# Consistency of Bohmian mechanics

• I
I have a free particle with a Gaussian wave function ##\psi(x,0) = N \exp\left(-\frac{x^2}{2}\right)##. After time evolution with ##H=\frac{p^2}{2}##, the wave function is given by ##\psi(x,t)=N\left(\frac{1}{1+i t}\right)^{3/2} \exp \left(-\frac{x^2}{2 (1+i t)}\right)##. The time evolution of a Bohmian particle is given by ##\dot x(t) = \mathrm{Im} \frac{\nabla \psi}{\psi} = \frac{t x(t)}{1+t^2}##, which is solved by ##x(t) = x_0\sqrt{1+t^2}##.

Now my question: If I measure the particle at some positive location (##x(0) =x_0 > 0##) at time ##t=0##, then Bohmian time evolution predicts that its position will always stay on the positive side of the real line, because ##x(t) > 0## for all ##t##. However, if I perform a quantum measurement, the wave function instead collapses to ##\psi(x,0) = \delta(x-x_0)##, which evolves into a Gaussian again. Thus the probability to find the particle in the negative half axis at time ##t>0## is non-zero according to quantum mechanics. This contradicts the previous Bohmian result.

How does Bohmian mechanics deal with this issue?

atyy

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Demystifier
Gold Member
The particle can move to the left part of the wave packet during the measurement. That's because during the measuement the Hamiltonian is more complicated than simply ##p^2/2## so the Bohmian equations that you have written above are not correct during the measurement.

Can you show me how to calculate that? Thanks!

Demystifier
Gold Member
Can you show me how to calculate that? Thanks!
It's much easier to draw than to calculate, see the attached drawing.

The picture shows the evolution of wave function (black) and the Bohmian trajectory (blue) with time. The non-shadowed region shows the region where the wave function is not zero. At the beginning we have one wide wave packet, at intermediate times (the time at which the wave function collapse happens) we have two narrow wave packets, and finally at late times we have two wide wave packets. The particle is initially in the right half of the wave packet, but finally it is in the left half of the wave packet on the right.

EDIT: Sorry for the blue color that cannot be easily distinguished from the black. I did not have a red pencil, but I hope you get the idea.

Spinnor, atyy and Minnesota Joe
A. Neumaier
2019 Award
It's much easier to draw than to calculate, see the attached drawing.
But one needs calculations to justify the drawing....

Demystifier
Gold Member
But one needs calculations to justify the drawing....
In principle yes, but if one understands the idea conveyed by the drawing, then it's not necessary if one is satisfied with intuitive qualitative understanding. Those who are not satisfied are welcome to perform a full calculation by themselves. It's not an easy calculation and I have no intention to do it here.

Thank you very much for the drawing! There are still some things I don't understand: I assume the two white regions stand for the resulting branches after collapse? But in a position measurement, we should have infinitely many branches, one for each real number ##x_0## and all of them shoud evolve into Gaussians again, so I can't imagine what the appropriate drawing would look like. Also, the black region seems to indicate that the particle can't ever cross it again, so while it may be able to go back a little, it still can't go back beyond a certain boundary. Is that right?

I still would be really interested in a calculation. It needn't be a full calculation, but someone must have studied similar situations in a toy model. I just don't know where to look for in the literature. It would also be nice if you could tell me just the Hamiltonian ##H## that I need to look at and I can see if I can solve it for myself. Thanks again!

Demystifier
Gold Member
Thank you very much for the drawing! There are still some things I don't understand: I assume the two white regions stand for the resulting branches after collapse?
Yes.

But in a position measurement, we should have infinitely many branches, one for each real number ##x_0## and all of them shoud evolve into Gaussians again,
If the position measurement is perfectly precise, you are right. But a realistic measurement is not perfectly precise, so in reality there are only a finite number of narrow branches. To illustrate my point, I have drawn the simplest nontrivial case with only two branches.

so I can't imagine what the appropriate drawing would look like.
I hope you can imagine what the drawing would look like for a finite number of branches, say 10.

Also, the black region seems to indicate that the particle can't ever cross it again, so while it
may be able to go back a little, it still can't go back beyond a certain boundary. Is that right?
Right.

