Entanglement swapping and Bohmian mechanics

[PeterDonis]

what we need is a description of the unitary operator that is realized by the BSM/SSM

If I discard the concerns I raised in my last two posts, and just try to compute this from the evolution given in the Ma paper, I get more confused still.

The evolution given in the Ma paper, schematically, describes an operator $U$ such that

$$
U \ket{\Phi^+} = i \left( \ket{HV}_{bb} - \ket{HV}_{cc} \right) / \sqrt{2}
$$

$$
U \ket{\Phi^-} = i \left( \ket{HH}_{bc} - \ket{VV}_{bc} \right) / \sqrt{2}
$$

(I have left out the double primes on the b and c subscripts since everything refers to the same pair of output channels.)

Since $\ket{HH} = \ket{\Phi^+} + \ket{\Phi^-}$ and $\ket{VV} = \ket{\Phi^+} - \ket{\Phi^-}$, by linearity, we should have

$$
U \ket{HH} = U \ket{\Phi^+} + U \ket{\Phi^-} = i \left( \ket{HV}_{bb} - \ket{HV}_{cc} + \ket{HH}_{bc} - \ket{VV}_{bc} \right) / \sqrt{2}
$$

$$
U \ket{VV} = U \ket{\Phi^+} - U \ket{\Phi^-} = i \left( \ket{HV}_{bb} - \ket{HV}_{cc} - \ket{HH}_{bc} + \ket{VV}_{bc} \right) / \sqrt{2}
$$

But this means that if, for example, we have both detectors in b or both detectors in c firing, which is the condition the paper gives for saying that photons 2 and 3 were swapped into the state $\Phi^+$, we can't actually say that, because they could have been in the states $\ket{HH}$ or $\ket{VV}$ and still produced either of those results, since both results have a nonzero amplitude in the states $U \ket{HH}$ and $U \ket{VV}$, just as they do in the state $U \ket{\Phi^+}$.

So at this point I think the paper must be leaving out some crucial elements either of the experiment or of the logic they are using.

 

[PeterDonis]

an operator $U$

To be clear, the kets that $U$ operates on are kets at the input channels to the BSM/SSM (the ones labeled b and c with no primes in the Ma paper), and what it produces are kets at the final output channels, the ones labeled b'' and c'', for which I omitted the double primes in my previous post.

 

[Demystifier]

Presumably the Hilbert space changed too.

No.

 

[PeterDonis]

at this point I think the paper must be leaving out some crucial elements either of the experiment or of the logic they are using.

One possible item: I assumed in my previous posts that the operator $U$ realized by the Bell state analyzer was unitary. But it might not be. If the analyzer is intended to be a projector, that is, a standard "measurement" operator that projects any arbitrary state onto its measurement basis, then treating it as unitary, at least the way I did, is not valid. I am trying to find references that discuss in more detail the theory behind these things.

That said, if the analyzer is indeed intended to be a projector, it still seems weird that the evolution given in the Ma paper would have Bell states going into the device instead of coming out.

 

[PeterDonis]

Do we have experiments that show that, in the latter case, no swap can occur?

Note that, if this...

the analyzer is intended to be a projector, that is, a standard "measurement" operator that projects any arbitrary state onto its measurement basis

...is correct, then we would not necessarily expect that no swap would occur if we passed photons 2 and 3 through vertical polarizers before the BSM. On some runs, of course, one or both of those photons would be absorbed and we would get no swap. But if both made it through, and the BSM is a projector, it could stiill project them into a Bell state, which means it could still execute a swap.

I wonder how much deeper this rabbit hole goes...

 

[PeterDonis]

It is not entangled under the decomposition (1), but it is entangled under many other decompositions. In particular, it is entangled under decomposition (2)

Note, though, that by "entangled" here you mean entanglement between the two different subsets of photons you picked out (AB-CD or AD-BC). But each of those subsets can be further subdivided into subsystems consisting of single photons, and there are always pairs of photons that are entangled according to that further subdivision, even in the case where you've picked out the 2-photon subsets in such a way that there is no entanglement between them (because each subset consists of a pair of photons that are maximally entangled with each other, which leaves no room for entanglement with anything else).

All this is worth keeping in mind, but I don't think that by itself it resolves any concerns about the counterintuitiveness of entanglement swapping experiments.

 

[PeterDonis]

If I have to choose one of those, I choose i).

