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. The book tells us that this shift is given by
$$\Delta \lambda = 2 \lambda_0 \frac{(p_0 + P) c}{E – Pc} \sin^2(\theta / 2)$$
where $p_0 = h / \lambda_0$ is the momentum of the incident photon, $\lambda_0$ is the original wavelength of the photon (before scattering), $\theta$ is the photon scattering angle, and $E = \sqrt{m^2 c^4 + P^2 c^2}$ is the initial energy of the electron.
In proving this, I started in the same way as in the derivation for "stationary electron" – conservation of momentum and energy along each axis. Suppose the photon and the electron are both moving initially along the $x$-axis. Then momentum conservation gives, for the $x$– and $y$– axes respectively $-$
$$(h \nu_0 / c) + P = (h \nu / c) \cos\theta + p \cos \phi$$
$$0 = (h \nu / c) \sin \theta – p \sin \phi$$
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, multiplying both equations by $c$,
$$p c \cos \phi = h \nu_0 + P c – h \nu \cos \theta$$
$$p c \sin \phi = h \nu \sin \theta$$
Squaring both and adding gives $-$
$$p^2 c^2 = (h \nu_0 + P c – h \nu \cos \theta)^2 + (h \nu \sin \theta)^2$$
$$\Rightarrow p^2 c^2 = (h \nu_0)^2 + P^2 c^2 + 2 h \nu_0 P c + (h \nu)^2 – 2 (h \nu_0 + P c)(h \nu \cos \theta)…..(1)$$
Now we come to energy conservation: The total energy of the electron is given by
$$T + E = \sqrt{m^2 c^4 + p^2 c^2}$$
where $T$ is the energy given to the electron by the photon, which is essentially $h (\nu – \nu_0)$.
$$\Rightarrow T^2 + m^2 c^4 + P^2 c^2 + 2 T E = m^2 c^4 + p^2 c^2$$
$$\Rightarrow (p^2 – P^2) c^2 = (h \nu_0 – h \nu)^2 + 2 (h \nu_0 – h\nu) E$$
$$= (h \nu_0)^2 + (h \nu)^2 – 2 h^2 \nu_0 \nu + 2 (h \nu_0 – h \nu) E…..(2)$$
Therefore from equations $(1)$ and $(2)$, we get $-$
$$(h \nu_0)^2 + (h \nu)^2 + 2 h \nu_0 P c – 2 (h \nu_0 + P c) (h \nu \cos \theta) = (h \nu_0)^2 + (h \nu)^2 – 2 h^2 \nu_0 \nu + 2(h \nu_0 – h \nu) E$$
which after a little bit of algebra becomes $-$
$$h^2 \nu_0 \nu (2 \sin^2 (\theta /2)) + P c h \nu \cos \theta = h \nu_0 (E – P c) – h \nu E.$$
Here I am stuck since this does not give me the required expression for the shift – just transforming frequencies to the corresponding wavelength does not give me the required result. Where have I gone wrong?
Best Answer
The mistake is in the "little bit of algebra" at the end where it is transformed into an undesired and probably incorrect form. First note that: $$ \Delta \lambda = \lambda- \lambda_0 = \frac{h}{p} - \frac{h}{p_{0}} = h\big( \frac{{p_0}-p}{pp_{0}}\big), $$ so we will first try to get all terms that contain $p_0-p$ to the LHS. I will rewrite your last correct step using $pc=h\nu,\ p_{0}c=h\nu_{0}$. Then, after dividing both sides by $2c$, \begin{align} E(p_0-p) &= (p_0p+p_0P-p(p_0+P)\cos\theta)c\\ &=(p(p_0+P)(1-\cos\theta) + (p_0-p)P)c. \end{align} After rearranging and using $1-\cos \theta =2\sin^2\big(\frac{\theta}{2}\big)$, $$ (E-Pc)(p_0-p)=2p(p_0+P)c\sin^2\Big(\frac{\theta}{2}\Big), $$ therefore, after dividing both sides by $pp_0$ and substituting $\Delta \lambda$, $$ \Delta \lambda=\frac{2h}{p_0}\frac{(p_0+P)c}{E-Pc}\sin^2\Big(\frac{\theta}{2}\Big) = 2\lambda_0 \frac{(p_0+P)c}{E-Pc}\sin^2\Big(\frac{\theta}{2}\Big). $$