Radiation Interacting with Free Electrons Problem: Difference between revisions
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The transition rates are: | The transition rates are: | ||
<math> \Gamma^{abs}_{i \rightarrow f} = \frac{2 \pi}{\hbar}|<\psi_{f}|V^{abs}|\psi_{i}>|^{2}\delta(E_{f} - E_{i} - \hbar\omega) </math> | |||
<math> \Gamma^{ems}_{i \rightarrow f} = \frac{2 \pi}{\hbar}|<\psi_{f}|V^{ems}|\psi_{i}>|^{2}\delta(E_{f} - E_{i} - \hbar\omega) </math> | |||
where, | |||
V = |
Latest revision as of 23:30, 11 April 2010
Queston
Show that free electrons can neither emit nor absorb photons.
Solution
Let us initialy assume that it is posible for free electrons to emit and absorb photons. We know that the wave funtion for a free electorn is a plane wave, so our initial and final wave functions are:
The transition rates are:
where,
V =