Polarization Entanglement, following Bohmian ideas and/or orthodox QM

[DrChinese]

This question follows recent several threads on Bohmian approach to entanglement concepts, and referencing Norsen also.

We have our usual polarization entangled (|HH>+|VV> photon pair A and B going to Alice and Bob. We measure A and it emerges from the V> port of a Polarizing Beam Splitter. (Let's ignore consideration of anything that might be hypothetically emerging from the H> port in some interpretations.) We know with certainty that the B photon, were it to be measured on the H/V basis, would be seen to be V> also.

How would the Bohmian (or anyone really) describe the difference (if any) between A and B before they arrive at their respective detectors? What statements might you make about their respective states, wave functions in terms of spin/polarization? Would you call them V> polarized, in a pure state? Both? Neither?

Does the A photon need to be actually detected before there is any change to the state or wave function (or anything else) of B? (As best as I can tell from Norsen, A simply emerging from the PBS should be enough to influence B.)

Edited to add: Not sure I was clear; although I am interested in the Bohmian perspective, I would also equally appreciate anyone's thoughts from the perspective of standard QM, Copenhagen, MWI, etc.

 

[DrChinese]

I might guess that in Standard QM: The A photon is in a pure V> polarization state and the B photon is still in an entangled state even though an H/V measurement would yield V>.

And I might naively guess that in Bohmian Mechanics (after consulting Norsen, figure 6): The A photon is now in a pure vertical polarization state and the B photon is also in a pure vertical polarization state.

But I am really not sure. Anyone?

 

[anuttarasammyak]

Hello.

We have our usual polarization entangled (|HH>+|VV> photon pair A and B going to Alice and Bob.

I understand that A and B entangle so that they have same |X>. Alice observes |X>. She anticipates that Bob should observe |X> also but it is not sure, e.g.,
-He forgets to do it.
-His apparatus is out of service or in service but mal-tuned
-Photon going to Bob is destroyed or invoved in new entanglement with unknown body
Alice can confirm/decline her anticipation by light speed contact from/to Bob after planned time of observation.

In this sense orthodox QM wave function and its BM correspondent should share that
they encode the information available to the observer and provide predictions for possible measurement outcomes.

 

[Matterwave]

I might guess that in Standard QM: The A photon is in a pure V> polarization state and the B photon is still in an entangled state even though an H/V measurement would yield V>.

In standard QM, once A is measured to emerge from the V polarization port then B will be in state $|V\rangle$, but I think you know this. I'm not sure why you would say that B is still entangled even after A has been measured.

And I might naively guess that in Bohmian Mechanics (after consulting Norsen, figure 6): The A photon is now in a pure vertical polarization state and the B photon is also in a pure vertical polarization state.

For Bohmian mechanics, you should not say "A photon is now in a pure vertical polarization state". Closer would be "the wave function of A is in a pure vertical polarization state". These are not the same sentences.

In Bohmian mechanics, the position-basis is preferred since it is only to particle positions (and velocities) that BM gives ontological status. As has been emphasized many many times now, the wave function of a particle is not the state of the particle in BM. The wave function is used to calculate the probability current and density in the guidance equation (eq. 2 of the paper you cited).

Unrelated -- if you are going to be following along a paper about spin, why not ask about spin? Why ask about polarization? Is there something about polarization that is special to you?

 

[DrChinese]

1. In standard QM, once A is measured to emerge from the V polarization port then B will be in state $|V\rangle$, but I think you know this. I'm not sure why you would say that B is still entangled even after A has been measured.

2. For Bohmian mechanics, you should not say "A photon is now in a pure vertical polarization state". Closer would be "the wave function of A is in a pure vertical polarization state". These are not the same sentences.

In Bohmian mechanics, the position-basis is preferred since it is only to particle positions (and velocities) that BM gives ontological status. As has been emphasized many many times now, the wave function of a particle is not the state of the particle in BM. The wave function is used to calculate the probability current and density in the guidance equation (eq. 2 of the paper you cited).

