Consider the vector current and axial vector current like $$j^{a\mu} = \bar{\psi}\gamma^{\mu}\frac{\tau^{a}}{2}\psi,$$ $$j^{a\mu}_5 = \bar{\psi}\gamma^{\mu}\gamma_5\frac{\tau^a}{2}\psi,$$ where $\tau^a$ is pauli matrices. Then the charges defined by $$Q^a = \int{d^3x j^{a0}(x, t)},\space \space \space Q^a_5 = \int{d^3x j_5^{a0}}.$$ I am having trouble proving the following commutation relations: $$[Q^a, Q^b] = i\epsilon^{abc}Q^c$$ $$[Q^a, Q_5^b] = i\epsilon^{abc}Q_5^c$$ $$[Q_5^a, Q_5^b] = i\epsilon^{abc}Q^c$$ I tried \begin{eqnarray}i\epsilon^{abc}Q^c &=& i\epsilon^{abc}\int{d^3x\bar{\psi}\gamma^{0}\frac{\tau^{c}}{2}\psi} = \int{d^3x\bar{\psi}\gamma^0\biggr[\frac{\tau^a}{2}, \frac{\tau^b}{2}\biggl]\psi} = \int{d^3x\bar{\psi}\gamma^0\frac{\tau^a}{2}\frac{\tau^b}{2}\psi} - \int{d^3x\bar{\psi}\gamma^0\frac{\tau^b}{2}\frac{\tau^a}{2}\psi}, \end{eqnarray} but I think this is not the correct way. How can I prove these relations?

  • $\begingroup$ Delete what you tried and write down explicitly the left hand side of each of your three relations: try the simplest first one: it is a double integral of the commutator of two fermion bilinears. Do the fermion commutators, obtaining delta functions collapsing one of the integrals and the group matrices into an epsilon contracted on a matrix. This is almost certainly a homework problem, or it should be. $\endgroup$ Commented Nov 25, 2019 at 20:16

1 Answer 1


Let us simplify the charges a little bit

\begin{equation} Q^a = \int \overline{\psi} \gamma ^0 \frac{\tau ^a}{2} \psi \mathrm{d}^3 x = \int \psi ^{\dagger} \gamma ^0 \gamma ^0 \frac{\tau ^a}{2} \psi \mathrm{d}^3 x = \int \psi ^{\dagger} \frac{\tau ^a}{2} \psi \mathrm{d}^3 x , \end{equation}

and similarly $Q_5^a = \int \psi ^{\dagger} \gamma _5 \frac{\tau ^a}{2} \psi \mathrm{d}^3 x$.

In the derivation, we need to use the following trick

\begin{equation} \left[ A B , C D \right] = A \left\{ B , C \right\} D - C \left\{ D , A \right\} B , \end{equation}

when $\left[ A , C \right] = \left[ B , D \right] = 0$ or $\left\{ A , C \right\} = \left\{ B , D \right\} = 0$.

Now knowing that $\left\{ \psi _i \left( x \right) , \psi _j \left( y \right) \right\} = \left\{ \psi _i ^{\dagger} \left( x \right) , \psi _j ^{\dagger} \left( y \right) \right\} = 0$, we have

\begin{equation} \begin{split} \left[ Q^a , Q^b \right] & = \iint \left[ \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) , \psi _k^{\dagger} \left( y \right) \frac{\tau ^b_{k l}}{2} \psi _l \left( y \right) \right] \mathrm{d}^3{x} \mathrm{d}^3{y} \\ & = \iint \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \left\{ \psi _j \left( x \right) , \psi _k^{\dagger} \left( y \right) \right\} \frac{\tau ^b_{k l}}{2} \psi _l \left( y \right) \mathrm{d}^3{x} \mathrm{d}^3{y} - \iint \psi _k^{\dagger} \left( y \right) \frac{\tau ^b_{k l}}{2} \left\{ \psi _l \left( y \right) , \psi _i^{\dagger} \left( x \right) \right\} \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} \mathrm{d}^3{y} \\ & = \iint \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \delta _{j k} \delta ^{\left( 3 \right)} \left( \vec{x} - \vec{y} \right) \frac{\tau ^b_{k l}}{2} \psi _l \left( y \right) \mathrm{d}^3{x} \mathrm{d}^3{y} - \iint \psi _k^{\dagger} \left( y \right) \frac{\tau ^b_{k l}}{2} \delta _{l i} \delta ^{\left( 3 \right)} \left( \vec{x} - \vec{y} \right) \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} \mathrm{d}^3{y} , \\ \end{split} \end{equation}

apply the $\delta$ functions

\begin{equation} \begin{split} \left[ Q^a , Q^b \right] & = \int \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \frac{\tau ^b_{j l}}{2} \psi _l \left( x \right) \mathrm{d}^3{x} - \int \psi _k^{\dagger} \left( x \right) \frac{\tau ^b_{k i}}{2} \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} \\ & = \int \left( \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \frac{\tau ^b_{j l}}{2} \psi _l \left( x \right) - \psi _k^{\dagger} \left( x \right) \frac{\tau ^b_{k i}}{2} \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \right) \mathrm{d}^3{x} , \\ \end{split} \end{equation}

rename the dummy indices

\begin{equation} \begin{split} \left[ Q^a , Q^b \right] & = \int \psi _i^{\dagger} \left( x \right) \left[ \frac{\tau ^a}{2} , \frac{\tau ^b}{2} \right]_{i j} \psi _j \left( x \right) \mathrm{d}^3{x} \\ & = \int \psi _i^{\dagger} \left( x \right) \imath \epsilon ^{a b c} \frac{\tau ^c_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} = \imath \epsilon ^{a b c} Q^c . \\ \end{split} \end{equation}

Same for $Q_5^a$.


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