My first ever question on stackexchange! Sorry if it is clumsy...

I am trying to follow the computation of $\mu \rightarrow e \gamma$ in Cheng and Li and am confused about the second and third ligns in equation (13.97):

$S_1 \rightarrow \tilde{S}_1 = (p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u_\mu \right] 2 m_\mu \left[2(1-\alpha_1)^2 + (2\alpha _1-1)\alpha _2\right]\\ S_2 \rightarrow \tilde{S}_2 = -k^2(p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u_\mu \right] (m_\mu /M^2) \times \left[(3\alpha _2-1)+\left[2\alpha_1^2-\alpha_1+\alpha_2(2\alpha_1-1/2)\right]\right]\\ S_3 \rightarrow \tilde{S}_3 = -k^2(p\cdot \varepsilon )\left[\bar{u}_e(1+\gamma _5) u_\mu \right] (m_\mu /M^2) \left[2\alpha_1^2+\alpha_1+\alpha_2(2\alpha_1-1/2)\right]$


$S_1 = \Gamma ^{\mu \nu}N_{\mu \nu}\\ S_2 = (k^\lambda \Gamma ^\mu _\lambda)(k^\nu N_{\mu \nu})/M^2\\ S_3 = \left[(k+q)^\lambda \Gamma ^\mu _\lambda \right]\left[(k+q)^\nu N_{\mu \nu}\right]/M^2$


$\Gamma _{\alpha \beta} = \left[(2k\cdot \varepsilon) g_{\alpha \beta} -(k+2q)_\beta \varepsilon _\alpha - (k-q)_\alpha \varepsilon _\beta\right] $

(this being equation (13.81))


$N_{\mu \nu} =\bar{u}_e(p-q)\gamma _\mu ((p+k)\cdot \gamma ) \gamma _\nu (1-\gamma _5)u_\mu(p)$

(equation (13.88)).

In the calculation $S_i\rightarrow \tilde{S}_i$ we shift the integration variable $k$ to $k-\alpha_1 p - \alpha_2 q$ and discard all terms that are not $\sim (p\cdot \varepsilon)$.

Whereas I managed to obtain the expression for $\tilde{S}_1$ (with a minus sign, but apparently that's alright...), I do not understand what to do with the many momenta in the calculation of $\tilde{S}_2$ (and $\tilde{S}_3$): Why do Cheng and Li only get (or keep?) terms $\sim k^2$? I would have thought we would get a whole bunch of terms $\sim m_\mu ^2$ from the combinations $\sim p^2$ and $\sim p\cdot q$. (The mass of the electron is neglected here.)

I would be very grateful for some tips... Thank you in advance!

  • $\begingroup$ Is it just a matter of $M \gg m_\mu$? But the W-boson is not on-shell... $\endgroup$
    – user188104
    Mar 15, 2018 at 19:34
  • $\begingroup$ I have worked out the quadratic part for general $\xi$ for $S_3$. But not surprisingly my answer is different from the book. There is a overall sign (this seems a recurrent theme for this section) and one more sign in the $\alpha_1$ factor. The computation is 6 pages in pdf format (all details are included). If you are still interested and may want to check it, I will be happy to post it as an answer. $\endgroup$ Jun 20, 2019 at 10:25
  • $\begingroup$ Dear Oбжорoв, That would be great! I haven't looked at this calculation for over a year, but I always had it in the back of my mind. I will go home now and look for my notes.... I guess your answer would be useful to many studying QFT? $\endgroup$
    – user188104
    Jun 24, 2019 at 17:39
  • $\begingroup$ I have posted the calculation of $S_3$ as an answer. Feel free to check if you find any errors. If it is Ok, pls accept the answer and upvote it. If it is correct, I can also post the calculation for $S_2$ and $S_1$. Good luck checking ! $\endgroup$ Jun 25, 2019 at 7:28

2 Answers 2


I recently went through this calculation, but only working in the $\xi = 1$ gauge. I would first note that I also got a minus sign relative to Cheng and Li, for every single diagram, so it all worked out in the end.

