I was reading Fermi's 1932 paper "Quantum Theory of Radiation". I was able to understand the paper until this particular derivation (which was excluded).

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$u_m,u_n$ are the eigen-functions of the electron which was initially solved using the unperturbed hamiltonian.

How to derive the highlighted-equation, using the left hand side?

I know that $\hat{p}=-i\hbar \nabla$ in three dimensions. But I never quite understood how to arrive at the energies $E_m$ and $E_n$.


1 Answer 1


This is likely using the fact that $[\hat{H},\vec{r}] = -i\frac{\hbar}{m}\vec{p}$, which you can verify by direct computation. (Here, I am using $\vec{r}$ for the position vector, replacing $X$ in the paper.) By replacing the momentum operator with the commutator, you can expand out the commutator and act to the left/right with the Hamiltonian, which brings out the energy factors. Simplifying will yield the result.


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