# Matrix element squared for up + antidown -> photon + W boson? [closed]

I am in trouble… I am a master student (first year) in theoretical particle physics and I have been working for several months on the calculation of the total cross-section at leading order for the process:

$$u + \overline{d} \rightarrow W^{+} + \gamma$$

I performed the whole calculation by hand and a couple of weeks ago I obtained my final answer. However, my professor has actually proven that one term in my total matrix element squared cannot be correct. After spending weeks on checking this term, I really cannot find the mistake. My question is whether someone would like to check my calculation or do this calculation by computer software e.g. Mathematica or Form. (I do not have much experience with calculating such quantities by a computer, neither do I have time to figure out how that could be done. That’s why I chose to do everything by hand.)

The problem is as follows:

There are three tree level Feynman diagrams for this process: $M_1$, $M_2$ and $M_3$. In order to calculate the cross-section, one first needs to calculate the total matrix element squared: $M_{tot}^2$. I performed this calculation piece by piece, i.e. I calculated $M_1 \cdot M_1^*$, $M_2 \cdot M_2^*$, $M_3 \cdot M_3^*$, $M_1 \cdot M_2^*$, $M_1 \cdot M_3^*$ and $M_2 \cdot M_3^*$. (All by hand, which was rather tedious…)

According to my professor all the terms of my $M_{tot}^2$ seem reasonable and could, in principle, be correct. Except for one term! $M_3 \cdot M_3^*$, namely, has a different energy behavior (my third Feynman diagram is the one with the $W^+$ propagator). All the other terms ($M_1 \cdot M_1^*$, $M_2 \cdot M_2^*$, $M_1 \cdot M_2^*$, $M_1 \cdot M_3^*$ and $M_2 \cdot M_3^*$) approach a constant as $E \to \infty$, whereas $M_3 \cdot M_3^*$ contains a term that goes with $E^2$ for large $E$. My professor has pointed out (and also proven) that this energy behavior cannot be correct and that $M_3 \cdot M_3^*$ should also approach a constant for $E \to \infty$. (This by the way implies that the cross-section must approach zero for $E \to \infty$.)

I defined the momenta as follows:

\begin{align*} p_1 = \begin{pmatrix} E_1 \\ 0 \\ 0 \\ E_1 \end{pmatrix} \,, \qquad p_2 = \begin{pmatrix} E_1 \\ 0 \\ 0 \\ -E_1 \end{pmatrix} \end{align*}

and for the outgoing particles:

\begin{align*} p_3 = \begin{pmatrix} E_3 \\ E_3 \sin\theta \cos\phi \\ E_3 \sin\theta \sin\phi \\ E_3 \cos\theta \end{pmatrix} \,, \qquad p_4 = \begin{pmatrix} \sqrt{E_3^2 + m_{\mathrm{w}}^2} \\ -E_3 \sin\theta \cos\phi \\ -E_3 \sin\theta \sin\phi \\ -E_3 \cos\theta \end{pmatrix} \,. \end{align*}

Some conventions: 1, 2, 3 and 4 correspond with $\overline{d}$, $u$, $\gamma$ and $W^+$ respectively; $E \equiv 2 \,E_1$ and the quark masses are neglected, i.e. $p_1^2 = p_2^2 = 0$.

The Feynman rules that I used for $M_3$:

\begin{align*} \mathrm{incoming \; \overline{d}:}& \quad \sqrt{\hbar} \,\overline{v_{(1)}} \\ \mathrm{incoming \; u:}& \quad \sqrt{\hbar} \,u_{(2)} \\ \mathrm{outgoing \; \gamma:}& \quad \sqrt{\hbar} \,\epsilon^{\mu\,*}_{(3)} \\ \mathrm{outgoing \; W^+:}& \quad \sqrt{\hbar} \,\epsilon^{\nu\,*}_{(4)} \\ \mathrm{propagator \; W^+:}& \quad -\frac{i \hbar \left( g^{\alpha\beta} - q^\alpha q^\beta / m_{\mathrm{w}}^2 \right)}{q^2 - m_{\mathrm{w}}^2} \\ \mathrm{vertex \; u \,\overline{d} \,W^+:}& \quad \frac{i g_{\mathrm{w}}}{\hbar} \left( 1+\gamma^5 \right) \gamma_\alpha \\ \mathrm{vertex \; W^+ \,W^+ \,\gamma:}& \quad \frac{iQ_{\mathrm{w}}}{\hbar} \left( g_{\nu\beta} (q + p_4)_\mu + g_{\mu\nu} (p_3 - p_4)_\beta - g_{\beta\mu} (q + p_3)_\nu \right) \end{align*}

