Initial-value problem for Bohmian mechanics

[A. Neumaier]

Does Bohmian mechanics have a mathematically well-defined initial-value problem with unique solution for given initial data?

The right hand side of the guiding equation has singularities at all configuration space positions where $\psi$ vanishes. Thus the particle dynamics breaks down.

Thus the dynamics of Bohmian mechanics looks ill-defined.

 

[Demystifier]

This is a removable singularity, which is most easily seen by writing $\psi$ in the polar form $\psi=Re^{iS}$, where $R$ and $S$ are real functions.

 

[DarMM]

There is the paper by Berndl et al the seems to prove global existence and uniqueness of solutions of the joint system of both the Schrödinger and Guiding equation:
https://projecteuclid.org/euclid.cmp/1104274916

I haven't gone through it personally.

 

[A. Neumaier]

There is the paper by Berndl et al the seems to prove global existence and uniqueness of solutions of the joint system of both the Schrödinger and Guiding equation:
https://projecteuclid.org/euclid.cmp/1104274916

... almost surely only. I need to read the details.

This is a removable singularity, which is most easily seen by writing $\psi$ in the polar form $\psi=Re^{iS}$, where $R$ and $S$ are real functions.

Where can I find the resulting dynamical equations for R and S?

 

[Demystifier]

Where can I find the resulting dynamical equations for $R$ and $S$?

Here: https://en.wikipedia.org/wiki/Quantum_potential

 

[A. Neumaier]

Here: https://en.wikipedia.org/wiki/Quantum_potential

But this shows that your transformation does not remove the singularity. It is now in the definition of Q.

 

[Demystifier]

But this shows that your transformation does not remove the singularity. It is now in the definition of Q.

You misunderstood me. I meant "removable singularity" in the sense of theory of analytic functions. For instance, the function
$$\frac{{\rm sin}x}{x}$$
has a singularity at $x=0$, but it is a removable singularity. For that purpose, it is completely irrelevant what are the equations of motion for $R$ and $S$. Writing $\psi$ in terms of $R$ and $S$ does not replace solving the Schrodinger equation for $\psi$, but is just a calculus trick to see that the singularity in the guiding equation is a removable singularity in the sense of theory of analytic functions. If you wish, you can see that the singularity is removable without ever introducing $R$ and $S$.

 

[A. Neumaier]

You misunderstood me. I meant "removable singularity" in the sense of theory of analytic functions. If you wish, you can see that the singularity is removable without ever introducing $R$ and $S$.

But the expression defining Q is not of this kind. For example if $R=x^2$ then $Q=const*\frac{2}{x^2}$ which has a pole at $x=0$. This is not removable.

 

[DarMM]

Okay I worked my way through this paper, it's a much easier read. Pretty interesting proof, nice idea:
https://arxiv.org/abs/math-ph/0406030

Again it is "almost sure" existence and uniqueness, but for a much broader class of models, e.g. other spins.

 

[Demystifier]

But the expression defining Q is not of this kind. For example if $R=x^2$ then $Q=const*\frac{2}{x^2}$ which has a pole at $x=0$. This is not removable.

I agree, the singularity in $Q$ is not removable. But you originally asked about singularity in the guiding equation.

Think of it this way. Suppose that $\psi$ is given, but that the Schrodinger equation is not known. Without the Schrodinger equation there would be no natural way to define $Q$. Nevertheless, one could still solve the guiding equation, and one could still write $\psi=Re^{iS}$.

 

[maline]

I have never understood the point in defining a "quantum potential". The first-order guiding equation gives all of the desired dynamics and is very simple and intuitive. Writing a second order equation using a complicated potential seems perverse to me, quite apart from issues with singularities.

 

[Demystifier]

I have never understood the point in defining a "quantum potential". The first-order guiding equation gives all of the desired dynamics and is very simple and intuitive. Writing a second order equation using a complicated potential seems perverse to me, quite apart from issues with singularities.

Exactly! The only purpose of quantum potential is to make equations look more similar to classical mechanics. But this similarity can be misleading.

 

[A. Neumaier]

I agree, the singularity in $Q$ is not removable. But you originally asked about singularity in the guiding equation.

Think of it this way. Suppose that $\psi$ is given, but that the Schrodinger equation is not known. Without the Schrodinger equation there would be no natural way to define $Q$. Nevertheless, one could still solve the guiding equation, and one could still write $\psi=Re^{iS}$.

But this substitution does not remove the singularity (division by zero $\psi$( in the guiding equation. It only replaces them with a singularity in the differential equation of S. Note that the polar decomposition is ambiguous at zeros of $\psi$; this is the reason why S is not well-defined when crossing a zero.

 

[Demystifier]

But this substitution does not remove the singularity (division by zero $\psi$( in the guiding equation. It only replaces them with a singularity in the differential equation of S. Note that the polar decomposition is ambiguous at zeros of $\psi$; this is the reason why S is not well-defined when crossing a zero.

A quote from https://plato.stanford.edu/entries/qm-bohm/ addresses the problem more appropriately:
"Since the denominator on the right hand side of the guiding equation vanishes at the nodes of ψ, global existence and uniqueness for the Bohmian dynamics is a nontrivial matter. It is proven in Berndl, Dürr, et al. 1995 and in Teufel and Tumulka 2005."
The cited papers are mathematically rigorous, so they should satisfy you.

 

[maline]

But this substitution does not remove the singularity (division by zero $\psi$( in the guiding equation. It only replaces them with a singularity in the differential equation of S. Note that the polar decomposition is ambiguous at zeros of $\psi$; this is the reason why S is not well-defined when crossing a zero.