When you shift the integration variable (and this shift will be different if you assigned your Feynman Parameters differently) the numerator takes on new terms which are either scalar $(\sim p \cdot \varepsilon )$ or vector $(\sim \gamma \cdot \varepsilon)$ under Lorentz transformations. Integrating over your new $k^\mu$ will not change the Lorentz transformation properties of these terms so, since they found that $$T = A \overline{u}_e(p-q)(1+\gamma_5)(2p\cdot \varepsilon-m_\mu \gamma \cdot \varepsilon) u_\mu(p)$$ (assuming $m_e \rightarrow 0$) (this is $(13.79)$) they know to keep only the scalar terms (which pleseantly happen to be proportional to $p\cdot \varepsilon$). Also all the terms which were odd under $k^\mu \rightarrow - k^\mu$ simply cancel from the integrations.

Unfortunately, working in $\xi =1$, I didn't need to calculate the $S_2$ or $S_3$ terms so I can't give you any more specific advice other than to keep staring at it. Understanding why the above scalar - vector decomposition is so clever took a lot of headscratching on my part and I can only hope I didn't convince myself of something that was completely false, but I made it to the correct answer in the end.


This is a very long answer, but it consists of two parts: first we explain the general method of how to calculate this. Next we work out the nitty gritty details. Readers only interested in the method may contend themselves with the first part. The result of the second part however differs from Cheng & Li's result. So any feedback on the calculation is more than welcome.


We work out $\boldsymbol{S_3}$ in terms of the shifted variable $\ell = k-\alpha_1 p - \alpha_2q$ \begin{align} M^2 \boldsymbol{S_3} =& (k+q)^\lambda \Gamma_\lambda^{\; \nu} (k+q)^\mu \boldsymbol{N}_{\mu\nu} \nonumber\\ =& \big(\ell- \alpha_1 p+ (1-\alpha_2) q \big)^\lambda \times \big[ 2 (\ell - \alpha_1p -\alpha_2 q)\cdot \epsilon\, \delta^\nu_\lambda - \big(\ell -\alpha_1p + (2-\alpha_2) q\big)^\nu \epsilon_\lambda\nonumber\\ & - \big(\ell -\alpha_1p -(1+\alpha_2) q \big)_\lambda \epsilon^\nu \big] \times \big(\ell -\alpha_1 p +(1-\alpha_2)q \big)^\mu \nonumber\\ &\times \big(\ell+(1-\alpha_1) p -\alpha_2 q\big)^\sigma \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \qquad \mathrm{[Eq 1]} \end{align} This is a messy expression, but let us keep our goal in mind. Our final expression must be of the form $p\cdot \epsilon \, \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p)$ so we can focus on obtaining these terms and can neglect everything that does not look like it. Identifying contributions of the form $p\cdot \epsilon$ is easy, but how do we get $\bar{u}_e (p-q) (1+\gamma^5) u_\mu(p)$. First we need to use the Clifford algebra to reduce the three $\gamma$ matrices to one. We will then have expressions that contain terms of the form $\bar{u}_e (p-q) \not{\! p} (1-\gamma^5) u_\mu(p)$ and $\bar{u}_e (p-q) \not{\! q} (1-\gamma^5) u_\mu(p)$. We can rewrite these as terms of the form $\bar{u}_e (p-q) (\not{\! p}-\not{\! q}) (1-\gamma^5) u_\mu(p)$ and $\bar{u}_e (p-q) \not{\! p} (1-\gamma^5) u_\mu(p)$. The first term then vanishes by the Dirac equation $\bar{u}_e (p-q) (\not{\! p}-\not{\! q}) =0$ in the limit of massless electrons. For the second term, we flip the $\not{\! p}$ through the $1-\gamma^5$, giving the desired $1+\gamma^5$, and use the Dirac equation $\not{\! p} u_\mu(p) = m_\mu u_\mu (p)$. This will then give something of the form $m_\mu \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p)$, which is what we need. As a final point we only need to work out terms in $\ell^2$ and $\ell^0$ as other terms are zero by symmetry considerations of the $\ell$-integral. Mind you, even with all these simplifications it is still a mess!

