Consistency of Bohmian mechanics

In summary: Thank you for your interest! I am happy to share some of the literature that may be of help. However, a full calculation is beyond the scope of this answer. If you are interested in a calculation, you are welcome to try and do it yourself. It's not an easy calculation and I have no intention to do it here.
  • #36
Elias1960, just pretend that ##x## is already a macroscopic pointer observable which can be read off without disturbances, not a particle. It's only for illustrative purposes, because nothing changes in the argument if you include more variables.
 
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  • #37
Thors10 said:
Elias1960, just pretend that ##x## is already a macroscopic pointer observable which can be read off without disturbances, not a particle. It's only for illustrative purposes, because nothing changes in the argument if you include more variables.
Whatever, you have to consider the full system, with measurement devices, if you want something following the Schroedinger equation. For this full system, the standard equivalence theorem holds. If you want to see how the collapse happens, you have to look at the subsystem during the measurement, and its wave function is defined by the full wave function (following the Schroedinger equation) and the trajectory of the measurement device. This leads, applied to a particular trajectory of the measurement device, to a particular collapse trajectory.
For me, it looks like you have not yet understood the whole picture.
 
  • #38
Elias1960: Even though I don't think it's necessary, I did this already in post 28, so we can move on to the real problem I'm interested in.

The problem I see is the following. There are two types of variables needed in BM:
1. Microscopic variables, which can only be measured by interaction with a measurement apparatus. BM claims to be able to derive an apparent collapse for the subsystem that describes these variables. Alright, I won't question this anymore, if we can then agree to move on to the problem I really care about.
2. Macroscopic pointer variables, which can be read off directly from the device. Gaining knowledge about them doesn't require interaction. Instead we can just learn the approximate value of these variables.

The problem is that gaining knowledge and updating the probability distribution for variables of the second kind in BM is not in agreement with the the collapse of the wave function in QM. The literature in BM doesn't discuss this problem at all as far as I can tell. If you disagree, please point me to a paper or book that treats this.
 
  • #39
Thors10 said:
My question concerns the collapse of the full system's wave function, not the subsystem.
You are right that in BM there is no collapse of the full system, including the measuring apparatus. But that's not much different from QM, because the standard textbook QM usually does not contain a collapse of the full system. That's because in QM textbooks the measurement apparatus is usually not represented by a wave function at all. Bohr, for instance, argued that measurement apparatus is classical. This does not contradict existing experiments because the wave function of a measurement apparatus (or any other system with many effective degrees of freedom) cannot be determined experimentally. In this sense, QM and BM have the same measurable predictions. They have some differences on how they treat the full system, but these differences cannot be measured.
 
  • #40
Thors10 said:
The problem is that gaining knowledge and updating the probability distribution for variables of the second kind in BM is not in agreement with the the collapse of the wave function in QM. The literature in BM doesn't discuss this problem at all as far as I can tell. If you disagree, please point me to a paper or book that treats this.
The global wave function remains uncollapsed. So, it contains all the empty worlds which we already know to be irrelevant given that we have direct access to the trajectories. Schroedinger's cat's wave function remains to be a superposition of the living and the dead cat. But what we see is the trajectory of the cat, not the wave function. The collapse is something we have only if we reduce the global wave function to an effective one for some subsystem, using the trajectory we see. That's all.
 
  • #41
Thors10 said:
2. Macroscopic pointer variables, which can be read off directly from the device. Gaining knowledge about them doesn't require interaction.
That's wrong, gaining knowledge about them requires interaction too. But since they are macroscopic, effects of those interactions can be approximated with classical physics.
 
  • #42
Demystifier:
Right, standard QM doesn't need to include the apparatus in the quantum system. But if we include the apparatus in the quantum system, standard textbook QM does need the collapse, because decoherence only gives us density matrix containing the incoherent mixture of all possible resulting wave functions. We need to collapse the full systems wave function to obtain a definite result. That's what the measurement problem is about. On the other hand, BM needs to update the probability distribution, because probabilities are supposed to be only due to ignorance. It is critical that these different processes of updating are in agreement, because otherwise they will lead to different predictions.

