# Why does the term $i\epsilon^{\alpha\mu\beta\nu}p_{\alpha}q_{\beta}$ vanish in the squared amplitude of the top decay?

I would like to calculate the squared of the amplitude of a top-decay $$t-> Wb$$. I know already the solution, but I can't get there: I get stuck with the Levi-Civita-Tensor on my way. Can you help me, please?

The amplitude is

$$M=\frac{g}{\sqrt{2}}\overline{u}(q)\gamma^{\mu}(\frac{1-\gamma^5}{2})u(p)\epsilon_{\mu}^*(k)$$

with q=Four-vector of bottom quark;
p=four-vector of top quark;
k=four-vector of W-boson

When I square it I get (mass of bottom quark neglected):

$$\frac{1}{2}\sum_{spins}|M|^2=\frac{g^2}{4}[\overline{u}(q)\gamma^{\mu}(\frac{1-\gamma^5}{2})u(p)\epsilon_{\mu}^*(k)][\overline{u}(q)\gamma^{\nu}(\frac{1-\gamma^5}{2})u(p)\epsilon_{\nu}^*(k)]^*$$

$$=\frac{g^2}{4}Tr[(p_{\alpha}\gamma^{\alpha}+m_t)\gamma^{\mu}(\frac{1-\gamma^5}{2})q_{\beta}\gamma^{\beta}\gamma^{\nu}(\frac{1-\gamma^5}{2})](-g_{\mu\nu}+\frac{k_{\mu}k_{\nu}}{m_W^2})$$

$$=\frac{g^2}{8}p_{\alpha}q_{\beta}Tr[\gamma^{\alpha}\gamma^{\mu}\gamma^{\beta}\gamma^{\nu}-\gamma^{\alpha}\gamma^{\mu}\gamma^{\beta}\gamma^{\nu}\gamma^5)(-g_{\mu\nu}+\frac{k_{\mu}k_{\nu}}{m_W^2})$$

$$=\frac{4g^2}{8}[p^{\mu}q^{\nu}+p^{\nu}q^{\mu}-(p \cdot q)g^{\mu\nu}+i\epsilon^{\alpha\mu\beta\nu}p_{\alpha}q_{\beta})(-g_{\mu\nu}+\frac{k_{\mu}k_{\nu}}{m_W^2})$$

Now the book says:

$$\frac{1}{2}\sum_{spins}|M|^2=\frac{4g^2}{8}[p^{\mu}q^{\nu}+p^{\nu}q^{\mu}-(p \cdot q)g^{\mu\nu})(-g_{\mu\nu}+\frac{k_{\mu}k_{\nu}}{m_W^2})$$

Why does the term $$i\epsilon^{\alpha\mu\beta\nu}p_{\alpha}q_{\beta}$$ vanish? I just can't find any answer.

In your expression, you have a term of the form $$I = \epsilon^{\alpha \mu \beta \nu} S_{\mu\nu}$$ where $$S_{\mu\nu} = - g_{\mu\nu} + k_\mu k_\nu / m_W^2$$ is a symmetric tensor. The contraction of an antisymmetric tensor with a symmetric one is always zero, because $$I = \epsilon^{\alpha \nu \beta \mu} S_{\nu\mu} = - \epsilon^{\alpha \mu \beta \nu} S_{\mu\nu} = - I$$ where in the first step we renamed the indices $$\mu \leftrightarrow \nu$$, and in the second step we used the symmetry and antisymmetry properties.