Undergrad Consistency of Bohmian mechanics

[Demystifier]

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.

 

[Thors10]

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.

 

[Thors10]

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.

 

[vanhees71]

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.

 

[Thors10]

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.

 

[A. Neumaier]

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.

 

[kith]

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.

 

[Thors10]

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.

 

[kith]

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.

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.

 

[kith]

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.

 

[Thors10]

Well, I didn't want to discuss this issue initially in the first place ;-) I was just forced to, because Elias1960 claimed that standard QM can be done without collapse, attempting to circumvent my question regarding the consistency of BM.I don't agree that wave function collapse is in contradiction with unitary evolution. One may also interpret it as no evolution at all, but rather epistemically as a quantum version of Bayesian updating. One would do the same in classical probability theory after gaining new information. In quantum theory it's just counterintuitive, because it has some strange side effects.

If you interpret the wave function epistemically, you don't need a further external device to justify the collapse of the combined particle/pointer system.

But even if you disagree, the question remains how to calculate the joint probability I asked for within QM. I would be happy if you could do this without collapse. The only thing I insist on is that you don't include additional unphysical variables that record the intermediate measurement result until $t=t_2$. In particular, we have a lab with only one measurement device at our disposal. In the typical institute with only little funding, that's not uncommon. ;-)

 

[Thors10]

Hmm, I don't see what you gain by making the example more complicated. Of course I was lazy and modeled the device by only one variable, but how does including additional variables fix the problem? At $t=t_1$, the state will be a superposition of the many possible measurement results and we must make sure that only the actually measured one survives, because otherwise QM gives us no tools to discern them after further time evolution, which is why computing the joint probability I'm asking for seems impossible to me.

Can you give an example that you find acceptable and a formula for calculating that joint probability?

 

[Thors10]

Or let me formulate the question in an equivalent way: Let's say we already have the complicated wave function of the full system after all the measurements and after only unitary time evolution: $\psi(\mathbf{X},t_2)$. How do I extract the information $P(y(t_1) \in A_1 \wedge y(t_2) \in A_2)$ from this wave function? My only constraint is that the intermediate measurement result at $t_1 < t_2$ has not been recorded into some auxiliary variable. The measurement has been performed, but the result was discarded before $t=t_2$. The experimenter was a bit sleepy and just forgot to write it down.

 

[kith]

If you interpret the wave function epistemically, you don't need a further external device to justify the collapse of the combined particle/pointer system.

You are right, I misunderstood your scenario. I retract my claim from #88 that if you are right, the measurement device cannot be described by QM. The external "thing" which makes the collapse sensible is the observer herself. So essentially, we are in the Wigner's friend scenario. Intermediate \chis occur in the description of the friend but not in Wigner's description.

In the typical institute with only little funding, that's not uncommon. ;-)

 

[Thors10]

Well, but I don't see how that helps. What if Wigner's friend already forgot the result of the $t_1$ measurement at $t=t_2$? Introducing Wigner's friend is just another way of evading my constraint that the intermediate measurement result at $t_1$ shouldn't be recorded in an auxiliary variable. I'm quite sure that that the joint probability I keep asking for can't be computed from the final wave function alone if no intermediate information has been recorded in an auxiliary variable.

Let me ask in a more general way: It's clear that unitary evolution doesn't only produce new information. It must also forget some old information, because there isn't enough space to store all newly gained information in the wave function. 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?

 

[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]

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".

 

[Thors10]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

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

 

[A. Neumaier]

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]

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]

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]

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]

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]

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]

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 let's 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]

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]

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.