Wavefunction from probability, OR current from charge density?

[jjustinn]

So I bugged the folks in General Physics about the latter form of the question a while back, and got some rather unconvincing "can't be done" replies. To state the problem specifically (and my motivation):

Let's say that the probability density of finding a particle at any place/time is given (ρ(r, t)) -- from that, shouldn't we be able to get *a* wave function? Of course, it will probably not be unique (certainly not up to a phase), but it seems like it might be something that would be useful to do.

More useful in my mind -- though basically the same problem (see below for justification) -- would be to derive the (say, electric) current given the time-dependence of the global (electric) charge density ρ(r, t). This is easy if you are dealing with point particles, where Ji = ρvi, where vi is the (four-) velocity of particle i (or if you had a "velocity field", v(r, t), it would just be J = ρv)...but in general, such a field (or breakdown into individual particles) won't exist - and even if it did, in those particular cases, you could derive the current from the charge density, provided that no partcles are adjacent:
It's not covariant, but you could take the gradient of the change in the field as time increases -- e.g. Ji = (ρ,t),i (here i is the coordinate -- e.g. x/y/z).

For example, take a single point charge in 1D moving to the right at 1 unit/s: ρ(x, t) = δ(x-t). So here's the charge (integrated around a single point to get rid of the delta) at various times/places:

t | x | ρ | ρ,t
0 | 0 | 1 | 0 (assume it started there)
0 | 1 | 0 | 0
1 | 0 | 0 | -1 (charge left)
1 | 1 | 1 | 1 (charge appeared)

So, the "gradient" of ρ,t at t=1 points from low to high, or in the direction the charge is moving, proportional to the speed it's moving and to the strength of the charge -- e.g., it's the current, up to a scalar factor.

However, if you instead had a line of charge on the x-axis and the charge was moving along that axis, ρ,t would be 0, so it breaks down.

HOWEVER, it still seems like the time-dependence of the charge density -- perhaps with suitable boundary conditions -- should be sufficient to determine the current.
To get back to my claim that this was the same as finding a wave function for a given probability density, there I'm appealing to standard QM -- say, unquantified Dirac electron theory -- where the wave function is directly tied to the current via J = ψ∇ψ* - ψ*∇ψ (pulled from the wikipedia's probability current article -- it may not hold exactly for the Dirac equation, but IIRC it's similar -- but in either case, it's not important).

So, I started working on the problem of expressing a given density (probability or charge) as the norm squared of a wave function: e.g. Given ρ(r, t) find ψ(r, t) such that ψ*ψ = ρ, *AND* it gives a nontrivial current (ψ∇ψ* - ψ*∇ψ). I found that it would have the general form of

ψ = ε + i√(ρ + ε^2)

And ε ≠ 0, because then j = 0. I found some other interesting things, like if you restrict ε,x = 0 you get j = ψ,x ε...but as far as something I could use, I got nowhere.

However, I did get a fresh breath of hope from that probability current article : there it mentions that when ψ is a plane wave exp(i[kx - ωt]), ρ is constant, and j turns out to be ρ*hk/m, but momentum = hk, and v = momentum / m, sp j = ρv for a plane wave -- and *everyone* knows that any function can be expressed as a superposition of plane waves! So my problem was solved! That is, until I sat down to try to actually do it--here's where I mention I never got past the intro physics courses in college. So after an hour or so of transforming, twiddling/fiddling and inverse-transforming, I began to wonder again if I wasn't on a fool's errand, and decided to try you crazy-smart folks in the QM, in hopes that someone can look at this and say "oh yeah, everyone knows that's (impossible/trivial/useless/all of those)".

So...any thoughts?

Thanks -
Justin

 

[SergioPL]

"So, the "gradient" of ρ,t at t=1 points from low to high, or in the direction the charge is moving, proportional to the speed it's moving and to the strength of the charge -- e.g., it's the current, up to a scalar factor.

However, if you instead had a line of charge on the x-axis and the charge was moving along that axis, ρ,t would be 0, so it breaks down."

J contains the current of density probability and therefore is the current itself.

The gradient of J controls the variation of probability density in one point, that means, if the current that comes to a point i a determinated moment is bigger that the current that exits, then the "charge density" in that point will increase, on the other hand it will decrease.

∂p/∂t=∇·j


Sergio

 

[vanhees71]

If you have a probability distribution for position, you are far from having the full information about the particle's state.

One way to make objective predictions about the state under the constraint of given information is the maximum-entropy principle, in the context of quantum theory also called the Shannon-Jaynes principle. You find this idea in Jaynes's famous paper,

E. T. Jaynes, Phys. Rev. 106, 620–630 (1957)

So we look for the most objective estimate for the statistical operator, [\itex]\hat{R}[/itex], given the position-probability distribution

[tex]\rho(x)=\langle{x}|\hat{R}|x \rangle.[/tex]

The criterion is, to make the von Neumann entropy,

[itex]S=-\mathrm{Tr} (\hat{R} \ln \hat{R}),[/itex]

maximal under the constraint that the position-probability distribution is [itex]\rho(x)[/itex]. Introducing a lagrange multiplyer for this constraint and doing the variation for the corresponding entropy functional, leads after some algebra to the conclusion

[tex]\hat{R}=\int \mathrm{d}^3 x \rho(\vec{x}) |\vec{x} \rangle \langle \vec{x}|.[/tex]

 

[jjustinn]

J contains the current of density probability and therefore is the current itself.

The gradient of J controls the variation of probability density in one point, that means, if the current that comes to a point i a determinated moment is bigger that the current that exits, then the "charge density" in that point will increase, on the other hand it will decrease.

