# Full calculation of B meson mixing amplitude

I am trying to calculate B mixing in the Standard Model (in preparation to go beyond the SM). I have no trouble doing the gamma matrix algebra etc. but the loop integral keeps tripping me up. In my calculation I have $$\int \frac{d^4 k}{(2\pi)^4} \frac{k^2}{(k^2-m_1^2) (k^2 - m_2^2) (k^2- M_W^2)^2}$$ I know about Feynman parametrization etc. but the result I get does not comply with what I find in the Literature. Unfortunately basically all calculations simply say "we calculate by standard methods" and have a function $$S(x_t) = \frac{4x_t - 11 x_t^2 + x_t^3}{4(1-x_t)^2} - \frac{3x_t^3 \ln x_t}{2(1-x_t)^3}$$ with $x_t = m_t^2 / M_W^2$ if $m_1 = m_2 = m_t$. This is not directly the result of evaluating the above integral though, since I have at least a factor of $1/M_W^2$ that is in the integral, but not included in S.

Where can I find a full calculation of the box diagram and what is the exact relation of the loop integrals to the functions $S$?

I do know about the general 1, 2, 3 and 4 point functions, generally called $A_0, B_0, B_1, \dots$. The $S$ is different from $D(0, 0, 0, m_1, m_2, M_W, M_W)$ though!

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It is a standard integral calculable via Feynman parametrization etc. Note that the integral at the top is dimensionless - energy to the sixth over energy to the sixth - and so is $S$. So if you have some extra dimensionful factors, your result is even dimensionally wrong and it should be trivial to find where you introduced the mistake. – Luboš Motl Jul 25 '13 at 14:34
@LubošMotl: you are mistaken. The integral has dimension $E^{-2}$, as the W propagator is squared! – Neuneck Jul 26 '13 at 8:39

Alternatively one could use the correct W propagator $$\frac{-i\big(g^{\mu \nu} - 1 \frac{k^\mu k^\nu}{k^2}\big)}{k^2 - M_W^2 + i \epsilon}$$ instead of the abbreviated one I used $$\frac{-i g^{\mu \nu}}{k^2 - M_W^2 - i \epsilon}$$ in order to have less diagrams but a more difficult integral to solve.
No, the actual goldstone bosons. (Those who are not physical degrees of freedom. Their Feynman rule includes a $k^\mu$ in the coupling to the fermions, so that they reproduce the $k^\mu k^\nu$ by which my two propagators differ. – Neuneck Jul 26 '13 at 14:03
Yeah that's what I meant (the goldstones which make 3/4 of the Higgs - should have been more specific there). So you're noticing that the form of the propagators is gauge dependent. You have a choice of unitary gauge with the physical projection operator for massive vector particles $g^{\mu\nu}-k^\mu k^\nu /k^2$ or some other gauge where the pieces come from different places. Just have to be extra careful to be consistent working out the Feynman rules for whatever gauge you're in. – Michael Brown Jul 26 '13 at 14:13