Question 7.4: Calculation of QCD Sum-Rule Graphs with Dimensional Regulari......
Calculation of QCD Sum-Rule Graphs with Dimensional Regularization
Calculate the first integral in (7.259) using the techniques of dimensional regularization introduced in Sect. 4.3.
\begin{aligned} -\mathrm{i} S(x, y ; A)= & \frac{1}{2 \pi^2} \frac{(x-y) \cdot \gamma}{(x-y)^4}-\frac{g G^{\varrho \mu}(0)}{4 \pi^4} \\ & \times \frac{1}{2} \int \mathrm{d}^4 z \frac{\not x-\not}{(x-z)^4} z_{\varrho} \gamma_\mu \frac{\not z-\not y}{(z-y)^4} \\ + & \frac{g^2 G^{\varrho \mu}(0) G^{\varrho^{\prime} \mu^{\prime}}(0)}{8 \pi^6} \\ & \times \frac{1}{4} \int \mathrm{d}^4 z_1 \mathrm{~d}^4 z_2 \frac{\not x-k_1}{\left(x-z_1\right)^4}\left(z_1\right)_{\varrho} \gamma_\mu \frac{\not k_1-k_2}{\left(z_1-z_2\right)^4} \\ & \times\left(z_2\right)_{\varrho^{\prime}} \gamma_{\mu^{\prime}} \frac{\not k_2-\not y}{\left(z_2-y\right)^4} \\ & +\ldots . & (7.259) \end{aligned}
We first introduce the usual Feynman parameters:
\begin{aligned} I & =\int \mathrm{d}^{4} z \frac{(\not x-\not z) \gamma_{\mu}(\not z-\not y) z_{\varrho}}{(x-z)^{4}(z-y)^{4}} \\ & =\Gamma(4) \int \mathrm{d} u u(1-u) \int \mathrm{d}^{4} z \frac{(\not x-\not z) \gamma_{\mu}(\not z-\not y) z_{\varrho}}{\left[u(x-z)^{2}+(1-u)(z-y)^{2}\right]^{4}} . & (1) \end{aligned}
Then, as usual, we bring the denominator into the quadratic normal form:
\begin{aligned} & u(x-z)^{2}+(1-u)(z-y)^{2} \\ & \quad=z^{2}-2 z[u x+(1-u) y]+u x^{2}+(1-u) y^{2} \\ & \quad=\{z-[u x+(1-u) y]\}^{2}+u(1-u) x^{2}+u(1-u) y^{2}-2 u(1-u) x \cdot y \\ & \quad=\{z-[u x+(1-u) y]\}^{2}+u(1-u)(x-y)^{2} . & (2) \end{aligned}
Substituting z \rightarrow z+u x+(1-u) y gives
\begin{aligned} I= & \Gamma(4) \int \mathrm{d} u u(1-u) \int \mathrm{d}^{4} z \frac{[(1-u)(\not x-\not y)-\not z] \gamma_{\mu}}{\left[z^{2}+u(1-u)(x-y)^{2}\right]^{4}} \\ & \times[\not z+u(\not x-\not y)]\left[z_{\varrho}+u x_{\varrho}+(1-u) y_{\varrho}\right] . & (3) \end{aligned}
Only the terms even in z contribute to the integral, and z_{\varrho} z_{\alpha} contribute z^{2} g_{\varrho \alpha} / 4 :
\begin{aligned} I= & 6 \int\limits_{0}^{1} \mathrm{~d} u~ u(1-u) \int \mathrm{d}^{4} z \frac{(1-u) u(\not x-\not y) \gamma_{\mu}(\not x-\not y)}{\left[z^{2}+u(1-u)(x-y)^{2}\right]^{4}} \\ & \times\left[u x_{\varrho}+(1-u) y_{\varrho}\right]+6 \int\limits_{0}^{1} \mathrm{~d} u~ u(1-u) \\ & \times \int \frac{\mathrm{d}^{4} z}{\left[z^{2}+u(1-u)(x-y)^{2}\right]^{4}}\left\{\frac{1}{2} z^{2} \gamma_{\mu}\left[u x_{\varrho}+(1-u) y_{\varrho}\right]\right. \\ & \left.+\frac{1}{4} z^{2}(1-u)(\not x-\not y) \gamma_{\mu} \gamma_{\varrho}-\frac{1}{4} z^{2} \gamma_{\varrho} \gamma_{\mu} u(\not x-\not y)\right\} . & (4) \end{aligned}
The nominator of the second integral is abbreviated by writing z^{2} \cdot f(u). We now integrate using (4.97) and (4.99):