I would choose ii). I would think that would be the natural choice for any advocate of Bohmian mechanics. The natural Bohmian account seems to me to be that the effective collapse due to the pre-measurement of two of the photons "steers" the other two, via the effective wave function/pilot wave, into the appropriate output channels to register a swap that is consistent with the pre-measured photon results. This is basically the account I gave in my earlier post (and have been elaborating in follow-up posts).

 

[PeterDonis]

From the data you cannot tell if anything was done on B/C first or on A/D. But your/BM explanations are different!

I think this is an unavoidable aspect (or drawback, if you want to see it that way) of having to give a forward in time account of such experiments, which is what BM has to give since it has straightforward forward in time dynamics. All interpretations have drawbacks.

I think one might be able to use considerations of time translation symmetry to argue that the explanations for different time orders, although they appear different on the surface, are in fact equivalent. By "time translation symmetry" I mean, in the context of these experiments, that putting a photon into a time delay line, and doing nothing else to it, doesn't change its state (at least as long as we assume it doesn't get delayed so long that coherence is lost). The Megidish paper uses this kind of reasoning when it blithely writes down states that entangle photons existing at different times, and thereby obtains a wave function that is formally the same as the one you get from a straightforward entanglement swap experiment where all the photons coexist and there is no delayed choice.

 

[DrChinese]

Indeed, looking at this and the diagram of the experimental setup (Figure 2 in the paper) makes me even more confused, because the evolution starts before BS1, i.e., the evolution assumes that photons 2 and 3 are in one of the Bell states before BS1--but then what puts them into those states? The only things in the experiment before BS1 are the preparations of the initial entangled pairs (1 and 2, 3 and 4), the time delay, and a half wave and quarter wave plate, which as far as I can tell are the same for both photons, so they should not change the phase relationship between them. What in all this can put photons 2 and 3 into a Bell state?

The Ma experimental diagram deviates from the standard setup for technical reasons related to the optional fast choice of BSM or SSM. So it is unusually complicated, and those extensions make it very difficult to discuss here. But here is the explanation they provide showing the |Φ+〉 evolution. It's hiding in the Supplementary section, page 14.

|Φ+〉 = (|𝐻𝐻〉bc + |𝑉𝑉〉bc)/√2
BS1 ---> 𝑖 (|𝐻𝐻〉b'b' + |𝑉𝑉〉b'b' + |𝐻𝐻〉c'c' + |𝑉𝑉〉c'c'))/2
EWPs & EOMs ---> (|𝑅𝑅〉b'b' − |𝐿𝐿〉b'b' + |𝐿𝐿〉c'c' − |𝑅𝑅〉c'c')/2
BS2 ---> 𝑖(|𝐻𝐻〉b''b'' − |𝐻𝐻〉c''c'')/√2

Note: every one of the states above are entangled.

I would strongly recommend looking at these other experimental designs that are "standard" swapping arrangements, since trying to decipher the quirks of the Ma experiment won't really relate to our discussion:

Hensen et al, Figure 1d: The "fibre-based beam splitter (FBS)" is responsible for the swap.

Megidish et al, Figure 2: The "projecting PBS" creates the required overlap and indistinguishability; and the swap occurs. The subsequent 2 PBSs and 4 single photon detectors serve to identify (where possible) the specific Bell state that the swap led to.

Ultimately: The overlapping 2 and 3 photons go into a beam splitter of some type within a narrow time window; they are either both transmitted or both reflected; and the output photons now lack distinguishing characteristics as to which source pair they came from. This is the swap. The remaining apparatus is to identify the swap. As stated previously, all indistinguishable emerging pairs lead to a swap to one of the 4 Bell states, though some lack identifying characteristics as to which one.

 

[DrChinese]

Note that, if this...


...is correct, then we would not necessarily expect that no swap would occur if we passed photons 2 and 3 through vertical polarizers before the BSM. On some runs, of course, one or both of those photons would be absorbed and we would get no swap. But if both made it through, and the BSM is a projector, it could stiill project them into a Bell state, which means it could still execute a swap.

I wonder how much deeper this rabbit hole goes...

LOL, it's a rabbit hole...

Yes, we now start to get into some areas in which I have seen very little literature. I will be starting a separate thread to speak precisely to the accepted theory/experiment - at least what I know, so don't laugh - about this.

You can bet your bottom dollar that I will try to show that those vertical polarizers you mention will prevent a swap.

