It has been noted here$\! { \, }^{\text(1, 2)}$, for instance, that

$$\mathbf{F} = \frac{d}{dt}\!\!\biggl[ \, \mathbf{p} \, \biggr]$$

is true in all contexts.

Likewise, in notable contexts it is apparently true that

$$\mathbf{F} = - \nabla \Phi := - \frac{d}{d\mathbf{r}}\!\!\biggl[ \, \Phi \, \biggr].$$

Is this, in a nutshell, a sufficient and valid justification for setting (in the corresponding suitable contexts)
the momentum operator as$\! { \, }^{\text(3)}$

$$\mathbf{\hat p } \propto -i \nabla := -i\frac{d}{d\mathbf{r}}$$

and setting the (potential) energy operator as$\! { \, }^{\text(4)}$

$$\hat \Phi \propto i\frac{d}{dt}$$

and both with the same constant of proportionality, $\hbar$, whereby

$$\mathbf{\hat F} = \frac{d}{dt}\biggl[-i\hbar\frac{d}{d\mathbf{r}}\biggr] = -\frac{d}{d\mathbf{r}} \biggl[i\hbar\frac{d}{dt}\biggr] \sim \frac{d^2}{dt \, d\mathbf{r}} = \frac{d^2}{d\mathbf{r} \, dt}$$


EDIT (related merely to formalities):

(${ \, }^{\text 1}$: Please note that the assertion to be noted here had been expressed explicitly in the form

$F = \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}$ is true in all contexts.

However, since it appears admissible to take note of an assertion without strictly quoting and clinging to its original literal expression (as had been tacitly presumed already in the initial statement of my question, and as it seems to be thus confirmed) I'd like, as far as unambiguously possible, to express the operation of "differentiation" consistently by using (a form of) Leibniz's notation.)

(${ \, }^{\text 2}$: Please note that the question whose notable answer to which reference has been made above had been tagged (primarily) as has "https://physics.stackexchange.com/questions/tagged/newtonian-mechanics".)

($ { \, }^{\text 3}$: Please note that the indicated expression of the momentum operator is explicitly stated there as

${\bf \hat p } = -i \hbar \nabla$


In one spatial dimension this becomes: $\hat{p}=\hat{p}_x=-i\hbar{\partial \over \partial x}$,

where the nabla symbol ($\nabla$) is related to http://en.wikipedia.org/wiki/Directional_derivative#Notation .)

(${ \, }^{\text 4}$: Please note that the indicated expression of the energy operator is explicitly stated there as

$\hat{E} = i\hbar\frac{\partial }{\partial t}$.



1 Answer 1


Note that forces are usually not well defined in a quantum mechanical context. Most forces (i.e. conservative, non-uniform force fields) are just ways of expressing position-dependent changes in momentum. For this to be a meaningful concept, we need position and momentum to be simultaneously well defined; and in quantum mechanics, they are not.

If you put together $\mathbf{F} = \mathbf{\dot{p}}$ and $\mathbf{F} = -\nabla\Phi(\mathbf{x})$, you get the expression $\mathbf{\dot{p}} = -\nabla\Phi(\mathbf{x})$. The left-hand side of this equation describes momentum, while the right-hand side is a function of position. But Heisenbergs uncertainty principle $\Delta p_i \Delta x_i \geq \hbar/2$ implies that position and momentum cannot be simultaneously well-defined, consequently the two sides of the equation are not simultaneously well-defined. This implies that this equation is invalid in quantum mechanics. (The connection to classical mechanics is basically that if you take the expectation value of both sides of the equation, it also holds in quantum mechanics. Google Ehrenfests theorem for more information about this.)

As for justifying the expressions for energy and momentum operators, perhaps my answer here is somewhat relevant?

  • $\begingroup$ jabirali: Thanks for answering. Unfortunately it caught me in the middle of editing my question (which I did on formal grounds only). Also, unfortunately, I cannot at all address the contents of your answer at the moment; but I hope to be able to do that within the next 24 h. $\endgroup$
    – user12262
    Nov 9, 2013 at 9:12
  • $\begingroup$ "This implies that this equation is invalid in quantum mechanics". At least in Heinseberg representation, it is correct, as an equation between operators. For instance for the quantum harmonic oscillator, you have : $ \mathbf {\dot P}(t) = -m \omega^2 \mathbf {X}(t) $. See also this previous answer $\endgroup$
    – Trimok
    Nov 9, 2013 at 9:55
  • $\begingroup$ I was implicitly thinking in the Schroedinger picture, where the equation doesn't hold neither for operators ($\mathbf{p}$ is time-independent, so $\mathbf{\dot{p}} = 0$) nor for eigenvalues (not simultaneously well-defined). I stand corrected :) $\endgroup$
    – jabirali
    Nov 9, 2013 at 10:33
  • $\begingroup$ jabirali: "[...] perhaps my answer here ..." and there (final section): "Technically, we say that the Hamiltonian is the generator of time translations, and momentum is the generator of space translations." -- Yes, thanks, that seems relevant to what I'm trying to get at, namely: Is the expression $\mathbf{\dot{p}} = -\nabla\Phi(\mathbf{x})$ useful or even required for figuring out which (if any) generator (of "time translations", or "space translations", or "rotations", or who knows what) to call/associate with "momentum, p" ? $\endgroup$
    – user12262
    Nov 9, 2013 at 23:39

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