# The Fifth Gamma Matrix

This is regarding $\gamma^5$, the fifth gamma matrix in quantum field theory. I know its defining properties, namely,

$$\gamma^5= -i\gamma^0 \gamma^1 \gamma^2 \gamma^3$$

with $\{\gamma^5,\gamma^{\mu}\}=0$ and $(\gamma^5)^2=1$, but so far all I've used these for is to help prove some gamma matrix identities.

It seems obvious to me that $\gamma^5$ is dependent on our representation (e.g. Weyl or Dirac), so my original idea that it's useful because of invariance on choice of basis seems wrong.

What do we like about $\gamma^5$?

• In the full Standard model, certain Feynman rules incorporate $\gamma^5$, e.g. in the charged current interaction vertex. In addition, the fifth gamma matrix can be used to construct axial vectors and pseudo-scalars with spinors. Mar 31, 2014 at 10:51
• So it's just a notation to denote the product of the gamma matrices? Because this product just so happens to be useful? Mar 31, 2014 at 11:21

One reason why $\gamma^5$ is important is because it corresponds to a symmetry of the massless Dirac Lagrangian: $$\mathcal{L} = i \bar{\psi} \gamma^\mu \partial_\mu \psi$$ and the global transformation: $$\psi \to \psi' = e^{i \alpha \gamma^5} \psi \tag{1}$$ Then using the identity: $$\{ \gamma^5,\gamma^\mu \} = 0$$ we can show: $$\gamma^\mu e^{i \alpha \gamma^5} = e^{-i \alpha \gamma^5} \gamma^\mu$$ Therefore, it is easy to see that: $$\bar{\psi} \to \bar{\psi}' = \bar{\psi} e^{ i \alpha \gamma^5}$$ Thus, we see that the transformation expressed by equation $(1)$ is a symmetry of the massless Dirac Lagrangian: $$\mathcal{L} \to \mathcal{L}' = \mathcal{L}$$ which is known as the axial symmetry. This symmetry will play an important role in the Standard Model and its anomalies. I'm sure you will study this at some point in the future.

Edit:

As innisfree mentions in the comments, using the $\gamma^5$ matrices, we can form the Lorentz invariant projection operators: $$P_{\pm} = \frac{1}{2} ( \mathbb{1} \pm \gamma^5)$$ such that: $$P_+ P_+ = P_+$$ $$P_- P_- = P_-$$ $$P_+ P_- = P_- P_+ = 0$$ (Note that the above three equations are valid regardless of the representation used.) In the chiral representation, they act on the Weyl spinor as: $$P_+ \psi = \begin{pmatrix} 0 \\ \psi_R \end{pmatrix}$$ $$P_- \psi = \begin{pmatrix} \psi_L \\ 0 \end{pmatrix}$$ In addition, for an any representation of the Clifford algebra, we use $\gamma^5$ to define the left- and right-handed parts of the field $\psi$: $$\psi_R = P_+ \psi$$ $$\psi_L = P_- \psi$$

• true, but surely most useful feature is that it is a projection operator that gives the chiral parts of a spinor? Mar 31, 2014 at 11:42
• @innisfree Yeah I agree, but this is (often) one of the first things you learn when you come across the $\gamma^5$ matrix (and so I kind of assumed the OP has already seen this). I thought it would interesting for the OP to see this symmetry Mar 31, 2014 at 11:45
• @Hunter Thanks for the edit. You were right, I have seen this property of the fifth gamma matrix (it appears in any QFT text) but not in as much detail. The symmetry was also very helpful as I hadn't realised\seen it before Mar 31, 2014 at 12:03
• @user13223423 happy to hear that. I didn't pay much attention to the axial symmetry the first time I came across it. But once you will study the SM, you'll realize it is very important. (Personally, I think anomalies are fascinating). Mar 31, 2014 at 12:09

In the standard model, and more generally in quantum field theory, the fifth gamma matrix has several uses. The charge current interaction between $\psi_{u,d}$, $\psi_{d,u}$ and a $W^{\pm}_\mu$ carries a factor of

$$V = \frac{ig}{\sqrt{2}}\gamma_\mu \frac{1-\gamma^5}{2}$$

In addition, the fifth gamma matrix can be used to construct Lagrangians with pseudo-scalars, e.g. $\bar{\psi}\gamma^5 \psi$, or axial vectors such as $\bar{\psi}\gamma^5 \gamma^\mu \psi$.