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In trying to calculate the Møller scattering cross section, I arrived at the following term$^1$: $$\frac{e^4}{(p_3-p_1)^4}\bar u(p_3)\gamma^\mu u(p_1)\bar u(p_4)\gamma_\mu u(p_2)\bar u(p_2)\gamma_\nu u(p_4)\bar u(p_1)\gamma^\nu u(p_3),$$ where $p_1,p_2$ are the incoming momenta and $p_3,p_4$ are the outgoing momenta. In QFT in a nutshell, Zee separates this term like this$^2$: $$\frac{e^4}{(p_3-p_1)^4}[\bar u(p_3)\gamma^\mu u(p_1)\bar u(p_1)\gamma^\nu u(p_3)][\bar u(p_4)\gamma_\mu u(p_2)\bar u(p_2)\gamma_\nu u(p_4)].$$ I don't understand how he arrived at this. What relations am I missing?

$^1$ Spin polarizations have been omitted.

$^2$ The original notation is $p_1,p_2$ for incoming momenta, $P_1,P_2$ for outgoing momenta and $k\equiv(p_3-p_1)$.

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  • $\begingroup$ "I don't understand how he arrived at this." By the same principle that allows you to write $AB=(A)(B)$ for any pair of complex numbers $A$ and $B$. $\endgroup$
    – AccidentalFourierTransform
    Commented Oct 17, 2017 at 19:02
  • $\begingroup$ $\bar{u}$ is a row vector, $u$ is a column vector, and in-between you have matrices, everything with dimension 4. Make sense now? $\endgroup$
    – user154997
    Commented Oct 18, 2017 at 16:55
  • $\begingroup$ @LucJ.Bourhis Got it! $\bar u\gamma^\mu u$ is a scalar, so the above expression can be separated as 4 scalars and these can be rearranged to Zee's expression. Thank you! $\endgroup$
    – user140561
    Commented Oct 18, 2017 at 17:56

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As Luc points in the comments, $\bar u, u$ are row and column vectors, so $\bar u \gamma^\mu u$ is a scalar. Thus the expression may be arranged this way: $$\frac{e^4}{(p_3-p_1)^4}\underbrace{[\bar u(p_3)\gamma^\mu u(p_1)]}_1\underbrace{[\bar u(p_4)\gamma_\mu u(p_2)]}_2\underbrace{[\bar u(p_2)\gamma_\nu u(p_4)]}_3\underbrace{[\bar u(p_1)\gamma^\nu u(p_3)]}_4.$$ Changing the order from $1234$ to $1423$ gives Zee's result.

Edit: As @JamalS points out, $\bar u, u $ carry spinor indices and $\gamma^\mu$ a Lorentz index, so $\bar u\gamma^\mu u$ is a scalar with respect to spinor indices only.

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    $\begingroup$ $u$ and $\bar u$ are spinors, so they carry $\alpha,\beta$ indices, just as when we write $\gamma^\mu$ we mean $(\gamma^\mu)_{\alpha\beta}$. Two spinors with one $\gamma^\mu$ soaks up the two spinor indices, but we still have the Lorentz index. So while you can swap them safely, strictly speaking the factor on its own is not a scalar unless you are more specific with regards to what it is a scalar with respect to. $\endgroup$
    – JamalS
    Commented Oct 18, 2017 at 18:10
  • $\begingroup$ @JamalS You're right. I edited the answer to be more precise. Thank you. $\endgroup$
    – user140561
    Commented Oct 18, 2017 at 18:20

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