PBS, would end up on the right hand side of that PBS - as per this image: http://physics.illinois.edu/people/kwiat/Interaction-Free-Measurements_files/current.gif (bottom PBS)

 

[atyy]

I did. However by adding more than what I have in my set-up complicates following what would happen from what is written in the link you provided.

I'm looking in Zeilinger's book, where at the first PBS a |V> polarized photon would be transmitted through, and at the 2nd PBS, would end up on the right hand side of that PBS - as per this image: http://physics.illinois.edu/people/kwiat/Interaction-Free-Measurements_files/current.gif (bottom PBS)

I thought that in Zeilinger's book the two PBSs are not oriented the same way. The first PBS splits a 45° beam into two, so the first PBS is V/H oriented. The output of the second PBS is a pure 45° beam on one side, so I think it is 45°/135° oriented.

 

[StevieTNZ]

It appears that they are both orientated in the H/V basis. The 2nd PBS just recombines the two beams (into one) from the 1st PBS.

 

[atyy]

It appears that they are both orientated in the H/V basis. The 2nd PBS just recombines the two beams (into one) from the 1st PBS.

Yes, you are right. I found an explanation by googling, which also answers your question about what happens if the initial beam has a different polarization.

http://www.users.csbsju.edu/~frioux/photon/MZ-Polarization.pdf

So that means in the entangled case, my guess above was wrong. In the entangle cased, when we look at only one photon of a pair, we have a mix of 45° and 135° photons. Since this arrangement always preserves the orientation, but changes the direction of the input beam, the output beam will exit in the same direction as the pure 45° beam, but it will contain a mix of 45° and 135° photons.

 

[StevieTNZ]

Totally confused now. In the PDF you provided, it says |H> or |V> polarized photon will be found in the same output path.

But if this diagram is correct: http://physics.illinois.edu/people/kwiat/Interaction-Free-Measurements_files/current.gif (bottom PBS, the alignment of the PBS is the same as my set-up), a |V> polarized photon coming in from the top towards the 2nd PBS (which it would if it is transmitted at the 1st PBS in my set-up) would end up to the right of the PBS, not below it.

 

[StevieTNZ]

Yeah, I'm confused. I'm getting conflicting information. In Zeilinger's book, the PBS orientation is the same as that in the PDF you provide. But in that PDF it says horizontally polarized photons are transmitted. In Zeilinger's book it says transmitted photons are vertically polarized.

 

[atyy]

Yeah, I'm confused. I'm getting conflicting information. In Zeilinger's book, the PBS orientation is the same as that in the PDF you provide. But in that PDF it says horizontally polarized photons are transmitted. In Zeilinger's book it says transmitted photons are vertically polarized.

I think it doesn't matter. The set up is such that if component x is transmitted the first time, then it will be transmitted the second time; and if component y is reflected the first time, then it will be reflected the second time.

 

[naima]

I realize that there is something that i do not understand.
Take a Michelson interferometer
look at fig 3 a) Suppose that this beam splitter is a PBS. If an horizontal photon goes thru the PBS when it comes back from the mirror it will pass again thru the BPS and will not be detected?

.

 

[naima]

You can see what happens in an interferometer setup when you follow the Jones matrix along the paths.
A PBS has two input channels and two output channels.
It receives a J1 matrix from the left and a J2 matrix from the other channel.
the two outputs are
$$\left( \begin{array}{cc} 1 & 0 \\0& 0 \end{array} \right) J1 + \left( \begin{array}{cc} 0 & 0 \\0& 1\end{array} \right) J2$$.
and
$$\left( \begin{array}{cc} 0 & 0 \\0& 1 \end{array} \right) J1 + \left( \begin{array}{cc} 1 & 0 \\0& 0 \end{array} \right) J2$$.
on the other path
The fist PBS receive J1 = Id and J2 = 0
The Jones Matrix of a mirror is
$$\left( \begin{array}{cc} 1 & 0 \\0& -1 \end{array} \right)$$

So we multiply (by the left) by this matrix and we get the inputs of the second PBS.
This will give you the final result.
You said that the link i gave you was complcated. You can easily follow this method along each of tjhe 8 paths;