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? $\endgroup$ – BioPhysicist May 28 '18 at 2:45
  • $\begingroup$ @AaronStevens edited; they should be $\phi$s $\endgroup$ – Dwagg May 29 '18 at 11:44

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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