Elias1960:
The global wave function in BM remains uncollapsed, right. But nevertheless, the probability distribution must be updated, because probabilities are only due to ignorance in BM. If we gain knowledge, our ignorance reduces and the probability distribution must reflect that.

Demystifier:
What does the pointer observable need to interact with during the gaining of knowledge about it?
 
  • #43
Thors10 said:
Elias1960:
The global wave function in BM remains uncollapsed, right. But nevertheless, the probability distribution must be updated, because probabilities are only due to ignorance in BM. If we gain knowledge, our ignorance reduces and the probability distribution must reflect that.
But it has to be updated because we see a measurement result. We see it, it is, therefore, not in the quantum part, it is described not by the wave function, but by the trajectory. There is no point in using a wave function once the real information about the trajectory is available. The part which will be described by a wave function is what remains.
 
  • #44
What is your point? Do you disagree that we need to update our probability distribution once we learned something about the system? I'm just talking about standard Bayesian updating. BM tells us that our particles are initially distributed according to ##\left|\psi(\mathbf{X},0)\right|^2##. After gaining knowledge, this distribution can no longer be maintained. The new probabilities will be given by ##P'(X)=P(X|A)## and the probability distribution function of ##P'## will be given by ##\frac{1}{P(A)}\chi_A(\mathbf{X})\left|\psi(\mathbf{X},0)\right|^2##.

My point is that the time evolution of this new probability density disagrees with the time evolved probability we get from QM and thus BM and QM make different predictions.

In BM, the time evolution of the updated probability density is given by ##\frac{1}{P(A)}\chi_A(E_t^{-1}(\mathbf{X}))\left|\psi(E_t^{-1}(\mathbf{X}),0)\right|^2 = \frac{1}{P(A)}\chi_{E_t(A)}(\mathbf{X})\left|\psi(\mathbf{X},t)\right|^2##.

In QM, the time evolution of the updated probability density is given by ##\frac{1}{P(A)}\left|(U(t)\chi_A({-})\psi({-},0))(\mathbf{X})\right|^2## and this is incompatible with the Bohmian formula.

If BM and QM are supposed to be equivalent, then these two formulas must necessarily be equal. There just isn't any other way. So the way I see it after this discussion so far is that BM and QM are really inequivalent theories (which may occasionally make similar predictions).
 
  • #45
Just don't malign Bohmian Rhpsody.
 
  • #46
Just to make it completely clear:

I consider a system that starts with an initial wave function ##\psi(\mathbf{X},0)##. Then I perform a position measurement at time ##t=0## and find the system in ##\mathbf{X} \in A##. Then I evolve the system to time ##t##. I look at the evolution of the initial probability density of this process in both BM and QM.

In BM, this works as follows:
##\left|\psi(\mathbf{X},0)\right|^2 \rightarrow \frac{1}{P(A)}\chi_{A}(\mathbf{X})\left|\psi(\mathbf{X},0)\right|^2 \rightarrow \frac{1}{P(A)}\chi_{E_t(A)}(\mathbf{X})\left|\psi(\mathbf{X},t)\right|^2##

Orthodox QM has the following evolution:
##\left|\psi(\mathbf{X},0)\right|^2 \rightarrow \frac{1}{P(A)}\left|\chi_{A}(\mathbf{X})\psi(\mathbf{X},0)\right|^2 \rightarrow \frac{1}{P(A)}\left|\left(U(t)\chi_{A}({-})\psi({-},0)\right)(\mathbf{X})\right|^2##

The math is correct and it's also the correct application of the BM and QM formalisms. So the question to BM is whether these formulas agree. But it's clear that they don't in general, because there are counterexamples. Hence, BM and QM are not equivalent at the level of full systems.
 