Let us start with the terms quadratic in $\ell$. We will have terms of the form $\ell^\alpha \ell^\beta p^\gamma p^\delta$ and $\ell^\alpha \ell^\beta p^\gamma q^\delta$. We can ignore the terms of the form $\ell^\alpha \ell^\beta q^\gamma q^\delta$ because they don't have a $p$ and hence cannot end up as $\epsilon \cdot p$. We can then replace $\ell^\alpha \ell^\beta$ by $(\ell^2/4) g^{\alpha \beta}$. If we multiply out [Eq 1] we get six different combinations of $\ell^\alpha \ell^\beta$. Each can come with one of three combinations $p^\gamma p^\delta, p^\gamma q^\delta$ or a $q^\gamma p^\delta$ (the order of the $p$ and $q$ is important as they are contracted with $\gamma$ matrices). So we need to work out a total of 18 terms of the form $g^{\alpha \beta} f^\gamma g^\delta \epsilon^\eta \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p)$. Where the indices $\alpha, \beta, \gamma, \delta$ and $\eta$ are some permutation of $\mu, \sigma, \nu$ and two of these indices are contracted. Also $(f,g)$ denotes $(p,p), (p,q)$ or $(q,p)$. It so happens that each such combination will be of the form $c m_\mu \epsilon\cdot p \bar{u}_e(p-q) (1+\gamma^5) u_\mu(p)$ and we need to determine the factor $c$ for each such combination.


Let's get cracking. We will ignore any unnecessary symbols and ignore any terms that don't lead to a contribution of the desired form. When we are ignoring anything we will use the symbol $\leadsto$ in stead of the equal sign. Note that we will not write the spinors $ \bar{u}_e (p-q) $ and $ u_\mu(p) $. They will be understood to be there. We will also write $\gamma^\pm = 1\pm \gamma^5$. Recall that with our abbreviations if we can get something of the form $\cdots \not{\! p} \gamma^-$ then this is equal to $\cdots \gamma^+ \not{\! p} = \cdots m_\mu \gamma^+$. We will also repetitively use the fact that $\not{\! p}^2 = p^2$.

  1. $\epsilon^\nu p^\mu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = \epsilon^\nu p^2 \gamma^\nu \gamma^- = m_\mu^2 \epsilon \cdot \gamma \gamma^- \leadsto 0$ as this cannot be brought into a form containing $\epsilon\cdot p$;

  2. $\epsilon^\nu p^\mu q^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = \epsilon^\nu (p-q) ^\mu q^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- + \epsilon^\nu q^\mu q^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- \leadsto 0$. The first term is zero because of the zero electron mass and the second term doesn't have a $p$ left to contract with the polarisation $\epsilon$;

  3. $\epsilon^\nu q^\mu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = \epsilon^\nu (q-p) ^\mu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- + \epsilon^\nu p^\mu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- \leadsto 0$. Due to $m_e=0$ and to the fact that we have already seen that the $\epsilon^\nu p^\mu p^\sigma$ combination doesn't contribute;

  4. $g^{\mu\nu} \epsilon_\rho p^\rho p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^-$. We first work out the general case because it will be useful for two more permutations. \begin{align*} g^{\mu\nu} \epsilon_\rho f ^\rho g^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- &= g^{\mu\nu} \epsilon_\rho f ^\rho g^\sigma \gamma_\mu (-\gamma_\nu \gamma_\sigma + 2 g_{\nu\sigma}) \gamma^- = -\epsilon_\rho f^\rho g^\sigma \gamma^\mu \gamma_\mu \gamma_\sigma \gamma^- + 2 \epsilon_\rho f^\rho g^\sigma \gamma_\sigma \gamma^- \nonumber\\ &= -4 \epsilon \cdot f \not{\! g} \gamma^- +2 \epsilon \cdot f \not{\! g} \gamma^- = -2 \epsilon \cdot f \not{\! g} \gamma^- \end{align*} So $g^{\mu\nu} \epsilon_\rho p^\rho p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = -2 \epsilon \cdot p \not{\! p} \gamma^- = -2 m_\mu \epsilon \cdot p \gamma^+$;

  5. $g^{\mu\nu} \epsilon_\rho p^\rho q^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = -2 \epsilon \cdot p \not{\! q} \gamma^- = -2 \epsilon \cdot p (\not{\! q}-\not{\! p}) \gamma^- - 2 \epsilon \cdot p \not{\! p}\gamma^- \leadsto -2 m_u \epsilon \cdot p \gamma^+$;

  6. $g^{\mu\nu} \epsilon_\rho q^\rho p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = -2 \epsilon \cdot q \not{\! p} \gamma^- =-2 m_\mu \epsilon \cdot q \gamma^+ \leadsto 0$;

