0
$\begingroup$

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.

$\endgroup$
1
  • 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

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.