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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – perplexity
    Jun 4, 2013 at 12:58

2 Answers 2


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 $U(\vec x)$.

  • $\begingroup$ So, $\dfrac{d^2 \hat x}{dt^2} = \frac {\large -1}{\large \hbar^2} (\hat H^2\hat x - 2\hat H \hat x \hat H + \hat x \hat H^2)$ $\endgroup$
    – Trimok
    Jun 4, 2013 at 19:39
  • $\begingroup$ Isn't $\dfrac{i}{\hbar}[\hat U, \frac{\hat P}{m}] = \frac{U (\vec x)}{m} \nabla - \frac{\nabla U(\vec x)}{m}$ ? $\endgroup$
    – asmaier
    Jun 4, 2013 at 20:34
  • $\begingroup$ @asmaier, by the product rule, there's another term that cancels the first. $\endgroup$ Jun 4, 2013 at 20:47
  • $\begingroup$ 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 $\psi$? $\endgroup$
    – user12262
    Jun 4, 2013 at 21:05
  • 1
    $\begingroup$ @user12262, I think that your $\hat a$ is just the acceleration operator defined in my answer. $\endgroup$ Jun 4, 2013 at 21:28

The operator corresponding to the variable ė, and the microquotient of the operator ê itself to time, the meaning of the two is not the same. To get the former, Landau's quantum mechanics gives a formula(it appears in the chapter 2, energy and momentum); As for the latter, it is necessary to calculate under the Heisenberg picture. In fact, under the Schrödinger picture, operators do not evolve over time, so there is no microquotient of operators to time.

  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
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    Jun 22, 2023 at 18:26

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