9.4 Coupling to matter fermions

The coupling of the Yang-Mills field from Section 9.3 to matter fields can be achieved by a gauging and minimal coupling procedure that is similar to the case of electromagnetism, see Section 7.1. The role that is played by the electric charge \(q\) in the context of QED is now played by a linear representation \(\rho : G\to \mathsf {GL}(V)\) (in the sense of Definition 9.6) of the Lie group \(G\) underlying the Yang-Mills theory. To simplify our presentation, we again assume that \(G\) is either the unitary group \(\mathsf {U}(n)\) or the special unitary group \(\mathsf {SU}(n)\) and we further assume that \(\rho \) is a unitary representation, e.g. the trivial or the defining representation from Example 9.7.

The first type of matter field we discuss is given by a Dirac field

on the \(d\)-dimensional Minkowski spacetime \((\bbR ^d,\eta )\) that takes values in a unitary representation \(\rho \) of \(G\) on a complex vector space \(V\). We demand that this field transforms under gauge transformations \(U:\bbR ^d\to G\,,~x\mapsto U(x)\), according to the given representation, i.e.

We shall again suppress most of the time the argument \(x\in \bbR ^d\) and simply write this as \(T_U\Psi = \rho (U)\,\Psi \), but it is important to keep in mind that \(U(x)\) is a local (i.e. \(x\)-dependent) transformation. From this transformation behavior we deduce that the Dirac adjoint field \(\overline {\Psi } :=\Psi ^\dagger \,(\ii \,\gamma ^0)\) transforms according to

\begin{flalign} T_U\overline {\Psi }\,:=\,\overline {T_U\Psi } = \big (\rho (U)\,\Psi \big )^\dagger \,(\ii \,\gamma ^0) = \Psi ^\dagger \,\rho (U^{-1})\,(\ii \,\gamma ^0) =\Psi ^\dagger \,(\ii \,\gamma ^0)\,\rho (U^{-1}) = \overline {\Psi }\,\rho (U^{-1})\quad , \end{flalign} where in the third step we used that \(\rho \) is a unitary representation, hence \(\rho (U)^\dagger = \big (\rho (U)\big )^{-1} = \rho (U^{-1})\). In the fourth step we used that \(\rho (U^{-1})\,(\ii \,\gamma ^0)= (\ii \,\gamma ^0)\,\rho (U^{-1})\) commute because they act on different tensor factors of \(\bbC ^N\otimes V\). (When one is more pedantic, one could write \(\mathrm {id}\otimes \rho (U^{-1})\) and \((\ii \,\gamma ^0)\otimes \mathrm {id}\) for these linear maps on \(\bbC ^N\otimes V\), but this would lead to a horrible notation.) Using the terminology of Remark 9.9, this means that the Dirac adjoint \(\overline {\Psi }\) transforms in the dual representation of \(\rho \). From this discussion it follows that the Dirac inner product is gauge invariant

\begin{flalign} \overline {\Psi }\,\chi \,\longmapsto \, \big (T_U\overline {\Psi }\big )\,\big (T_U\chi \big ) = \overline {\Psi }\,\rho (U^{-1})\,\rho (U)\,\chi = \overline {\Psi }\,\chi \quad , \end{flalign} which is a useful observation that will later allow us to write down a mass term in the action.

Similarly to the case of electrodynamics, defining a gauge invariant kinetic term for \(\Psi \) is more problematic because, being a local transformation \(U(x)\), the gauge transformations do not commute with partial derivatives, i.e. \(\partial _{\mu }\big (\rho (U(x))\,\Psi (x)\big ) \neq \rho (U(x))\,\partial _\mu \Psi (x)\). The solution of this problem lies again in introducing a gauge covariant derivative, which in the current non-Abelian case is given by

where \(A_\mu \) is the Yang-Mills field and \(\underline {\rho }\) is the Lie algebra representation induced by \(\rho \), see Example 9.11. One can show by a direct calculation that, under a combined gauge transformation on \(\Psi \) and \(A_\mu \) according to (9.36) and (9.48), the gauge covariant derivative transforms as

\begin{flalign} T_U\,:\,D_\mu \Psi (x)\,\longmapsto \, \rho \big (U(x)\big )\,D_\mu \Psi (x)\quad . \end{flalign} For completeness, I will spell out this calculation, suppressing again the argument \(x\in \bbR ^d\) to improve readability

