In order to compute the scattering cross section for Møller scattering, one needs the amplitudes for both the $t$- and the $u$-channel. Since the cross section is proportional to $|\mathcal{M}|^2$, the matrix element squared, and since there are two channels, we will get mixed terms. One of these terms is, after averaging over the possible spin states:

$$ -\frac{e^4}{4 t u}\mathop{Tr}\left[ (p_3\!\!\!\!\!/+m)\gamma^\mu(p_1\!\!\!\!\!/+m)\gamma^\beta(p_4\!\!\!\!\!/+m)\gamma_\mu(p_2\!\!\!\!\!/+m)\gamma_\beta \right], $$ where $e$ is the electron charge, $m$ the electron mass, $p_1,p_3$ the momenta of the incoming electrons, $p_2,p_4$ the momenta of the outgoing ones and $\gamma$ the $\gamma$-matrices. $\mathop{Tr}$ stands for the trace of an operator. $t$ and $u$ are Mandelstam variables.

Now, one could simply expand the expression I have to take the trace of but this will lead to an exploding amount of terms. So, is there any smart way of calculating this trace without getting lost in cumbersome calculations?


3 Answers 3


Typically, one uses the fact that the trace of an odd number of gamma matrices vanishes in $d$ dimensions other than select odd $d$. This allows you to see, by inspection really, the contributing terms. It usually amounts to dropping some mass contributions, for example. In your case, there is more simplication to be made first by anti commuting the $\gamma^{\mu}$ and $\gamma^{\beta}$ so as to e.g rewrite the pair $\gamma^{\mu}\gamma_{\mu} = d1$. This then leaves you with a trace of at most four gamma matrices, which is relatively simple and whose evaluation is explicitly given in the literature.


One way to reduce the number of terms is to set $m=0$.

For $E\gg m$ and using the approximation $m=0$ the result for the trace calculation is

$$ -8s^2 $$

This result (obtained by computer) indicates the magnitude of simplification that occurs.

Complete derivation follows.

In center of mass coordinates the momentum vectors for Moller scattering are $$ p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\quad p_2=\begin{pmatrix}E\\0\\0\\-p\end{pmatrix}\quad p_3=\begin{pmatrix} E\\ p\sin\theta\cos\phi\\ p\sin\theta\sin\phi\\ p\cos\theta \end{pmatrix} \quad p_4=\begin{pmatrix} E\\ -p\sin\theta\cos\phi\\ -p\sin\theta\sin\phi\\ -p\cos\theta \end{pmatrix} $$

where $p=\sqrt{E^2-m^2}$. The spinors are \begin{gather*} u_{11}=\begin{pmatrix}E+m\\0\\p\\0\end{pmatrix}\quad u_{21}=\begin{pmatrix}E+m\\0\\-p\\0\end{pmatrix}\quad u_{31}=\begin{pmatrix}E+m\\0\\p_3^z\\p_3^x+ip_3^y\end{pmatrix}\quad u_{41}=\begin{pmatrix}E+m\\0\\p_4^z\\p_4^x+ip_4^y\end{pmatrix} \\ u_{12}=\begin{pmatrix}0\\E+m\\0\\-p\end{pmatrix}\quad u_{22}=\begin{pmatrix}0\\E+m\\0\\p\end{pmatrix}\quad u_{32}=\begin{pmatrix}0\\E+m\\p_3^x-ip_3^y\\-p_3^z\end{pmatrix}\quad u_{42}=\begin{pmatrix}0\\E+m\\p_4^x-ip_4^y\\-p_4^z\end{pmatrix} \end{gather*}

The last digit in a spinor subscript is 1 for spin up and 2 for spin down. Note that the spinors are not individually normalized. Instead, a combined spinor normalization constant $N=(E+m)^4$ will be used where needed.

This is the probability density for Moller scattering. The formula is from Feynman diagrams. $$ |\mathcal{M}(s_1,s_2,s_3,s_4)|^2=\frac{e^4}{N} \left| \frac{1}{t}(\bar{u}_3\gamma^\mu u_1)(\bar{u}_4\gamma_\mu u_2) -\frac{1}{u}(\bar{u}_4\gamma^\nu u_1)(\bar{u}_3\gamma_\nu u_2) \right|^2 $$

Symbol $s_j$ selects the spin (up or down) of spinor $j$. Symbol $e$ is electron charge. Symbols $t$ and $u$ are Mandelstam variables $t=(p_1-p_3)^2$ and $u=(p_1-p_4)^2$.

Let \begin{equation*} a_1=(\bar{u}_3\gamma^\mu u_1)(\bar{u}_4\gamma_\mu u_2) \qquad a_2=(\bar{u}_4\gamma^\nu u_1)(\bar{u}_3\gamma_\nu u_2) \end{equation*}

Then \begin{align*} |\mathcal{M}(s_1,s_2,s_3,s_4)|^2 &= \frac{e^4}{N} \left|\frac{a_1}{t} - \frac{a_2}{u}\right|^2\\ &= \frac{e^4}{N} \left(\frac{a_1}{t} - \frac{a_2}{u}\right)\left(\frac{a_1}{t} - \frac{a_2}{u}\right)^*\\ &= \frac{e^4}{N} \left( \frac{a_1a_1^*}{t^2} - \frac{a_1a_2^*}{tu} - \frac{a_1^*a_2}{tu} + \frac{a_2a_2^*}{u^2} \right) \end{align*}