I still would be really interested in a calculation. It needn't be a full calculation, but someone must have studied similar situations in a toy model. I just don't know where to look for in the literature. It would also be nice if you could tell me just the Hamiltonian ##H## that I need to look at and I can see if I can solve it for myself. Thanks again!
For some modeling of quantum measurements in Bohmian mechanics see e.g. http://de.arxiv.org/abs/1610.01138

But if the particle can't cross the forbidden region, this still doesn't solve the problem that QM predicts a non-zero probability for the particle to be found behind the forbidden region. Also, the resulting wave-function after the collapse looks nothing like a Gaussian, which is what the QM prediction would be. I still can't reconcile the drawing with the QM predictions, even if I imagine an imperfect measurement with finitely many branches. After the collapse, the wave-function should look at least somewhat like a Gaussian centered at ##x_0##, even in an imperfect measurement and there shouldn't be any branches.

Thank you for the paper, I will have a look at it!

Demystifier
Gold Member
But if the particle can't cross the forbidden region, this still doesn't solve the problem that QM predicts a non-zero probability for the particle to be found behind the forbidden region.
When I say that the particle can't cross the forbidden region, I really mean that probability for crossing it is negligibly small. The white region is e.g. the ##5\sigma## region of the Gaussian.

Also, the resulting wave-function after the collapse looks nothing like a Gaussian, which is what the QM prediction would be. I still can't reconcile the drawing with the QM predictions, even if I imagine an imperfect measurement with finitely many branches. After the collapse, the wave-function should look at least somewhat like a Gaussian centered at ##x_0##, even in an imperfect measurement and there shouldn't be any branches.
In my drawing the wave function after the collapse looks like two Gaussians with a negligible overlap. In more general case, it looks like ##N## Gaussians with a negligible overlap. But the main point of Bohmian mechanics is that only one of those Gaussians is filled with a particle, while all the other Gaussians are empty. This means that all the other Gaussians do not have influence (or more precisely, that their influence is negligible) on the particle, so are irrelevant. In other words, the particle behaves as if only one Gaussian (the non-empty one) existed, which for all practical purposes looks as if a collapse of the wave function happened. That is how Bohmian mechanics explains the appearance of collapse (in agreement with standard QM) without the actual collapse.

Oh I see, so the forbidden region is not really forbidden, just very unlikely. That would indeed resolve the initial problem. However, I don't agree that the wave-function after the collapse can still be a superposition of Gaussians. One can imagine a dynamics that will later squeeze all the Gaussians together by turning on a certain potential term at some later time (e.g. ##V(x,t) = x^2\theta(t-t_1)##), so all of them will become relevant again, while in the QM collapse, the other Gaussians will be gone forever. BM and QM seem to make different predictions in this scenario.

Demystifier
Gold Member
Oh I see, so the forbidden region is not really forbidden, just very unlikely. That would indeed resolve the initial problem.
Good.

However, I don't agree that the wave-function after the collapse can still be a superposition of Gaussians. One can imagine a dynamics that will later squeeze all the Gaussians together by turning on a certain potential term at some later time (e.g. ##V(x,t) = x^2\theta(t-t_1)##), so all of them will become relevant again, while in the QM collapse, the other Gaussians will be gone forever. BM and QM seem to make different predictions in this scenario.
The Gaussians (that is, the appearance of collapse) are stable because they are entangled with the wave function of the measuring apparatus (and of the environment). Technically, this phenomenon in the literature is known as decoherence. Since the number of degrees of freedom associated with the apparatus (and the environment) is enormous, a typical time needed to bring the Gaussians together is many orders of magnitude longer that the age of the Universe. So due to decoherence (on which you may want to read more) BM and QM make the same predictions.

But the difference to decoherence is that in decoherence, the state of the particle evolves into a mixed state, while your drawing depicts a superposition and the interference terms will become relevant if the Gaussians collide (e.g. in an ##x^2## potential).

If you consider a system with decoherence and learn something e.g. about the pointer variable at ##t_0##, then the same effect as in my initial post occurs: The conditional probability density of the whole system at time ##t>t_0## (given the information at ##t_0##) won't coincide with the probability density that comes from the unitary dynamics (##\left|\psi(\{x\},t)\right|^2##). The difference is just that we now look at a very high-dimensional system.