\begin{aligned} \int \mathrm{d}^d k \frac{1}{\left(k^2+2 k \cdot p+m^2\right)^\alpha}= & \frac{\mathrm{i} \pi^{d / 2}}{\left(m^2-p^2\right)^{\alpha-d / 2}} \frac{\Gamma(\alpha-d / 2)}{\Gamma(\alpha)} . & (4.97)\\ \int \mathrm{d}^d k \frac{k^2}{\left(k^2+2 k \cdot p+m^2\right)^\alpha}= & \frac{\mathrm{i} \pi^{d / 2}}{\left(m^2-p^2\right)^{\alpha-d / 2}} \frac{1}{\Gamma(\alpha)}\left[\Gamma\left(\alpha-\frac{d}{2}\right) p^2\right. \left.+\Gamma\left(\alpha-1-\frac{d}{2}\right) \frac{d}{2}\left(m^2-p^2\right)\right] . & (4.99) \end{aligned}
\begin{aligned} I= & 6 \int\limits_{0}^{1} \mathrm{~d} u u(1-u)\left\{\left[u(1-u)(x-y)^{2}\right]^{-2}\right. \\ & \times u(1-u)(\not x-\not y) \gamma_{\mu}(\not x-\not y)\left[u x_{\varrho}+(1-u) y_{\varrho}\right] \frac{\mathrm{i} \pi^{2} \Gamma(2)}{\Gamma(4)} \\ & \left.+f(u) \frac{\mathrm{i} \pi^{2} \Gamma(1)}{\Gamma(4)} 2\left[u(1-u)(x-y)^{2}\right]^{-1}\right\} \\ = & \mathrm{i} \pi^{2} \frac{1}{(x-y)^{4}}\left[\int\limits_{0}^{1} \mathrm{~d} u(\not x-\not y) \gamma_{\mu}(\not x-\not y)\left[u x_{\varrho}+(1-u) y_{\varrho}\right]\right. \\ & +2 \int\limits_{0}^{1} \mathrm{~d} u\left\{\frac{1}{2} \gamma_{\mu}\left[u x_{\varrho}+(1-u) y_{\varrho}\right]+\frac{1}{4}(1-u)(\not x-\not y) \gamma_{\mu} \gamma_{\varrho}\right. \\ & \left.\left.-\frac{1}{4} \gamma_{\varrho} \gamma_{\mu} u(\not x-\not y)\right\}(x-y)^{2}\right] . & (5) \end{aligned}
Both the integral \int u \mathrm{~d} u and \int(1-u) \mathrm{d} u give 1 / 2
\begin{aligned} I= & \frac{\mathrm{i} \pi^{2}}{2} \frac{1}{(x-y)^{4}}\left\{(\not x-\not y) \gamma_{\mu}(\not x-\not y)(x+y)_{\varrho}+\gamma_{\mu}\left(x_{\varrho}+y_{\varrho}\right)(x-y)^{2}\right. \\ & \left.+\frac{1}{2}\left[( \not x-\not y) \gamma_{\mu} \gamma_{\varrho}-\gamma_{\varrho} \gamma_{\mu}(\not x-\not y)\right](x-y)^{2}\right\} \\ = & \frac{\mathrm{i} \pi^{2}}{4} \frac{1}{(x-y)^{2}}\left[(\not x-\not y) \gamma_{\mu} \gamma_{\varrho}-\gamma_{\varrho} \gamma_{\mu}( \not x-\not y)\right] \\ & +\frac{\mathrm{i} \pi^{2}}{2} \frac{1}{(x-y)^{4}}\left[2(x-y)_{\mu}(\not x-\not y)(x+y)_{\varrho}\right] \\ = & \frac{\mathrm{i} \pi^{2}}{4} \frac{1}{(x-y)^{2}}\left[(\not x-\not y) \gamma_{\mu} \gamma_{\varrho}-\gamma_{\varrho} \gamma_{\mu}(\not x-\not y)\right] \\ & +\mathrm{i} \pi^{2} \frac{1}{(x-y)^{4}}\left[x_{\mu} x_{\varrho}+x_{\mu} y_{\varrho}-y_{\mu} x_{\varrho}-y_{\mu} y_{\varrho}\right](\not x-\not y) . & (6) \end{aligned}
This still has to be multiplied by G_{\varrho \mu}(0), which is antisymmetric in \varrho and \mu, leading to
\begin{aligned} \Delta S(x, y ; A)= & -\frac{\mathrm{i} g}{32 \pi^{2}} \frac{1}{(x-y)^{2}}\left[(\not x-\not y) \gamma_{\mu} \gamma_{\varrho}-\gamma_{\varrho} \gamma_{\mu}(\not x-\not y)\right] G^{\varrho \mu}(0) \\ & -\frac{\mathrm{i} g}{8 \pi^{2}} \frac{1}{(x-y)^{2}} 2 x_{\mu} y_{\varrho}(\not x-\not y) G^{\varrho \mu}(0) . & (7) \end{aligned}