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In the book “Gauge theory of elementary particles-Cheng and Li, section 13.3, the $\mu\rightarrow e\gamma$ decay amplitude is calculated in the $R_{\xi}$ gauge. Regarding this derivation, I'm stuck with a couple of questions.

(i) I think this calculation would have been easier in the unitary gauge since in that case we will not have loops with unphysical Higgs. I can't understand why is it not calculated in the unitary gauge. I think, this amplitude should give finite and identical result in unitary gauge as well. Isn't it?

Are there any other useful references where diagrams such as $\mu\rightarrow e\gamma$, $\mu\rightarrow e\nu\bar{\nu}$ etc are computed in reasonable detail?

(ii) Although it is explained in one statement, it is not clear to me how the contributions from four diagrams of figure 13.6(e) cancel from similar terms appearing in diagrams 13.6(a)-(d).

(iii)Why did they say that “we need to concentrate on the $p\cdot \epsilon$ term in Eqn. 13.79. What did they left out the $\gamma\cdot\epsilon$ term? My guess is that it is the $\gamma\cdot\epsilon$ term from the diagrams 13.6(a)-(d), is cancelled by contributions from the four diagrams of 13.6(e). But is this cancellation so trivial to see that one need not calculate them at all?

EDIT: The Lorentz invariant transition amplitude is proportional to $T(\mu\rightarrow e\gamma)\sim \bar{u}_e\sigma_{\mu\nu}(A+B\gamma_5)u_\mu$. Then he argues that neglecting the electron mass i.e., $m_e=0$ leads to $A\approx B$. How does that work out? In another account, when we do not throw the electron mass the amplitude is proportional to $T(\mu\rightarrow e\gamma)\sim \bar{u_{e}} i \sigma_{\mu\nu}q^\nu[m_e(1-\gamma_5)+m_\mu(1+\gamma_5)]u_\mu$. With $m_e=0$, though, the Lorentz invariant transition amplitude from Cheng and Li can be reproduced. I couldn't prove the second proportionality either.

I used the fact that the operators $\sigma_{\mu\nu}$ and $\sigma_{\mu\nu}\gamma_5$ connects the left-chiral state with the right-chiral one i.e., $\bar{\psi}\sigma_{\mu\nu}\psi=\overline{\psi_R}\sigma_{\mu\nu}\psi_L+\overline{\psi_L}\sigma_{\mu\nu}\psi_R$ and same for $\sigma_{\mu\nu}\gamma_5$. Using these I proceeded upto the step $$T(\mu\rightarrow e\gamma)\sim A\overline{u_{eL}}\sigma_{\mu\nu}u_{\mu R}+A\overline{u_{eR}}\sigma_{\mu\nu}u_{\mu L}+B\overline{u_{eL}}\sigma_{\mu\nu}\gamma_5 u_{\mu R}+B\overline{u_{eR}}\sigma_{\mu\nu}\gamma_5 u_{\mu L}$$

But it is not clear to me that how this to reduce to an expression of the form $T(\mu\rightarrow e\gamma)\sim i \sigma_{\mu\nu}q^\nu[m_e(1-\gamma_5)+m_\mu(1+\gamma_5)]$ or in the $m_e=0$ limit $T(\mu\rightarrow e\gamma)\sim \bar{u_{e}}i \sigma_{\mu\nu}q^\nu[m_\mu(1+\gamma_5)]u_\mu$?