 

[Morbert]

Yes, we now start to get into some areas in which I have seen very little literature. I will be starting a separate thread to speak precisely to the accepted theory/experiment - at least what I know, so don't laugh - about this.

I may save my responses to @PeterDonis for that thread as they are about the experiment in general rather than a Bohmian accounting.

 

[PeterDonis]

here is the explanation they provide showing the |Φ+〉 evolution.

That's precisely the evolution that's confusing me. If the BSM is supposed to be a projector, i.e., it takes an arbitrary pair of photon 2 & 3 states as input and projects them onto one of the four Bell states, then I would expect to see Bell states coming out of the evolution, i.e., at the end. But instead what I see is Bell states going in, i.e., at the start, and at the end I see the states that are actually detected in order to distinguish one Bell state from another. Where in the experiment is the projection done? How do I get from an arbitrary 2-photon state to one of the four Bell states? The evolution doesn't show that.

I'll look at the other papers you reference to see if they address this.

 

[PeterDonis]

The overlapping 2 and 3 photons go into a beam splitter of some type within a narrow time window; they are either both transmitted or both reflected; and the output photons now lack distinguishing characteristics as to which source pair they came from. This is the swap.

Yes, I get that, and in the Ma setup, that (or at least the start of it) is supposed to be BS1. But the evolutions in the paper, that you wrote down in your post, show Bell states going into BS1, i.e., before photons 2 and 3 even reach it. That's the part I can't understand.

 

[PeterDonis]

Hensen et al, Figure 1d: The "fibre-based beam splitter (FBS)" is responsible for the swap.

Megidish et al, Figure 2: The "projecting PBS" creates the required overlap and indistinguishability; and the swap occurs. The subsequent 2 PBSs and 4 single photon detectors serve to identify (where possible) the specific Bell state that the swap led to.

Both of these look much more like what you describe: a beam splitter (FBS or PBS), and Bell states coming out of it, not going in. The Megidish paper says they post-select for runs where the photons come out of the PBS by different ports, and their choice of whether or not to swap is whether or not they set things up for indistinguishability as the photons come out. That all makes sense to me. But it's different from what the Ma paper describes.

 

[DrChinese]

Can you be more specific?

"Why certainly, my boy!"-Bugs Bunny

Norsen (Fig. 6) says* that an |Up> (or V> or H> in our terminology, specific label doesn't matter) detection for entangled particle labeled 1 (explicitly occurring first) casts remote particle 2 into a |Down> state even absent a measurement on that particle.

"...particle 1 (on the right) encounters its SGz device first; the particle is found to be spin-up (solid trajectories) ... depending on the initial z-coordinate of the particle. If particle 1 goes up, the collapse suffered by the CWF [conditional wave function] of particle 2 (on the left) causes it to go down (solid trajectories) regardless of its initial z-coordinate. ... [Otherwise the] result of measuring the z-spin of particle 2 is then determined by the initial z-coordinate of particle 2. The difference between the two scenarios exemplifies the contextuality of the pilot-wave theory [Bohmian Mechanics] (since the result of measuring the spin of particle 2 depends not only on the initial state but on whether or not the spin of particle 1 is measured jointly) and also its nonlocality (since the choice of whether or not to measure particle 1 could be made at space-like relativistic separation from the measurement of particle 2)."

It says the same thing in the accompanying text: measurement of particle 1 changes particle 2 nonlocally, measurement on particle 2 is not required to occur for that change. Norsen: "The non-locality is thus clear: ... the outcome of the measurement on particle 2 depends on what Alice chooses to do in the vicinity of particle 1."

There really isn't any wiggle room on this. He is describing the Bohmian nonlocal action at a distance as it applies to spin (or polarization). This is a physical change to the z-spin of particle 2. The underlying details of the mechanics does not affect his pronouncement. And for the purposes Norsen is applying this example, it accurately reproduces traditional QM spin statistics.

---

So going back to this thread, and considering Norsen's as fair reference (which hopefully @Demystifier concurs with):

Bohmian description:
With an initially entangled photon pair 1 and 2 in state |Ψ−〉12: A |V> state outcome on particle 1 causes/leads to/is associated with an immediate change to distant particle 2 into a pure |H> state.

Now, please tell me you accept this without any further qualification.


*Quotes are edited slightly for brevity/clarity, you can always read the entire thing via the link.

