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- Homework Statement
- Show that the Feynman amplitude for Compton scattering ##\mathcal{M} = \mathcal{M}_a + \mathcal{M}_b## is gauge invariant while the individual contributions ##\mathcal{M}_a## and ##\mathcal{M}_b## are not, by considering the gauge transformations

$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \lambda k^{\mu}, \ \ \ \ \varepsilon^{\mu} (\vec k_f) \rightarrow \varepsilon^{\mu} (\vec k_f) + \lambda' k^{'\mu}$$

This is exercise 8.7 in Mandl & Shaw

- Relevant Equations
- Please see below

Show that the Feynman amplitude for Compton scattering ##\mathcal{M} = \mathcal{M}_a + \mathcal{M}_b## is gauge invariant while the individual contributions ##\mathcal{M}_a## and ##\mathcal{M}_b## are not, by considering the gauge transformations

$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \lambda k^{\mu}, \ \ \ \ \varepsilon^{\mu} (\vec k_f) \rightarrow \varepsilon^{\mu} (\vec k_f) + \lambda' k^{'\mu}$$

The Feynman amplitudes for Compton Scattering by electrons are (let us ignore helicity and polarization indices for simplicity)

\begin{equation*}

\mathcal{M}_a = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i)

\end{equation*}

\begin{equation*}

\mathcal{M}_b = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i)

\end{equation*}

Given the gauge transformations

$$\varepsilon\!\!\!/ (\vec k_i) \rightarrow \varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/, \ \ \ \ \varepsilon\!\!\!/ (\vec k_f) \rightarrow \varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/'$$

We get

\begin{align*}

\mathcal{M}'_b &= -e^2 \bar u(\vec p_f) \left(\varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/ \right) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \left(\varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/' \right) u(\vec p_i) \\

&= -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i) \\

&- \lambda \lambda' e^2 \bar u(\vec p_f) k\!\!\!/ \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/' u(\vec p_i) \\

&-\lambda' e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/' u(\vec p_i) \\

&-\lambda e^2 \bar u(\vec p_f) \lambda k\!\!\!/ \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i) \\

\end{align*}

Analogously

\begin{align*}

\mathcal{M}'_a &= -e^2 \bar u(\vec p_f) \left(\varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/' \right) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \left(\varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/ \right) u(\vec p_i) \\

&= -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i) \\

&- \lambda \lambda' e^2 \bar u(\vec p_f) k\!\!\!/' \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/ u(\vec p_i) \\

&-\lambda' e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/ u(\vec p_i) \\

&-\lambda' e^2 \bar u(\vec p_f) \lambda k\!\!\!/' \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i) \\

\end{align*}

OK. I am not sure if what follows is correct but there we go. Next we should use Ward's identity but I am not sure how to apply it in this problem. I thought of ##k\!\!\!/ =0## and ##k\!\!\!/' =0## but that is not correct, as it would imply that the individual Feynman amplitudes are invariant by themselves, which is stated to not be the case.

Any hint is appreciated.

Thank you

$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \lambda k^{\mu}, \ \ \ \ \varepsilon^{\mu} (\vec k_f) \rightarrow \varepsilon^{\mu} (\vec k_f) + \lambda' k^{'\mu}$$

The Feynman amplitudes for Compton Scattering by electrons are (let us ignore helicity and polarization indices for simplicity)

\begin{equation*}

\mathcal{M}_a = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i)

\end{equation*}

\begin{equation*}

\mathcal{M}_b = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i)

\end{equation*}

Given the gauge transformations

$$\varepsilon\!\!\!/ (\vec k_i) \rightarrow \varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/, \ \ \ \ \varepsilon\!\!\!/ (\vec k_f) \rightarrow \varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/'$$

We get

\begin{align*}

\mathcal{M}'_b &= -e^2 \bar u(\vec p_f) \left(\varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/ \right) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \left(\varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/' \right) u(\vec p_i) \\

&= -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i) \\

&- \lambda \lambda' e^2 \bar u(\vec p_f) k\!\!\!/ \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/' u(\vec p_i) \\

&-\lambda' e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/' u(\vec p_i) \\

&-\lambda e^2 \bar u(\vec p_f) \lambda k\!\!\!/ \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i) \\

\end{align*}

Analogously

\begin{align*}

\mathcal{M}'_a &= -e^2 \bar u(\vec p_f) \left(\varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/' \right) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \left(\varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/ \right) u(\vec p_i) \\

&= -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i) \\

&- \lambda \lambda' e^2 \bar u(\vec p_f) k\!\!\!/' \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/ u(\vec p_i) \\

&-\lambda' e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/ u(\vec p_i) \\

&-\lambda' e^2 \bar u(\vec p_f) \lambda k\!\!\!/' \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i) \\

\end{align*}

OK. I am not sure if what follows is correct but there we go. Next we should use Ward's identity but I am not sure how to apply it in this problem. I thought of ##k\!\!\!/ =0## and ##k\!\!\!/' =0## but that is not correct, as it would imply that the individual Feynman amplitudes are invariant by themselves, which is stated to not be the case.

Any hint is appreciated.

Thank you