  7. $g^{\nu\sigma} \epsilon_\rho p^\rho p^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^-$. Again we start with the general case. $ g^{\nu\sigma} \epsilon_\rho f^\rho g^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = \epsilon_\rho f^\rho g^\mu \gamma_\mu \gamma^\nu \gamma_\nu \gamma^- = 4 \epsilon\cdot f \not{\! g}\gamma^-$. Thus $g^{\nu\sigma} \epsilon_\rho p^\rho p^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = 4 \epsilon\cdot p \not{\! p}\gamma^- = 4 m_u \epsilon \cdot p \gamma^+$;

  8. $g^{\nu\sigma} \epsilon_\rho p^\rho q^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = 4 \epsilon\cdot p \not{\! q}\gamma^- = 4 \epsilon\cdot p (\not{\! q}- \not{\! p}) \gamma^- + 4 \epsilon\cdot p \not{\! p}\gamma^- = 4 m_u \epsilon \cdot p \gamma^+$;

  9. $g^{\nu\sigma} \epsilon_\rho q^\rho p^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = 4 \epsilon\cdot q \not{\! p}\gamma^- = 4 m_\mu \epsilon\cdot q \gamma^+ \leadsto 0$

  10. $\epsilon^\mu p^\nu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = p^2 \epsilon\cdot \gamma \leadsto 0$

  11. $\epsilon^\mu p^\nu q^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^-$. This requires a little bit more work: \begin{align*} \epsilon^\mu p^\nu q^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- &= \epsilon^\mu q^\sigma \gamma_\mu \gamma_\sigma \not{\! p} \gamma^- = m_\mu \epsilon^\mu q^\sigma \gamma_\mu \gamma_\sigma \gamma^+ = m_\mu \epsilon^\mu q^\sigma (-\gamma_\sigma \gamma_\mu + 2 g_{\mu\sigma}) \gamma^+ \nonumber\\ &= -m_\mu \epsilon^\mu \not{\! q} \gamma_\mu \gamma^+ + 2 m_\mu \epsilon\cdot q \gamma^+ \leadsto -m_\mu \epsilon^\mu (\not{\! q} -\not{\! p}) \gamma_\mu \gamma^+ -m_\mu \epsilon^\mu \not{\! p} \gamma_\mu \gamma^+ \nonumber\\ & \leadsto m_\mu \epsilon^\mu p^\sigma \gamma_\mu \gamma_\sigma \gamma^+ - 2 m_\mu \epsilon\cdot p \gamma^+ \leadsto - 2 m_\mu \epsilon\cdot p \gamma^+ \end{align*}

  12. $\epsilon^\mu q^\nu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^-$. This also requires a bit more work: \begin{align*} \epsilon^\mu q^\nu p^\sigma \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- & = -\epsilon^\mu q^\nu \gamma_\mu\gamma_\nu \not{\! p} \gamma^- + 2 \epsilon^\mu q\cdot p \gamma_\mu \gamma^- \leadsto -m_\mu \epsilon^\mu q^\nu \gamma_\mu\gamma_\nu \gamma^+ \nonumber\\ & = m_\mu \epsilon^\mu \not{\! q} \gamma_\mu \gamma^+ - 2 m_\mu \epsilon \cdot q \gamma^+ \leadsto m_\mu \epsilon^\mu (\not{\! q} -\not{\! p}) \gamma_\mu \gamma^+ + m_\mu \epsilon^\mu \not{\! p} \gamma_\mu \gamma^+ \nonumber\\ & =m_\mu \epsilon^\mu p^\sigma \gamma_\sigma \gamma_\mu \gamma^+ = - m_\mu \epsilon \cdot \gamma \not{\! p} \gamma^+ + 2 m_\mu \epsilon\cdot p \gamma^+ \leadsto 2 m_\mu \epsilon\cdot p \gamma^+ \end{align*}

  13. $\epsilon^\sigma p^\nu p^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = - p^2 \epsilon \cdot \gamma + 2 \epsilon \cdot p \not{\! p} \gamma^- = 2 m_\mu \epsilon\cdot p \gamma^+$;