\begin{flalign} \nn T_U\big (D_\mu \Psi \big )&= \partial _\mu \big (\rho (U)\,\Psi \big ) + \underline {\rho }\Big (U\,A_\mu U^{-1} + U\,\partial _\mu U^{-1}\Big )\,\rho (U)\,\Psi \\ \nn &=\rho (U)\,\partial _{\mu }\Psi + \big (\partial _\mu \rho (U)\big )\,\Psi + \rho (U)\,\underline {\rho }(A_\mu )\,\Psi + \rho (U)\,\big (\partial _\mu \rho (U^{-1})\big )\,\rho (U)\,\Psi \\ &=\rho (U)\,\partial _{\mu }\Psi + \rho (U)\,\underline {\rho }(A_\mu )\,\Psi = \rho (U)\,D_\mu \Psi \quad .\label {eqn:YMcovdercheck} \end{flalign} In the second step we used that \(\underline {\rho }(U\,A_\mu \,U^{-1}) = \rho (U)\,\underline {\rho }(A_\mu )\,\rho (U^{-1})\) and \(\underline {\rho }(U\,\partial _\mu U^{-1}) = \rho (U)\,\partial _\mu \rho (U^{-1})\), which follows from our construction of the Lie algebra representation \(\underline {\rho }\) in Example 9.11. The last step uses the Leibniz rule in order to write \(\big (\partial _\mu \rho (U^{-1})\big )\,\rho (U) = \partial _\mu \big (\rho (U^{-1})\,\rho (U)\big ) - \rho (U^{-1})\,\partial _\mu \rho (U) = \partial _\mu \mathrm {id} - \rho (U^{-1})\,\partial _\mu \rho (U)= -\rho (U^{-1})\,\partial _\mu \rho (U) \).

These are now all the ingredients we need to write down the action functional

for a massive Dirac field taking values in a unitary representation \(\rho \) of \(G\). This action is gauge invariant under the combined gauge transformations

\begin{flalign} \label {eqn:YMguagecombinedDirac} T_{U}\Psi = \rho (U)\,\Psi ~~,\quad T_U\overline {\Psi } = \overline {\Psi }\,\rho (U^{-1})~~,\quad (T_U A)_\mu = U\,A_\mu U^{-1} + U\,\partial _\mu U^{-1} \end{flalign} on all the fields. We observe that there is a cubic interaction term, which upon quantization will lead to a \(3\)-valent interaction vertex in the Feynman rules that is similar to the photon-Dirac-anti-Dirac vertex of QED from Section 7.2. The physical details of this interaction will of course depend on the choice of representation \(\rho \); in particular, for the trivial representation we would have that \(\underline {\rho _{\mathrm {triv}}} = 0\) is zero, hence this term disappears in this very special case. (This is analogous to the fact that the photon does couple trivially to particles of charge \(q=0\).) To obtain a full theory that describes both the Dirac and the Yang-Mills field, we simply add the two actions (9.43) and (9.54) and define

This action is of course gauge invariant under the combined gauge transformations (9.55) because the two individual summands are gauge invariant.

Chiral fermions in \(d=4\) spacetime dimensions: In what follows we consider the \(d{=}4\)-dimensional Minkowski spacetime \((\bbR ^4,\eta )\). An important physical feature of the standard model of particle physics (which is defined in \(d=4\) spacetime dimensions) is that the left and right-handed chiral components of Dirac spinors participate differently in the interactions mediated by the gauge bosons. From the representation theoretic perspective established in this section, this means that the left-handed Weyl fields \(\Psi _L\) live in a different representation of the Lie group than the right-handed Weyl fields \(\Psi _R\). (See around Eqn. (5.33) for the definition of left/right-handed Weyl spinors.) We take this into account by writing

\begin{flalign} \Psi _{L/R} \,:\,\bbR ^4\longrightarrow \bbC ^2\otimes V_{L/R}~,~~x\longmapsto \Psi _{L/R}(x) \end{flalign} for a left/right-handed Weyl field taking values in a unitary representation \(\rho _{L/R}\) of the Lie group \(G\) on a complex vector space \(V_{L/R}\). (In general, these representations will be different!) The corresponding gauge transformations from (9.48) then read as

\begin{flalign} \big (T_U\Psi _{L/R}\big )(x)\, =\,\rho _{L/R}\big (U(x)\big )\,\Psi _{L/R}(x) \end{flalign} and the gauge covariant derivatives from (9.51) are given by

