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Commutation relation between charges in current algebra

Physics Asked by Seal on August 4, 2021

Consider the vector current and axial vector current like
$$j^{amu} = bar{psi}gamma^{mu}frac{tau^{a}}{2}psi,$$
$$j^{amu}_5 = bar{psi}gamma^{mu}gamma_5frac{tau^a}{2}psi,$$
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:
$$[Q^a, Q^b] = iepsilon^{abc}Q^c$$
$$[Q^a, Q_5^b] = iepsilon^{abc}Q_5^c$$
$$[Q_5^a, Q_5^b] = iepsilon^{abc}Q^c$$
I tried
begin{eqnarray}iepsilon^{abc}Q^c &=& iepsilon^{abc}int{d^3xbar{psi}gamma^{0}frac{tau^{c}}{2}psi} = int{d^3xbar{psi}gamma^0biggr[frac{tau^a}{2}, frac{tau^b}{2}biggl]psi} = int{d^3xbar{psi}gamma^0frac{tau^a}{2}frac{tau^b}{2}psi} – int{d^3xbar{psi}gamma^0frac{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?

One Answer

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 $Q_5^a$.

Answered by zyy on August 4, 2021

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