  • #47
Thors10 said:
Demystifier:
Right, standard QM doesn't need to include the apparatus in the quantum system. But if we include the apparatus in the quantum system, standard textbook QM does need the collapse, because decoherence only gives us density matrix containing the incoherent mixture of all possible resulting wave functions. We need to collapse the full systems wave function to obtain a definite result. That's what the measurement problem is about. On the other hand, BM needs to update the probability distribution, because probabilities are supposed to be only due to ignorance.
So far I agree.

Thors10 said:
It is critical that these different processes of updating are in agreement, because otherwise they will lead to different predictions.
I agree that the two approaches make different "predictions" here, but I do not think that this difference can be measured in practice. Or if you think that it can be measured in practice, can you propose a specific experiment how that could be done?

Thors10 said:
Demystifier:
What does the pointer observable need to interact with during the gaining of knowledge about it?
Maybe I am missing the true motivation for this question, but obviously it must interact with the measured system. A nontrivial question is precisely how they should interact. For a model of an appropriate interaction see http://de.arxiv.org/abs/1406.5535 Sec. 6.1.
 
  • #48
Thors10 said:
Hence, BM and QM are not equivalent at the level of full systems.
I agree, but the claim is that this difference cannot be measured in practice. In principle the difference could be seen if one could prepare a true macroscopic Schrodinger cat state, but nobody has done it so far. In existing experiments it is possible to prepare a mesoscopic Schrodinger "cat", which shows that there is no collapse at the mesoscopic level, which can be taken as an experimental indication that the Bohmian interpretation could be closer to truth than the collapse interpretation. But it is consistent with a collapse interpretation too if one claims that the collapse only happens at the macroscopic level, and not on the mesoscopic one.
 
  • #49
Well, in order for the different predictions to be measured in practice, one would need an actual prediction of BM in the frist place. However, I don't think any experimentally relevant prediction of BM has ever been computed at all. There is only the claim that BM will make the same predictions as QM and therefore we should just compute the predictions in QM. But that's not what needs to be done if we want to test BM. We need a full BM computation that we could compare to experiment. Otherwise it's just a test of QM, not of BM.
Regarding the pointer observable: I said that these observables can be read off directly from the device, unlike the observables of the particles. Of course the pointer variable interacts with the particle, but in order to gain knowledge about it, I don't need to include yet another interaction term with some additional system.

You said: "That's wrong, gaining knowledge about them requires interaction too. But since they are macroscopic, effects of those interactions can be approximated with classical physics."

I asked you what these classical interactions would be, because certainly you can't mean the quantum interactions with the particle? They are not classical at all.

My point is just: In order to measure the position of a particle in BM, I'm told that I can't just take its Bohmian trajectory. Instead I must couple the particle to some measurement device and instead measure the position of the pointer variable. So the pointer variable is of a different kind: I don't need to couple it to yet another bigger system in order to gain knowledge about it. I can just take the actual value of its Bohmian trajectory.
 
  • #50
"I agree, but the claim is that this difference cannot be measured in practice."
Well that's the problem I'm having. The BM community seems to shy away from making predictions with their theory. I'm trying to understand the (in)equivalence by looking at some actual calculations, but I'm just told that they are not sufficient. Then again, nobody seems to have done a sufficient calculation at all, so I don't understand the confidence in the claim. I'm just interested in some quantitative computations in BM, but I only ever get loose arguments. And those arguments never suffice to get some quantitative results out of them that could in principle be compared to experiments.
 
  • #51
Thors10 said:
"I agree, but the claim is that this difference cannot be measured in practice."
Well that's the problem I'm having. The BM community seems to shy away from making predictions with their theory. I'm trying to understand the (in)equivalence by looking at some actual calculations, but I'm just told that they are not sufficient. Then again, nobody seems to have done a sufficient calculation at all, so I don't understand the confidence in the claim. I'm just interested in some quantitative computations in BM, but I only ever get loose arguments. And those arguments never suffice to get some quantitative results out of them that could in principle be compared to experiments.
Well, some authors do try to make new measurable predictions out of BM, but in my opinion such attempts are misguided. See my "Bohmian mechanics for instrumentalists" (linked in my signature below), Sec. 4.4.
 
