I've been trying to follow through this derivation of the total squared matrix element for Compton scattering. We have two first-order diagrams:

The two lowest-order Feynman diagrams for Compton Scattering (a) s-channel and (b) t-channel

Using Feynman rules, the matrix elements for each diagram are:

$$\mathcal{M}_{1} =\left(\overline{u}^{\left( s^{\prime }\right)}\left( p^{\prime }\right)\left( ie\gamma ^{\nu }\right) \epsilon _{\nu }^{\ast }\right)\left[\frac{{\not}p +{\not}k \ +m}{( p+k)^{2} -m^{2}}\right]\left( \epsilon _{\mu }\left( ie\gamma ^{\mu }\right) u^{( s)}( p)\right)$$

$$\mathcal{M}_{2} =\left(\overline{u}^{\left( s^{\prime }\right)}\left( p^{\prime }\right)\left( ie\gamma ^{\nu }\right) \epsilon _{\nu } \ \right)\left[\frac{{\not}p -{\not}k \ +m}{( p-k)^{2} -m^{2}}\right]\left( \epsilon _{\mu }^{\ast }\left( ie\gamma ^{\mu }\right) u^{( s)}( p)\right)$$

Now when the author solved for $| \mathcal{M}_{1}| ^{2}$, he put all the polarization vectors at the beginning.

$$| \mathcal{M}_{1}| ^{2} =\tfrac{e^{4}}{4\left( s-m_{e}^{2}\right)^{2}}\sum\limits _{s,s^{\prime }}\sum\limits _{r,r^{\prime }} \epsilon _{\nu }^{r^{\prime } \ast } \epsilon _{\mu }^{r\ast } \epsilon _{\rho }^{r\ast } \epsilon _{\sigma }^{r^{\prime } \ast }\left[\overline{u}^{\left( s^{\prime }\right)} \gamma ^{\nu }\left({\not}p +{\not}k \ +m_{e}\right) \gamma ^{\mu } u^{( s)}\right]$$

$$\times \left[\overline{u}^{( s)} \gamma ^{\rho }\left({\not}p +{\not}k \ +m_{e}\right) \gamma ^{\sigma } u^{\left( s^{\prime }\right)}\right]$$

Why is it valid to move all polarization vectors to the front? Do they commute with all the other factors in the equations above?

Other derivations (this and this) I found on the internet went through this exact same step and I'm confused why this is valid.

The way I understand it is that the first part of the matrix element $ \left(\overline{u}^{\left( s^{\prime }\right)}\left( p^{\prime }\right)\left( ie\gamma ^{\nu }\right) \epsilon _{\nu }^{\ast }\right) $ is calculated as follows:

$ \overline{u}^{\left( s^{\prime }\right)}\left( p^{\prime }\right)$ is a $1\times 4$ row vector, $\left( ie\gamma ^{\nu }\right) $ is a $4\times 4$ matrix, and $\epsilon _{\nu }^{\ast } $ is a $ 4\times 1$ column vector, and multiplying these three gives us a $1 \times 1$ constant, but since we're actually summing this for $\nu = 0$ to $3$, this actually becomes a $1\times 4$ row vector $j^\nu$.

Same goes for the outgoing part of the matrix element $\left( \epsilon _{\mu }\left( ie\gamma ^{\mu }\right) u^{( s)}( p)\right)$, which then becomes a $ 4\times 1$ vector $j^\mu$.

Then we multiply these two vectors: $j^\nu \left[\frac{{\not}p +{\not}k \ +m}{( p+k)^{2} -m^{2}}\right] j^\mu$, where the middle factor is a $ 4\times 4$ matrix. Is this understanding correct? It bothers me that if we "factor out" the $\epsilon$'s, the dimensions of the remaining factors would not be compatible for matrix multiplication.

  • 1
    $\begingroup$ They don't have Dirac indices and so are just Lorentz vectors. You can certainly keep them there if you want but you'll obtain the same expression ultimately. $\endgroup$
    – Triatticus
    Commented Dec 28, 2022 at 11:42

1 Answer 1


Let's look at $\gamma^\nu \epsilon_\nu^\ast$. According to the Einstein's summation convention it means:

$$\overbrace{\gamma^\nu \epsilon_\nu^\ast}^{4\times 4}\equiv\sum_{\nu=0,\ldots, 3} \gamma^\nu \epsilon_\nu^\ast\equiv \overbrace{\gamma^0 \epsilon^\ast_0}^{4\times 4} +\overbrace{\gamma^1 \epsilon^\ast_1}^{4\times 4} +\overbrace{\gamma^2 \epsilon^\ast_2}^{4\times4} +\overbrace{\gamma^3 \epsilon^\ast_3}^{4\times 4}$$

Each $\epsilon_0, \ldots, \epsilon_3 $ is a number not even a vector, well each is a component of a 4-vector, whereas each $\gamma_0,\ldots, \gamma_3$ is a $4\times 4$-matrix. Therefore $\gamma^\nu \epsilon_\nu^\ast$ is a 4x4 matrix, not more and not less. The same is true for $\epsilon_\mu \gamma^\mu$.

In total we have schematically:

$$\bar{u} \gamma^\nu \epsilon_\nu^\ast [ \ldots ] \epsilon_\mu \gamma^\mu u \quad\text{so we have}\quad \sum_\nu \sum_\mu\overbrace{(1\times 4)}^{\bar{u}}\overbrace{4\times 4}^{\gamma^\nu \epsilon_\nu^\ast } \overbrace{[ 4\times 4]}^{\text{propagator}}\overbrace{4\times 4}^{\epsilon_\mu \gamma^\mu }\overbrace{(4\times 1)}^{u}= 1\times 1$$

Nothing changes if the $\epsilon$'s are put in front of the matrix product since the $4\times 4$ matrix $\gamma^\nu \epsilon_\nu^\ast$ remains $4\times 4$ if it is only a $\gamma$ inside the full matrix product and the same is true for $\epsilon_\mu \gamma^\mu$. Or write out the full product in components, taking a component for instance $\epsilon_0$ in front of the matrix product in one of the summands or keeping it inside does not change anything as $\epsilon_0$ is just a number.


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