# Consistency of Bohmian mechanics

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kith
So let's say unitary evolution has just forgotten the position of some particle/pointer ##x_i## at time ##t_2##, even though it has been measured a while ago. How do I calculate from the final wave function ##\psi(\mathbf{X},t_2)## a probability for some information that once was available at ##t_1##, but has been forgotten in the meantime?
From Wigner's point of view, there hasn't been an intermediate measurement in the sense of QM and thus there's no information which was available at time $t_1$. In Wigner's description, the state at $t_1$ is a superposition which includes "friend has observed the particle to be in $A_1$" and "friend has observed the particle to be not in $A_1$".

Just to be clear that we are talking about the same situation: the friend is the observer of your scenario and Wigner is an additional external observer.

PeterDonis
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It's clear that unitary evolution doesn't only produce new information. It must also forget some old information
This is wrong. Unitary evolution is reversible so it exactly preserves information.

there isn't enough space to store all newly gained information in the wave function
There is no need to since there is no "newly gained information".

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kith:

Well, then we agree, because that's exactly what I'm trying to say. Without intermediate collapse, it is impossible to compute such joint and conditional probabilities and hence QM without collapse doesn't work. With collapse, you can compute the joint probability also in this setting. (Wigner can just ask his friend at ##t=t_1## and collapse his wave function)

Maybe you will now introduce some further observer who observes Wigner. My response would be: Let observer ##i+1## ask observer ##i## what they measured at time ##t_1##. Then let observer ##i## forget their result before ##t_2##. So you get an infinite chain of observers who must collapse their wave functions, yet there is no hypothetical full system which accurately describes the system without collapse. (Of course, this scenario is pathological, but as a gedankenexperiment, it shows that there is a conceptual problem if you drop the collapse axiom without replacement.)

PeterDonis:

That's not what I meant by information. When we talk about apparent collapse, some variable needs to become and remain almost classical if we want to perform the collapse at the end of the calculation. That's what I meant by "newly gained information." It's called deferred measurement. But you can't make all observables almost classical.

I agree that unitary evolution can be inverted and we can use this to compute the joint probability I asked for. But in order to do that, you will end up with an expression again, where you inserted a ##\chi## inbetween two unitary evolutions: First you evolve back, then you insert ##\chi## and then you evolve forward again. So again, you end up with a projection at an intermediate time.

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PeterDonis
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That's not what I meant by information.
Then you need to either give a reference for what you mean by "information", since you can't just make up its meaning for yourself, or use a different term. Or, you could recognize that the context in which you are using the term "information" is not unitary evolution; see below.

When we talk about apparent collapse, some variable needs to become and remain almost classical if we want to perform the collapse at the end of the calculation. That's what I meant by "newly gained information."
That has nothing to do with unitary evolution, since collapse is not unitary. When you apply collapse, mathematically, you are stopping unitary evolution and using a different mathematical process, which is not unitary and not reversible and does not preserve information, in the standard sense of "information" that I was using.

I agree that unitary evolution can be inverted and we can use this to compute the joint probability I asked for. But in order to do that, you will end up with an expression again, where you inserted a χ\chi inbetween two unitary evolutions: First you evolve back, then you insert χ\chi and then you evolve forward again. So again, you end up with a projection at an intermediate time.
I have no idea what you are talking about here.

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PeterDonis
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PeterDonis
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Thread reopened. Two overly argumentative posters have been thread banned and some overly argumentative posts have been deleted.

A. Neumaier
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What precisely is the experiment you want to describe? Without explanation it's not clear what you precisely want to measure.
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 one compute that without inserting ##\chi_{A_2}## in between the unitary evolution operators?
The standard way to measure your probability is to let in each round a particle pass two consecutive slits at ##A_1## and ##A_2##, with a screen at the end. Then one records which fraction of the in-particles appeared at the screen. But this involves three measurements, two projective ones at the slits and a final one - which would be redundant if impacts on the slit material were recorded rather than (as usual) discarded.
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 in between). 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
Please explain how an experimenter can measure your probability with a single measurement device, in sufficient detail to make it suggestive.

