I am trying to calculate the decay rate of the $W^-$ boson to a charged lepton and the corresponding antineutrino.

I denote the four momentum of the $W$ boson with $q = (M_W, \vec{0})$. The momenta of the outgoing leptons are given by $p_1 = (M_W/2, 0, 0, M_W/2)$ and $p_2 = (M_W/2, 0, 0, -M_W/2)$. For simplicity, we assume that the leptons are massless.

We have the following matrix element: $$ \mathcal{M} = \bar{u}(p_1) \left(-i \frac{g_w}{\sqrt{2}} \gamma^\mu \frac{1-\gamma^5}{2}\right)v(p_2)\epsilon_\mu(q)\,.$$ Here, $\bar{u}$ and $v$ denote the spinors and $\epsilon_\mu$ is the polarisation vector of the real $W$ boson.

Now, we derive the matrix element squared, averaged over the initial polarisations and summed over the final state fermion spins. We also use that $P_L P_L = P_L$ with $P_L = (1-\gamma^5)/2$ as the projector for left-handed chirality. \begin{align} |\mathcal{M}|^2 &= \frac{g_w^2}{2}\frac{1}{3} \sum_\mathrm{pol} \epsilon_\mu(q) \epsilon^*_\nu(q) \bar{u}(p_1) \sum_\mathrm{spins} \bar{u}(p_1) \gamma^\mu \frac{1-\gamma^5}{2} v(p2) \times \bar{v}(p_2) \gamma^\nu u(p1)\\ &= \frac{g_w^2}{12} \left( -g_{\mu \nu} + \frac{q_\mu q_\nu}{M_W^2}\right)\mathrm{Tr}[\displaystyle{\not}{p_1}\gamma^\mu\displaystyle{\not}{p_2}\gamma^\nu] \end{align} Where we made use of Casimir's trick and executed the sum over the three physical polarisation states of a massive spin-1 gauge boson. Additionally, we made use of the fact that any terms proportional to $\gamma^5$ will vanish as the trace includes an epsilon symbol, which yields zero when contracted with the symmetric polarisation sum.

Up until now, I understand what happened. However, I struggle to evaluate this expression.

First question: Is it allowed to move the metric tensor and the four-vectors inside the trace? If not, why not? This would greatly simplify the calculation. I tried this and found for the first term: $$- g_{\mu \nu} \mathrm{Tr}[\displaystyle{\not}{p_1}\gamma^\mu\displaystyle{\not}{p_2}\gamma^\nu] = - \mathrm{Tr}[\displaystyle{\not}{p_1}\gamma^\mu\displaystyle{\not}{p_2}\gamma_\mu] = - \mathrm{Tr}[4 p_1\cdotp_2 I_4] = - 16 p_1 \cdot p_2\,.$$ However, this does not look right as it conflicts with the direct calculation using the identity $\mathrm{Tr}[\displaystyle{\not}{a}\gamma^\mu\displaystyle{\not}{b}\gamma^\nu] = 4(a^\mu b^\nu + a^\nu b^\mu - g^{\mu \nu} a \cdot b)$, giving $$ - g_{\mu \nu} \mathrm{Tr}[\displaystyle{\not}{p_1}\gamma^\mu\displaystyle{\not}{p_2}\gamma^\nu] = - g_{\mu \nu}\cdot4(p_1^\mu p_2^\nu + p_1^\nu p_2^\mu - g^{\mu \nu} p_1\cdotp_2) = 8p_1\cdotp_2\,.$$ Can you please help me understand what I am doing wrong/inconsistently?

Concerning the second term, I am also having problems. Let us try to evaluate it using the same identity as before and attach the two $q$ four-vectors: $$ q_\mu q_\nu \cdot 4(p_1^\mu p_2^\nu + p_1^\nu p_2^\mu - g^{\mu \nu} p_1\cdotp_2) = 4[2(p_1\cdot q)(p_2\cdot q) - q^2(p_1\cdot p_2)]\,.$$ I calculated the four-vector products explicitly and obtained: \begin{align} q^2 &= M_W^2\\ q \cdot p_1 &= q \cdot p_2 = M_W^2/2\\ p_1 \cdot p_2 &= M_W/2 \cdot M_W/2 - M_W/2 \cdot (-M_W/2) = M_W^2/2 \end{align} Plugging this in gives $$4[2(p_1\cdot q)(p_2\cdot q) - q^2(p_1\cdot p_2)] = 4[2 \cdot M_W^2/2 \cdot M_W^2/2 - M_W^2 \cdot M_W^2/2] = 4M_W^4 [1/2 - 1/2] = 0\,.$$ This is really surprising to me, I would not have expected this term to vanish as this was also the not the case in similar calculations in textbooks like Peskin-Schroeder. Can you help me out? What did I do wrong?


1 Answer 1


Hi and welcome to PSE.

  1. Regarding your first question, yes, it is totally ok to move the metric tensor and the four-vectors inside the trace, since the trace is a trace with repsect to the $\gamma$ matrices, with the former tensor and four-vectors being "scalar" quantities with respect to these matrices. If you want, you can contract the Lorentz indices and substitute the $\gamma$ matrices and see it yourself. In your attempt, you first try to contract the Lorentz indices and then take the trance. Your attempt should look like something of the form $$- g_{\mu \nu} \mathrm{Tr}[\displaystyle{\not}{p_1}\gamma^\mu\displaystyle{\not}{p_2}\gamma^\nu] = -\mathrm{Tr}[\displaystyle{\not}{p_1}\gamma^\mu\displaystyle{\not}{p_2}\gamma_\mu] = \mathrm{Tr}[\displaystyle{\not}{p_1}2\displaystyle{\not}{p_2}] = 2\mathrm{Tr}[\displaystyle{\not}{p_1}\displaystyle{\not}{p_2}]=2\times4(p_1\cdot p_2) $$ which is the desired result. Look at Gamma matrices identities for the identities used here.

  2. Regarding your second question, I personally do not see anything wrong, I do not know if I have taken a careful look. I would suggest checking your conventions from your textbook and if they are what you are using, then you are correct.

  • 1
    $\begingroup$ Indeed, I was wrong to assume that gamma matrices and slashed four-vectors commute. After all, another gamma matrix is hidden in the slashed vector! I still cannot see a mistake in the derivation of the second term and still believe that it is equal to zero. That way, I can obtain $|\mathcal{M}|^2 = \frac{g_w^2}{12}(8p_1 \cdot p_2) = \frac{g_w^2 M_W^2}{3}$, which is the result from the literature for the $W$ boson decay to two massless fermions. Thank you so much! $\endgroup$
    – JLS
    Commented Aug 3, 2023 at 15:00

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