  • #52
How could such predictions be misguided? Either BM and QM can be shown to be equivalent. This seems not to be the case. Or we need to understand their difference quantitatively, in order to test which of them is correct. Without quantitative computations, BM can't be said to be compatible with experiments, yet.

I have looked at the section of that paper, but I still don't understand why we should allow BM to get away with not making quantitative predictions? Every physical theory must do that at some point. It's not a "trap" as you call it in that paper. It's just how science works.
 
  • #53
Thors10 said:
How could such predictions be misguided? Either BM and QM can be shown to be equivalent. This seems not to be the case. Or we need to understand their difference quantitatively, in order to test which of them is correct. Without quantitative computations, BM can't be said to be compatible with experiments, yet.

I have looked at the section of that paper, but I still don't understand why we should allow BM to get away with not making quantitative predictions? Every physical theory must do that at some point. It's not a "trap" as you call it in that paper. It's just how science works.
Obviously you have not understood my argument in the paper, so let me rephrase it. If practically measurable differences between two theories exist, then they should be pointed out. But if they do not exist, then honest scientists should not pretend that they exist just for the sake of looking "more scientific". I argue in the paper that they do not exist, and that authors who claim the opposite either misunderstand BM or pretend that they exist for the sake of looking more scientific. If you still don't understand it, then I don't know how to put it differently.

But note that latter in the paper I do make a new generic measurable prediction out of BM, but on an entirely different level. It's in Sec. 5.2.
 
  • #54
Demystifier:
"I argue in the paper that they do not exist"
Well, but I don't see any quantitative argument in the paper. You make a lot of approximations and you don't quantify how good these approximations are. You would have to work in an actual model to calculate such deviations. Only then can we test whether BM gives the same predictions as QM up to the measurement uncertainty of our state of the art experiments. Also you don't study how quickly the deviations will evolve. Maybe they are small at first but will explode after a short time interval. All of this must be checked. Just claiming it, isn't enough.

Let me give an example:
If I stack two spheres exactly on top of each other, then classical mechanics tells us that they will stay in this state forever. Now if I fail to find the right spot and make a small deviation, then you would argue that these deviations are small and because of this I should still expect the spheres to remain almost motionless. But the truth is that this situation is unstable and the deviations will quickly explode and the system will collapse. One can study this instability using math and it's very important in order to make the correct predictions. If we don't know quantitatively how the deviations will manifest themselves, the theory is not predictive at all.
 
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  • #55
Thors10 said:
Demystifier:
"I argue in the paper that they do not exist"
Well, but I don't see any quantitative argument in the paper. You make a lot of approximations and you don't quantify how good these approximations are. You would have to work in an actual model to calculate such deviations. Only then can we test whether BM gives the same predictions as QM up to the measurement uncertainty of our state of the art experiments. Also you don't study how quickly the deviations will evolve. Maybe they are small at first but will explode after a short time interval. All of this must be checked. Just claiming it, isn't enough.
In principle you are right. But I have seen a lot of papers in which calculations of that sort have been done explicitly, so I think I can say I have a good intuition about which effects can be significant and which can't. It's not a proof that I am right, but all physicists have some intuition about some effects that makes them fairly confident in ignoring some effects due to being negligible. In practice, it's impossible to check everything in a single paper. Theoretical physics is, among other things, an art of making approximations based on experience. The point of this paper is to convey to other physicists my intuition about this stuff, not to make proofs. The proofs, or at least more quantitative arguments, can be found in references I cited. In this paper I am trying to explain the big picture, not the details. I am showing the woods, not the trees. Anyone who wants to see more details can read the cited references and make some detailed calculations by himself.
 