Thus:

$$\mathscr{M}_3 = \,\frac{i \hbar g_{\mathrm{w}} Q_{\mathrm{w}}}{(p_1 + p_2)^2 - m_{\mathrm{w}}^2} \,\overline{v_{(1)}} \left( 1+\gamma^5 \right) \gamma_\alpha \left[ g^{\alpha\beta} - \frac{(p_1 + p_2)^\alpha (p_1 + p_2)^\beta}{m_{\mathrm{w}}^2} \right] \epsilon^{\nu\,*}_{(4)} \times \left[ g_{\nu\beta} (2 \,p_4 + p_3)_\mu + g_{\mu\nu} (p_3 - p_4)_\beta - g_{\beta\mu} (2 \,p_3 + p_4)_\nu \right] \epsilon^{\mu\,*}_{(3)} \,u_{(2)}$$

I obtained for $\langle \mathscr{M}_3 \mathscr{M}_3^* \rangle$:

$$\langle \mathscr{M}_3 \mathscr{M}_3^* \rangle = \,\frac{1}{36} \sum_{S,C} \mathscr{M}_3 \mathscr{M}_3^* = \,\frac{2 \,\hbar^2 g_{\mathrm{w}}^2 Q_{\mathrm{w}}^2}{9} \left[ \frac{E^4 \left( E^2 - 5 \,m_{\mathrm{w}}^2 \right)}{m_{\mathrm{w}}^2 \left( E^2 - m_{\mathrm{w}}^2 \right) ^2} + \sin^2\theta \right] \,,$$

which cannot be correct, unfortunately... A very important thing is that I have chosen to work in a specific gauge! When you sum over photon polarization vectors, some gauge vector, r, enters the calculation:

$$\sum_{s_3} \epsilon^\mu_{(3)} \,\epsilon^{\nu\,*}_{(3)} = -g^{\mu\nu} + \frac{p_3^\mu r^\nu + r^\mu p_3^\nu}{p_3 \cdot r} \,,$$

where $r$ (the gauge vector) is an arbitrary massless vector that is not parallel to $p_3$. By making the following choice

$$r = \begin{pmatrix} 0 \\ 1 \\ i \\ 0 \end{pmatrix} \,,$$

a lot of terms will drop out in the calculation. All inner products containing $r$ appear in one the following forms:

\begin{align*} \frac{p_1 \cdot r}{p_3 \cdot r} =& \,0 \,, \qquad \frac{p_1 \cdot r}{p_3 \cdot r} = 0 \,, \qquad \frac{p_3 \cdot r}{p_3 \cdot r} = 1 \,, \qquad \frac{p_4 \cdot r}{p_3 \cdot r} = -1 \,. \end{align*}

Hence by choosing for this specific gauge we can save ourselves a lot of time! Of course, once you have made this explicit choice for the gauge vector you will have to stick it! By the way, by choosing this gauge the effect of longitudinally polarized W bosons drops out!

For W bosons we have the following identity for a sum over polarization vectors:

\begin{align*} \sum_{s_4} \epsilon^\mu_{(4)} \,\epsilon^{\nu\,*}_{(4)} = -g^{\mu\nu} + \frac{p_4^\mu p_4^\nu}{m_{\mathrm{w}}^2} \end{align*}

Would someone be willing to either check/calculate $\langle \mathscr{M}_3 \mathscr{M}_3^* \rangle$? Perhaps someone has already a computer code for calculating such quantities... In principle I am only interested in the answer and not necessarily in the calculation. Again, it is very important that the calculation of $\langle \mathscr{M}_3 \mathscr{M}_3^* \rangle$ is performed in the gauge that I have chosen, as all the other partial matrix elements were also calculated in this specific gauge.

I realize that I ask a lot… In order to increase your motivation for this check/calculation, I award the person that provides me with a satisfying analytical expression for $\langle \mathscr{M}_3 \mathscr{M}_3^* \rangle$ with EUR 20, besides eternal glory, of course! :) Not being able to finish this calculation is really frustrating and therefore your help would be greatly appreciated. If you have any questions, please ask me.

Thanks!

-
 Hi TopQuark38, and welcome to Physics Stack Exchange! This is a site for conceptual questions, not really a place to find people to check your work or do it for you, especially not for financial compensation. If you can edit your question to ask about a particular concept that you suspect may be giving you trouble, I'll be happy to reopen it. – David Zaslavsky♦ Jun 14 '12 at 20:42 You are possibly finding a cross section that is growing too fast because you are using a massive propagator for the weak boson without the Higgs unitarization that comes from properly embedding the thing in the standard model. Try to do the calculation with a massless W boson and see if you get a reasonable result. I will vote to reopen, but you should rewrite the question to focus on the evaluation of the problematic term. Your professor is probably concerned about the Froissart bound. – Ron Maimon Jun 15 '12 at 2:03 One way to make progress is to calculate the thing without the photon--- this is just a standard weak decay matrix element, and it should reduce to the Fermi theory answer. I didn't go over your detailed calculation, however. – Ron Maimon Jun 15 '12 at 2:04

## closed as too localized by David Zaslavsky♦Jun 14 '12 at 20:37

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