What happens when entangled photons are split into different paths?

[vanhees71]

Seems to be some disagreement on whether the PBS breaks entanglement or not, but let's just assume the detectors also interact before the polarizer in Path A. So if the PBS doesn't break entanglement the detectors at least will.

You can check this for yourself. You just need to think about what the PBS does. It's (thinking about an idealized lossless device again) another unitary transformation, when acting on the polarization states. What it does is simply to entangle the polarization states H/V to the photon momentum: One state goes through the other is reflected to 90 degrees, i.e., you have
$$\hat{U}_{\text{PBS}} |H, \vec{p}_{\text{in}} \rangle = |H,\vec{p}_{\text{refl}} \rangle,$$ $$\hat{U}_{\text{PBS}} |V, \vec{p}_{\text{in}} \rangle = |V,\vec{p}_{\text{in}} \rangle.$$
So the two-photon state when one photon went through the polarizer and the other through the HWP and the PBS is $-|D \rangle \otimes |V,\vec{p}_{\text{in}}$ (which is not an entangled state, because it's a product state). Since in the direction $\vec{p}_{\text{in}}$ if Detector A detects the its photon then for sure Detector B2 clicks.

 

[DrChinese]

1. This may not be a material point, but my expectation is that we start with an H/V superposition explicitly, not A/D. I believe I have a method of creating entangled pairs in this way, and have come up with a way of validating that prior to the experiment. It's all beyond the scope of this question though, so if we can let's just assume that the entangled photons are generated such that we know one is H and the other is V (though don't know which is which).

2. Seems to be some disagreement on whether the PBS breaks entanglement or not, but let's just assume the detectors also interact before the polarizer in Path A. So if the PBS doesn't break entanglement the detectors at least will.

3. It doesn't seem to give the same result if we analyze from B's perspective though; I suppose that's what I'm struggling with. Photon B starts as H/V, encounters the HWP and becomes A/D, is routed 50% towards Detectors B1/B2. So B1 registers half A and half D, B2 also registers half A and half D...

4. So basically by evaluating a given Photon B that has reached either Detector B1 or B2 we can't say whether it started as H or V, because either state could have led to detection at either B detector. But, now that it's been detected at Detectors B1 or B2 the polarization state is defined; it's no longer in superposition. But, if Photon B started as H or V, as we expected, ...

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.

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.

3. This analysis is acceptable so far. Keep in mind, however, that the B stream A/D pairing now maps its entanglement to the A stream H/V pairing. Which also means: B stream H/V pairing now maps its entanglement to the A stream A/D pairing. This is critical to understand for the next step. And this is why your assumption is a problem.

4. When the rotated B stream H/V basis arrives at the H/V PBS, the matching A stream entanglement is on the A/D basis. So when those A photons pass the D oriented filter, they are being matched only to the B stream that goes one way at the PBS. Again, here we are examining from the perspective that the B stream is the controlling perspective (as if ordering mattered). So the PBS is serving to "cause" collapse of the A stream into its A/D entanglement.

Again, in this analysis (as in the analysis from A's perspective from my prior post): The B stream started life "as if" it was A/D polarized (of course it was really in a superposition of all polarization states). Naturally that means the A stream started life "as if" it was similarly polarized.

----------

So we conclude: your invalid assumption that the entangled photon pair is entangled at a preferred angle (H/V) - but not otherwise - cannot be used to yield a correct analysis of a quantum mechanical setup. If you use it anyway, you will come to a conclusion that is at odds with what actually happens, i.e. that somehow ordering matters (it doesn't per theory and experiment).

 

[Nugatory]

This may not be a material point, but my expectation is that we start with an H/V superposition explicitly, not A/D. I believe I have a method of creating entangled pairs in this way, and have come up with a way of validating that prior to the experiment.

You have made some mistake here, as the A/D superposition and the H/V superposition are the same state, just written as the sum of two vectors in different ways. I can write 12=7+5 and 12=8+4, but it does not follow that there are two kinds of cartons of a dozen eggs, the 7+5 kind and the 8+4 kind. Likewise $|A\rangle+|D\rangle$ and $|H\rangle+|V\rangle$ aren't different states, just the same state written as the sum of two vectors in different ways.

 

[PeterDonis]

the A/D superposition and the H/V superposition are the same state

To be clear: the states (ignoring normalization factors) $\ket{AD} - \ket{DA}$ and $\ket{HV} - \ket{VH}$ are the same entangled state (the "singlet" state), expressed in two different bases. This is the initial state in the experiment. ("Entangled" is a better term than "superposition" because, as I think I've already pointed out in this thread, the latter is basis dependent but the former is not.)

But after the HWP acts, the state is a different entangled state, $\ket{AH}+ \ket{DV}$ (again ignoring the normalization factor).

 

[PeterDonis]

$|A\rangle+|D\rangle$ and $|H\rangle+|V\rangle$ aren't different states

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.

 

[Nugatory]

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.

 

[vanhees71]

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.

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.

 

[DrClaude]

@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.

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.

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."

 

[JamesPaylow]

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).

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.

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

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?

 

[DrChinese]

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).

 

[PeterDonis]

Moderator's note: A separate discussion about terminology has been moved to a new thread in the interpretations subforum:

https://www.physicsforums.com/threads/quantum-nonlocality-and-spooky-action-at-a-distance.1056028/

 

[PeterDonis]

Moderator's note: Another separate discussion about how the MWI accounts for experiments like this one has been moved to another new thread in the interpretations subforum:

https://www.physicsforums.com/threads/mwi-and-the-entangled-photon-experiment.1059053/