Consider the process $e^-\rightarrow e^-\gamma$ depicted in the following Feynman diagram.

Feynman diagram

The spin-averaged amplitude with linearly polarised photons is $$\overline{|M|^2}=8\pi\alpha\left(-g^{\mu\nu}+\epsilon^\mu_+\epsilon^\nu_-+\epsilon^\mu_-\epsilon^\nu_+\right)\left(p_\mu p^\prime_\nu+p_\nu p^\prime_\mu-g_{\mu\nu}pp^\prime\right),$$ where the polarisation sum for massless vector bosons was used in terms of the unphysical polarisation vectors $\epsilon_\pm=\frac{1}{\sqrt{2}}\left(\epsilon_t\pm\epsilon_L\right)$, $\epsilon_t=(1,\mathbf{0}), \epsilon_L=\frac{1}{|\mathbf{k}|}(0,\mathbf{k})$.

Now, by the Ward identity the parts of the amplitude of the form $\epsilon^\mu_\pm M_\mu$ should vanish since $\epsilon_+\parallel k$. However, since $\epsilon_+\epsilon_-=1$ I find that instead $$\overline{|M|^2}=16\pi\alpha\left(\epsilon^\mu_+p\epsilon^\nu_-p^\prime+\epsilon^\mu_-p\epsilon^\nu_+p^\prime\right).$$

I know that in general the Ward identity does not hold for individual diagrams but only for the sum of all relevant ones. However, the above diagram is the only one to this order in $\alpha$.

Why does the term $\left(\epsilon^\mu_+\epsilon^\nu_-+\epsilon^\mu_-\epsilon^\nu_+\right)\left(p_\mu p^\prime_\nu+p_\nu p^\prime_\mu-g_{\mu\nu}pp^\prime\right)$ not vanish?


1 Answer 1


It is because the Ward identity applies to S-matrix elements, for which particles must be on mass shell. A simple calculation shows that with only one vertex, energy-momentum conservation is not consistent with the mass shell condition for all particles.


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