3. Unrelated -- if you are going to be following along a paper about spin, why not ask about spin? Why ask about polarization? Is there something about polarization that is special to you?

Great comments! Thanks!!

1. There is a technical reason for the distinction (is B still entangled or not) that few people are aware of. If B's wave function (your preferred term) is in a pure vertical polarization state i.e. V>, it is not able to participate in an entanglement swap. Were that the case, the Delayed Choice version of swapping would not work. See 3. below for explanation. Since Delayed Choice versions work, it must not be in a V> wave function. At least I think, assuming everything said in 3. below is correct. I'm asking if anyone knows more on this than I have read.

2. Thanks. I admit I'm not sure how this distinction figures into experimental situations and expectation values, but I will try to use appropriate phrasing.

3. Yes, there are several things specific to polarization entanglement when applied to entanglement swapping scenarios. I am unable to formulate a suitable parallel to spin 1/2 systems. Virtually every swapping scenario I've seen uses photons for the BSM regardless of the particle type for the Alice/Bob measurements.

Again, there is the issue pointed out in 1 regarding swapping protocol. The other is unique to Type I PDC entanglement. That method requires 2 PDC crystals oriented perpendicular. One crystal produces only V> polarized pairs. These are entangled, but not on the polarization basis such pairs will not generate a swap*. The other crystal produces only H> polarized pairs.

But when the H/V output cones are overlapped properly, an entangled superposition results due to indistinguishability of the source. So the B photons must (this is the question) still be polarization entangled on some level (I don't know how) even after the A photon is measured.

Keep in mind that in the entanglement swapping protocol, generally: the Bell state measurement (on photons B/C or 2/3 depending on label) is made on a different basis than the Alice/Bob measurements. If Alice and Bob measure H/V, the Bell state measurement (BSM) will be 1/0. But if the B/C photons already possess a pure V> wave function, those won't be able to relate back to the A/D photons in any way.

*How do I know that this is true? Where is a good reference? I've never found one, I assume because the answer is considered "obvious" (which it isn't). And I could be wrong, there could be other issues I am unaware of. However: Why would any experimenter go through the time and effort to overlap the output cones of the V> and H> crystals if V> crystals alone would do the trick? The only reasonable answer is you need both to obtain entanglement suitable for swapping.

 

[Morbert]

How would the Bohmian (or anyone really) describe the difference (if any) between A and B before they arrive at their respective detectors? What statements might you make about their respective states, wave functions in terms of spin/polarization? Would you call them V> polarized, in a pure state? Both? Neither?

The difference in ordinary QM would be in theirs states. If the state of A+B before they arrive at their respective detectors is $\ket{\Phi^+}_\mathrm{AB}$ then the state of A before it arrives is $\mathrm{tr}_\mathrm{B}\ket{\Phi^+}_\mathrm{AB}$ and the state of B before it arrives is $\mathrm{tr}_\mathrm{A}\ket{\Phi^+}_\mathrm{AB}$. Their states have the same form, but pertain to different degrees of freedom.

 

[DrChinese]

The difference in ordinary QM would be in theirs states. If the state of A+B before they arrive at their respective detectors is $\ket{\Phi^+}_\mathrm{AB}$ then the state of A before it arrives is $\mathrm{tr}_\mathrm{B}\ket{\Phi^+}_\mathrm{AB}$ and the state of B before it arrives is $\mathrm{tr}_\mathrm{A}\ket{\Phi^+}_\mathrm{AB}$. Their states have the same form, but pertain to different degrees of freedom.

1. And your answer is the same after A has passed through the Polarizing Beam Splitter?

2. What about after A has arrived at its detector?

Thanks!

 

[Morbert]

1. And your answer is the same after A has passed through the Polarizing Beam Splitter?

2. What about after A has arrived at its detector?

Thanks!

