# Computing box diagrams with non-vanishing external momenta

I'm trying to explicitly compute the following box diagram in the Feynman-t'Hooft gauge: If I neglect the impulsion of the $s$ quark, then the final amplitude is given by

$$\mathcal{A} \propto [\bar{s}(0) \gamma_\alpha \gamma_\delta \gamma_\mu P_L b(p_b)] \cdot [\bar{u}(p_1) \gamma^\mu \gamma^\Delta \gamma^\alpha P_L v(p_2)]\cdot I^\delta_\Delta,$$ where

$$I^{\delta\Delta}=\int \frac{\mathrm{d}^4k}{(2\pi)^4} \frac{k^\delta(k-p_2)^\Delta}{[k^2-m_t^2][(k-p_b)^2-m_W^2][k^2-m_W^2][(k-p_2)^2-m_W^2]}.$$

If I consider vanishing external momenta $p_b$ and $p_2$, then it is easy to compute this integral and express the final amplitude in terms of $m_t$ and $m_W$. However, when I carry out this computation in the general case, I don't know how to simplify the final amplitude. More precisely, I find some expressions like

$$\mathcal{A}\propto [\bar{s}(0) \gamma_\mu p_2^\mu P_L b(p_b)] \cdot [\bar{u}(p_1) P_L v(p_2)]+\dots$$ and I don't know how to get rid of the impulsion $p_2$ in the left part of the r.h.s of the equation. I can't use Dirac's equation, because $p_2$ is in the "wrong" current and the conservation of 4-momenta is completely useless, because this would just replace $p_2$ by $p_1$.

Could you give some hints about how to simplify this expression? Some references about the effect of external momenta in box diagrams might also be helpful.

• Maybe this ref could be useful. – Trimok Sep 27 '14 at 12:10
• @Trimok Thank you for the reference! It seems that they have neglected the problematic terms when they assumed that $p_1$ and $p_4$ are small (eq. 5a). – Melquíades Sep 29 '14 at 19:58

Anyway, if you perform the standard procedure for calculating this integral you will find an expression which is a function of all the possible external momenta and the metric tensor as well. Then you can carry the limit $p_s \to 0$ you will find the formula which are you looking for.
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