 

[PeterDonis]

Yes, we now start to get into some areas in which I have seen very little literature. I will be starting a separate thread to speak precisely to the accepted theory/experiment - at least what I know, so don't laugh - about this.

Ok. I'll be interested to see any other experimental references you can give.

You can bet your bottom dollar that I will try to show that those vertical polarizers you mention will prevent a swap.

From what @Morbert posted in #40:

if 1 and 4 are measured first (say, in the H,V basis), we get

$$\begin{equation*}\begin{aligned}\rho_{1234} = \frac{1}{8}\Big( &\left[HH\right]_{14}\left[\Phi^+-\Phi^-\right]_{23} + \left[VV\right]_{14}\left[\Phi^++\Phi^-\right]_{23} \\ &+ \left[HV\right]_{14}\left[\Psi^+-\Psi^-\right]_{23} + \left[VH\right]_{14}\left[\Psi^++\Psi^-\right]_{23}\Big)\end{aligned}\end{equation*}$$

Simple algebra gives (leaving out the normalization factors):

$$
\rho_{1234} = \left[HH\right]_{14}\left[VV\right]_{23} + \left[VV\right]_{14}\left[HH\right]_{23} + \left[HV\right]_{14}\left[VH\right]_{23} + \left[VH\right]_{14}\left[HV\right]_{23}
$$

If we know photons 2 and 3 are both vertically polarized, all but one of the above terms go away, and we are left with:

$$
\rho_{1234} = \left[HH\right]_{14}\left[VV\right]_{23}
$$

which of course can be rewritten as

$$
\rho_{1234} = \left[HH\right]_{14}\left[\Phi^+ - \Phi^-\right]_{23}
$$

So we can still execute a swap, but only into one of those two Bell states, and only if the photon 1 and 4 measurement results were both $H$. At least, that's what the math seems to say. (Indeed, we could have just read this off of what I quoted from @Morbert, if we observe that the only photon 1 & 4 state that is consistent with photons 2 and 3 being vertically polarized is $\left[HH\right]_{14}$, so that's the only term in his equation that can survive. But I thought it might be helpful to go the longer way around.)

This is also consistent with my previous posts where the rule seemed to be that as long as there was a nonzero amplitude for photons 2 and 3 to be parallel in any basis, we can swap into $\Phi^\pm$. (And as long as there is a nonzero amplitude for photons 2 and 3 to be antiparallel in any basis, we can swap into $\Psi^\pm$. Here that is not possible because there is zero amplitude for the photons to be parallel in the H-V basis, as I commented in an earlier post.)

 

[DrChinese]

1. Both of these look much more like what you describe: a beam splitter (FBS or PBS), and Bell states coming out of it, not going in.

2. The Megidish paper says they post-select for runs where the photons come out of the PBS by different ports, and their choice of whether or not to swap is whether or not they set things up for indistinguishability as the photons come out. That all makes sense to me. But it's different from what the Ma paper describes.

1. Well of course the 2 and 3 photons going in are in the |Ψ−〉 state, per their initial association with 1 and 4 respectively. They assume you know that, and I know you do.

2. In Megidish, the choice of whether to swap or not is accomplished before the 2 and 3 photons go into the Projecting PBS. A delay line introduces temporal distinguishability.

I agree that if you wade into Ma's BSM black box, you are taking your life into your hands. But obviously, eventually it makes sense. It just depends on how much time you want to spend on it.

 

[PeterDonis]

of course the 2 and 3 photons going in are in the |Ψ−〉 state

Not photons 2 and 3, no. If we go by how they were initially prepared, photon 2 is in a singlet state with photon 1, and photon 3 is in a singlet state with photon 4. In that state, photons 2 and 3 are only classically correlated; they are not in a singlet state with each other.

If we take into account that photons 1 and 4 are already measured, we get what @Morbert posted in #40, and I quoted in #106 just now. Which is still not photons 2 and 3 being in the singlet state. It's a mixture of linear combinations of Bell states. They are still only classically correlated.

2. In Megidish, the choice of whether to swap or not is accomplished before the 2 and 3 photons go into the Projecting PBS. A delay line introduces temporal distinguishability.

Yes, the indistinguishability has to be set up beforehand (by not selecting the delay line). But the swap itself does not happen until the photons come out of the PBS, as I understand their paper.

 

[PeterDonis]

I agree that if you wade into Ma's BSM black box, you are taking your life into your hands. But obviously, eventually it makes sense. It just depends on how much time you want to spend on it.