The state of A before it passes through the PBS would be $\mathrm{tr}_\mathrm{B}\ket{\Phi^+}_\mathrm{AB} = \frac{1}{2}(\ket{H}\bra{H}+\ket{V}\bra{V})$. The state of A after is passes through the PBS, but before it is detected, would be $\frac{1}{2}(\ket{H,h}\bra{H,h}+\ket{V,v}\bra{V,v})$ where $\{h,v\}$ are path degrees of freedom.

The state of A after it is detected would collapse to H or V or more correctly the pointer state of the detector would be the H or V pointer state.

 

[Morbert]

1. And your answer is the same after A has passed through the Polarizing Beam Splitter?

2. What about after A has arrived at its detector?

Thanks!

I forgot about B.

Before A is detected, the state of B is the usual $\mathrm{tr}_\mathrm{A}\ket{\Phi^+}_\mathrm{AB}$. After A is detected and the measurement outcome V is obtained, then B is in the state $\ket{V}$. Though there is some subtlety here as Bob, before learning of Alice's results, can still model B with $\mathrm{tr}_\mathrm{A}\ket{\Phi^+}_\mathrm{AB}$. See section 7 here: https://arxiv.org/pdf/1308.5290

 

[DrChinese]

The state of A before it passes through the PBS would be $\mathrm{tr}_\mathrm{B}\ket{\Phi^+}_\mathrm{AB} = \frac{1}{2}(\ket{H}\bra{H}+\ket{V}\bra{V})$. The state of A after is passes through the PBS, but before it is detected, would be $\frac{1}{2}(\ket{H,h}\bra{H,h}+\ket{V,v}\bra{V,v})$ where $\{h,v\}$ are path degrees of freedom.

The state of A after it is detected would collapse to H or V or more correctly the pointer state of the detector would be the H or V pointer state.

Not disputing this characterization, just want to be sure I hear what you are saying.

The state of A does not change upon passing through the PBS and emerging from the V> port. It changes upon actual (irreversible) detection.

 

[DrChinese]

I forgot about B.

Before A is detected, the state of B is the usual $\mathrm{tr}_\mathrm{A}\ket{\Phi^+}_\mathrm{AB}$. After A is detected and the measurement outcome V is obtained, then B is in the state $\ket{V}$. Though there is some subtlety here as Bob, before learning of Alice's results, can still model B with $\mathrm{tr}_\mathrm{A}\ket{\Phi^+}_\mathrm{AB}$. See section 7 here: https://arxiv.org/pdf/1308.5290

Again, not disputing your characterization, just want to be sure I hear what you are saying.

After A's detection as V>: B is in state V>, which is different than its earlier state of $\frac{1}{2}(\ket{H}\bra{H}+\ket{V}\bra{V})$

 

[Matterwave]

I will try to keep my response as narrow as possible to avoid saying wrong things.

There is a technical reason for the distinction (is B still entangled or not) that few people are aware of.

You mean post measurement of A right?

If B's wave function (your preferred term) is in a pure vertical polarization state i.e. V>, it is not able to participate in an entanglement swap.

By postulate in standard QM, the mathematical operation which corresponds to the physical measurement of particle $A$, which gives the resulting eigenvalue $v$ corresponding to state $|V\rangle_A$, when the particle started in the entangled state:

$$|\Psi\rangle_{AB}=\frac{1}{\sqrt{2}}(|H\rangle_A \otimes |H\rangle_B + |V\rangle_A \otimes |V\rangle_B)$$

Is to act on that state $\Psi$ via the projection operator $P_{A,v}=|V\rangle_A\langle V|_A \otimes I_B$ where $I_B$ is the identity operator in the $B$ particle space and then normalize the resulting state.

If you act with this operator on the state given and then normalize the result, assuming orthonormality of basis states $H$ and $V$, you get $|\Psi'\rangle_{AB}=|V\rangle_A\otimes |V\rangle_B$ which is by definition a product state and not an entangled state.

Were that the case, the Delayed Choice version of swapping would not work.

Detailed discussion of entanglement swapping might be best suited for the other thread. What I wrote here in this current post (#12) was simply about my understanding of standard QM.