Maybe I should explain my question a bit better: In BM, we have an initial probability density ##\rho(\mathbf{X})=\left|\psi(\mathbf{X})\right|^2## on the configuration variables ##\mathbf{X}##. The time evolution of these variables is given by ##\dot{\mathbf{X}}(t)=\mathrm{Im}\frac{\nabla U(t)\psi}{U(t)\psi}## (##U(t)## is the time evolution of QM). If we solve these differential equations, we get a map ##E_t## from the initial positions ##\mathbf{X}(0)## to the positions ##\mathbf{X}(t)##, i.e. ##\mathbf{X}(t) = E_t(\mathbf{X}(0))##. In my example above, we would have ##\mathbf{X}=x## and ##E_t(x) = x\sqrt{1+t^2}##.

Now there seems to be a difference between how BM and QM deal with learning new information:
1. In QM, we would just collapse the wave function.
2. in BM, we must instead update the probability distribution. If we habe learned that ##\mathbf{X}(t)\in A##, then the probability distribution on the initial configurations at ##t=0## (##\rho(\mathbf{X})##) gets updated to ##\rho'(\mathbf{X}) = \frac{\chi_{E_t^{-1}(A)}(\mathbf{X})\rho(\mathbf{X})}{\int\chi_{E_t^{-1}(A)}(\mathbf{X})\rho(\mathbf{X})\mathrm{d}\mathbf{X}}##. In my example above, ##\rho'(x) = 2\theta(x)N\exp(-\frac{x^2}{2})##, if we learn that the particle is on the positive side at ##t=0##.

My problem is that these don't coincide. Now you're saying decoherence helps, so we include the apparatus and the environment into the description (##\mathbf{X} = (x,a,e)##) and then the probability distribution ##\rho'(x)=\int\rho'(x,a,e)\mathrm{d} a\mathrm{d} e## for the particle ought to behave like in a quantum collapse. Maybe this is true, but it still certainly won't apply to the distribution ##\rho'(x,a,e)## on all variables, so BM still can't reproduce the QM dynamics on the full system.

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It's much easier to draw than to calculate, see the attached drawing.
The drawing is misleading because it ignores a key point: During and after the measurement, you have to draw a picture in higher dimensional space, not only the particle itself counts, but also the measurement result.

But the point that a particle at the left has to remain on the left side holds only in the one-dimensional case, given that trajectories cannot cross each other. In two dimensions, there is no such property. The trajectory of the particle measured on the right side will not cross the trajectory of the particle measured on the left side even if it goes left and the left particle goes right, simply because they travel are on different roads defined by the different measurement results.

In the three-dimensional situation, the calculation is analogous and you get ##\vec{x}(t) =\vec{x}_0\sqrt{1+t^2}##, so the particle always stays in the same octant. It's not a problem of the one-dimensional case only.

Now there seems to be a difference between how BM and QM deal with learning new information:
1. In QM, we would just collapse the wave function.
2. in BM, we must instead update the probability distribution.
No, in BM we have either to collapse the wave function too, or we have to use the bigger wave function which includes also the measurement instrument. And you have to use it taking into account as the Bohmian position of the quantum system, as the Bohmian position of the measurement device. Given that the trajectory of the measurement instrument is simply visible, the two possibilities are connected by the formula for the effective wave function of a subsystem:

$$\psi_{sys}(q_{sys},t) = \psi(q_{sys},q_{dev}(t), t)$$

Once you see the trajectory of the measurement device, it makes no sense to use the big wave function for it, and one uses the effective one of the quantum part alone. But this is also the collapsing one.

Demystifier
Gold Member
The drawing is misleading because it ignores a key point: During and after the measurement, you have to draw a picture in higher dimensional space, not only the particle itself counts, but also the measurement result.
I agree, but still the drawing does help to develop intuition to a certain extent.

I now understand that in BM, the collapse of a subsystem is conjectured to arise from decoherence. I think there is no agreement whether this works even among mainstream quantum physicists. But even if it did, this still wouldn't address my main problem:

My issue is that after a measurement, one needs to update the probability distribution of the full system in both BM and QM. In QM, one just collapses the wave function, but in BM, one has to use the conditional probability ##P'(X) = P(X|M)## as the updated probability distribution, where ##M## is the event that the particle (or the pointer position) has been measured somewhere. Now my calculation shows that the updated probability distribution is no longer given by ##\left|\psi\right|^2## and its evolution is no longer given by the Schrödinger equation. So BM really is a different theory that makes different predictions than QM. One may not see this on the level of a subsystem, but it's certainly true for the full system.

