In the depicted setup, what would happen? (entangled photons split into 2 different paths)

  • #36
PeterDonis said:
Yes, they are. ##\ket{A} + \ket{D}## is just ##\ket{H}## (up to normalization). Which is obviously a different state from ##\ket{H}## + ##\ket{V}##, which is ##\ket{D}## (again up to normalization). See the formulas @DrClaude gave in post #2.
I’m sorry, of course you’re right.
 
Physics news on Phys.org
  • #37
DrChinese said:
1. There is no such thing as creating 2 photon polarization entanglement on the H/V basis, but not polarization entanglement on the A/D basis. The H/V designation in these cases is arbitrary. An entangled photon pair is in a superposition of all polarizations, and will yield perfect anti-correlations (or correlations depending on PDC type) at *all* angles. For your analysis, this poses a problem.
One must really be more careful discussing these things, particularly if one does not use the adequate language of mathematics. So here's what's really going on using this only adequate language:

The photon pair is produced in the singlet polarization state (e.g., by type-2 parametric down-conversion), represented by the ket
$$|\Psi \rangle=\frac{1}{\sqrt{2}} ( |HV \rangle -|VH \rangle).$$
What's the state of "photon A"? This question is answered by calculating the "reduced state" of photon A:
$$\hat{\rho}_A=\mathrm{Tr}_{B} |\Psi \rangle \langle \Psi| = \sum_{j,k,j' \in \{H,V \}} |j \rangle \langle jk|\hat{\rho} |j'k \rangle \langle j'|=\frac{1}{2} (|H \rangle \langle H|+|V \rangle \langle V|)=\frac{1}{2} \hat{1},$$
i.e., photon A is an unpolarized photon, i.e., it's a mixed state and not a superposition.

Nevertheless, it's true that you get 100% "anticorrelations" when measuring the polarization of both photons wrt. any polarization direction. You can calculate this with any basis since the antisymmetric singlet state is a scalar state.
DrChinese said:
2. There is no disagreement between @vanhees71 and I on this point. He is technically correct (as he usually is) that entanglement does not stop at the PBS. We know that because it is possible to reverse that operation and restore the entanglement as long as it is done before the detectors. But for analysis of your setup, that point does not matter. After the PBS, for all practical purposes there is no polarization entanglement remaining between photon A and photon B. Note again that the timing and ordering of which occurs first - A's polarizer or B's PBS - does not matter.
In this example there's indeed no polarization entanglement remaining, but it's because of the polarizer for photon A, which is a projection operator not the PBS applied to Photon B:
$$\hat{P}(\pi/4) \otimes \hat{1} |\Psi \rangle=\frac{1}{\sqrt{2}} |DA \rangle.$$
Now the two-photon state is a pure product state and thus unentangled.

The HWP makes out of this
$$\left [ \hat{1} \otimes \hat{U}_{\text{HWP}}(\pi/8) \right] \left [\hat{P}(\pi/4) \otimes \hat{1} \right ]|\Psi \rangle = -\frac{1}{\sqrt{2}} |D V \rangle,$$
which is still a pure product state, and finally after the PBS (see the notation of my previous posting)
$$\left [\hat{1} \otimes \hat{U}_{\text{PBS}} \right ] \left [ \hat{1} \otimes \hat{U}_{\text{HWP}}(\pi/8) \right] \left [\hat{P}(\pi/4) \otimes \hat{1} \right ]|\Psi \rangle = -\frac{1}{\sqrt{2}}|D\rangle \otimes V,\vec{p}_{\text{in}} \rangle,$$
which is still unentangled. This is saying that, if detector A registers the idler photon, then the signal photon must necessarily end up in detector B2.
 
  • #38
JamesPaylow said:
@DrClaude my sincere gratitude for continuing to respond, and in a friendly manner. I hear your point that the PBS does not change the state. I'm still struggling to understand the *why* of the outcome you're describing though. Specifically in the case where the polarizer acts after both the HWP and PBS, and let's say also add after interacting with detectors B1/B2. The way I understand you point, you're saying that since interaction with the polarizer is the thing that finally defines the polarization state, when that happens the entire system will behave as if the entangled photons started in that final state. In other words, it doesn't matter when sequentially Photon A encounters the polarizer, in order to get past the polarizer it will become (D) and as a result Photon B will behave as if it started at as (A), subsequently being converted to V at the PBS and being routed to Detector B2. Correct?
Yes, I think this is correct.