  14. $\epsilon^\sigma p^\nu q^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^-$. We find \begin{align*} \epsilon^\sigma p^\nu q^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- &= \epsilon^\sigma q^\mu \gamma_\mu \gamma_\sigma \not{\! p} \gamma^- = m_\mu \epsilon^\sigma q^\mu \gamma_\mu \gamma_\sigma \gamma^+ = m_\mu \not{\! q} \epsilon^\sigma \gamma_\sigma \gamma^+\nonumber\\ & = m_\mu (\not{\! q}-\not{\! p}) \epsilon^\sigma \gamma_\sigma \gamma^++ m_\mu \not{\! p} \epsilon^\sigma \gamma_\sigma \gamma^+ \leadsto -m_\mu p^\sigma \epsilon^\sigma \gamma^\sigma \gamma^\mu +2 m_\mu p\cdot \epsilon \gamma^+\nonumber\\ & = 2 m_\mu \epsilon\cdot p \gamma^+ \end{align*}

  15. $\epsilon^\sigma q^\nu p^\mu \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = \epsilon^\sigma q^\nu (\not{\! p} -\not{\! q} ) \gamma_\sigma \gamma_\nu \gamma^- - \epsilon^\sigma q^\nu \not{\! q} \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- \leadsto 0 $

  16. $g^{\mu\sigma} \epsilon_\rho p^\nu p^\rho \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^-$. Again we start from the general case: $ g^{\mu\sigma} \epsilon_\rho f^\nu g^\rho \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = 4 \epsilon_\rho f^\nu q^\rho \gamma_\nu\gamma^- = 4 \epsilon\cdot g \not{\! f} \gamma^-$. Thus $g^{\mu\sigma} \epsilon_\rho p^\nu p^\rho \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = 4 \epsilon\cdot p \not{\! p} \gamma^- = 4 m_\mu \epsilon\cdot p \gamma^+$;

  17. $g^{\mu\sigma} \epsilon_\rho p^\nu q^\rho \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- = 4 \epsilon\cdot q \not{\! p} \gamma^-\leadsto0$;

  18. $g^{\mu\sigma} \epsilon_\rho q^\nu p^\rho \gamma_\mu \gamma_\sigma \gamma_\nu \gamma^- =4 \epsilon\cdot p \not{\! q} \gamma^- = 4 \epsilon\cdot p (\not{\! q}-\not{\! p}) \gamma^- + 4 \epsilon\cdot p \not{\! p} \gamma^- = 4 m_\mu \epsilon\cdot p \gamma^+$.

These results are summarised below.

$\epsilon^\nu p^\mu p^\sigma \leadsto 0; \;\; \epsilon^\nu p^\mu q^\sigma \leadsto 0; \;\;\epsilon^\nu q^\mu p^\sigma \leadsto 0;$

$g^{\mu\nu}\epsilon_\rho p^\rho p^\sigma \leadsto -2; \;\;g^{\mu\nu}\epsilon_\rho p^\rho q^\sigma \leadsto -2; \;\;g^{\mu\nu}\epsilon_\rho p^\rho q^\sigma \leadsto -2; $

$g^{\nu \sigma}\epsilon_\rho p^\rho p^\mu \leadsto +4; \;\; g^{\nu \sigma}\epsilon_\rho p^\rho q^\mu \leadsto +4; \;\;g^{\nu \sigma}\epsilon_\rho q^\rho p^\mu \leadsto 0$

$\epsilon^\mu p^\nu p^\sigma \leadsto 0; \;\;\epsilon^\mu p^\nu q^\sigma \leadsto -2; \;\; \epsilon^\mu q^\nu p^\sigma \leadsto +2$

$\epsilon^\sigma p^\nu p^\mu \leadsto +2; \;\;\epsilon^\sigma p^\nu q^\mu \leadsto +2; \;\; \;\;\epsilon^\sigma q^\nu p^\mu \leadsto 0$

$g^{\mu \sigma}\epsilon_\rho p^\nu p^\rho \leadsto +4;\;\; g^{\mu \sigma}\epsilon_\rho p^\nu q^\rho \leadsto 0;\;\; g^{\mu \sigma}\epsilon_\rho q^\nu p^\rho \leadsto +4$

