Where $T$ is the energy given to the electron by the photon, which is essentially $h (\nu - \nu_0)$. Now we come to energy conservation: The total energy of the electron is given by $$p c \cos \phi = h \nu_0 + P c - h \nu \cos \theta$$ $$p c \sin \phi = h \nu \sin \theta$$ Where $\nu$ is the frequency of the electron after scattering, $p$ is the momentum of the electron after scattering, and $\phi$ is the electron scattering angle. Then momentum conservation gives, for the $x$- and $y$- axes respectively $-$ Suppose the photon and the electron are both moving initially along the $x$-axis. In proving this, I started in the same way as in the derivation for "stationary electron" - conservation of momentum and energy along each axis. $$\Delta \lambda = 2 \lambda_0 \frac$ is the initial energy of the electron. The book tells us that this shift is given by In a problem from Bransden and Joachain's Quantum Mechanics, it is asked to calculate the Compton wavelength shift, but the electron is now moving, with a momentum $P$, in the same direction as the approaching photon.
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