Ok, you're right that there are singularities. But these are not important because they occur at points where the particle has zero probability of being. The guiding equation only needs to be well-defined at the particular point where it is relevant.

But I guess one should wish to prove that dynamics are well-defined even in a hypothetical non-equilibrium situation, where the Born rule does not apply...

 

[A. Neumaier]

A quote from https://plato.stanford.edu/entries/qm-bohm/ addresses the problem more appropriately:
"Since the denominator on the right hand side of the guiding equation vanishes at the nodes of ψ, global existence and uniqueness for the Bohmian dynamics is a nontrivial matter. It is proven in Berndl, Dürr, et al. 1995 and in Teufel and Tumulka 2005."
The cited papers are mathematically rigorous, so they should satisfy you.

Maybe; I haven't studied them yet. These are also the papers cited by DarMM. They don't prove global existence and uniqueness but only almost sure global existence and uniqueness, which makes a difference when claiming mathematical rigor.

 

[A. Neumaier]

they occur at points where the particle has zero probability of being

This means nothing.

In a continuum, a particle has zero probability to be in any particular position! Probabilities are positive only in domains with a nonempty interior.

 

[Demystifier]

They don't prove global existence and uniqueness but only almost sure global existence and uniqueness, which makes a difference when claiming mathematical rigor.

I think that they show if for almost all solutions, where "almost all" means all except for a set of solutions of measure zero. A set of measure zero is, mathematically, quite close to not existing at all.

For example, what is the probability that the random real number in the range [0,10] is exactly equal to $\pi$? The probability is zero.

 

[A. Neumaier]

A set of measure zero is, mathematically, quite close to not existing at all.

For example, what is the probability that the random real number in the range [0,10] is exactly equal to pipipi? The probability is zero.

So is the probability that the random number is exactly 1. We always observe such ''events with zero probability''. They obviously exist!

 

[Demystifier]

So is the probability that the random number is exactly 1. We always observe such ''events with zero probability''. They obviously exist!

No, we never observe them, because we never measure with perfect precision. For instance, experiments cannot distinguish 1 from 1.00000000000000000000001

 

[A. Neumaier]

No, we never observe them, because we never measure with perfect precision. For instance, experiments cannot distinguish 1 from 1.00000000000000000000001

This does not help. All experimental results are given as rational numbers (which have measure zero in the set of reals).

Thus measurement results have zero probability of occurring, and according to your arguments, experimental physics is close to not existing at all.

 

[maline]

This means nothing.

In a continuum, a particle has zero probability to be in any particular position! Probabilities are positive only in domains with a nonempty interior.

Of course, I meant zero probability density. I'm not sure whether we can interpret "probability zero because the set has measure zero" as "cannot happen", but if the probability density goes to zero at the point of interest then I'm more confident...

 

[Demystifier]

Does Bohmian mechanics have a mathematically well-defined initial-value problem with unique solution for given initial data?

The right hand side of the guiding equation has singularities at all configuration space positions where $\psi$ vanishes. Thus the particle dynamics breaks down.

Thus the dynamics of Bohmian mechanics looks ill-defined.

Note that singularities (more precisely, fixed points) of a similar kind appear also in classical mechanics, so one could say that classical mechanics is also ill-defined.

For an example, consider a Newtonian particle moving in one dimension, with the trajectory $x(t)$. Let the potential be $V(x)=-\kappa x^{3/2}$, where $\kappa$ is a positive constant. One would expect that the initial condition $x(0)$ and $\dot{x}(0)$ defines a unique solution of the Newton equation. However, for the initial condition $x(0)=\dot{x}(0)=0$, the solution is not unique. (I leave it as an exercise for the readers to show it.)

For a philosophical discussion of this see https://www.pitt.edu/~jdnorton/papers/DomePSA2006.pdf

For a mathematical discussion, in terms of conditions under which ordinary differential equations have unique solutions, see
V.I. Arnold, Ordinary Differential Equations, Secs. 2.1-2.3.

 

[A. Neumaier]

Note that singularities (more precisely, fixed points) of a similar kind appear also in classical mechanics, so one could say that classical mechanics is also ill-defined.

For an example, consider a Newtonian particle moving in one dimension, with the trajectory $x(t)$. Let the potential be $V(x)=-\kappa x^{3/2}$, where $\kappa$ is a positive constant.

This happens only for those systems where the potential has a gradient that is not Lipschitz continuous. For example, in gravitation, trajectories stop to exist at collisions. One understands this as limitations in the modeling!

But for Bohmian mechanics, the singular behavior is present for every system!

 

[Demystifier]

But for Bohmian mechanics, the singular behavior is present for every system!

That's not true. For instance, it does not appear for a Gaussian wave function or for a stationary state of the hydrogen atom. In fact, wave functions for which it appears are probably not physical.

 

[A. Neumaier]

That's not true. For instance, it does not appear for a Gaussian wave function or for a stationary state of the hydrogen atom. In fact, wave functions for which it appears are probably not physical.

It appears to me rather that wave functions for which it does not appear are not physical:

A Gaussian wave function will not stay Gaussian under the Bohmian dynamics, unless the Hamiltonian is quadratic (no interactions).

In a stationary state of the hydrogen atom, the Bohmian particle stands still at whatever position it happens to have. This contradicts quantum equilibrium, which is an integral assumption of the Bohmian picture.

Excited states of the hydrogen atom all have points $x$ where $\psi(x)$ vanishes.

 

[martinbn]

@Demystifier It would help if you gave a specific toy example, where all the calculation can be done. Say particle in a box or anything like it.