Let us now go back to the expression [Eq 1]. We split the calculation in three depending on whether we take the first, second or third term between the square brackets. Let us start with the first term, we thus have \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_1 =& 2 \epsilon_\rho \big(\ell- \alpha_1 p+ (1-\alpha_2) q \big)^\nu \times (\ell - \alpha_1p -\alpha_2 q)^\rho \big(\ell -\alpha_1 p +(1-\alpha_2)q \big)^\mu \nonumber\\ &\times \big(\ell+(1-\alpha_1) p -\alpha_2 q\big)^\sigma \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \end{align} Extracting the $\ell^2$ terms we find \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_1^{\ell^2} = 2\epsilon_\rho \times \Big[& \ell^\nu \ell^\rho \big( -\alpha_1 p + (1-\alpha_2)q \big)^\mu \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma\nonumber\\ + & \ell^\nu \ell^\mu \big( -\alpha_1 p -\alpha_2 q \big)^\rho \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & \ell^\nu \ell^\sigma \big( -\alpha_1 p -\alpha_2 q \big)^\rho \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & \ell^\rho \ell^\mu \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\nu \big((1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & \ell^\rho \ell^\sigma \big( -\alpha_1 p +(1-\alpha_2) q \big)^\nu \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & \ell^\mu \ell^\sigma \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\nu \big( -\alpha_1 p -\alpha_2 q \big)^\rho \Big] \nonumber\\ &\times \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \end{align}
Using $\ell^{\alpha\beta} = (\ell^2/4) g^{\alpha \beta}$ this gives \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_1^{\ell^2} = \frac{\ell^2}{2} \times \Big[& \epsilon^\nu \big( -\alpha_1 p + (1-\alpha_2)q \big)^\mu \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma\nonumber\\ + & g^{\mu\nu} \epsilon_\rho \big( -\alpha_1 p -\alpha_2 q \big)^\rho \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & g^{\nu\sigma} \epsilon_\rho \big( -\alpha_1 p -\alpha_2 q \big)^\rho \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & \epsilon^\mu \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\nu \big((1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & \epsilon^\sigma \big( -\alpha_1 p +(1-\alpha_2) q \big)^\nu \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & g^{\mu\sigma} \epsilon_\rho \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\nu \big( -\alpha_1 p -\alpha_2 q \big)^\rho \Big]\nonumber\\ &\times \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \label{eq:ssbqjjjje} \end{align}
It is now just a matter of expanding this and picking out all the non-zero combinations from the above table, multiplied by the appropriate factors in each combination: \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_1^{\ell^2}& = m_\mu \ell^2 \epsilon\cdot p \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p) \nonumber\\ &\times \Big[ 0 + 0 + 0 -(-\alpha_1)(1-\alpha_1) - (-\alpha_1)(-\alpha_2) + 0 \nonumber\\ &+ 2(-\alpha_1)(-\alpha_1) +2 (-\alpha_1)(1-\alpha_2) + 0 + 0- (-\alpha_1)(-\alpha_2) + (1-\alpha_2)(1-\alpha_1) \nonumber\\ & + (-\alpha_1)(-\alpha_1) + (-\alpha_1)(1-\alpha_2) +0 +2(-\alpha_1)(-\alpha_1) + 0 + 2(1-\alpha_2)(-\alpha_1) \Big] \nonumber\\ &= m_\mu \ell^2 \epsilon\cdot p \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p) (1-\alpha_1-\alpha_2) \big(1+\alpha_1(1-2-1-2)\big) \end{align} and so our final result for the quadratic part of this term is \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_1^{\ell^2}& = m_\mu \ell^2 (1-\alpha_1-\alpha_2) (1-4\alpha_1) \epsilon\cdot p \,\bar{u}_e (p-q) (1+\gamma^5) u_\mu(p) \end{align}

