# Chiral Symmetry and Charge Algebra

I'm following Section 5.1 in Cheng and Li's Particle Physics book and I am having trouble reproducing some of the commutation relations.

The axial current is given by $$(J_A^a)^\mu = \bar{\psi}_\alpha\gamma^\mu\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\gamma_5\psi_\beta$$ where $$\lambda^a$$ are the Gell-Mann matrices. The associated charge is then $$Q^{5a} = \int d^3 x\ \bar{\psi}_{\alpha}(x)\gamma^0\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\gamma_5\psi_\beta(x)$$ The 'normal' current is given by $$(J^a)^\mu = \bar{\psi}_\alpha\gamma^\mu\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\psi_\beta$$ and has associated charge $$Q^{a} = \int d^3 x\ \bar{\psi}_{\alpha}(x)\gamma^0\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\psi_\beta(x)$$ I am trying to show that $$[Q^a,Q^{5b}] = if^{abc}Q^{5c}$$ where $$f^{abc}$$ are the structure constants of $$SU(3)$$. My work so far is, $$\begin{equation} \begin{split} [Q^a,Q^{5b}] &= \int d^3x\ d^3 y\ \left[\bar{\psi}_\alpha(x)\gamma^0 (T^a)_{\alpha\beta}\psi_\beta(x),\bar{\psi}_{\gamma}(y)\gamma^0(T^b)_{\gamma\delta}\gamma_5\psi_\delta(y)\right]\\[0.25cm] &=\int d^3x\ d^3 y\ (T^a)_{\alpha\beta}(T^b)_{\gamma\delta} \left[\psi_\alpha^\dagger(x) \psi_\beta(x),\psi_{\gamma}^\dagger(y)\gamma_5\psi_\delta(y)\right] \end{split} \end{equation}$$ Using, $$[A B, C D]=A\left\{B, C\right\} D - AC\left\{B,D\right\} + \left\{A,C\right\}DB - C\left\{A,D\right\}B$$ and the normal anti-commutation relations for fermions, $$\begin{multline} \left[\psi_\alpha^\dagger(x) \psi_\beta(x),\psi_{\gamma}^\dagger(y)\gamma_5\psi_\delta(y)\right] =\\ \psi_\alpha^\dagger(x)\left\{\psi_\beta(x),\psi_\gamma^\dagger(y)\right\}\gamma_5\psi_\delta(y) -\psi_\alpha^\dagger(x)\psi_\gamma^\dagger(y)\left\{\psi_\beta(x),\gamma_5\psi_\delta(y)\right\} -\psi_\gamma^\dagger(y)\left\{\psi_\alpha^\dagger(x),\gamma_5\psi_\delta(y) \right\}\psi_\beta(x) \end{multline}$$ The first term is consistent what with I expect. However, I don't know how to deal with the anti-commutators that contain $$\gamma_5$$.

Any help would be appreciated!

• The $\gamma_5$s are just numbers: numerical coefficients. So, give them indices, and take them out of the anticommutators. Only two anticommutators are nontrivial. – Cosmas Zachos Feb 5 '20 at 17:25

This relation will be useful for you:

$$[\psi^\dagger(t,\textbf x) \Gamma^A \lambda^a \psi(t,\textbf x), \psi^\dagger(t,\textbf y) \Gamma^B \lambda^b \psi(t,\textbf y)] = \delta^{(3)}(\textbf x-\textbf y) \psi^\dagger(t,\textbf x)[\Gamma^A \lambda^a, \Gamma^B \lambda^b]\psi(t,\textbf y) \tag{1}$$

$$\Gamma^A \to 1,\quad \Gamma^B\to^5,\quad \lambda^a \to T^a,\quad \lambda^b\to T^b \tag{2}$$

Now instead of using the commutator relation you wrote, I'd use

$$[A B, C D] = A[B,C]D + [A,C]BD + CA[B,D] + C[A,D]B \tag{3}$$

and we see that $$A$$ with $$D$$, as well aas $$B$$ with $$C$$ won't contribute, since the commutator of a Dirac matrix with a Gell-Mann matrix is zero. Therefore,

$$[\Gamma^A \lambda^a, \Gamma^B \lambda^b] = [\Gamma^A, \Gamma^B]\lambda^a\lambda^b + \Gamma^B\Gamma^A [\lambda^a,\lambda^b] \tag{4}$$

Finally, $$[1,\gamma^5]=0$$ and $$[T^a, T^b]=\text{i} f^{abc}T^c$$.

As mentioned in a comment, this can be proven by writing the product of spinors and Dirac/Gell-Mann matrices in index notation:

$$\psi^\dagger \Gamma \lambda \psi = \psi^\dagger_{\alpha, i} \Gamma_{\alpha\beta} \lambda_{ij}\psi_{\beta, j} \tag{5}$$

By doing this, you can pull the $$\Gamma$$'s and $$\lambda$$'s out of the anti-commutator, as if they were just scalar numbers (which, in index notation, they are!).