LOL.

 

[PeterDonis]

measurement of particle 1 changes particle 2 nonlocally, measurement on particle 2 is not required to occur for that change.

We went over this in the previous thread. I can't answer for what Norsen meant, but as I pointed out in the other thread, the wave function is not the particle in BM. It's part of the Hamiltonian. So when we get an effective collapse of the wave function because we measured particle 1, which is what Norsen is describing, we nonlocally change particle 2's Hamiltonian, in a way that will end up steering it into the "down" output arm when we measure its spin, since particle 1's spin was up. That doesn't change particle 2 itself (its position); it just changes the Hamiltonian that will govern particle 2's motion from that instant on.

Now to go back to what I said that you agreed with, but you said Norsen said otherwise. My case (1) there was that photons 1 and 4 have been measured but no operation has yet been done on photons 2 and 3. By "operation" I meant either a measurement, or some other kind of interaction like going through a half-wave or quarter-wave plate, or more to the point a vertical polarizer, since that's the operation in my case (2) that I was contrasting with. I did not mean just freely traveling and the effective wave function changing because of an entangled particle being measured.

The point I was trying to make was that, just as a matter of making correct predictions with the basic math of QM (as opposed to applying any interpretation, Bohmian or otherwise), we might have to distinguish the two cases I described. But that depends on whether putting photons 2 and 3 through a vertical polarizer before the swap device, vs. not, actually makes a difference in experiments. Based on my post #106, I think it should--not to prevent a swap altogether, but to restrict it so it can only occur if photons 1 and 4 both give $H$ as the result of their polarization measurements. Any other combination of photon 1 and 4 measurement results should mean a swap can't happen if photons 2 and 3 are put through vertical polarizers--whereas if they were not, a swap could happen in those cases.

 

[DrChinese]

Not photons 2 and 3, no. If we go by how they were initially prepared, photon 2 is in a singlet state with photon 1, and photon 3 is in a singlet state with photon 4. In that state, photons 2 and 3 are only classically correlated; they are not in a singlet state with each other.

That's what I said. Photon 2 (should you think of it as entangled with 1) is still in the |Ψ−〉 state going into the PBS. I certainly didn't mean that going into the PBS, photons 2 and 3 are entangled with each other. They are simply going in more or less in the same state. (They come out entangled in 1 of 4 Bell states though.)

 

[PeterDonis]

the outcome of the measurement on particle 2 depends on what Alice chooses to do in the vicinity of particle 1."

...This is a physical change to the z-spin of particle 2.

Again, we went over this in the other thread, but to recap:

Note Norsen's careful phrasing: he does not say the z-spin of particle 2 changes. He only says the outcome of the mesaurement on particle 2 depends on the particle 1 measurement. He's phrasing it that way because, as I said in post #110 just now, and as I said in the other thread, the wave function is not the particle. So the fact that we had an effective collapse in particle 2's wave function does not mean its spin changed, or anything else about it. Indeed, we can't even define its spin, according to BM, until it's measured--after all, Bob could choose to measure its spin in some other basis than the z basis. We can say that the spin part of particle 2's wave function changed, but that's not "a physical change to the z-spin of particle 2", because "the z-spin of particle 2" isn't even a well-defined property at all unless and until we measure particle 2's spin in the z basis. (See Norsen's discussion of contextuality in section V of the paper for more on this.) All we can say is that, as I said in post #110, the effective collapse due to measuring particle 1 changes the Hamiltonian that will govern particle 2's motion from that instant on.

Now of course it's really a matter of words, not physics, whether you want to call all this a "physical change" to particle 2. If you insist on doing so, I can't stop you. But I don't think you can put those words in Norsen's mouth. More importantly, even if we agree to call it a "physical change", it's clearly a different kind of physical change from a measurement, or passing through a half-wave plate or a beam splitter or a polarizer or a Stern-Gerlach magnet. We don't want to lump all those things together into one category because they're not. We have to differentiate between them because they are different. That was the point I was trying to make with my cases (1) and (2) in that earlier post.

 

[PeterDonis]

Photon 2 (should you think of it as entangled with 1) is still in the |Ψ−〉 state going into the PBS. I certainly didn't mean that going into the PBS, photons 2 and 3 are entangled with each other.

Then, in the case where photons 1 and 4 have already been measured, would you agree that photon 2 is not in the singlet state going into the PBS? That's the case we're discussing, after all.