Let us now turn to the second term between brackets in [Eq 1]. It is \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_2=& - \epsilon_\lambda \big(\ell- \alpha_1 p+ (1-\alpha_2) q \big)^\lambda \times \big(\ell -\alpha_1p +(2-\alpha_2) q\big)^\nu \big(\ell -\alpha_1 p +(1-\alpha_2)q \big)^\mu \nonumber\\ &\times \big(\ell+(1-\alpha_1) p -\alpha_2 q\big)^\sigma \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \end{align} We extract the terms in $\ell^2$: \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_2^{\ell^2} = -\epsilon_\lambda \times \Big[& \ell^\lambda \ell^\nu \big( -\alpha_1 p + (1-\alpha_2)q \big)^\mu \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma\nonumber\\ + & \ell^\lambda \ell^\mu \big( -\alpha_1 p +(2-\alpha_2) q \big)^\nu \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & \ell^\lambda \ell^\sigma \big( -\alpha_1 p +(2-\alpha_2) q \big)^\nu \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & \ell^\nu \ell^\mu \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\lambda \big((1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & \ell^\nu \ell^\sigma \big( -\alpha_1 p +(1-\alpha_2) q \big)^\lambda \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & \ell^\mu \ell^\sigma \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\lambda \big( -\alpha_1 p+(2 -\alpha_2) q \big)^\nu \Big] \nonumber\\ &\times \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \end{align}
Using $\ell^{\alpha\beta} = (\ell^2/4) g^{\alpha \beta}$ this gives \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_2^{\ell^2} = -\frac{\ell^2}{4} \times \Big[& \epsilon^\nu\big( -\alpha_1 p + (1-\alpha_2)q \big)^\mu \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma\nonumber\\ + &\epsilon^\mu\big( -\alpha_1 p +(2-\alpha_2) q \big)^\nu \big( (1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & \epsilon^\sigma \big( -\alpha_1 p +(2-\alpha_2) q \big)^\nu \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & g^{\mu\nu}\epsilon_\lambda \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\lambda \big((1-\alpha_1) p -\alpha_2 q \big)^\sigma \nonumber\\ + & g^{\nu\sigma}\epsilon_\lambda \big( -\alpha_1 p +(1-\alpha_2) q \big)^\lambda \big( -\alpha_1 p +(1-\alpha_2) q \big)^\mu \nonumber\\ + & g^{\mu\sigma}\epsilon_\lambda \big( -\alpha_1 p+(1 -\alpha_2) q \big)^\lambda \big( -\alpha_1 p+(2 -\alpha_2) q \big)^\nu \Big] \nonumber\\ &\times \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \label{eq:ssbqihhh3492} \end{align}
We pick up the appropriate pieces from the table \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_2^{\ell^2}& = -\frac{m_\mu \ell^2}{2} \epsilon\cdot p \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p) \nonumber\\ & \times \Big[ 0+0+0 +0 - (-\alpha_1)(-\alpha_2) + (2-\alpha_2)(1-\alpha_1)\nonumber\\ & +(-\alpha_1)(-\alpha_1) + (-\alpha_1)(1-\alpha_2) + 0 -(-\alpha_1)(1-\alpha_1) - (-\alpha_1)(-\alpha_2 )+0 \nonumber\\ & 2(-\alpha_1)(-\alpha_1) + 2(-\alpha_1)(1-\alpha_2) +0 +2 (-\alpha_1)(-\alpha_1) + 2(-\alpha_1)(2-\alpha_2) +0 \Big]\nonumber\\ &= -\frac{m_\mu \ell^2}{2} \epsilon\cdot p \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p) \nonumber\\ &\times \Big[ 2(1-\alpha_1-\alpha_2) + \alpha_2 -\alpha_1 (1-\alpha_1-\alpha_2) + \alpha_1 (1-\alpha_1-\alpha_2) \nonumber\\ & \qquad -2 \alpha_1 (1-\alpha_1-\alpha_2) -2 \alpha_1 (1-\alpha_1-\alpha_2) -2\alpha_1 (1-\alpha_1-\alpha_2) -2 \alpha_1 \Big] \nonumber\\ &= -\frac{m_\mu \ell^2}{2} \epsilon\cdot p \bar{u}_e (p-q) (1+\gamma^5) u_\mu(p) \Big[(1-\alpha_1 -\alpha_2)(2 - 2\alpha_1 -2 \alpha_1 ) + \alpha_2 - 2 \alpha_1\Big] \end{align} and so our final result for the quadratic part of this term is \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_2^{\ell^2}& = -m_\mu \ell^2 \epsilon\cdot p \big[ (1-\alpha_1-\alpha_2)(1-2\alpha_1) - \alpha_1 +\frac{\alpha_2}{2}\big] \nonumber\\ &\bar{u}_e (p-q)(1+\gamma^5) u_\mu(p) \end{align}

The third term of [Eq 1] is \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_3=& - \epsilon^\nu \big(\ell- \alpha_1 p+ (1-\alpha_2) q \big)^\lambda \big(\ell -\alpha_1p -(1+\alpha_2) q \big)_\lambda \big(\ell -\alpha_1 p +(1-\alpha_2)q \big)^\mu \nonumber\\ &\times \big(\ell+(1-\alpha_1) p -\alpha_2 q\big)^\sigma \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \end{align} The terms quadratic in $\ell^2$ all have a $\epsilon^\nu$ coefficient, but by our table all the combinations $\epsilon^\nu f^\mu g^\sigma$ vanish and so we simply find that \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)_3^{\ell^2}& = 0 \end{align}

