Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am trying to understand the structure of the fermions in non-abelian gauge theories. Disclaimer: my question might be very trivial (I suspect the answer could simply be "a change of basis"), but I would be grateful is someone could shed some light if there's something deeper lurking around the corner.

Let's consider the Yang-Mills Lagrangian

$$ \mathcal{L} = -\frac{1}{4} (F_{\mu \nu}^a)^2 + \bar{\psi}(i \gamma^\mu D_\mu -m)\psi $$

where $ D_\mu = \mathbf{1} \partial_\mu - i g A_\mu^a t^a$, then $\psi$ needs to have both Dirac ($\mu$) and colour (a) degrees of freedom. I get confused when I change the regular approach to the problem and I am not sure if it is a real issue or if I'm just overcomplicating/overthinking this.

For instance, let's consider $SU(N)$ YM where the generators $t^a$ are $N \times N$ matrices, and there are $N^2-1$ of them. Therefore, the covariant derivative $D_\mu$ is an $N \times N$ matrix, and the index $a$ above runs over $N^2-1$ value.

When we start by contracting $D_\mu$ with the $\gamma$-matrices,

$$ (\gamma^\mu D_\mu)_{ij} = \delta_{ij} \gamma^\mu \partial_\mu - i g (\gamma^\mu A_\mu^a) (t^a)_{ij} $$

one gets an $N \times N$ matrix of $4 \times 4$ matrices. The corresponding $4N$-component object this matrix acts upon is the $N$ Dirac spinors arranged in a column.

However, notice that if we start in the following way instead:

$$ (D_\mu \psi)_i = \partial_\mu \psi_i - i g A_\mu^a (t^a)_{ij} \psi_j $$

we get that the covariant derivative acts on $\Psi \equiv (\psi_1, \cdots, \psi_N)$. For all we know, $\Psi$ has no spinor structure since we haven't contracted with the gamma matrices yet.

Contracting with the $\gamma^\mu$, one gets a $4 \times 4$ matrix of $N \times N$ matrices that encodes the same information as before. This time, it would seem that we only have one Dirac spinor, where each component is a $N$-valued singlet.

However, it seems that the matrix $\gamma^\mu D_\mu$ and the $4N$-component spinor look different, even though all we did was to change the order in which we constructed things.

In the first case, we get $N$ spinors corresponding to the $N$ colors of the adjoint representation. In $SU(3)$, this would be like saying we effectively have 3 spinors which correspond to the red/blue/green colors as $(\psi_R, \psi_B, \psi_G)$. In the second case, this identification fails since we only have one big complicated object.

What went wrong? Is this difference simply a change of basis for $\psi$? Is there something relevant we can learn from looking at the YM Lagrangian in these two different ways?

Also, in the first case, when we get $N$ spinors, I am confused about their signification. I always assumed that in QCD, $\psi$ would correspond to a quark, which is a fermion by itself. Does it mean that quarks are fermions that can be described by fermionic fields/degrees of freedom that we call color?

share|improve this question

2 Answers 2

In a sense, you are right. It's the same information, put in a different order. Try to do that with a simpler case, for instance two two-dimensional vector space, one with index $\alpha=1,2$, the other with index $a=1,2$. Then any vector of this now four-dimensional vector space can be written either $\psi_{\alpha,a}$ or $\psi'_{a,\alpha}$ or even $v_i$ with $i=1,2,3,4$. Then the matrix $\sigma$ and $S$ acting respectively on the space $\alpha$ and $a$ can associated to create $\gamma=\sigma \otimes S$ or $\gamma'=S\otimes \sigma$ acting respectively on $\psi_{\alpha,a}$ and $\psi'_{a,\alpha}$, or any other complicated 4 by 4 matrix acting $v$ depending on how we associate a given $i=1,2,3,4$ to a set $(\alpha,a)=(1,1), (1,2),(2,1),(2,2)$.

The choice to associate $i$ to $(\alpha,a)$ is free, but some choices are more natural than others. In you case, we are used to this of the system as $N$ fermions, described by spinors, which corresponds to $1\otimes \gamma_\mu$, because we think of the well defined spinors in interaction through the gauge field. But this is just a convenient representation.

share|improve this answer

You may write :

$(\gamma^\mu D_\mu) (\psi_i)_k = [\delta_i^j \partial_\mu - i g A_\mu^a (t_a)_{i}^j] ~~ (\gamma^\mu)^k_l ~~(\psi_j)_l \tag{1}$

Here $i,j$ are in $1..N$, and $k,l$ are in $1..4$, $(\psi_i)_k$ is the k-th ($1 \leq k \leq 4$) component of the i-th ($1 \leq i \leq N$) spinor.

We could use the notation $\psi_{j~l} = (\psi_j)_l$, now we see that $[\delta_i^j \partial_\mu - i g A_\mu^a (t_a)_{i}^j]$ is acting on the first indice of $\psi_{j~l}$, while $(\gamma^\mu)^k_l$ is acting on the second indice of $\psi_{j~l}$, so we could use a tensorial compact notation, with $\psi$ representing the $\psi_{{j~l}}$ :

$(\gamma^\mu D_\mu) ~\psi = ([\mathbb{Id}~ \partial_\mu- i g A_\mu^a t_a] \otimes \tag{2}\gamma^\mu) ~\psi $

Of course, it is not very much useful for practical calculus, but it is, at least, a view on the whole structure.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.