I'm not sure I follow -- nowhere did I mention the "gradient of the probability current (density)" -- that wouldn't even make sense, since J is a vector quantity (ignoring the "vector gradient"). However, you did mention the continuity equation right after that:

∂p/∂t=∇·j

...which would mean that ∇(ρ,t) is actually ∇(∇.j) -- the "vector laplacian" of the current...so I guess that shows that my attempt at a derivation is at least problematic (unless the current is an eigenfunction of the laplacian - Δj = kj for some scalar k).

However, stating the continuity equation like that, it does make it seem like given ρ,t you *should* be able to get j -- isn't there some theorem that given the divergence of a vector field everywhere, the vector field itself can be determined (obviously not uniquely without more conditions)?

 

[jjustinn]

If you have a probability distribution for position, you are far from having the full information about the particle's state.

Sure, definitely: but two things -- first, this isn't necessarily restricted to a single particle (or even any definite number of particles); second, I'm not looking for the complete description of the "state" (whatever that may mean here) -- the target is simply the current, which seems a quantity derived from the time-dependence of the corresponding charge density (SI be damned).

I probably messed up saying "wavefunction", since that would imply more than just the probability amplitude density for position, which is all I meant to imply. I was looking at the "wave function" as just a sort of formal complex "square root" of the (position) probability density (though from the Fourier transform, one could also get the "momentum" [frequency] representation, which *would* seem to be sufficient for a classical 'phase space' description of the system's complete state ).

The rest of your post went far over my head, but this is the second time in the past week (in wildly different contexts) I've been pointed to work by Jaynes, so I'll have to check that out.

 

[Jano L.]

Let's say that the probability density of finding a particle at any place/time is given (ρ(r, t)) -- from that, shouldn't we be able to get *a* wave function?

jjustinn, have a look on p. 169 of this Bohm's paper:
https://www.nd.edu/~dhoward1/Bohm%20HV-I%20Phys%20Rev%201952.pdf

He decomposes the wavefunction into product [itex]Re^{i\varphi}[/itex] and derives equation for both R and [itex]\varphi[/itex]. If you know the function [itex]\rho(r,t)[/itex] for all r and all t, you can plug [itex]R=\sqrt{\rho}[/itex] into his equation 4 for the phase [itex]\varphi[/itex]. Then you can use the equation to find out the phase [itex]\varphi(r,t)[/itex], provided that you know the gradient of the phase [itex] \nabla \varphi [/itex] at the initial time. When you have [itex]\varphi[/itex], you can calculate [itex]\psi[/itex] and the current j as well.vanhees71, your equations look nice and novel but I am afraid they are not correct. Can you give or provide link to their derivation?

The problem is that, if |x> is meant to be the usual symbol that fulfills [itex]<x'|x>=\delta(x-x')[/itex], the symbol |x><x| in your definition of [itex]\hat R[/itex] has no meaning, since when this R is used to calculate [itex]\rho[/itex] according to your first formula, it leads to multiplication of two same delta distributions, which is not defined.

 

[jjustinn]

jjustinn, have a look on p. 169 of this Bohm's paper:
https://www.nd.edu/~dhoward1/Bohm%20HV-I%20Phys%20Rev%201952.pdf

He decomposes the wavefunction into product [itex]Re^{i\varphi}[/itex] and derives equation for both R and [itex]\varphi[/itex]. If you know the function [itex]\rho(r,t)[/itex] for all r and all t, you can plug [itex]R=\sqrt{\rho}[/itex] into his equation 4 for the phase [itex]\varphi[/itex]. Then you can use the equation to find out the phase [itex]\varphi(r,t)[/itex], provided that you know the gradient of the phase [itex] \nabla \varphi [/itex] at the initial time. When you have [itex]\varphi[/itex], you can calculate [itex]\psi[/itex] and the current j as well.

VERY interesting, and totally contrary to everything I've seen yet. Decomposing ψ in polar form should have been obvious, since if ψ = Rexp(iS/h), (R/S real) then ρ = R*R, so it's just a matter of finding "the" phase (e.g. One of many that would probably work).

I find equation 5 in the paper to be particularly interesting:

ρ,t + div(ρ∇S/m)

That's obviously the continuity equation, in a particularly familiar form:

q,t + div(qv), where qv is the "current" of q, J = qv...and the J in eq. 5 is even in that form, if we take ∇S to be (non-relativistic) momentum -- that is, mass times velocity,p = mv...so it breaks down to ρ,t + div(ρv).

Now, I'm going to see what happens if I equate the two different definitions of current (ψ∇ψ* - ψ*∇ψ)...hopefully it's not a tautology. It seems like stating it this way might make VanHees' basic point that having ρ(r, t) for all r/t isn't the whole story, since that tells nothing about the phase...but that paper at least opens up a lot of other ways to find it (Hamilton-Jacobi theory for one)...

I'll post again if I find anything of interest (to me, anyway)...any further comments, hints or random thoughts would of course be appreciated.

 

[jjustinn]

So, it turns out that if ψ = √ρ exp(iS), then J= ρ∇S = ψ∇ψ* - ψ*∇ψ *is* a tautology, so I'm still left with fundamentally the same number of unknowns (though now, at least I know what that unknown is: the phase). In that Bohm paper, it is further restricted via the Schrodinger equations, but that won't work for the case of a general "conserved charge".

I think I could probably narrow it down a bit by assuming that ρ is incompressible or some other similar constraints.

I still haven't tried by idea of trying to solve it in momentum-space, nor done any serious investigation into the significance of the gradient of the phase being the "velocity" (which sounds familiar from the wave/Hamilton-Jacobi sections of Spivak's mechanics)...so there are a few avenues open before I consider it a dead end.