SU(N) Yang-Mills Theory and Scalar Fields

Source: Perimeter Scholars International, Quantum Field Theory II, Tutorial 8; 11 worked parts.

Question prompts paraphrased from the official Perimeter PSI Online sheet; the equations and notation follow the source.

1 Coupling to Scalar Fields

Let $\phi_i(x)$ be a scalar field in a representation $r$ of $SU(N)$, with $i=1,\ldots,\dim r$.

Determine the dimension of the group $SU(N)$.

An $N\times N$ complex matrix has $2N^2$ real dimensions. The hermitian condition, $U^*_{ik}U_{jk} = \delta_{ij}$, is $N^2$ constraints. In addition, $\det U =1$ gives an extra constraint. Therefore, $\dim SU(N) = N^2 -1$.
Determine the dimension of the fundamental representation of $SU(N)$.

The fundamental representation of $SU(N)$ is $N \times N$ complex matrices, which has dimension $N$.
The adjoint action is defined by \[\operatorname{Ad}_g(T^a)\equiv gT^ag^{-1}.\tag{1.1}\] Thus there is a $\dim G\times\dim G$ matrix $C(g)$ such that \[gT^ag^{-1}=C^{ab}(g)T^b.\tag{1.2}\] Expand $g$ and $C(g)$ infinitesimally and use \[[T_i,T_j]=if_{ijk}T_k\tag{1.3}\] to derive the adjoint generators \[(T_i)_{jk}=if_{ijk}.\tag{1.4}\]

\[\begin{aligned} \operatorname{Ad}(g)Y = (I + \epsilon X)Y(I-\epsilon X) = I + \epsilon[X,Y] = (I + \epsilon \operatorname{ad}(X))Y \implies \operatorname{ad}(X)Y = [X,Y]. \end{aligned}\tag{1.5}\] Taking $X = T^a, Y=T^b$, we have \[\operatorname{ad}(T^a)^{bc}T^c = if^{abc}T^c \implies \operatorname{ad}(T^a)^{bc} =if^{abc}.\tag{1.6}\]
Determine the dimension of the adjoint representation of $SU(N)$.

As we have seen above, the adjoint representation of the Lie algebra is a $\dim G \times \dim G$ matrix. Therefore, \[\dim \operatorname{adj}(su(N)) = \dim SU(N).\tag{1.7}\]

Now consider the local transformation \[\phi_i(x)\longmapsto g_{ij}(x)\phi_j(x), \qquad g(x)=\exp\!\bigl[i\alpha_m(x)T_r^m\bigr],\tag{1.8}\] where $T_r^m$ are the $su(N)$ generators in the representation $r$ and the $\alpha_m(x)$ are spacetime-dependent group parameters. Require the covariant derivative to transform as \[D_\mu\phi_i(x)\longmapsto g_{ij}(x)D_\mu\phi_j(x),\tag{1.9}\] and define, with matrix indices suppressed, \[D_\mu\phi=\partial_\mu\phi-ieA_\mu\phi, \qquad A_\mu(x)=A_\mu^m(x)T_r^m.\tag{1.10}\]

Derive the finite gauge-transformation law of $A_\mu(x)$ required by the covariance of $D_\mu\phi$.

\[\begin{aligned} D_\mu \phi \to D_\mu' g \phi = (\partial_\mu g)\phi + g \partial_\mu \phi - ie A_\mu' g\phi = g(\partial_\mu \phi - ie A_\mu \phi), \end{aligned}\tag{1.11}\] which implies \[A_\mu' = g A_\mu g^{-1} - \frac{i}{e}(\partial_\mu g) g^{-1}.\tag{1.12}\]
For infinitesimal $\alpha_m(x)$, show that \[\delta A_\mu^a=\frac{1}{e}D_\mu\alpha_a, \qquad D_\mu\alpha=\partial_\mu\alpha+ie[\alpha,A_\mu].\tag{1.13}\]

\[\begin{aligned} A_\mu^aR(T^a) & \to (I + i \alpha_bR(T^b))A_\mu^a R(T^a)(I - i \alpha_bR(T^b)) + \frac{1}{e}(\partial_\mu \alpha_b)R(T^b)(I + i \alpha_a R(T^a))\\ &=A_\mu^aR(T^a) + i\alpha_b[R(T^b),R(T^a)]A_\mu^a+\frac{1}{e}(\partial_\mu \alpha_b)R(T^b)\\ &=A_\mu^aR(T^a) + \frac{1}{e}D_\mu\alpha, \end{aligned}\tag{1.14}\] from which we can recognise $\alpha$ as the adjoint representation of $su(N)$.
Prove that the Lagrangian \[\mathcal{L} =-(D_\mu\phi)^\dagger D^\mu\phi -m^2\phi^\dagger\phi -\frac{\lambda}{2}(\phi^\dagger\phi)^2\tag{1.15}\] is invariant under the local scalar transformation and the gauge-field transformation found above.

It can be easily seen through the transformation of $\phi$ and $D_\mu \phi$.
Extract every Feynman rule from this Lagrangian that contains at least one scalar line.

Expanding the covariant derivative, we have \[|D_\mu\phi|^2 = |\partial_\mu\phi|^2 + ieA_\mu(\phi^\dagger\partial_\mu \phi - \phi \partial_\mu \phi^\dagger) + e^2A_\mu^2|\phi|^2,\tag{1.16}\] and the Feynman rules can be read off.
Apply Noether's theorem to obtain the conserved currents.

\[j^a_\mu = \frac{\partial \mathcal{L}}{\partial(\partial^\mu \phi)}\delta\phi^a + \frac{\partial \mathcal{L}}{\partial(\partial^\mu \phi^\dagger)}\delta\phi^{\dagger a}= i (D_\mu\phi)^\dagger R(T^a)\phi - i\phi^\dagger R(T^a)(D_\mu \phi).\tag{1.17}\]

2 Zero Modes of the Yang--Mills Lagrangian

Take the unfixed Yang--Mills Lagrangian \[\mathcal{L} =-\frac{1}{2g^2}\operatorname{Tr}(F^{\mu\nu}F_{\mu\nu}),\tag{2.1}\] with the trace over adjoint indices and arbitrary $N$ in $SU(N)$.

Expand the $F^2$ term, isolate the quadratic terms containing two derivatives, and write them as \[\int d^4x\,A_\mu^a(x)D_x^{\mu\nu}A_\nu^a(x).\tag{2.2}\]

\[\begin{aligned} S &= -\frac{1}{4g}\int d^dx F_{\mu\nu}^a F^{a\mu\nu}\\ &=-\frac{1}{2g}\int d^dx A_\nu(-\partial^2\eta^{\mu\nu} + \partial^\mu\partial^\nu)A_\mu + \cdots, \end{aligned}\tag{2.3}\] where the interacting terms are suppressed.
Show that there is a nonzero field $B_\mu$ satisfying \[D_x^{\mu\nu}B_\nu(x)=0.\tag{2.4}\] Hint: regard $D^{\mu\nu}$ as an integral kernel and inspect its momentum-space representation. Explain why this zero mode obstructs inversion of the quadratic operator before gauge fixing.

It is easy to see that any field of the form $\partial_\mu \omega$ is the kernel of the operator $D^{\mu\nu}_a$. Therefore, $D^{\mu\nu}_a$ is clearly not invertible and hence we need the Faddeev-Popov procedure.