- #1
metricspace
- 2
- 0
- Homework Statement
- Ok
- Relevant Equations
- Ok
I am trying to compute the following loop integral:
$$
\require{cancel}
\displaystyle
\begin{align}
I= \int \frac{d^4k}{(2\pi)^4}\bar u(p')\frac{[k^2k^\mu - (k\cdot p)\cancel{k}\gamma^\mu -(k\cdot p')\gamma^\mu\cancel{k}]}{[k^2-M^2+i\epsilon][(k-p)^2-m^2+i\epsilon][(k-p')^2-m^2+i\epsilon]}u(p),
\end{align}
$$
with 2 different methods.
First Method: Rewrite the numerator and cancel the denominator. The numerator is
$$
\begin{align}
N^\mu &= k^2k^\mu - (k\cdot p)\cancel{k}\gamma^\mu -(k\cdot p')\gamma^\mu\cancel{k}
\nonumber \\
&= k^2k^\mu - \frac{1}{2}(k^2-[(k-p)^2-m^2])k_\nu \gamma^\nu\gamma^\mu -\frac{1}{2}(k^2-[(k-p')^2-m^2])k_\nu\gamma^\mu\gamma^\nu
\nonumber \\
&= \frac{1}{2}[(k-p)^2-m^2]k_\nu \gamma^\nu\gamma^\mu +\frac{1}{2}[(k-p')^2-m^2]k_\nu\gamma^\mu\gamma^\nu.
\end{align}
$$
Thus,
$$
\begin{align}
\require{cancel}
\displaystyle
I =& \frac{1}{2}\int\frac{d^4k}{(2\pi)^4}\bar u(p') \frac{1}{[k^2-M^2+i\epsilon]} \left( \frac{k_\nu}{[(k-p')^2-m^2+i\epsilon]}\gamma^\nu\gamma^\mu + \frac{k_\nu}{[(k-p)^2-m^2+i\epsilon]}\gamma^\mu\gamma^\nu \right) u(p)
\nonumber \\
&= \frac{1}{2}A\bar u(p')(p'_\nu\gamma^\nu\gamma^\mu + p_\nu\gamma^\mu\gamma^\nu) u(p)
\nonumber \\
&= Am \bar u(p') \gamma^\mu u(p),
\end{align}
$$
where ##A## is a scalar.
Second Method: Feynman Parametrization. The denominator becomes:
$$
\require{cancel}
\displaystyle
\begin{align}
[k^2-M^2+i\epsilon]^{-1}[(k-p)^2-m^2+i\epsilon]^{-1}[(k-p')^2-m^2+i\epsilon]^{-1}=
\int_0^1dx\int_0^{1-x}dy \frac{2}{[l^2-\Delta^2+i\epsilon]^3},
\end{align}
$$
where ##l^\mu=(k-xp-yp')^\mu##, ##\Delta^2=(x+y)^2m^2+(1-x-y)M^2-xyq^2##, and, ##q^\mu=(p'-p)^\mu##.
Next, we perform the shift in the numerator:
$$
\begin{align}
N^\mu = k^2k^\mu - (k\cdot p)\cancel{k}\gamma^\mu -(k\cdot p')\gamma^\mu\cancel{k} = A_1 \bar u(p') \gamma^\mu u(p) + A_2 \bar u(p') \frac{i\sigma^{\mu\nu}q_\nu}{2m} u(p),
\end{align}
$$
where ##A_1## and ##A_2## are scalars. This method clearly gives a different result.
The first method gives the correct result (the result should not contain a ##i\sigma^{\mu\nu}q_\nu/2m## term), so the problem must be related with the Feynman Parametrization somehow.
The denominator ##[(k-p)^2-m^2+i\epsilon]=[k^2-2(k\cdot p)+i\epsilon]## (and the other one with ##p'##) could be problematic when ##k\to 0##, however there are other terms in the numerator (which I did not included here) that give the correct result with the same denominator and using Feynman Parametrization, i.e., using the second method. On the other hand these same terms (which give the correct result) are UV finite but the one in question is not. Is the problem in the IR or UV or both?