Demystifier
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I don't assume a specific version of collapse.
Well, you argue that in Bohmian mechanics we don't have collapse, but an update of knowledge. But in standard QM, the collapse is also often interpreted as an update of knowledge. So it seems that you assume collapse that cannot be interpreted as an update of knowledge. Indeed, by the collapse you mean a collapse of the wave function of the whole universe (and not merely a collapse of the wave function of a small subsystem), which cannot be interpreted as an update of knowledge because in practice nobody knows the wave function of the whole universe anyway. So you do assume a specific version of the collapse postulate which is not generally accepted within standard QM.

vanhees71
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The standard way to measure your probability is to let in each round a particle pass two consecutive slits at ##A_1## and ##A_2##, with a screen at the end. Then one records which fraction of the in-particles appeared at the screen. But this involves three measurements, two projective ones at the slits and a final one - which would be redundant if impacts on the slit material were recorded rather than (as usual) discarded.

Please explain how an experimenter can measure your probability with a single measurement device, in sufficient detail to make it suggestive.
You have to distinguish two different things.

If you want to meausure the position probabilities for the particles prepared at ##t=0## in the state ##|\psi_0 \rangle## at two times ##t_1,t_2>0## you have to first calculate ##|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi_0 \rangle## and then you have ##W(t,\vec{x})=|\psi(t,\vec{x})|^2## with ##\psi(t,\vec{x})=\langle \vec{x}|\psi(t,\vec{x})##. To measure ##W(t,\vec{x})## you have to repeat the entire setup sufficiently often and first measure the particle's position at ##t=t_1## and then another ensemble measuring at ##t=t_2##.

If you want what you describe in the quoted post, you have to take the two slits into account in the Hamiltonian, and then you see that you measure a different probability distribution for the outcomes of these measurement devices since the time evolution of the state is affected by the presence of the slits.

You always need the context when describing a measurement, including relevant measurement devices which can influence the measurement outcomes of other devices in the experimental setup.

A. Neumaier
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If you want what you describe in the quoted post, you have to take the two slits into account in the Hamiltonian, and then you see that you measure a different probability distribution for the outcomes of these measurement devices since the time evolution of the state is affected by the presence of the slits.
This is precisely what the formula of @Thors10 quoted in post #82 does. Thus it is his intended setup. The only questionable thing in his description is that he claims he can do it by repeatedly measuring a single particle twice with a single detector. So I inquired about that.

vanhees71
vanhees71
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That's a third variant then, because then you'd need to register the passing of the particle at each of the slits, i.e., you have to let it interact with something else, e.g., a photon, and then again you get something else. That's why I asked for the specific experimental setup he has in mind.

This is the issue of "contextuality" of quantum theory, which is also heavily debated in foundation circles, and the resolution of any quibbles usually are of course to precisely describe the physical setup of the experiment.

A nice classical example of this is v. Weizsäcker's consideration of the Heisenberg microscope gedankenexperiment, where you observe an electron in the double-slit experiment through scattering of a single photon, which you let go through a lense and register it either in the focal plane for measuring its momentum or in the plane determined by the lense equation for meausuring the electron's position. Repeating this in the first case you get a double-slit interference pattern in the second case you get the incoherent single-slit superposition. So you need the complete (!) context, i.e., the full setup to describe the outcome though the interaction necessary for measurement (i.e., the photon scattering with the electron in the one or the other slit) is the same in both setups.

A. Neumaier
2019 Award
That's a third variant then, because then you'd need to register the passing of the particle at each of the slits,
The two filters containing the consecutive slits automatically register those particles that don't pass (increasing its energy slightly - whether this is actually recorded or not does not make any change in the final result), while the final screen registers the particle that passes. Thus nothing special needs to be done. The final probability is the ratio of in-particles produced by the source (which can be determined by appropriate calibration) to out-particles (which are registered by the final screen) and should agree under perfect conditions with the integral given by Thors10.
So you need the complete (!) context, i.e., the full setup to describe the outcome
No. Since the intervals specified by Thors10 for the two slits are wide, diffraction - which would matter for narrow slits - can be neglected. Thus a straightforward analysis is adequate.

vanhees71
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Ok, I still don't understand the precise setup you have in mind. if you have two slits having an effect on the particles to be meausered you modify the dynamics, i.e., the Hamiltonian and thus change the probabilities compared to the situation of freely moving particles. Then it also makes a difference whether you register the particle (or at least imposing something that in principle lets you know about the particle running through either slit or not. This has been discussed already by Bohr and Einstein in their famous debates like Einstein's box apparently determining position and momentum of a particle both precisely, but counterargued rightly by Bohr taking into account the uncertainties of the box itself. I don't think that you can avoid contextuality that easily!

