I am confused by formula (17.88) and (17.89) on page 578 in P&S. They are computing the matrix element for electron splitting ($e^-\rightarrow e^-+\gamma$) in the massless limit.

They call $z$ the fraction of energy of the initial electron that is carried off by the photon and $p_\perp$ the transverse momentum of the photon.

To order $p_\perp^2$ the momenta are then as follows:

Initial electron: $p = (p,0,0,p)$

Emitted photon: $q= (zp, p_\perp,0 ,zp - \frac{p_\perp^2}{2zp})$

Final electron: $k = ((1-z) p, -p_\perp, 0, (1-z) p + \frac{p_\perp^2}{2zp})$

They then proceed to compute the matrix elements for given helicities and find e.g. for left-handed electrons in the chiral representation of $\gamma$-matrices:

$$ i\mathcal{M} = ie \sqrt{2(1-z)p}\sqrt{2p} \xi^\dagger(k) \sigma^I \xi(p) \epsilon_T^{*I}(q) $$

With $\xi$ the $1\times 2$ spinor. For the left-handed initial electron we have $$ \xi(p) = \begin{pmatrix} 0\\1 \end{pmatrix} $$

But then they say that $$ \xi(k) = \begin{pmatrix} p_\perp/2(1-z)p\\1 \end{pmatrix} $$ and that the polarisation vectors for the photons are $$ \epsilon_L ^{*i}(q) = \frac{1}{\sqrt{2}} (1,i,-\frac{p_\perp}{zp}) \qquad \text{and} \qquad \epsilon_R ^{*i}(q) = \frac{1}{\sqrt{2}} (1,-i,-\frac{p_\perp}{zp}) $$

This is my question: where do the expressions for $\xi(k), \epsilon_L ^{*i}(q)$ and $\epsilon_R ^{*i}(q)$ come from?


A left-handed electron moving the 3 direction has the 2-component spinor (0,1), and a left-handed photon moving in the 3 direction has the polarization vector $\epsilon^* = \frac{(1,i, 0)}{\sqrt{2}}$. However, in this problem, the outgoing electron and photon are not moving exactly in the 3 direction. They are moving at a small angle to this direction specified by the transverse momentum $p_{perp}$. If you rotate the polarization spinor and the polarization vector into the directions of motion of the particles, working to 1st order in $p_{perp}$, you will find the expressions given above and in eqs. (17.88) and (17.89) of our book.

  • $\begingroup$ Welcome to the den, Mike! Searching for your name in the search window nets plenty of product support challenges. $\endgroup$ – Cosmas Zachos Apr 18 at 14:27
  • $\begingroup$ Thank you so much for taking the time to answer this question. Brilliant book, by the way. I thoroughly enjoyed it, sweat and all included. $\endgroup$ – Oбжорoв Apr 19 at 9:51

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