The expected probability density $\langle|\mathcal{M}|^2\rangle$ is computed by summing $|\mathcal{M}|^2$ over all spin states and dividing by the number of inbound states. There are four inbound states. \begin{align*} \langle|\mathcal{M}|^2\rangle &= \frac{1}{4}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2 |\mathcal{M}(s_1,s_2,s_3,s_4)|^2\\ &= \frac{e^4}{4}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2 \frac{1}{N}\left( \frac{a_1a_1^*}{t^2}-\frac{a_1a_2^*}{tu}-\frac{a_1^*a_2}{tu}+\frac{a_2a_2^*}{u^2} \right) \end{align*}

Use the Casimir trick to replace sums over spins with matrix products. \begin{align*} f_{11}&=\frac{1}{N}\sum_{\rm spins}a_1a_1^*= \mathop{\rm Tr}\left( (\not p_3+m)\gamma^\mu(\not p_1+m)\gamma^\nu \right) \mathop{\rm Tr}\left( (\not p_4+m)\gamma_\mu(\not p_2+m)\gamma_\nu \right) \\ f_{12}&=\frac{1}{N}\sum_{\rm spins}a_1a_2^*= \mathop{\rm Tr}\left( (\not p_3+m)\gamma^\mu(\not p_1+m)\gamma^\nu (\not p_4+m)\gamma_\mu(\not p_2+m)\gamma_\nu \right) \\ f_{22}&=\frac{1}{N}\sum_{\rm spins}a_2a_2^*= \mathop{\rm Tr}\left( (\not p_4+m)\gamma^\mu(\not p_1+m)\gamma^\nu \right) \mathop{\rm Tr}\left( (\not p_3+m)\gamma_\mu(\not p_2+m)\gamma_\nu \right) \end{align*}

Hence \begin{equation*} \langle|\mathcal{M}|^2\rangle=\frac{e^4}{4} \left( \frac{f_{11}}{t^2}-\frac{f_{12}}{tu}-\frac{f_{12}^*}{tu}+\frac{f_{22}}{u^2} \right) \end{equation*}

Here is another way to compute probability densities. \begin{align*} f_{11}&= 32 (p_1\cdot p_2) (p_3\cdot p_4) + 32 (p_1\cdot p_4) (p_2\cdot p_3) - 32 m^2 (p_1\cdot p_3) - 32 m^2 (p_2\cdot p_4) + 64 m^4 \\ f_{12}&= -32 (p_1\cdot p_2) (p_3\cdot p_4) + 16 m^2 (p_1\cdot p_2) + 16 m^2 (p_1\cdot p_3) + 16 m^2 (p_1\cdot p_4) \\ &\phantom{=}\qquad{} + 16 m^2 (p_2\cdot p_3) + 16 m^2 (p_2\cdot p_4) + 16 m^2 (p_3\cdot p_4) - 32 m^4 \\ f_{22}&= 32 (p_1\cdot p_2) (p_3\cdot p_4) + 32 (p_1\cdot p_3) (p_2\cdot p_4) - 32 m^2 (p_1\cdot p_4) - 32 m^2 (p_2\cdot p_3) + 64 m^4 \end{align*}

In Mandelstam variables $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$, $u=(p_1-p_4)^2$ the formulas are \begin{align*} f_{11} &= 8 s^2 + 8 u^2 - 64 s m^2 - 64 u m^2 + 192 m^4 \\ f_{12} &= -8 s^2 + 64 s m^2 - 96 m^4 \\ f_{22} &= 8 s^2 + 8 t^2 - 64 s m^2 - 64 t m^2 + 192 m^4 \end{align*}

When $E\gg m$ a useful approximation is to set $m=0$ and obtain \begin{align*} f_{11}&=8s^2+8u^2\\ f_{12}&=-8s^2\\ f_{22}&=8s^2+8t^2 \end{align*}

For $m=0$ the Mandelstam variables are \begin{align*} s&=4E^2 \\ t&=-2E^2(1-\cos\theta)=-4E^2\sin^2(\theta/2) \\ u&=-2E^2(1+\cos\theta)=-4E^2\cos^2(\theta/2) \end{align*}

The corresponding expected probability density is \begin{align*} \langle|\mathcal{M}|^2\rangle &=\frac{e^4}{4} \left( \frac{8s^2+8u^2}{t^2}+\frac{16s^2}{tu}+\frac{8s^2+8t^2}{u^2} \right) \\ &=2e^4 \left( \frac{s^2+u^2}{t^2}+\frac{2s^2}{tu}+\frac{s^2+t^2}{u^2} \right) \\ &=2e^4 \left( \frac{1+\cos^4(\theta/2)}{\sin^4(\theta/2)} +\frac{8}{\sin^2\theta} +\frac{1+\sin^4(\theta/2)}{\cos^4(\theta/2)} \right) \end{align*}


I would use the following identities, which are obtained by following CAF's approach:

$$\gamma^\mu \gamma^\alpha \gamma_\mu = -2\gamma^\alpha$$ $$\gamma^\mu \gamma^\alpha \gamma^\beta \gamma_\mu = 4g^{\alpha\beta}$$

$$\gamma^\mu \gamma^\alpha \gamma^\beta \gamma^\delta \gamma_\mu = -2\gamma^\delta\gamma^\beta\gamma^\alpha$$


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