- #1
ur22
Homework Statement
I am currently working on an exercise list where I need to calculate the second functional derivative with respect to Grassmann valued fields.
$$
\dfrac{\overrightarrow{\delta}}{\delta \psi_{\alpha} (-p)} \left( \int_{x} \widetilde{\bar{\psi}}_{\mu} (x) i \partial_{s}^{\mu \nu} \widetilde{\psi}_{\nu} (x) \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} \\
$$
where the ##\psi## are euclidean Grassmann valued Dirac spinors. The tilde denotes Spinors in position space and the index ##s## stands for Feynman slashes
$$
a_{s} = \gamma^{\mu} a_{\mu}
$$
Homework Equations
I used
$$
\int_{x} \equiv \int\mathrm{d}^{4}x \\
\int_{p} \equiv \int \dfrac{\mathrm{d}^{4}p}{(2 \pi)^{4}}
$$
The convolution is defined as:
$$
\left( F \ast G \right) (p) \equiv \int_{- \infty}^{+ \infty} F(\lambda) G(p - \lambda) \mathrm{d} \lambda
$$
The Attempt at a Solution
My calculation reads:
$$
\dfrac{\overrightarrow{\delta}}{\delta \psi_{\alpha} (-p)} \left( \int_{x} \widetilde{\bar{\psi}}_{\mu} (x) i \partial_{s}^{\mu \nu} \widetilde{\psi}_{\nu} (x) \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
\dfrac{\overrightarrow{\delta}}{\delta \psi_{\alpha} (-p)} \left( \int_{x} \int_{p'} e^{-ip'x} \left[ -\int d \lambda \, \bar{\psi}_{\mu} (\lambda) (p_{s}'^{\mu \nu} - \lambda_{s}^{\mu \nu}) \psi_{\nu}(p' - \lambda) \right] \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
\left( \int_{x} \int_{p'} e^{-ip'x} \left[ \int d \lambda \, \bar{\psi}_{\mu} (\lambda) (p_{s}'^{\mu \nu} - \lambda_{s}^{\mu \nu}) \delta^{\alpha}_{\nu} \delta (\lambda - (p' + p)) \right] \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
\left( \int_{x} \int_{p'} e^{-ip'x} \bar{\psi}_{\mu} (p' + p) (-p_{s}^{\mu \alpha}) \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
- \int_{x} \int_{p'} e^{-ip'x} \delta^{\beta}_{\mu} \delta (p'+p+q) \, p_{s}^{\mu \alpha} = \\
%
- \dfrac{1}{(2 \pi)^4} \int_{x} e^{i (p+q) x} \, p_{s}^{\beta \alpha} = \\
%
- p_{s}^{\beta \alpha} \delta (p+q) = \\
%
- (p_{s}^{T})^{\alpha \beta} \delta (p+q)
$$However, the solution of the exercise reads:
$$
- p_{s}^{\alpha \beta} (2 \pi)^{4} \delta (p-q)
$$
The factor ##(2 \pi)^{4}## might be due to different conventions regarding the Fourier Transform, but what's the deal with the different arguments in the delta function and the transposition of ##p_s##?Best regards