They are simply going in more or less in the same state.

I agree with this. I just don't see how that "same state" is a singlet state in the case under discussion.

 

[PeterDonis]

Bohmian description:
With an initially entangled photon pair 1 and 2 in state |Ψ−〉12: A |V> state outcome on particle 1 causes/leads to/is associated with an immediate change to distant particle 2 into a pure |H> state.

Now, please tell me you accept this without any further qualification.

Can't quite, for the reason I expounded in #110 and #112. I will agree that the wave function of particle 2 is effectively collapsed into the $\ket{H}$ state in the case you describe.

 

[DrChinese]

We went over this in the previous thread. I can't answer for what Norsen meant, but as I pointed out in the other thread, the wave function is not the particle in BM...

Surprise surprise surprise.-Gomer Pyle.

And here we are again sir. You ask for backup for Norsen's position, and I provided it. And where does that lead us? To move on? NO. To say "good job DrC with the requested specifics, I see your point about the Bohmian position". NO.

I asked specifically for you to accept my Norsen quote - YOU ASKED FOR IT - and move on. And you do the opposite, simply keep the increasingly argumentative nature of your posts going. See for instance your #112 where you parse Norsen's words to death. Or your incredibly stubborn #114 where you parse my words to death.

What you are talking about in this snippet is a different subject, and not germane to this thread. We're talking about exactly what Norsen means and says, it's an authoritative proxy for the position of Bohmian Mechanics. If he isn't good enough, tell me why not. I am not going down your diversionary path. Norsen is talking about instantaneous nonlocal influences, that's what Bohmian mechanics is all about.

PS This is your usual cue to close a thread for "moderation".

 

[DrChinese]

Then, in the case where photons 1 and 4 have already been measured, would you agree that photon 2 is not in the singlet state going into the PBS? That's the case we're discussing, after all.

I can't agree because this is the question I am asking here. It seems in that in the Bohmian view, photon 2 is not in the singlet state going into the PBS. At least according to Norsen.

For the orthodox view, I'm not sure. I haven't seen literature on this exact point, and you never provide any relevant support to your statements (as you haven't in this thread other than references I already supplied). But if you have something authoritative that answers this point specifically, then please share it. I don't think the answer appears in any of my swapping references.

 

[PeterDonis]

I asked specifically for you to accept my Norsen quote - YOU ASKED FOR IT - and move on.

I'm not saying you quoted Norsen incorrectly. I'm saying you added words that aren't Norsen's, they're yours:

This is a physical change to the z-spin of particle 2.

I do not agree that Norsen said this. Those aren't his words. Those are your words. Those are your interpretation of what Norsen said, and I don't agree with your interpretation.

If you want me to stop being "argumentative", you need to stop putting words in the mouths of the references you quote, that are not theirs, but yours. If you want to quote them, fine, quote them, leave their words just as they are, and move on. If you want to express your own opinions as your own opinions, fine, do that. But you won't do that. You keep adding your own words, expressing your own opinions, to the words of the references that you quote, and then behaving as if the reference said those words and they're not your words, not your opinion, but "established science". And if I don't agree that that's what your references said, I'm going to say so. That's not going to change.

PS This is your usual cue to close a thread for "moderation".

I'm not going to take any moderation action whatever in this thread. I've reported your post so other moderators can evaluate and decide if they want to do anything.

 

[PeterDonis]

It seems in that in the Bohmian view, photon 2 is not in the singlet state going into the PBS. At least according to Norsen.

Norsen's view, which I understand to be the standard Bohmian view, is that photon 2's wave function effectively collapses when photon 1 is measured, so its wave function is no longer effectively the singlet state, yes.

For the orthodox view, I'm not sure.

I think it depends on what you consider to be "the orthodox view".

The textbook I have handiest is Ballentine, and his Chapter 9 discusses the general topic of how to treat information about one measurement result when computing probabilities for others, particularly section 9.6 on joint and conditional probabilities. His general approach looks similar to the Bohmian notion of effective collapse, i.e., to compute the probability of results for measurement B conditioned on a particular result for measurement A, you use the effectively collapsed wave function that corresponds to that result for measurement A.

Ballentine uses the ensemble interpretation, but my understanding is that the above approach would be considered at least fairly "orthodox". It's also the approach I used in post #106, drawing on @Morbert's post #40, when looking at the effect of putting photons 2 and 3 through a vertical polarizer.