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

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


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.

  • $\begingroup$ What are the $u$ terms? $\endgroup$ May 28, 2018 at 2:45
  • $\begingroup$ @AaronStevens edited; they should be $\phi$s $\endgroup$
    – Dwagg
    May 29, 2018 at 11:44

1 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 $\nu_{nm}X_{nm} = iP_{nm}$.


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