  • #56
Alright, then thanks for the discussion, I think it enhanced my understanding of BM quite a bit. Initially I thought that BM is really an exact interpretation of QM and gives the exact same results and I was just missing some basic argument. Now that I understand the difference, I think one should see this inequivalence as a chance, because it makes the question of interpretations amenable to experimental tests rather than a matter of opinion. I think the most important task for the BM community is to come up with some general methods of quantifying these deviations, so the question, which interpretation is right, can ultimately be answered by experiments, just like with Bell's theorem, which made some deep interpretational questions accessible to experimental physics. This could be really exciting.
 
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  • #57
Thors10 said:
Alright, then thanks for the discussion, I think it enhanced my understanding of BM quite a bit. Initially I thought that BM is really an exact interpretation of QM and gives the exact same results and I was just missing some basic argument. Now that I understand the difference, I think one should see this inequivalence as a chance, because it makes the question of interpretations amenable to experimental tests rather than a matter of opinion. I think the most important task for the BM community is to come up with some general methods of quantifying these deviations, so the question, which interpretation is right, can ultimately be answered by experiments, just like with Bell's theorem, which made some deep interpretational questions accessible to experimental physics. This could be really exciting.
In principle, I agree. In fact, in my younger days I was trying myself to make such a measurable distinction, see http://de.arxiv.org/abs/quant-ph/0406173 . But later, with more experience, I realized that my attempt (as well as attempts of many others) was too naive.
 
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  • #58
Here's another question for you: If you think one can do QM without intermediate collapse, then how do you calculate probabilities such as ##P(\mathbf{X}(t_1) \in A_1 \wedge \mathbf{X}(t_2) \in A_2)##? In orthodox QM, you would just calculate ##\int \left|(\chi_{A_2}U(t_2-t_1)\chi_{A_1}U(t_1)\psi)(\mathbf{X})\right|^2 \mathrm{d}\mathbf{X}##, but you can't do that without collapse, because the Heisenberg position observables ##\hat{x}(t)## don't commute for different ##t## (not even if they are pointer positions), so one can't perform the calculation at the end of the time evolution. One has to insert the ##\chi##'s at intermediate times.
 
  • #59
Thors10 said:
Here's another question for you: If you think one can do QM without intermediate collapse, then how do you calculate probabilities such as ##P(\mathbf{X}(t_1) \in A_1 \wedge \mathbf{X}(t_2) \in A_2)##? In orthodox QM, you would just calculate ##\int \left|(\chi_{A_2}U(t_2-t_1)\chi_{A_1}U(t_1)\psi)(\mathbf{X})\right|^2 \mathrm{d}\mathbf{X}##, but you can't do that without collapse, because the Heisenberg position observables ##\hat{x}(t)## don't commute for different ##t## (not even if they are pointer positions), so one can't perform the calculation at the end of the time evolution. One has to insert the ##\chi##'s at intermediate times.
Counterquestion: How do you measure it? To compare your probability with observation, you have to have information about what was at ##t_1## as well as at ##t_2##. How is this done in QM? Very simple, you measure ##A_1## at ##t_1##, and care about that the measurement result is preserved. So, you start with
##\psi_1(a_1) \psi_2(a_2)\psi(q,0)##. Independent variables defining the states of the two measurement devices and the state of the system, and independent initial wave functions. And physics which makes the interaction between them zero except for the time of the measurements. At ##t_1## this will become a superposition
$$\sum_i c_i \psi^i_1(a_1,t_1) \psi_2(a_2,t_1)\psi_i(q,t_1).$$
After this, the first measurement result will be preserved, thus, no more interaction with all the other things, and trivial evolution of ##a_1## itself. Thus, the future evolution of the ##\psi^i_1(a_1,t)## will be trivial after ##t_1##.
At ##t_2## this will become a superposition
$$\sum_i c_i \psi^i_1(a_1,t_1) \sum_j c_{ij} \psi^j_2(a_2,t_2)\psi_{ij}(q,t_2).$$

This is all yet quantum evolution without collapse, without any ##\chi## inserted. Then comes the final measurement of the positions ##a_1, a_2, q## which gives you ##\rho(a_1, a_2, q, t_2)##. It measures positions of different objects, they commute, no problem.
 
