Gauge fields -- why are they traceless hermitian?

Question

A gauge field is introduced in the theory to preserve local gauge invariance. And this field (matrix) is expanded in terms of the generators, which is possible because the gauge field is traceless hermitian.

Now why did we choose it as traceless Hermitian? What was the insight behind the choice that made us expand it in terms of the generators? I read somewhere that 'the gauge field belongs to the Lie algebra' and I tried to follow it up, but I can't make any sense out of what I read. Can anybody explain in clear, intuitive terms?

Answer 1

The elements of the gauge transformations belong to a gauge group. In physics, it's most typically $SU(N)$ (both the electroweak theory, with its $SU(2)$, and the QCD for quarks, $SU(3)$, use these $SU(N)$ groups; $U(1)$ we first learn in electromagnetism – but we must reinterpret the charge as the "hypercharge" when we study the electroweak theory – is the only extra addition we need for the Standard Model). It's a group of all complex $N\times N$ matrices $M$ that obey $$MM^\dagger=1, \quad \det M = 1$$ Note that $M^\dagger=(M^*)^T$ is the Hermitian conjugate; the first condition makes the matrix "unitary", therefore $U$. The determinant of a unitary matrix could be any complex number whose absolute value equals one. The second condition says that the determinant must be one and nothing else, that's the "special" or $S$ condition in $SU(N)$.

The gauge field transforms as $$ A_\mu \to M(A_\mu+ie\partial_\mu) M^\dagger$$ up to different conventions. That's needed for the covariant derivative $D_\mu$ to transform nicely. Forget about the complicated formula above. The point is that $A_\mu$ takes values in the Lie algebra of the Lie group.

In other words, you may imagine an infinitesimal transformation – infinitely close to the identity – in the gauge group, e.g. $SU(N)$. Assume $$ M = 1+i\epsilon G $$ The factor $\epsilon$ makes it infinitesimal, the factor of $i$ is a convention popular among physicists but omitted by mathematicians (physicists like things to be Hermitian, without $i$, they would have to be anti-Hermitian).

Here, $G$ is the kind of $N\times N$ matrix that the gauge field can have as a value.

Now, substitute this Ansatz for $M$ into the conditions $MM^\dagger=1,\det M=1$. You may neglect $\epsilon^2$ "very small" terms and the conditions become $$1+i\epsilon G - i\epsilon G^\dagger = 1, \quad \det(1+i\epsilon G) = 1$$ Mathematics implies that these conditions are equivalent to $$ G = G^\dagger, \quad {\rm Tr}(G) =0 .$$ To get the first one, I just subtracted $1$ from both sides and cancelled $i\epsilon$. To get the latter, I used the "sum of products over permutations" formula for the determinant and noticed that only the product of the diagonal entries contributes $O(\epsilon)$ terms and they're proportional to the sum of the diagonal entries, the trace.

At any rate, you should try to understand this maths and its conclusion is that the Hermiticity of the generator $G$ – matrices that are combined with various real coeffcients to get $A_\mu$ – is equivalent to the gauge group's being unitary; and the tracelessness is equivalent to the group's being "special" i.e. requiring the unit determinant.

It's perhaps useful to mention why $SU(N)$ is considered the "simplest" class of gauge groups. The $S$ has to be there because $U(N)$ isn't simple – it's pretty much isomorphic to $SU(N)\times U(1)$ where the two factors could be treated separately and we want to work with the smallest allowed pieces of gauge groups which are $SU(N)$ and $U(1)$. And $SU(N)$ is more "elementary" than $SO(N)$ or $USp(2N)$ because complex numbers are more fundamental in group theory (and physics) than real numbers or quaternions. In fact, the groups $SO(N)$ and $USp(2N)$ may be defined as $SU(N)$ with some "extra structure" (orientifolds) added which makes some natural group-theoretical analyses somewhat more convoluted than those for $SU(N)$. But one may still say that the Lie algebra for $SO(N)$ would be composed of antisymmetric real (or antisymmetric pure imaginary, depending on the conventions concerning factors of $i$) matrices, in analogy with the Hermitian matrices above; they're automatically traceless.

Answer 2

What we want to achieve is an invariance of the Lagrangian under a certain symmetry. These symmetries are described by Lie groups (if they are continuous).

Let's take QCD as a working example: We want our Lagrangian to be invariant under certain redefinitions of color, i.e. $$ \psi = \begin{pmatrix} q_r \\ q_b \\ q_g \end{pmatrix} \mapsto \psi' = \begin{pmatrix} q'_r \\ q'_b \\ q'_g \end{pmatrix} = \mathbf{U} \begin{pmatrix} q_r \\ q_b \\ q_g \end{pmatrix}$$ where $\mathbf U$ is a unitary matrix with unit derminant. That means it belongs to the group $SU(3)$. Lie groups like $SU(3)$ are differentiable manifolds, i.e. we can exand it as a kind of Taylor series around the unit element $$ \mathbf U = \exp\left( i g \alpha^a(x) T^a\right) = \mathbf 1 + i g \alpha^a(x) T^a + \dots $$ where I already made explicit that we are interested in redefining color locally, i.e. we can do a different definition at any point in spacetime.

In order for the Matrix $\mathbf U$ to be unitary, the $T^a$ who are the generators must be hermitian $$ \mathbf U^\dagger = \exp\left( -i g \alpha^a(x) (T^a)^\dagger \right) \overset{!}{=} \mathbf U^{-1} = \exp\left( -i g \alpha^a(x) T^a \right)$$ Moreover, using $\operatorname{det}(e^\mathbf A) = e^{ \operatorname{tr} \mathbf A} $ we find that $$\operatorname{det} \mathbf U = 1 \Leftrightarrow \operatorname{tr} T^a = 0$$

What has this to do with the gauge field?

Look at the kinetic term $$\bar \psi i \gamma^\mu \partial_\mu \psi \mapsto \bar \psi \mathbf U^\dagger i \gamma^\mu \partial_\mu \mathbf U \psi$$ To evaluate the partial, we must use the Leibnitz rule and using $\partial_\mu \mathbf U = \mathbf U * (i g \partial_\mu \alpha^a(x) (T^a)_{ij})$ we get $$ \bar \psi_i \mathbf U^\dagger \mathbf U i \gamma^\mu \partial_\mu \psi_i - \bar \psi_i \mathbf U^\dagger \mathbf U g \gamma^\mu \big(\partial_\mu \alpha^a(x)\big) T^a_{ij} \psi_j $$ where I did write out the color indices $(i, j)$ on the spinors and the generator, but left out those of $\mathbf U^\dagger_{ij} \mathbf U_{jk} = \delta_{ik}$, since they drop out anyway. In order to get rid of the second term we introduce a new field defined by its transformation property under a $SU(3)$ transformation $$A_\mu(x) = A^a_\mu T^a \mapsto \big(A^a_\mu - \partial_\mu \alpha^a(x)\big) T^a$$ just to cancel the new term we got from Leibnitz. That's why the gauge field must be a linear combination of generators - the term we want to kill is!

The gauge-kinetic term $$\bar \psi i \gamma^\mu (\partial_\mu + i g A_\mu) \psi$$ is actually invariant under the $SU(3)$ transformation on $\psi$ an $A_\mu$ simultaneousely. That's a nice exercise to carry out.