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  • $\begingroup$ (ii) and (iii). Earlier he uses gauge invariance to justify the form 13.76. So he is arguing that any terms which don't fit this form must cancel to satisfy gauge invariance. I haven't looked deeper into it than that. $\endgroup$ – Bruce Greetham Aug 28 '16 at 17:07
  • $\begingroup$ If you go through the steps, 13.76 is nothing but the equation 13.79 written in a different manner. The four diagrams of 13.6(e) resemble the term $\sim m_\mu \bar{u}_e\gamma\cdot\epsilon u_\mu$ in 13.79. So I don't understand the argument of not considering these diagrams. $\endgroup$ – SRS Aug 28 '16 at 17:27
  • $\begingroup$ Yes- having looked into it more I see your confusion. No its not clear. Sorry - I need an expert. $\endgroup$ – Bruce Greetham Aug 28 '16 at 18:36
  • $\begingroup$ I've tried in my answer to dissect his logic: I've probably got as far as you. $\endgroup$ – Bruce Greetham Aug 28 '16 at 19:07
  • $\begingroup$ It can not really be a tree diagram. $\endgroup$ – Roghan Arun May 14 at 14:07
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(i) It is not advisable to do loop calculations in unitary gauge due to blowing up of the propagator for W boson in ultraviolet limit as the loop involves integrating over momenta. Renormalizability(perturbative) is also obscured by using unitary gauge. So I doubt you will find any good reference on doing loop calculation using unitary gauge in this case.

(ii) A term like $\bar u\gamma_\mu u$ can be extracted from the loop diagrams (a)-(d) by noting the vertex where fermion line is present and fermion propagator contribute one gamma factor. There are three gammas coming with some of them multiplying with $\gamma_5$. some duality relations like $\epsilon_{abcd}\gamma^{bcd}=i\gamma_5\gamma_a$ and simple gamma technology will produce $\bar u\gamma_\mu u$ and $\bar u\gamma_{\mu\nu} u$ type term. [$\gamma_{\mu\nu}$ is same as your $\sigma_{\mu\nu}$(apart from some numerical factor probably)]

(iii) The term $\gamma\centerdot\epsilon$ does not cancel from the diagrams but is rejected on the reality condition of emitted photon. Simply saying the term violates gauge invariance. To see it explicitly, $\epsilon^\nu \bar u_e(1+\gamma_5)\gamma_\nu u_\mu$ with $\epsilon^\nu$ removed, operate with $q^\nu$ on $\bar u_e (1+\gamma_5)\gamma_\nu u_\mu$ and write q as q-p+p and then use Dirac equation in momentum space for muon and electron and you will see it does not vanish.

For the edit part- In the massless limit of electron, you can pick a certain helicity and the coupling of W boson is to left handed electron. With the bar form of electron spinor, it basically requires to multiply it with $1+\gamma_5$. This amounts to $(1+\gamma_5)(A+B\gamma_5)$ or $A(1+\gamma_5)+B(1+\gamma_5)$ as $\gamma_5$ commutes with $\sigma_{\mu\nu}$. So both term involving A and B corresponds to same matrix element in massless electron limit, So they can be taken to be equal .

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  • $\begingroup$ Your answer to number 3 is not correct. The reason is because there's only one parameter to be deterined, so one need only consider the p.eps term. p. eps does not satisfy the ward identity either, but 2 p.eps - m eps.gamma does. $\endgroup$ – thedoctar Jan 26 at 0:53
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To answer parts (ii) and (iii) follow the logic from equation 13.73:

In general it is possible to have 3 types of terms:
$q^\nu\sigma_{\lambda\nu}$
$\gamma_\lambda$
$q_\lambda$

Then he proves that only the first of these three are allowed in the final solution. This is how he arrives at 13.76.

As we know that we have to end up with the form of 13.76 we can ignore any diagrams or terms within diagrams which don't have this form.

Now for the confusing bit: Having justified dropping all terms of the $\gamma_\lambda$ type, he then uses Gordon decomposition in 13.79 to replace $i\sigma_{\lambda\nu}q^\nu\epsilon^\lambda$ by $(2p\centerdot\epsilon - m_\mu\gamma\centerdot\epsilon)$ which seems to bring a $\gamma_\lambda$ term back in. That seems to be the point that has got you confused.

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  • $\begingroup$ @ Bruce Greetham-Exactly. $\endgroup$ – SRS Aug 28 '16 at 19:31
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iii) He ignores the $\gamma.\epsilon$ term because he only needs to determine $A$, which can be read off the coefficient of the $p.\epsilon$ term.

Regarding your fourth question, $A+B\gamma^5=(A-B)P_L+(A+B)P_R$. Since the weak force only couples to the left-handed electron, we want $A=B$.

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