# Puzzled by $\mathrm p = -i\hbar \nabla = m \dot{\mathrm r}$?

I'm working on Physics of Atoms and Molecules by Bransden and Joachain. And I've come across the following statement, which I don't understand (p.195, ch. 4.3, "The dipole approximation")

... In terms of the momentum operator $\mathrm p = -i\hbar \nabla = m \dot{\mathrm r}$, we can also write $$\langle \psi_b |\mathrm p| \psi_a\rangle = m\langle \psi_b |\dot{\mathrm r}| \psi_a\rangle$$ ...

Here I'm puzzled by the operator $\dot{\mathrm r}$. First of all, I don't see how it makes any sense? $\mathrm r$ is just given by

$$\mathrm r = \begin{pmatrix} x \\ y\\ z \end{pmatrix}$$

in cartesian coordinates. So I don't see any time dependence here that would allow me to differentiate. Maybe their equality is to be taken in an average sense? For example is what they mean:

$$\langle \psi_b |\mathrm p| \psi_a\rangle = m\frac{\mathrm d}{\mathrm dt}\langle \psi_b |{\mathrm r}| \psi_a\rangle$$

? This would seem to make sense, since it would in particular give us $\langle p \rangle = m \frac{\mathrm d}{\mathrm dt} \langle r\rangle$, which is the classical expression for momentum.

If my guess is correct: how can I derive this equality?

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This is a Heisenberg picture construction. It is easiest if you learn Heisenberg's quantum mechanics first, then Schrodinger's. They are equivalent in a simple way, but the intuitions they give are different. – Ron Maimon Dec 23 '11 at 4:18

This is a Heisenberg picture/Schrodinger picture confusion. The operator "r" is not time dependent in the Schrodinger picture, but in the Heisenberg picture, it is. The time dependent version of the operator is the operator r(t) which has the property that

$$\langle\psi(0)|r(t)|psi(0)\rangle = \langle\psi(t)| r(0) |\psi(t)\rangle$$

For all choice of $\psi(0)$. It is a unitary transformation which shifts the time dependence from the state to the matrix.

$$|\psi(t)\rangle = e^{-iHt} |\psi(0)\rangle$$

so that

$$r(t) = e^{iHt} r(0) e^{-iHt}$$

and you can check that everything works formally.

Whenever you hear somebody talking about the time derivative of an operator, they are always talking in the Heisenberg picture, because otherwise you need explicit time dependence in the operator for this to be meaningful. Most modern people talk in the path integral, where you have both pictures simultaneously. Schrodinger picture is moving the boundary condition forward in time, Heisenberg picture is moving the insertion point of a local object forward in time.

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Thank you for answering and clearing this up! – Sam Dec 23 '11 at 14:28

$\frac{\mathrm d}{\mathrm dt} \langle r\rangle$ is equivalent to $\dot{\mathrm r}$, the over-dot is just Newton's notation for differentiation with respect to time. Unless I'm missing something more profound in your question?

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He means that the operator "r" is not time dependent in the Schrodinger picture. – Ron Maimon Dec 23 '11 at 4:10
Ah, gotcha. Thanks @RonMaimon – Mark Beadles Dec 23 '11 at 16:09