  • #60
Elias1960 said:
As far as I can see, whatever is described here as a difference between BM and QM appears as well in QM itself if we use different cuts.
I agree.
 
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  • #61
Thors10 said:
The collapse is a necessary ingredient of QM. All textbooks agree on that.
I guess you didn't see the textbook by Ballentine.

All textbooks (except perhaps Ballentine) agree that some version of collapse is necessary, but they do not agree on which version exactly that should be. Here you seem to assume one specific version of collapse, but that's not the only version that can be found in textbooks.
 
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  • #62
Well, the textbook by Ballentine may be an exception, but I think it is universally agreed that this part of the book is flawed.

I don't assume a specific version of collapse. I'm just asking: How do you calculate ##P(x(t_1) \in A_1 \wedge x(t_2) \in A_2)## in standard QM without inserting projections inbetween the unitary evolutions. I think it is impossible. Elias1960' proposal doesn't answer the question, because he calculates a completely different quantity. He assumes several measurement devices at the same time rather than a single one at different times. It's just an attempt to evade the problem, but surely one must be able to answer the original question. Standard QM can do it, no matter what version of collapse we assume.
 
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  • #63
I didn't defame you at any point. This is simply what I understood from your post and I still don't see the difference.

Anyway, the challenge still stands: Compute ##P(x(t_1) \in A_1 \wedge x(t_2) \in A_2)## without intermediate collapse. You haven't done this so far. You have computed a different quantity.
 
  • #64
What precisely is the experiment you want to describe? Without explanation it's not clear what you precisely want to meausure. Within the ensemble interpretation you can measure the probability distribution to find the particle around ##\vec{x}## at time ##t_1## or at time ##t_2##. The ensemble interpretation just applies to ensembles. Since you usually disturb the individual system in measuring ##\vec{x}## at ##t_1##, if you want to measure the probability distribution for ##\vec{x}## at a later time ##t_2## you have to prepare another ensemble and measure ##\vec{x}## at time ##t_2## rather than at time ##t_1##. However then I don't know the meaning of the logical and in your probability.
 
  • #65
Well, let me make it very precise:
I have a Hamiltonian ##H=\frac{p^2}{2}## and an initial state ##\psi_0(x)=\frac{1}{2\pi}e^{-\frac{x^2}{2}}##. What is the probability to find the particle in the set ##A_1 = [2,3]## at ##t_1 = 1## and in the set ##A_2 = [0,1]## at ##t_2 = 2##?

In standard QM, we would compute ##P(x(t_1)\in A_1 \wedge x(t_2)\in A_2) = \int \left|(\chi_{A_2}U(t_2-t_1)\chi_{A_2}U(t_1)\psi_0)(x)\right|^2\mathrm{d}x##.

But how can compute that without inserting ##\chi_{A_2}## inbetween the unitary evolution operators? I know one could in principle include two measurement devices into the description, one recording the position at ##t=t_1## and another one recording the position at ##t=t_2## and then measure the value of the corresponding pointers at ##t=t_2## (assuming the value of the first pointer hasn't changed inbetween). But that's not the question. It's just a trick to get around the actual problem. In fact, an experimenter can use the very same measurement device to measure the position at different times and he can even chose to discard the result of the first measurement completely without recording it at all. Since in this particular realistic setting, only one measurement device was used, it must be possible to describe this situation with a model that only contains one single measurement device. A model that includes two devices would just describe a different experimental setting. Anyway, in the experimental setup with one mesurement device, the measurements at ##t_1## and ##t_2## have really taken place, so I must be able to compute the probability ##P(x(t_1)\in A_1 \wedge x(t_2)\in A_2)## within the corresponding theory. You may replace the particle position ##x## by the corresponding pointer variable of the measurement device if you want to/need to. But you are not allowed to introduce two pointer variables, because this would correspond to a different experimental setting.
 
