I'm following the lecture notes by Timo Weigand on QFT.

On page 169, section 6.2 he is briefly touching on the non-abelian gauge symmetry in the SM.

The fundamental representation makes sense to me. For example, for $SU(3)$, we define the object or column vector with three component, suppressing spinor indices, $\psi(x)=(\psi_1, \psi_2, \psi_3)^T$. The fundamental representation, a $3*3$ matrix $V(h)$, acts on this column vector, with each component itself a Dirac spinor that has 4 components of complex numbers. Lagrangian density then is showed to remain invariant when the $\psi$ is multiplied by the $3*3$ matrix $V(h)$.

But the next step puzzles me a little, when we use the adjoint representation. Here instead of a 3 component column vector, we use a $3*3$ matrix $\psi(x)$, whose elements $\psi_{ij}(x)$ are complex numbers based on the definition given in Eqn.(6.32). $$\forall x: \psi(x) \equiv \psi_{i j}(x) \in\left\{\mathbb{C}^{N, N} \mid \psi^{\dagger}(x)=\psi(x), \operatorname{tr} \psi(x)=0\right\}$$

It is a $3*3$ matrix or representation of the $SU(3)$ Lie algebra.

But when in Eqn.(6.34) $$\mathcal{L}=\operatorname{tr} \bar{\psi}(i \partial-m) \psi \equiv \operatorname{tr}\left[\psi_{i j}(i \partial-m) \psi_{j l}\right] \equiv \psi_{i j}(i \partial-m) \psi_{j i} $$ the Lagranian density is defined as $tr(\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi)= \bar{ \psi_{ij}}(i\gamma^\mu\partial_\mu-m)\psi_{ji}$. Here $\psi$ is a $3*3$ matrix whose elements, $\psi_{ji}$, are not complext numbers but Dirac spinors.

In another word, is this what we mean by the spinor indices are suppressed in the adjoint representation? How can we think of $\psi(x)$ both as a $3*3$ matrix with components to be both complex numbers as defined in (6.32) and Dirac spinors as used in (6.34) that have themselves four components, each a complex number.


1 Answer 1


Perhaps this will help. Let $t^a$ be the generators of the Lie algebra is some representation. For example, this could be the adjoint representation as you're asking about and each $t^a$ will be a $3\times 3$ matrix.

Then any field $\phi$ valued in the Lie algebra may be written generically as $\phi=\phi_a t^a$. This is a common trick employed when dealing with gauge connections.

For the spinors, let $\alpha$ be the spinor index. Then we could write $$ \psi_\alpha=\psi_{\alpha a}t^a. $$ The fields $\psi_i$ would then be valued in the Lie algebra. If we wanted to suppress the mention of $t$, then we could define $$ \psi_{\alpha ij}=\psi_{\alpha a}(t^a)_{ij}. $$ Suppressing the spinor index means writing $\psi_{ij}$ and viewing each component as spinor-valued.


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