My issue is that after a measurement, one needs to update the probability distribution of the full system in both BM and QM. In QM, one just collapses the wave function, but in BM, one has to use the conditional probability ##P'(X) = P(X|M)## as the updated probability distribution, where ##M## is the event that the particle (or the pointer position) has been measured somewhere. Now my calculation shows that the updated probability distribution is no longer given by ##\left|\psi\right|^2## and its evolution is no longer given by the Schrödinger equation. So BM really is a different theory that makes different predictions than QM. One may not see this on the level of a subsystem, but it's certainly true for the full system.
No. Here is the formula which connects the wave function of the subsystem with the wave function of the full system inclusive measurement device:
$$\psi_{sys}(q_{sys},t) = \psi(q_{sys},q_{dev}(t), t)$$
Here ##\left|\psi\right|^2## defines the probability that the measurement device is at ##q_{dev}## and the system in ##q_{sys}##. The evolution of the full wave function follows the Schroedinger equation. But if there is some interaction, the reduced wave function does not follow the Schroedinger equation reduced to the subsystem during the interaction time.
I now understand that in BM, the collapse of a subsystem is conjectured to arise from decoherence.
The formula above works always. No decoherence is necessary. One needs decoherence only if one (1) wants to be sure that the "measurement" may not be reverted later so that one can effectively forget about ##\psi## and reduce the consideration to the subsystem ##\psi_{sys}##. (2) sees the result measured by the measurement device.

Well, I don't care about subsystems. My issue concerns the full system and I'm asking whether BM produces the same predictions for the full system as QM. The answer seems to be no, because for that to be true, the probability density of the updated probability distribution ##P'(X) = P(X|M)## is no longer given by ##\left|\psi\right|^2## anymore and hence there will be observables for which BM and QM make different predictions. In my example, the probability distribution after measurement would be ##2\theta(x)\left|\psi(x,t)\right|^2##, which is not equal to ##\left|\psi(x,t)\right|^2##, if we consider for the moment for illustrative purposes the possibility that the full system consists only of one particle.

Demystifier
Gold Member
In my example, the probability distribution after measurement would be ##2\theta(x)\left|\psi(x,t)\right|^2##, which is not equal to ##\left|\psi(x,t)\right|^2##, if we consider for the moment for illustrative purposes the possibility that the full system consists only of one particle.
I think your analysis is inconsistent, for if the full system consists of only one particle then there can be no measurement to begin with.

vanhees71
So what is a measurement in BM? If I understood you correctly, one should include a pointer variable, which can be read off without influencing the system. So we have a wave function ##\psi(x,y,t)##, where ##y## is the pointer variable. Now if I learn something about the pointer variable, e.g. that it is positive, then I have to update the probability distribution from ##\left|\psi(x,y,t)\right|^2## to ##N\theta(y)\left|\psi(x,y,t)\right|^2##, where ##N## is a normalization constant. The situation is completely identical, just more complicated. The difference is that I have just assumed my ##x## to be the pointer variable, which can be read off without influence.

Demystifier
Gold Member
Now there seems to be a difference between how BM and QM deal with learning new information:
1. In QM, we would just collapse the wave function.
2. in BM, we must instead update the probability distribution. If we habe learned that ##\mathbf{X}(t)\in A##, then the probability distribution on the initial configurations at ##t=0## (##\rho(\mathbf{X})##) gets updated to ##\rho'(\mathbf{X}) = \frac{\chi_{E_t^{-1}(A)}(\mathbf{X})\rho(\mathbf{X})}{\int\chi_{E_t^{-1}(A)}(\mathbf{X})\rho(\mathbf{X})\mathrm{d}\mathbf{X}}##. In my example above, ##\rho'(x) = 2\theta(x)N\exp(-\frac{x^2}{2})##, if we learn that the particle is on the positive side at ##t=0##.
The problem with your analysis is that you don't consider how the new information is learned. The new information is not told us by God. The new information is acquired through a process of measurement. The process of measurement can be described in terms of wave functions for the full system (namely, the measured system + the apparatus), and when this is taken into account (which your analysis doesn't) it turns out that QM and BM make the same measurable predictions.

Demystifier