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how do I calculate the interference of these two diagrams, where the fermions are electrons?

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The matrix element of the first diagram is calculated as $\frac{1}{(p_3+p_4)^2}\bar{u}(p_2)\gamma_\mu u(p_1) \bar{u}(p_3) \gamma^\mu v (p_4)$, the amplitude neglecting the electron mass is calculated as $\frac{1}{(p_3+p_4)^4} (\gamma p_2) \gamma_\mu (\gamma p_1) \gamma_\nu (\gamma p_3) \gamma^\mu (\gamma p_4) \gamma^\nu $, the second diagram is just calculated by exchanging $p_2$ and $p_3$.

When I attempt to calculate the interference, I can't just write a matrix element for it and multiply it with the complex conjugate since now there are not two distinct fermion lines but one large one. Therefore, none of the spinors can be commuted and I cannot properly combine spinors into momenta. My only idea would be to directly write the amplitude as $\frac{1}{(p_3+p_4)^2 (p_2+p_4)^2}(\gamma p_1)\gamma_\nu (\gamma p_2) \gamma^\mu (\gamma p_4) \gamma^\nu (\gamma p_3) \gamma_\mu$.

Is this correct and if not how do I calculate diagrams with longer fermion lines like this.

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  • $\begingroup$ Not sure what you mean by long fermion lines, you just need to keep applying the spin sum identities until the expression is of the form $\bar{u}(p_3) (\dots) u(p_3)$, this can then be viewed as $ \bar{u}_i Q_{ij} u_j$, so you can still do your fermion spin sum on this object and write it as $\mathrm{Tr}(Q \bar{u} u)$. $\endgroup$ – Triatticus Mar 10 at 14:44
  • $\begingroup$ If I just mulitply the matrix elements I cannot properly apply spin sum identities because the corresponding spinors are not next to each other. Only if I write the amplitude directly like I did in the end of the question I can combine the spinors. $\endgroup$ – Nik Mar 11 at 13:39
  • $\begingroup$ Only the spinors from $p_3$ are not next to one another, the rest appear in pairs. After contracting all the pairs you deal with the final pair formed by the $\bar{u}_3$ and $u_3$. $\endgroup$ – Triatticus Mar 11 at 15:10
  • $\begingroup$ The matrix element is $\frac{1}{(p_3+p_4)^2}\bar{u}(p_2)\gamma_\mu u(p_1) \bar{u}(p_3) \gamma^\mu v (p_4)$ and the cc of the crossed matrix element is $\frac{1}{(p_2+p_4)^2}\bar{v}(p_4)\gamma_\nu u(p_2) \bar{u}(p_1) \gamma^\nu u(p_3)$, so only the spinors from $p_4$ appears in pairs and the rest does not. $\endgroup$ – Nik Mar 11 at 15:26
  • $\begingroup$ Along with the answer below, don't forget that the amplitude involves the trace of that expression. $\endgroup$ – Triatticus Mar 11 at 15:46
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Let $q_1$ and $q_2$ be the exchanged photon's momentum of each diagram, and for sake of brevity I will abbreviate $u(p_a)=u_a$ so as to limit the expression length.

The first Diagram has a matrix element:

$$ \mathcal{M}_1 = \frac{i}{q_1^2}(\bar{u}_3\gamma_{\mu}v_4)(\bar{u}_2\gamma^{\mu}u_1) $$

The second:

$$ \mathcal{M}_2 = \frac{i}{q_2^2}(\bar{u}_2\gamma_{\mu}v_4)(\bar{u}_3\gamma^{\mu}u_1) $$ Then: $$ \frac14\sum_{\text{spins}}|\mathcal{M}_1\bar{\mathcal{M}}_2|= \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\gamma_{\mu}v_4)(\bar{u}_2\gamma^{\mu}u_1)(\bar{u}_1\gamma^{\nu}u_3)(\bar{v}_4\gamma_{\nu}u_2) $$

This is, I assume where you get stuck, so we do as normal and rearrange the terms to put pairs of spinors together:

$$ \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\gamma_{\mu}v_4)(\bar{u}_2\gamma^{\mu}u_1)(\bar{v}_4\gamma_{\mu}u_2)(\bar{u}_1\gamma^{\mu}u_3) \to \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\gamma_{\mu}v_4)(\bar{v}_4\gamma_{\nu}u_2)(\bar{u}_2\gamma^{\mu}u_1)(\bar{u}_1\gamma^{\nu}u_3) $$

Now just form all the easier spin sums you can to get:

$$ \frac{1}{4q_2^2q_1^2}\sum_{\text{spins}}(\bar{u}_3\underbrace{\gamma_{\mu}\not{p_4}\gamma_{\nu}\not{p_2}\gamma^{\mu}\not{p_1}\gamma^{\nu}}_{Q_{ij}}u_3) \to \frac{1}{4q_2^2q_1^2}\sum_{ij}\bar{u}_{3i} Q_{ij} u_{3j} = \frac{1}{4q_2^2q_1^2}\sum_{ij}Q_{ij}(u_3\bar{u}_3)_{ji} $$

This just leaves you with:

$$ \frac{1}{4q_2^2q_1^2}\mathrm{Tr}(Q\not{p_3}) = \frac{1}{4q_2^2q_1^2}\mathrm{Tr}(\gamma_{\mu}\not{p_4}\gamma_{\nu}\not{p_2}\gamma^{\mu}\not{p_1}\gamma^{\nu}\not{p_3}) $$

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  • $\begingroup$ Hi, I just forgot to write the trace, sry about that. The part I got stuck was indeed where you assumed that I got stuck. I thought the rearrangements you made were not possible because if we take both diagrams into account, all outgoing momenta are "connected", I will look into this in more detail to understand better how I can rearrange the terms. Thank you very much! $\endgroup$ – Nik Mar 11 at 19:44
  • $\begingroup$ It has to do with the spinor indices, much like how you can rearrange products with lorentz indices because the sum doesn't care, if you explicitly write the spinor indices you can see how rearrangements can take place. A good example can be found in Griffiths Intro to Particle physics. $\endgroup$ – Triatticus Mar 12 at 0:48

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