Consistency of Bohmian mechanics

[Thors10]

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?

 

[Demystifier]

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.

 

[Thors10]

Can you show me how to calculate that? Thanks!

 

[Demystifier]

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.

 

[A. Neumaier]

It's much easier to draw than to calculate, see the attached drawing.

But one needs calculations to justify the drawing...

 

[Demystifier]

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.

 

[Thors10]

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]

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

 

[Thors10]

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]

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.

 

[Thors10]

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]

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.

 

[Thors10]

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.

 

[Thors10]

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.

 

[Elias1960]

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.

 

[Thors10]

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.

 

[Elias1960]

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]

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.

 

[Thors10]

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.

 

[Elias1960]

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.

 

[Thors10]

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]

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.

 

[Thors10]

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]

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]

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.

That's not how it works. You must take into account the entanglement between the pointer variable and the measured system. See e.g. my "Bohmian mechanics for instrumentalists" linked in my signature below, especially Sec. 3.2.

 

[Thors10]

No that's not true. My analysis in that post contains this possibility. $\mathbf{X}$ is a placeholder for a whole set of coordinates. One of them may be the coordinate of the subsystem, another one may be the coordinate of the pointer variable and you may also include the environment. I explained it n the end of that post. For example $\mathbf{X} = (x,a,e)$ and if you now learn that the pointer variable $a$ is positive, you would have to update the probability distribution from $\left|\psi(x,a,e,t)\right|^2$ to $N\chi_{\mathbb{R}\times\mathbb{R}_+\times\mathbb{R}}(x,a,e)\left|\psi(x,a,e,t)\right|^2$. The interaction between $x$, $a$ and $e$ suring measurement is already contained in $\psi(x,a,e,t)$, but the updating of the probability distribution must be done nevertheless, because in BM, the probability distribution is supposed to be due to ignorance, so if I learn something, I have to update it.

 

[Thors10]

The entanglement between the pointer observable and the measured system is already contained in $\psi(x,y,t)$.

 

[Thors10]

Here's another example. Let $x$ be the measured system and $y$ be the pointer variable and let the interaction be given by $H=\frac{p_x^2}{2} + \frac{p_y^2}{2} + (x-y)^2$. We can solve this analytically again by going to center of mass coordinates $c = \frac{x+y}{2}$, $r=x-y$. The evolution of an initial wave function $\psi(x,y,0) = \frac{1}{2\pi}e^{-\frac{(x+y)^2}{8}}e^{-\frac{(x-y)^2}{2}}$ is then given by $\psi(x,y,t) = \frac{e^{-it}}{2\pi(1+\frac{it}{2})^{\frac{3}{2}}}e^{-\frac{(x+y)^2}{8(1+\frac{it}{2})}}e^{-\frac{(x-y)^2}{2}}$, because the center of mass Hamiltonian is a sum of a free center of mass movement and a harmonic oscillator. The interaction between the measured system $x$ and the pointer variable $y$ is already there. Now if I learn at $t=0$ that the pointer variable $y$ is positive for instance, I have to multiply the probability distribution $\left|\psi(x,y,t)\right|^2$ by $\theta(y)$ and normalize it. This new distribution is no longer given by $\left|\psi(x,y,t)\right|^2$ in the future.

 

[Demystifier]

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 $t=0$.

You didn't define what is $\chi_{E_t^{-1}(A)}$. I guess it's a characteristic function which is equal to 1 in $A$ and equal to 0 elsewhere. In other words, $\chi$ projects the wave function to the region $A$. But if so, then it coincides with QM (your item 1.) because the collapse is nothing but a projection of the wave function to $A$. Or perhaps by $\chi_{E_t^{-1}(A)}$ you mean something else?

 

[Demystifier]

Now if I learn at $t=0$ that the pointer variable $y$ is positive for instance, I have to multiply the probability distribution $\left|\psi(x,y,t)\right|^2$ by $\theta(y)$ and normalize it. This new distribution is no longer given by $\left|\psi(x,y,t)\right|^2$ in the future.

But standard QM also says that the new new distribution is no longer $\left|\psi(x,y,t)\right|^2$, due to the collapse.

 

[Thors10]

$E_t$ is the time evolution of the Bohmian hidden variables (from $t_0=0$ to $t$). In my example it is given by $E_t(x)=x\sqrt{1+t^2}$, so $E_t^{-1}(x) = \frac{x}{\sqrt{1+t^2}}$ and given $x(t)$, it just gives us $x(0)$. So if we learn something at time $t$, we compute what the initial conditions must have been through $E_t^{-1}$ and update the probability distribution on the initial conditions accordingly. In my example, $t$ was just $0$, so $E_0^{-1}(x) = x$ and $\chi_{E_0^{-1}(\mathbb{R}_+)}(x) = \chi_{\mathbb{R}_+}(x) = \theta(x)$.

 

[Thors10]

Yes, of course in standard QM we have to update the wave function due to collapse. The problem is that the updated Bohmian probability distribution of the full system is not equal to the collapsed and evolved QM wave function at later time. In this harmonic oscillator example, we would have to collapse the wave function to some function of $x$ times the delta distribution $\delta(y-y_0)$ and the Hamiltonian I wrote won't evolve this into the updated Bohmian distribution, because in BM the wave function of the full system evolves unitarily and the updating just restricts the possible initial conditions of the Bohmian variables.

 

[Demystifier]

Yes, of course in standard QM we have to update the wave function due to collapse. The problem is that the updated Bohmian probability distribution of the full system is not equal to the collapsed and evolved QM wave function at later time.

There are many texts in the literature that show that they are equal. It would help if you could take some of those texts and say what, in your opinion, is exactly wrong in this text.

In this harmonic oscillator example, we would have to collapse the wave function to some function of $x$ times the delta distribution $\delta(y-y_0)$ and the Hamiltonian I wrote won't evolve this into the updated Bohmian distribution, because in BM the wave function of the full system evolves unitarily and the updating just restricts the possible initial conditions of the Bohmian variables.

I don't quite follow your arguments, but note that the Hamiltonian you wrote is not an interaction that may serve the purpose of measurement. It's not that any interaction counts as measurement. A measurement is an interaction that splits the wave function into separated branches as in the drawing in my first post on this thread.

 

[Thors10]

What is conjectured in the literature is that BM reproduces the wave function collapse on the level of a subsystem. I'm not convinced by the arguments, because the calculations are too simplistic, but anyway, that's not what I care about. My question concerns the collapse of the full system's wave function, not the subsystem. I don't think this is discussed anywhere in the literature, but I would of course be interested if you could point me somewhere.

Well, as Elias1960 said, your drawing is misleading. One can achieve the split you want by turning the interaction on and off in some finite time interval. Nevertheless, the Bohmian updating won't agree with the quantum collapse. It's easy to see that this can only be true if the pointer variable commutes with the Hamiltonian and this is never the case, because then the pointer observable would be conserved. A measurement device with a constant pointer is broken.

 

[Elias1960]

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. ... 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.

That's funny, you claim not to care about subsystems but then give a wave function for the subsystem.

Once you consider a measurement in BM, the full system you have to consider is always the one which includes the measurement device too, so that it simply cannot be a one particle system.