Hi
I realize that this thread is very old, but it's new to me! Anyway, it seems to me that there are more questions here than answers (and some answers seemed to be misleading). So please allow me to first summarize the calculation in order to clarify the question, and then make a suggestion for the answer.
The Question:
Basically the question is how do the two box diagrams, which are equal and sum with a relative minus sign, not sum to zero? (The sum should actually give a function A(x_i,x_j) rather than zero.)
The calculation:
I've attached the relevant Feynman diagrams for the W boson box diagram which I'm going to discuss (see bottom of post). The reason I include it is because the original post linked to some diagrams which had a few mistakes.
Starting from the diagram on the left, and using the Feynman Gauge, it's easy to find,
<br />
\left(\frac{g}{\sqrt{2}}\right)^4<br />
\sum_{i,j=u,c,t}\lambda_i \lambda_j \int \frac{d^4 q}{(4\pi)^4}<br />
\,(\bar{d}\gamma^\mu P_L \frac{i(\not\! q+m_i)}{q^2-m_i^2}\gamma^\nu P_L s)\,<br />
(\bar{d}\gamma_\nu P_L \frac{i(\not\! q+m_j)}{q^2-m_j^2}\gamma_\mu P_L s)<br />
\left(\frac{-i}{q^2-M_W^2}\right)^2<br />
As the projection operators must match (i.e. P_L P_R =0), and recalling that P_L changes to P_R whenever it passes a \gamma^\mu, it is clear that the m_i and m_j terms are zero, leaving only the \not\! q terms;
<br />
\frac{g^4}{4}<br />
\sum_{i,j=u,c,t}\lambda_i \lambda_j \int \frac{d^4 q}{(4\pi)^4}<br />
\,\frac{q_\alpha q_\beta}{D}<br />
\,(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s)<br />
(\bar{d}\gamma_\nu \gamma^\beta\gamma_\mu P_L s)<br />
Here we have defined, \lambda_i=V^{}_{is}V^{*}_{id}, where V is the CKM matrix, and
<br />
\frac{1}{D}=<br />
\frac{1}{(q^2-M_W^2)^2}\frac{1}{q^2-m_i^2}\frac{1}{q^2-m_j^2}<br />
\,.<br />
The next step is to do the integration using,
<br />
I_{\alpha\beta}(i,j)\equiv \int d^4 q \, <br />
\frac{q_\alpha q_\beta}{(q^2-M_W^2)^2(q^2-m_i^2)(q^2-m_j^2)}<br />
= \frac{-i\pi^2}{4 M_W^2}A(x_i,\,x_j)\,g_{\alpha\beta}<br />
with x_i=m_i^2/M_W^2. The function A is defined by,
<br />
A(x_i,\,x_j) = \frac{J(x_i)-J(x_j)}{x_i-x_j} <br />
and
<br />
J(x_i)=\frac{1}{1-x_i}<br />
+ \frac{x_i^2 \,\ln x_i}{(1-x_i)^2}<br />
This is actually easy to prove, but I'll leave it out for now. Using this we now get
<br />
\frac{-i\,g^4\,\pi^2}{16 M_W^2(2\pi)^4}<br />
\sum_{i,j=u,c,t}\lambda_i \lambda_j <br />
\,A(x_i,\,x_j)<br />
\,(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s)<br />
\,(\bar{d}\gamma_\nu \gamma_\alpha\gamma_\mu P_L s)<br />
\,.<br />
Next we simplify the combination, (\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s)<br />
\,(\bar{d}\gamma_\nu \gamma_\alpha\gamma_\mu P_L s). All you need to do here is make use of the \gamma matrix
identity,
<br />
\gamma^\mu \gamma^\alpha \gamma^\nu<br />
= g^{\mu\alpha}\gamma^\nu + g^{\nu\alpha}\gamma^\mu <br />
- g^{\mu\nu}\gamma^\alpha - i\epsilon^{\mu\alpha\nu\beta}\gamma_5 \gamma_\beta<br />
to show that,
<br />
(\bar{d}\gamma^\mu \gamma^\alpha\gamma^\nu P_L s)<br />
\,(\bar{d}\gamma_\nu \gamma_\alpha\gamma_\mu P_L s)<br />
=<br />
4\,<br />
(\bar{d}\gamma^\mu P_L s)<br />
\,(\bar{d}\gamma_\mu P_L s)<br />
and then the result for the left diagram becomes,
<br />
-\frac{i\,g^4\,\pi^2}{4 M_W^2\,(2\pi)^4}<br />
\sum_{i,j=u,c,t}\lambda_i \lambda_j <br />
\,A(x_i,\,x_j)<br />
\,<br />
(\bar{d}\gamma^\mu P_L s)<br />
\,(\bar{d}\gamma_\mu P_L s)<br />
\,.<br />
Now if we do the same calculation for the box diagram on the right we actually get the same result. So how is it that we don't get zero when we add up the two diagrams?
My suggested answer:
The 4-quark operators which appear in the two diagrams aren't actually the same.
They connect different s-quarks to different d-quarks. In the left diagram it is the two initial(final) state quarks which are connected, while in the right diagram it is one from the initial state and one from the final state. In order to become the same 4-quark operator we need to use a Fierz rearrangement,
<br />
(\bar{d}_{1L} \gamma^\mu s_{2L})<br />
\,(\bar{d}_{3L} \gamma_\mu s_{4L})<br />
= -<br />
(\bar{d}_{1L} \gamma^\mu s_{4L})<br />
\,(\bar{d}_{3L} \gamma_\mu s_{2L})<br />
and hence introduce an extra minus sign, so that the total is just twice that for the first diagram given above.
Does anyone know if this is the correct solution to the problem?
