So we all know, or can look up Fermi's golden rule to be something like:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\sigma\left(E\right) \propto \rho\left(E\right) |\langle \psi_i | \mu | \psi_f\left(E\right) \rangle|^2[/tex]

Sigma is the cross section (or can be the transition rate as well), rho is the density of states of the final state and mu is the dipole operator (or other transition operator).

My question relates to the density of states, and what exactly are we to use as the "final" state.

Suppose we are calculating a photodetachment cross section for H and that the electron the S-angular momentum state ground state. Our final state is then a P-wave free electron state, as that is the only final state with a nonzero transition dipole matrix element.

In literature, the formulas for the cross section always use

[tex]\rho\left(E\right) \propto \sqrt{E}[/tex]

which one can derive from plane waves in a 3-D box (disregarding the fact that the positive energy solutions of the schrodinger equation aren't in fact plane waves).

But, if I limit myself to only l = 1 (e.g. P) angular momentum spherical waves, haven't I essentially put myself into a 1-D box? In that case

[tex] \rho\left(E\right) \propto 1/\sqrt{E}[/tex].

So which density of states should I use? Literature picks the 3-D box but I don't see why, since the distribution of l = 1 energy eigenvalues follows the 1-D box results.

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# Fermi's Golden Rule and Density of States

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