We are now in a position to extract the full quadratic factor of $\boldsymbol{S_3}$ \begin{align} \Big( &M^2 \boldsymbol{S_3}\Big)^{\ell^2} = m_\mu \ell^2 \epsilon\cdot p \bar{u}_e (p-q)(1+\gamma^5) u_\mu(p) \nonumber\\ &\times \big[(1-\alpha_1-\alpha_2)(1-4\alpha_1) - (1-\alpha_1-\alpha_2)(1-2\alpha_1) + \alpha_1 -\frac{\alpha_2}{2} \big] \end{align} And this gives our final result \begin{align} \Big(M^2 \boldsymbol{S_3}\Big)^{\ell^2}& = m_\mu \ell^2 \big[-\alpha_1+2 \alpha_1^2 + 2\alpha_1\alpha_2 -\frac{\alpha_2}{2} \big] \epsilon\cdot p\, \bar{u}_e (p-q)(1+\gamma^5) u_\mu(p) \end{align}

Note that this result differs from Cheng & Li in two ways. First there is an overall sign. Next the sign of the $\alpha_1$ term is different.

Now that we have worked out the terms of $\boldsymbol{S_3}$ quadratic in $\ell^2$, let us turn to the terms independent of $\ell$. These are \begin{align} \Big( M^2 \boldsymbol{S_3}\Big)^{\ell^0}=& \big(- \alpha_1 p+ (1-\alpha_2) q \big)^\lambda \times \big[ 2 ( - \alpha_1p -\alpha_2 q)\cdot \epsilon\, \delta^\nu_\lambda - \big(-\alpha_1p+ (2-\alpha_2) q\big)^\nu \epsilon_\lambda\nonumber\\ & - \big( -\alpha_1p -(1+\alpha_2) q \big)_\lambda \epsilon^\nu \big] \times \big( -\alpha_1 p +(1-\alpha_2)q \big)^\mu \nonumber\\ &\times \big(+(1-\alpha_1) p -\alpha_2 q\big)^\sigma \bar{u}_e (p-q) \gamma_\mu \gamma_\sigma \gamma_\nu (1-\gamma^5) u_\mu(p) \end{align} Keeping in mind that our end result must be proportional to $ \epsilon\cdot p\, \bar{u}_e (p-q)(1+\gamma^5) u_\mu(p)$, we can go through the same procedure as for the quadratic terms. We have a product of three factors linear in $p$ and $q$ and one factor linear in $p^\alpha \epsilon^\beta$ and $q^\alpha \epsilon^\beta$. One of the $p$'s will have to combine with the polarisation to give the $\epsilon \cdot p$. Another $p$ can be used to extract the muon mass $m_\mu$ as we did for the quadratic term. We are then left with the product of two terms linear in $p$ and $q$. These need to have contracted indices, so can only be a function of $p^2, q^2$ and $p\cdot q$. But $q^2=0$ and $p^2=2p\cdot q = m_\mu^2$, and so whatever the end result is we pick up another $m_\mu^2$ and hence our and result will be \begin{align} \Big( M^2 \boldsymbol{S_3}\Big)^{\ell^0}=& F_3(\alpha_1,\alpha_2) m_\mu^3 \epsilon\cdot p\, \bar{u}_e (p-q)(1+\gamma^5) u_\mu(p) \end{align} with $F_3(\alpha_1,\alpha_2)$ some complicated function. The $\ell^2$ terms will give a contribution in the loop integral of order $m_\mu/M^2$. The $\ell^0$ terms will give a contribution of an order $m_\mu^2/M^2$ higher and so these terms can be ignored as $m_\mu \ll M$.

  • $\begingroup$ I actually think this answer may be too long to be useful...? Or, at least, I think the first paragraph constitutes quite a reasonable answer by itself. If you really do want to keep the entire calculation in this post, I'd at least suggest explicitly noting that everything past the first paragraph is optional for all but the most interested readers and only fills in the details of what you explain in the first paragraph. $\endgroup$
    – David Z
    Jun 25, 2019 at 9:38
  • $\begingroup$ Point taken, but the usefulness of the answer is that there are a lot op typos in Cheng and Li and this answer at least enables one to check all the details. But I will edit and add a sentence at the beginning as you suggest. $\endgroup$ Jun 25, 2019 at 9:55

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