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I'm reading through Halzen and Martin Chapter 6.3 and I have a few questions about what they're doing. We're calculating the invariant amplitude for $e^-\mu^- \rightarrow e^-\mu^-$ scattering. The text is (up to some rearranging to make the question clearer) as follows:

It is convienent to separate the sums over the electron and muon spins by

$$\overline{|\mathfrak{R}|^2} = \frac{1}{(2s_A+1)(2s_B+1)}\sum_{\substack{all\ spin \\ states}}|\mathfrak{R}|^2 = \frac{e^4}{q^4} L_e^{\mu\nu}L_{\mu\nu}^{muon}$$

where $s_A$, $s_B$ are the spins of the incoming particles, and the tensor associated with the electron vertex is

$$L_e^{\mu\nu} = \frac{1}{2}\sum_{e\ spins}[\bar{u}(k')\gamma^\mu u(k)][\bar{u}(k')\gamma^\nu u(k)]^*$$

and with a similar expression for $L_{\mu\nu}^{muon}$.

It's not clear to me why the second equality in the first line is true. It sounds as if the authors know a priori that we can separate the sums by writing it as a contraction of two tensors. Is this always possible? I'm not sure how to verify this on my own, and the authors don't really do much to explain it.

The book continues:

$$L_e^{\mu\nu} = \frac{1}{2}\sum_{s'}\bar{u}_\alpha^{(s')}(k')\gamma_{\alpha\beta}^{\mu}\sum_{s}\bar{u}_\beta^{(s)}(k)\bar{u}_\gamma^{(s)}(k)\gamma_{\gamma\delta}^{\mu}\bar{u}_\delta^{(s')}(k')$$ $$=\frac{1}{2}\sum_{s'}\bar{u}_\delta^{(s')}(k')\bar{u}_\alpha^{(s')}(k')\gamma_{\alpha\beta}^{\mu}\sum_{s}\bar{u}_\beta^{(s)}(k)\bar{u}_\gamma^{(s)}(k)\gamma_{\gamma\delta}^{\mu}$$

justifying this repositioning of $\bar{u}_\delta$ from the back to the front of the equation by saying

The completeness relations allows the sums over both the initial and the final electron spins to be preformed. The repositioning of $\bar{u}_\delta$ makes this evident; it can be moved as the matrix character is recorded by the component.

I don't understand why we can reposition $\bar{u}_\delta$ like this. Does it commute with all of the other $u$'s and $\gamma$'s? The second sentence makes no sense to me at all. What is the matrix character?

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  • $\begingroup$ Parts of your question marked as quotes don't actually appear as such in the book. $\endgroup$ – Suzu Hirose Oct 14 '16 at 2:02
  • $\begingroup$ Yes, I've edited it to make it clear that I paraphrased by grabbing a few things from previous pages. Thanks. $\endgroup$ – Ecclesiastic Oct 14 '16 at 4:18
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It's not clear to me why the second equality in the first line is true. It sounds as if the authors know a priori that we can separate the sums by writing it as a contraction of two tensors. Is this always possible?

This comes from conservation of angular momentum. The spin before is the same as the spin after, so each different component can be separated.

I don't understand why we can reposition u¯δ like this.

The four components of the wavefunction are scalars and can be arbitrarily rearranged, you are just using commutativity of multiplication ($a\times b=b\times a$). They are only moving it to make the relation obvious.

Does it commute with all of the other u's and γ's?

Yes, of course it does, the four components of the wavefunction are scalars.

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  • $\begingroup$ >I don't understand why we can reposition u¯δ like this. $\endgroup$ – Ecclesiastic Oct 14 '16 at 4:50
  • $\begingroup$ Sorry, I clicked enter early. Okay, that makes sense now. I guess it just doesn't appear obvious when in the notation with spinor indices suppressed. $\endgroup$ – Ecclesiastic Oct 14 '16 at 4:52

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