A. Neumaier
2019 Award
Ok, I still don't understand the precise setup you have in mind. if you have two slits having an effect on the particles to be measured you modify the dynamics, i.e., the Hamiltonian and thus change the probabilities compared to the situation of freely moving particles.
In order that I understand you, please tell me the modified Hamiltonian for a free particle in a coherent state with fairly precise momentum that at time ##t=t_1## moves through an idealized, infinitely thin filter with a single wide slit.

I think it should have the time-dependent form ##H(t)=H_0## for ##t<t_1## and ##H(t)=H_1## for ##t>t_1##. It seems to me that both ##H_0## and ##H_1## are the free Hamiltonian. The effect of the filter is simply to change at ##t=t_1## the wave function by setting ##\psi(x)## to zero at positions outside the slit and rescaling.

If you agree, you can repeat this (if the slit is wide, the result is still an approximate coherent state) with a second filter passed at time ##t=t_2##, and get the formula of Thors10 from post #65.

If you don't agree, please specify your version of the story precisely enough that I can see how it differs from mine.

vanhees71
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I don't know that Hamiltonian. And also you again changed the setup. Now we have an infinitely thin filter. What is this filter supposed to do? If the filter is not moving, why should the Hamiltonian be time dependent?

I still do not know, which experiment you really have in mind. Maybe it's more like a cloud chamber, where you can follow a "track", and that's what's described in Mott's famous paper about ##\alpha## particles emitted from a radioactive source in a cloud chamber? Then you could make a movie showing the cloud track being formed, but then due to the vapor you don't observe a free particle but a particle interacting with the vapor atoms, and this is what gives a track. Again, the context about what precisely is how measured must be given to make sense of the question.

A. Neumaier
2019 Award
I don't know that Hamiltonian. And also you again changed the setup. Now we have an infinitely thin filter. What is this filter supposed to do?
It is a piece of metal with a hole in it. Idealized to be infinitely thin; otherwise I'd need to model the interaction of the electrons with the metal. I don't know of any discussion of an experiments with slits where the latter is done; hence the idealization. Then all the complicated stuff happens instantaneously and can be summarized by the projection. At least this is done in informal discussions when blending out partial beams (in Stern-Gerlach experiments, say).
If the filter is not moving, why should the Hamiltonian be time dependent?
You were claiming that the filters change the Hamiltonian:
if you have two slits having an effect on the particles to be measured you modify the dynamics, i.e., the Hamiltonian and thus change the probabilities compared to the situation of freely moving particles.
I was just trying to understand what you mean. Surely away from the filters the Hamiltonian is the free Hamiltonian, so the only way I could give meaning to your claim was to assume that you thought that the Hamiltonian is time-dependent with three constant pieces before, during and after the passage through the slit.
I still do not know, which experiment you really have in mind.
A very simple experiment. I have a source of electrons, two close and parallel (in this post only one) sheets of metal both with a big hole, and a screen parallel to the plates at the end. I want to know the probability that an electron emitted by the source is detected by the screen. Under the usual idealizations and the standard collapse assumption, this probability is given by the formula of Thors10 from post #65, with characteristic functions specified by the positions of the holes.

A. Neumaier
2019 Award
Thread reopened. Two overly argumentative posters have been thread banned and some overly argumentative posts have been deleted.
I hope that Thors10 is not banned from posting in this thread, since he is the originator of the thread, and the current discussion needs his active participation:
This is precisely what the formula of @Thors10 quoted in post #82 does. Thus it is his intended setup. The only questionable thing in his description is that he claims he can do it by repeatedly measuring a single particle twice with a single detector. So I inquired about that.

PeterDonis
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