  • #66
vanhees71 said:
What precisely is the experiment you want to describe? Without explanation it's not clear what you precisely want to meausure. Within the ensemble interpretation you can measure the probability distribution to find the particle around ##\vec{x}## at time ##t_1## or at time ##t_2##. The ensemble interpretation just applies to ensembles. Since you usually disturb the individual system in measuring ##\vec{x}## at ##t_1##, if you want to measure the probability distribution for ##\vec{x}## at a later time ##t_2## you have to prepare another ensemble and measure ##\vec{x}## at time ##t_2## rather than at time ##t_1##. However then I don't know the meaning of the logical and in your probability.
Mott's particle track analysis is about multiple position measurements of the same particle producing a single track.
 
  • #67
Thors10 said:
But how can compute that without inserting ##\chi_{A_2}## inbetween the unitary evolution operators?
In the Many Worlds interpretation, there are also no intermediate ##\chi##s. So I agree with the others here that your critique isn't specific to Bohmian mechanics.

If your argument that the calculation cannot be done without inserting intermediate ##\chi##s is correct, the measurement device itself cannot be described by QM. I.e. you are arguing that QM isn't a universal theory but has only a limited domain of application.
 
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  • #68
Well, but this part of the discussion is neither about MWI nor BM. It's about the question whether intermediate collapse is necessary in standard QM. My argument is that it is required, because otherwise, standard QM has no way to calculate the joint probability that I asked for.

Can you explain how my argument in #86 implies that the device itself cannot be described by QM? I don't understand your reasoning. If I have wave function collapse at my disposal (i.e. the ability to insert projectors at intermediate times), I see no problem in principle.

Here's a toy model: Let ##x## be the particle position and ##y## be the pointer position. Let ##H=\frac{p_x^2}{2} + \frac{p_y^2}{2} + (x-y)^2##. Of course that's not a realistic interaction between particle and device, but a more complicated interaction won't add anything of value to the discussion. I now just apply the formula ##P(y(t_1)\in A_1 \wedge y(t_2)\in A_2) = \int \left|(\chi_{\mathbb{R}\times A_2}U(t_2-t_1)\chi_{\mathbb{R}\times A_1}U(t_1)\psi_0)(x,y)\right|^2\mathrm{d}x\mathrm{d}y## to get the desired probability. However, I have still inserted a ##\chi## inbetween the unitary evolutions.

In fact, I would be happy to see a formula without intermediate collapse. The situation I have described (using the same device twice at different times) is clearly physically possible and is also done regularly in the lab. I just see no possibility to describe the situation within standard QM without intermediate collapse.
 
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  • #69
Thors10 said:
Well, but this part of the discussion is neither about MWI nor BM.
Well, it's titled "Consistency of Bohmian mechanics". ;-) If you are interested in something more general, it probably would be good to ask a mentor to edit the title.

Thors10 said:
Can you explain how my argument in #86 implies that the device itself cannot be described by QM? I don't understand your reasoning. If I have wave function collapse at my disposal (i.e. the ability to insert projectors at intermediate times), I see no problem in principle.
Wave function collapse is in contradiction with unitary evolution.

If your quantum system consists only of the particle, this is not a problem. During the measurement, unitary evolution of the particle needn't hold because there's an interaction with the external measurement apparatus which isn't included in the quantum description.

If your quantum system consists of the particle and the apparatus, you don't have intermediate wave function collapse at your disposal anymore. Unitary evolution doesn't predict intermediate unique outcomes but intermediate superposition states of the combined system apparatus+particle. You again need something external (like a measurement with a device which measures the state of the combined system) in order to justify using the collapse.
 
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  • #70
Thors10 said:
Here's a toy model: Let ##x## be the particle position and ##y## be the pointer position. Let ##H=\frac{p_x^2}{2} + \frac{p_y^2}{2} + (x-y)^2##. Of course that's not a realistic interaction between particle and device, but a more complicated interaction won't add anything of value to the discussion.
The essential thing isn't the interaction but that you are including only a small part of the measurement apparatus in your description. If you include the whole device, you don't have anything external anymore which allows you to use the collapse.
 

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