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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
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@LubošMotl: you are mistaken. The integral has dimension $E^{-2}$, as the W propagator is squared! –  Neuneck Jul 26 '13 at 8:39

1 Answer 1

It was pointed out to me by my advisor, that one has to include the appropriate Feynman diagrams with Goldstone bosons too in order to arrive at the proper result.

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.

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Just curious: what Goldstone bosons? I presume you mean the Higgs? It would be interesting to see this worked out if you ever get the chance to write it up. I can ask nicely. :) –  Michael Brown Jul 26 '13 at 8:57
    
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
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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

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