# Can one define an acceleration operator in quantum mechanics?

It seems most books about QM only talk about position and momentum operators. But isn't it also possible to define a acceleration operator?

I thought about doing it in the following way, starting from the definition of the momentum operator:

$\hat{p} = -i\hbar \frac{\partial }{\partial x}$

Then we define a velocity operator in analogy to classical mechanics by dividing momentum by the mass $m$

$\hat{v} = \frac{-i\hbar}{m} \frac{\partial }{\partial x}$

In classical mechanics acceleration is defined as the time derivative of the velocity, so my guess for an acceleration operator in QM would be

$\hat{a} = \frac{-i\hbar}{m} \frac{\partial }{\partial t} \frac{\partial }{\partial x}$

Is that the general correct definition of the acceleration operator in QM? How about relativistic quantum mechanics?

• Not a very good idea. Time is a parameter, not a variable, in Quantum Mechanics. If you are given a wave function $\psi(x)$ (or $\psi(p)$) you wouldn't know what to do to obtain the acceleration using your operator. You could infer some kind of average value from the history of your system, but you could not obtain eigenvalues/eigenvectors. So it is better to think in terms of the force, that you can obtain as (minus) the gradient of the potential divided by the mas. – perplexity Jun 4 '13 at 12:58

I think you might try approaching this in the Heisenberg picture.

The time derivative of the position operator is:

$$\dfrac{d \hat x}{dt} = \dfrac{i}{\hbar}[\hat H, \hat x]$$

which is a reasonable velocity operator. The time derivative of the velocity operator is then:

$$\dfrac{d^2 \hat x}{dt^2} = \dfrac{i}{\hbar}[\hat H, \dfrac{d \hat x}{dt}]$$

For example, consider a free particle so that $\hat H = \frac{\hat P^2}{2m}$. The velocity operator would then be $\frac{\hat P}{m}$. This certainly looks reasonable as it is of the form of the classical $\vec v = \frac{\vec p}{m}$ relationship.

But, note that the velocity operator commutes with this Hamiltonian so the commutator in the definition of the acceleration operator is 0. But that is what it must be since we're assuming the Hamiltonian of a free particle which means there is no force acting on it.

Now, consider a particle in a potential so that $\hat H = \frac{\hat P^2}{2m} + \hat U$. The velocity operator, for this system, is then $\frac{\hat P}{m} + \frac{i}{\hbar}[\hat U, \hat x]$.

Assuming the potential is not a function of momentum, the commutator is zero and the velocity operator is the same as for the free particle.

The acceleration operator is then $\dfrac{i}{\hbar}[\hat U, \frac{\hat P}{m}]$.

In the position basis, this operator is just $\frac{-\nabla U(\vec x)}{m}$ which looks like the acceleration of a classical particle of mass m in a potential given by $U(\vec x)$.

• So, $\dfrac{d^2 \hat x}{dt^2} = \frac {\large -1}{\large \hbar^2} (\hat H^2\hat x - 2\hat H \hat x \hat H + \hat x \hat H^2)$ – Trimok Jun 4 '13 at 19:39
• Isn't $\dfrac{i}{\hbar}[\hat U, \frac{\hat P}{m}] = \frac{U (\vec x)}{m} \nabla - \frac{\nabla U(\vec x)}{m}$ ? – asmaier Jun 4 '13 at 20:34
• @asmaier, by the product rule, there's another term that cancels the first. – Alfred Centauri Jun 4 '13 at 20:47
• So ... could $\langle \psi \, {\hat x} \, \psi \rangle$ be evaluated even if ${\hat U}$ and thus ${\hat H}$ were not (yet) known? Could $d^2/dt^2[ \langle \psi \, {\hat x} \, \psi \rangle ]$ then be determined? Might therefore even an operator ${\hat {\bf a}}$ be defined such that $\langle \psi \, {\hat {\bf a}} \, \psi \rangle := d^2/dt^2[ \langle \psi \, {\hat x} \, \psi \rangle ]$ for all $\psi$? – user12262 Jun 4 '13 at 21:05
• @user12262, I think that your $\hat a$ is just the acceleration operator defined in my answer. – Alfred Centauri Jun 4 '13 at 21:28