Again, there is the issue pointed out in 1 regarding swapping protocol. The other is unique to Type I PDC entanglement. That method requires 2 PDC crystals oriented perpendicular.

I would need to review the relevant literature to contribute further, so I will not make statements regarding these topics.

 

[Morbert]

Not disputing this characterization, just want to be sure I hear what you are saying.

The state of A does not change upon passing through the PBS and emerging from the V> port. It changes upon actual (irreversible) detection.

The less important issue: The state of A does change as it passes through the PBS

The state of A before it passes through the PBS would be $\mathrm{tr}_\mathrm{B}\ket{\Phi^+}_\mathrm{AB} = \frac{1}{2}(\ket{H}\bra{H}+\ket{V}\bra{V})$. The state of A after is passes through the PBS, but before it is detected, would be $\frac{1}{2}(\ket{H,h}\bra{H,h}+\ket{V,v}\bra{V,v})$ where $\{h,v\}$ are path degrees of freedom.

The polarization degree of freedom becomes correlated with the "which port" degree of freedom. But if we ignore this additional degree of freedom, it is fair to say the state of A's polarization does not change.

The more important issue: ordinary QM doesn't let us make a statement like "the photon emerged from the V> port" independent of measurement. Interpretations like consistent histories let you recover such intuitions, but ordinary QM only gives you states for managing expectations on measurement outcomes. Deeper statements must be made with respect to some interpretational framework.

Again, not disputing your characterization, just want to be sure I hear what you are saying.

After A's detection as V>: B is in state V>, which is different than its earlier state of $\frac{1}{2}(\ket{H}\bra{H}+\ket{V}\bra{V})$

Yes, a measurement on A with a recorded outcome lets us update our state for B. What this means will be interpretation-dependent though.

 

[DrChinese]

1. The state of A does change as it passes through the PBS

The polarization degree of freedom becomes correlated with the "which port" degree of freedom. But if we ignore this additional degree of freedom, it is fair to say the state of A's polarization does not change.

The more important issue: ordinary QM doesn't let us make a statement like "the photon emerged from the V> port" independent of measurement. Interpretations like consistent histories let you recover such intuitions, but ordinary QM only gives you states for managing expectations on measurement outcomes. Deeper statements must be made with respect to some interpretational framework.

2. Yes, a measurement on A with a recorded outcome lets us update our state for B. What this means will be interpretation-dependent though.

1. I follow this. Just was looking for a take on the "when" of A's state change: After the PBS before detection? Or only after detection. I think your comments are all appropriate. I'm also trying to compare this to the Bohmian view. Norsen says one thing, and Demystifier has a little different take because he follows the "predetermination" side of BM more strongly. Not sure this represents any actual difference between them or not.

Norsen: "It is thus meaningful already at this stage to speak of the actual outcome of the SGz-based measurement of the z-spin of particle 1.' [The diagram confirms the change occurs upon passing through the PBS or other spin splitting device.]


2. Good, you are saying the state for B can be update (per oQM). And going to the Bohmian interpretation:

Norsen: "The crucial point is now that the CWF for particle 2 – the thing that will determine how particle 2 behaves when it subsequently encounters its SGz device – depends on where particle 1 ended up." [I read everything Norsen says around the process to mean there is an instantaneous physical change, caused by the A movement, to remote B (per our example).]

Demystifier might have an issue with my use of the word "caused" as he mentions essentially: everything participates in the determination of particle 1's direction of movement, therefore there is no independent cause.

---

Hopefully I am not misrepresenting him, Norsen or yourself. But either way, "something" happens to B when A either goes through the PBS or is detected. In oQM, it need not be physical because that would be dependent on adopting an interpretation. But in BM, it seems to be physical. Either way, we would now label the state of B as being V> on that basis.

Interestingly: PDC entanglement will have A and B entangled as to frequency up to the point of detection, regardless of intermediate polarization measurements. This is somewhat dependent on the specific filters being used in the experiment.

$\omega_p = \omega_a + \omega_b$