Acceleration Operator

1. Feb 23, 2010

jfy4

Hello,

Has anyone heard of, or know of a quantum mechanical acceleration operator?
I thought I would take the time derivative of the momentum operator divided by mass but I'm not sure if that would be correct.

Any help would be nice.

2. Feb 23, 2010

meopemuk

Yes, your guess is pretty good for the non-relativistic case. More precisely, acceleration is defined as the second time derivative of position. Taking into account that in QM the time derivative of an observable is $$-i/\hbar$$ times commutator with the full Hamiltonian, you can also obtain

$$\mathbf{a} = \frac{d^2\mathbf{r}}{dt^2} = -\frac{1}{\hbar^2}[[\mathbf{r}, H], H]$$

Eugene.

3. Feb 24, 2010

SpectraCat

For the acceleration of a single massive particle, isn't it even simpler than that? Since (non-relativistic) acceleration is also the time derivative of the momentum, divided by the mass, can't you just follow the Ehrenfest theorem-style derivation as below? . We have H=K + V, but the momentum operator commutes with K, you are just left with:

$$\hat{a}=\frac{1}{m}\frac{d}{dt}\hat{p}=\frac{i}{m\hbar}\left[V,\hat{p}\right]$$

but since $$\hat{p}=-i\hbar\nabla$$, this simplifies to:

$$\hat{a}=\frac{1}{m}\frac{d}{dt}\hat{p}=-\frac{\nabla V}{m}$$

Assuming this is all correct, I would guess that this generalizes to the case of multiple particles in a rather straightforward way.

4. Feb 24, 2010

meopemuk

SpectraCat,

your formula is correct in the non-relativistic case and if the potential V does not depend on momentum.

Eugene.

5. Feb 24, 2010

SpectraCat

Don't those mean the same thing? Is it possible to have a non-relativistic potential that depends on momentum? I have seen cases where it is useful to write *effective* potentials as if they were functions of momentum, such as in the case of a centrifugal barrier for L>0 in a central potential, but I can't think of a case where there is an explicit dependence of the potential on momentum without considering relativistic effects. Am I missing something obvious?

6. Feb 24, 2010

meopemuk

You are right. All physical momentum-dependent potentials that I know are relativistic corrections (i.e., proportional to a power of 1/c). For example, the Darwin potential, which is the Hamiltonian form of the magnetic interaction. However, in principle, one can also imagine a momentum-dependent interaction that does not depend on 1/c.

Eugene.

7. Feb 24, 2010

jfy4

Thank you very much for your responses guys, they were great. Just to clarify, do you think it would be ok for me to write the acceleration operator like this:

$$\hat{a}=-\frac{i \hbar}{m}\frac{d}{dt}\nabla$$

Like i suggested then in the OP?

8. Feb 24, 2010

meopemuk

This notation could be formally correct, but it should be used with caution, as it can be misleading. Suppose we have a free particle described by the plane wave function $$\psi(x,t) = N \exp(ipx -iEt)$$. Naturally, it should be an eigenfuction of the acceleration operator with eigenvalue 0. However, if we formally apply your operator to $$\psi(x,t)$$ we get a non-zero result.

Eugene.

9. Feb 24, 2010

Hans de Vries

This is the in principle correct operator which one obtains via the
method given by meopemuk. With the operator also goes a method
of how to apply the operator into an observable.

The observed acceleration (which is the 4-acceleration actually)
should be real and should not contain anymore the phase information
of the wave function. The standard way of applying the operator is
(as long as $\tilde{\mathcal{O}}^i$ contains no higher as first order derivatives)

$$\left(\, \tilde{\mathcal{O}}^i~\hat{\psi} \,\right)~|\psi|^2 ~=~ \frac{1}{2}\left(\, \psi^* \tilde{\mathcal{O}}^i \psi ~-~ \psi\, \tilde{\mathcal{O}}^i \psi^*\,\right)$$

This won't work however if the operator contains higher order derivatives
as is the case of the acceleration operator. A way of applying the operator
which works for a general wave function is this one:

$$\mathtt{A}^i\,\psi^*\psi~=~ -\frac{i\hbar}{2m}\left[~\frac{\partial}{\partial t}\left(~ \frac{1}{\psi}\frac{\partial \psi }{\partial x^i} - \frac{1}{\psi^*}\frac{\partial \psi^* }{\partial x^i} ~\right)\right] \psi^*\psi$$

This gives you the 4-acceleration density. Integrating over the wave-
function will give you the acceleration (average) just like integrating over
the position operator turns a "position density" into the average position.

You can find the derivation and the description in the relevant chapter of
my book here: http://physics-quest.org/Book_Chapter_Klein_Gordon_operators.pdf

In section 10.7 you'll see meopemuk's commutator as equation (10.45),
your result as equation (10.47) and the expression above as (10.60)

Regards, Hans

Last edited: Feb 24, 2010
10. Feb 24, 2010

jfy4

wow, thank you very much. ill check it out.