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Ian Lovejoy
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Homework Statement
I'm working my way through Quantum Field Theory in a Nutshell by A. Zee. I'm religiously doing the exercises but since I'm doing it on my own (I'm not in school) I have no one to ask when I get stuck. Any hints would be appreciated.
The problem is IV.7.5, on page 253 of the book. The desire is to derive the anomaly using the triangle diagram. However this is not to be done using the standard shift of integration variable treatment (that is developed in the chapter itself). Instead, one is to start with the massive Fermion case, and use Lorentz invariance, Bose statistics, and vector current conservation. All of that part I get. But in the end, one still has to evaluate a Feynman integral, and that is where I am stuck.
Homework Equations
The equation one gets from the triangle diagram is:
[tex]
\Delta^{\lambda\mu\nu}(k_1,k_2) = (-1)i^3\int\frac{d^4p}{(2\pi)^4} tr(\gamma^\lambda\gamma^5\frac{1}{p\!\!\!/ - q\!\!\!/ - m}\gamma^\nu\frac{1}{p\!\!\!/ - k\!\!\!/_1 - m}\gamma^\mu\frac{1}{p\!\!\!/ - m}) + \{\mu, k_1 \leftrightarrow \nu, k_2\}
[/tex]
Also one can argue by Lorentz invariance that:
[tex]
\Delta^{\lambda\mu\nu}(k_1,k_2) =
\epsilon^{\lambda\mu\nu\sigma}k_{1\sigma}A_1 +
\epsilon^{\lambda\mu\nu\sigma}k_{2\sigma}A_2 +
\epsilon^{\lambda\nu\sigma\tau}k_{1\sigma}k_{2\tau}k^{1\mu}A_3 +
\epsilon^{\lambda\mu\sigma\tau}k_{1\sigma}k_{2\tau}k^{1\nu}A_4 +
\epsilon^{\mu\nu\sigma\tau}k_{1\sigma}k_{2\tau}k^{1\lambda}A_5
[/tex]
[tex]
+ \epsilon^{\lambda\nu\sigma\tau}k_{1\sigma}k_{2\tau}k^{2\mu}A_6 +
\epsilon^{\lambda\mu\sigma\tau}k_{1\sigma}k_{2\tau}k^{2\nu}A_7 +
\epsilon^{\mu\nu\sigma\tau}k_{1\sigma}k_{2\tau}k^{2\lambda}A_8
[/tex]
Where the [tex]A_i[/tex] are functions of the Lorentz scalars. Using Bose statistics one can show:
[tex]
A_2(k_1, k_2) = -A_1(k_2, k_1)
[/tex]
[tex]
A_6(k_1, k_2) = -A_4(k_2, k_1)
[/tex]
[tex]
A_7(k_1, k_2) = -A_3(k_2, k_1)
[/tex]
[tex]
A_8(k_1, k_2) = A_5(k_2, k_1)
[/tex]
And using vector current conservation you can show:
[tex]
A_2 = {k_1}^2A_3 + k_1\cdot k_2 A_6
[/tex]
[tex]
A_1 = k_1\cdot k_2 A_4 + {k_2}^2A_7
[/tex]
(continued in reply, having trouble with the length of this post)