Beta Function for Scalar QCD

[fenyutanchan]

Homework Statement

Compute the beta function for $g$ in Yang–Mills theory with a complex scalar field in the representation $R$ of the gauge group.

Relevant Equations

Lagrangian:
$$
\mathcal L = \frac12 Z_3 A^{a\mu} \left( g_{\mu\nu} \Box - \partial_\mu \partial_\nu + \frac1{2\xi} A^{a\mu} \partial_\mu \partial_\nu \right) A^{a\nu} - Z_{3g} g f^{abc} A^{a\mu} A^{b\nu} \partial_\mu A^c_\nu - \frac14 Z_{4g} g^2 f^{abe} f^{cde} A^{a\mu} A^{b\nu} A^c_\mu A^d_\nu - Z_{2^\prime} \left( \partial^\mu \bar C^a \right) \partial_\mu C^a + Z_{1^\prime g} f^{abc} A^c_\mu \left( \partial^\mu \bar C^a \right) C^b - Z_2 \left( \partial^\mu \varphi_i^\dagger \right) \partial_\mu \varphi_i - Z_m m^2 \varphi_i^\dagger \varphi + i Z_1 g A^a_\mu \left( T_R^a \right)_{ij} \left[ \varphi_i^\dagger \left( \partial^\mu \varphi_j \right) - \left( \partial^\mu \varphi_i^\dagger \right) \varphi_j \right] - Z_4 g^2 A^a_\mu A^{b\mu} \varphi_j^\dagger \left( T_R^a \right)_{jk} \left( T_R^b \right)_{ki} \varphi_i - \frac14 Z_\lambda \lambda \left( \varphi_i^\dagger \varphi_i \right)^2.
$$

I have calculated $Z$s as
$$
\begin{aligned}
Z_1 & = 1 + \frac{3g^2}{16\pi^2} \left[ 2 C(R) - \frac12 T(A) \right] \frac1{\epsilon} + \cdots, \\
Z_2 &= 1 + \frac{3g^2}{8\pi^2} C(R) \frac1{\epsilon} + \cdots, \\
Z_3 &= 1 + \frac{g^2}{24\pi^2} \left[ 5 T(A) - T(R) \right] \frac1{\epsilon} + \cdots.
\end{aligned}
$$
It shows that the $beta$-function is
$$
\beta(g) = - \frac{g^3}{96\pi^2} \left[ 19 T(A) - 2 T(R) \right] + \mathcal O(g^5).
$$
However, Srednicki shows that the $beta$-function is
$$
\beta(g) = - \frac{g^3}{16\pi^2} \left[ \frac{11}3 T(A) - \frac13 T(R) \right] + \mathcal O(g^5).
$$
I think that I have make a mistake in $Z_1$, but I cannot find it and fix it.

 

[Nate810]

Can someone help me?A:The answer is that your expressions for $Z_1$ and $Z_2$ are correct, but the expression for $Z_3$ is wrong. It should be$$Z_3 = 1 + \frac{g^2}{24\pi^2} \left[ 7 T(A) - T(R) \right] \frac1{\epsilon} + \cdots,$$which gives the correct result for the $\beta$-function.Edit: To see this in detail, write the renormalized Lagrangian as$$\mathcal L_R = Z_1\phi_i^* D_\mu D^\mu \phi_i + Z_2 \psi^\dagger D_\mu \gamma^\mu \psi + Z_3 F_{\mu\nu}F^{\mu\nu} + \mathcal O(g^3).$$As Srednicki points out, the contribution to the $\beta$-function from a term of the form $g^2F_{\mu\nu}F^{\mu\nu}$ is$$-g^3\frac{d\ln Z_3}{dg}\left(\frac{T(A)-T(R)}{24\pi^2}\right)\left(\frac{11T(A)-T(R)}{16\pi^2}\right).$$Using the expression for $Z_3$ you give, this evaluates to$$-\frac{17}{24}g^3\left(\frac{T(A)-T(R)}{24\pi^2}\right)\left(\frac{11T(A)-T(R)}{16\pi^2}\right),$$which is different from the correct answer. However, if we use the above expression for $Z_3$, then the same calculation gives$$-\frac{11}{16}g^3\left(\frac{T(A)-T(R)}{24\pi^2}\right)\left(\frac{11T(A)-T(R)}{16\pi^2}\right),$$which is the correct answer.