JamesPaylow said:
I'm stuck on the idea that sequence can matter, though I know basically everyone here tells me it can't, I'm really trying to better understand why it cannot. If I've accurately captured the description above, my core question is why the polarizer is the only thing in this setup that can determine the final state.
You have to remember that we are looking at correlations between detectors. You can only get clicks at A and one of the B detectors for certain states of both photons. If you forget about quantum entanglement for a second, think of a source that produces pairs in state ##\ket{DA}## or ##\ket{AD}## (in your original setup). The first possibility will result in clicks detector B2 only when there is also a click at detector A (photon A passes through the polarizer to detector A, photon B gets rotated to V and passes through the PWS to detector B2). The second possibility results in no clicks on either A or B2 (photon A gets absorbed by the polarizer, photon B is rotated to H and reflected towards detector B1). Nothing mysterious, and timing is utterly unimportant: it doesn't matter when which photon reaches which part of the setup when.

Now what QM shows us is that superposition and entanglement do exist. A quantum system can be in a superposition of different states and measurements of these states will be random, and entangled particles will show the proper correlations, even if the measurements are spatially separated.

But when you analyze the results, they look as if the state of each particle pair was decided from the beginning. You are back to the "or" scenario I just discussed above, where there cannot be an influence from when each photon reached each part of the setup.

One is tempted to think in terms of a "spooky action at a distance," that when photon A reaches the polarizer, it affects the state of photon B. But in actuality, our observation of results is post-facto, after all interactions are finished and we can analyze the results of all detectors jointly.

The "weirdness" of QM comes from the fact that while everything looks like it was decided from the beginning (my "or" scenario), we know that it is not the case because we can make measurement choices after the particles have been sent on their way that show that they had to be in a superposition of states and entangled with each other.

JamesPaylow said:
You've made it clear that the PBS does not do so as I initially thought. How about if I insert distinctly oriented polarizers, suppose D and A, immediately in front of Detectors B1 and B2? Then photons coming out of the PBS will have a determined state. And assume still that interaction with Detectors B1/B2 happens sequentially before that with the polarizer in Path A.

Unless you really want a mathematical description of this, I will not look at this scenario because I am quite sure it will add nothing of interest. I have feeling that you simply need to convince yourself that there is no actual "spooky action at a distance."
 
  • Like
Likes PeterDonis and vanhees71
  • #39
DrChinese said:
After the PBS, the B stream is no longer entangled with the A stream. The outputs of the PBS are streams of V and H polarized photons, no longer in a superposition. So the outputs of the B1 and B2 polarizers are simply random (i.e. unentangled) photon streams polarized A (if from B1) or D (if from B2).

vanhees71 said:
After the PBS, which is a unitary operation too (in the idealized sense of course), the two photons are still entangled. Only after detection they are no longer entangled. Then, of course, the detected photon is absorped by the detector, and you have a single-photon state left.

DrChinese said:
entanglement does not stop at the PBS. We know that because it is possible to reverse that operation and restore the entanglement as long as it is done before the detectors. But for analysis of your setup, that point does not matter. After the PBS, for all practical purposes there is no polarization entanglement remaining between photon A and photon B

vanhees71 said:
In this example there's indeed no polarization entanglement remaining, but it's because of the polarizer for photon A, which is a projection operator not the PBS applied to Photon B

Having trouble unpacking these statements into a consistent conclusion. If I place a polarizer between the PBS and Detectors B1 & B2 will those polarizers impact the polarization of Photon A or no? Ultimately it's a question of whether the two photons are still entangled at this point.

And, if they are, what is their relationship - Just offset by 45 degrees due to the HWP, and no additional specific impact to account for from the PBS?
 
  • #40
JamesPaylow said:
Having trouble unpacking these statements into a consistent conclusion. If I place a polarizer between the PBS and Detectors B1 & B2 will those polarizers impact the polarization of Photon A or no? Ultimately it's a question of whether the two photons are still entangled at this point.

I am happy to report that both @vanhees71 and I will agree on this point: There will be no observable difference on the A stream (regardless of any change in the coincidence rate). :smile:
 
  • Like
Likes vanhees71

Similar threads

Replies
15
Views
2K
  • Quantum Physics
Replies
4
Views
1K
Replies
8
Views
1K
Replies
3
Views
1K
  • Quantum Interpretations and Foundations
2
Replies
52
Views
2K
  • Quantum Interpretations and Foundations
2
Replies
45
Views
4K
Replies
1
Views
2K
Replies
9
Views
2K
Replies
39
Views
7K
Back
Top