I believe I didn't make any mistakes on the passages, so what am I missing?
$$
\require{cancel}
\displaystyle
\begin{align}
I= \int \frac{d^4k}{(2\pi)^4}\bar u(p')\frac{[k^2k^\mu - (k\cdot p)\cancel{k}\gamma^\mu -(k\cdot p')\gamma^\mu\cancel{k}]}{[k^2-M^2+i\epsilon][(k-p)^2-m^2+i\epsilon][(k-p')^2-m^2+i\epsilon]}u(p),
\end{align}
$$
with 2 different methods.
First Method: Rewrite the numerator and cancel the denominator. The numerator is
$$
\begin{align}
N^\mu &= k^2k^\mu - (k\cdot p)\cancel{k}\gamma^\mu -(k\cdot p')\gamma^\mu\cancel{k}
\nonumber \\
&= k^2k^\mu - \frac{1}{2}(k^2-[(k-p)^2-m^2])k_\nu \gamma^\nu\gamma^\mu -\frac{1}{2}(k^2-[(k-p')^2-m^2])k_\nu\gamma^\mu\gamma^\nu
\nonumber \\
&= \frac{1}{2}[(k-p)^2-m^2]k_\nu \gamma^\nu\gamma^\mu +\frac{1}{2}[(k-p')^2-m^2]k_\nu\gamma^\mu\gamma^\nu.
\end{align}
$$
Thus,
$$
\begin{align}
\require{cancel}
\displaystyle
I =& \frac{1}{2}\int\frac{d^4k}{(2\pi)^4}\bar u(p') \frac{1}{[k^2-M^2+i\epsilon]} \left( \frac{k_\nu}{[(k-p')^2-m^2+i\epsilon]}\gamma^\nu\gamma^\mu + \frac{k_\nu}{[(k-p)^2-m^2+i\epsilon]}\gamma^\mu\gamma^\nu \right) u(p)
\nonumber \\
&= \frac{1}{2}A\bar u(p')(p'_\nu\gamma^\nu\gamma^\mu + p_\nu\gamma^\mu\gamma^\nu) u(p)
\nonumber \\
&= Am \bar u(p') \gamma^\mu u(p),
\end{align}
$$
where ##A## is a scalar.
Second Method: Feynman Parametrization. The denominator becomes:
$$
\require{cancel}
\displaystyle
\begin{align}
[k^2-M^2+i\epsilon]^{-1}[(k-p)^2-m^2+i\epsilon]^{-1}[(k-p')^2-m^2+i\epsilon]^{-1}=
\int_0^1dx\int_0^{1-x}dy \frac{2}{[l^2-\Delta^2+i\epsilon]^3},
\end{align}
$$
where ##l^\mu=(k-xp-yp')^\mu##, ##\Delta^2=(x+y)^2m^2+(1-x-y)M^2-xyq^2##, and, ##q^\mu=(p'-p)^\mu##.
Next, we perform the shift in the numerator:
$$
\begin{align}
N^\mu = k^2k^\mu - (k\cdot p)\cancel{k}\gamma^\mu -(k\cdot p')\gamma^\mu\cancel{k} = A_1 \bar u(p') \gamma^\mu u(p) + A_2 \bar u(p') \frac{i\sigma^{\mu\nu}q_\nu}{2m} u(p),
\end{align}
$$
where ##A_1## and ##A_2## are scalars. This method clearly gives a different result.
The first method gives the correct result (the result should not contain a ##i\sigma^{\mu\nu}q_\nu/2m## term), so the problem must be related with the Feynman Parametrization somehow.
The denominator ##[(k-p)^2-m^2+i\epsilon]=[k^2-2(k\cdot p)+i\epsilon]## (and the other one with ##p'##) could be problematic when ##k\to 0##, however there are other terms in the numerator (which I did not included here) that give the correct result with the same denominator and using Feynman Parametrization, i.e., using the second method. On the other hand these same terms (which give the correct result) are UV finite but the one in question is not. Is the problem in the IR or UV or both?
I believe I didn't make any mistakes on the passages, so what am I missing?