What do momentum matrix elements in energy basis represent?

Question

Suppose we have a Hamiltonian $H$ with eigenfunctions $\phi_1(\mathbf{x}),\ldots, \phi_n(\mathbf{x})$ and eigenvalues $E_1,\ldots, E_n$. What, if anything, does the matrix element

$$P_{mn} = \langle \phi_m | P| \phi_n\rangle = -i\hbar\int \overline{\phi_m}(\mathbf{x})\vec\nabla\phi_n(\mathbf{x})\,\,d^3x $$

represent?

In the specific case of an electron in an atomic potential, my textbook (Quantum Theory of Radiation by E. Fermi) claims that this represents the momentum of the electron and it can immediately be proved that $$-i\hbar\int \overline{\phi_m}(\mathbf{x})\vec\nabla\phi_n(\mathbf{x})\,\,d^3x = -im\nu_{mn}X_{nm}$$ where $$\nu_{mn} = (E_m-E_n)/\hbar$$ and $$X_{mn} = X_{nm} = \int \phi_n(\mathbf{x})\,\mathbf{X}\, \phi_m(\mathbf{x}) \,d^3x$$ where $\mathbf{X}$ is the position vector operator. I don't follow his claim.

Answer 1

For single particle dynamics with Hamiltonian $H = \frac{p^2}{2m} + V(x)$, it is clear that $[H,x] = [\frac{p^2}{2m},x] = i\hbar\frac{p}{m}$. Then taking matrix elements of both sides (in the energy eigenbasis $|n>$ with $H|n> = E_n|n>$) gives $<n|Hx|m> - <n|xH|m> = i\hbar P_{nm}$. The left hand side is $(E_n-E_m)X_{nm}$ so $\nu_{nm}X_{nm} = iP_{nm}$.