\begin{flalign} D_\mu \Psi _{L/R}(x)\,=\,\partial _\mu \Psi _{L/R}(x) + \underline {\rho _{L/R}}\big (A_\mu (x)\big )\,\Psi _{L/R}(x)\quad . \end{flalign} This all looks pretty much the same as for Dirac fields, so what’s the problem? The problem arises when we want to write down an action functional similar to (9.54) for the chiral fields. To see this, let us recall the \(4\)-dimensional gamma-matrices from (5.25), which we shall write compactly in the following block matrix form

\begin{flalign} \gamma ^\mu = -\ii \,\begin{pmatrix} 0 & \sigma ^\mu \\ \overline {\sigma }^\mu & 0 \end {pmatrix}\quad , \end{flalign} where \(\sigma ^0=\overline {\sigma }^0=1_{2\times 2}\) is the \(2\times 2\) identity matrix and \(\sigma ^i = -\overline {\sigma }^{i}\) are the Pauli matrices. The Dirac inner product then reads in terms of the chiral components as

\begin{flalign} \overline {\Psi }\,\chi = \Psi ^\dagger \,(\ii \,\gamma ^0)\,\chi = \begin{pmatrix} \Psi _L^\dagger & \Psi _R^\dagger \end {pmatrix}\, \begin{pmatrix} 0 & 1_{2\times 2}\\ 1_{2\times 2} & 0 \end {pmatrix}\, \begin{pmatrix} \chi _L\\ \chi _R \end {pmatrix} = \Psi _L^\dagger \,\chi _R + \Psi _R^\dagger \,\chi _L\quad , \end{flalign} i.e. it mixes between the left and the right-handed chiral components. This means that this expression is in general not gauge invariant whenever the two representations \(\rho _{L/R}\) are different, which is however the characteristic feature of a chiral gauge theory. Explicitly, under a gauge transformation we have in general that

\begin{flalign} \overline {\Psi }\,\chi \,\longmapsto \,\Psi _L^\dagger \,\rho _L(U^{-1})\,\rho _{R}(U)\,\chi _R + \Psi _R^\dagger \,\rho _R(U^{-1})\,\rho _L(U)\,\chi _L\,\neq \, \overline {\Psi }\,\chi \quad . \end{flalign} So mass terms are problematic in chiral gauge theories, but what about kinetic terms? These are much better, because

\begin{flalign} \nn \overline {\Psi }\,\gamma ^\mu \,\chi &= \Psi ^\dagger \,(\ii \,\gamma ^0)\,\gamma ^\mu \,\chi = -\ii \, \begin{pmatrix} \Psi _L^\dagger & \Psi _R^\dagger \end {pmatrix}\, \begin{pmatrix} 0 & 1_{2\times 2}\\ 1_{2\times 2} & 0 \end {pmatrix}\, \begin{pmatrix} 0 & \sigma ^\mu \\ \overline {\sigma }^\mu & 0 \end {pmatrix}\, \begin{pmatrix} \chi _L\\ \chi _R \end {pmatrix} \\[6pt] &=-\ii \, \begin{pmatrix} \Psi _L^\dagger & \Psi _R^\dagger \end {pmatrix}\, \begin{pmatrix} \overline {\sigma }^\mu & 0\\ 0& \sigma ^\mu \end {pmatrix}\, \begin{pmatrix} \chi _L\\ \chi _R \end {pmatrix} = -\ii \,\Big (\Psi _L^\dagger \,\overline {\sigma }^\mu \,\chi _L + \Psi _R^\dagger \,\sigma ^\mu \,\chi _R \Big ) \end{flalign} does not mix between the left and the right-handed chiral components, hence it is gauge invariant. This allows us to define the gauge invariant action functional

that consists only of a kinetic term with gauge covariant derivative. The cubic interaction term in this action will lead again, upon quantization, to a \(3\)-valent interaction vertex in the Feynman rules. We can further add this to the Yang-Mills action (9.43) and obtain the action functional

that is gauge invariant under the combined gauge transformations

\begin{flalign} T_{U}\Psi _{L/R} &= \rho _{L/R}(U)\,\Psi _{L/R}~~,\quad \\ T_U\Psi _{L/R}^\dagger &= \Psi _{L/R}^\dagger \,\rho _{L/R}(U^{-1})~~,\quad \\ (T_U A)_\mu &= U\,A_\mu U^{-1} + U\,\partial _\mu U^{-1}\quad . \end{flalign} It is important to emphasize again that a standard mass term \(m\,\overline {\Psi }\Psi \) is forbidden in this action because it is in general not gauge invariant for \(\rho _L\neq \rho _R\). The standard model of particle physics comes with a clever way to nevertheless introduce a suitable mass term, which is through the Higgs mechanism for spontaneous gauge symmetry breaking.