Fermion Current Commutators in 2 dimensions

[EoinBrennan]

Homework Statement

Given the current: J^{\epsilon}_{0} (t,x) = \overline{\psi_{L}}(t,x + \epsilon) \gamma^{0} \psi_{L}(t,x - \epsilon) = \psi_{L}^{\dagger} (x + \epsilon) \psi_{L}(x - \epsilon) with \psi_{L} = \frac{1}{2} (1 - \gamma^{5}) \psi_{D}.

Use the canonical equal time commutation relations for fermions to compute the equal time commutator:
[J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)].

Homework Equations

Canonical equal time commutation relations:

\{\psi_{a} (x), \psi^{\dagger}_{b} (y)\} = i \delta^{3} (x - y) \delta_{a b}

\{\psi_{a} (x), \psi_{b} (y)\} = \{\psi^{\dagger}_{a} (x), \psi^{\dagger}_{b} (y)\} = 0

The Attempt at a Solution

So [J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] = \psi^{\dagger}_{L} (x + \epsilon) \psi_{L} (x - \epsilon) \psi^{\dagger}_{L} (y + \epsilon) \psi_{L} (y - \epsilon) - \psi^{\dagger}_{L} (y + \epsilon) \psi_{L} (y - \epsilon) \psi^{\dagger}_{L} (x + \epsilon) \psi_{L} (x - \epsilon)

From here I'm not sure what path to take.

\{ \psi_{L} (x - \epsilon), \psi^{\dagger}_{L} (y + \epsilon) \} = i \delta^{3} (x - y - 2 \epsilon) \\ \Rightarrow \psi_{L} (x - \epsilon) \psi^{\dagger}_{L} ( y + \epsilon) = i \delta^{3} (x - y - 2 \epsilon) - \psi^{\dagger}_{L} ( y + \epsilon) \psi_{L} (x - \epsilon)

Subbing this into the commutation relation gives

i.e. [J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] = \psi^{\dagger}_{L} (x + \epsilon) ( i \delta^{3} (x - y - 2 \epsilon) - \psi^{\dagger}_{L} ( y + \epsilon) \psi_{L} (x - \epsilon)) \psi_{L} (y - \epsilon) - \psi^{\dagger}_{L} (y + \epsilon) ( i \delta^{3} (y - x + 2 \epsilon) - \psi^{\dagger}_{L} ( x + \epsilon) \psi_{L} (y - \epsilon)) \psi_{L} (x - \epsilon)

With \psi^{\dagger}_{L} (x + \epsilon) \psi^{\dagger}_{L} (y + \epsilon) = \frac{1}{2} \{ \psi^{\dagger}_{L} (x + \epsilon), \psi^{\dagger}_{L} (y + \epsilon) \} = 0, etc.

So now I have

[J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] = \psi^{\dagger}_{L} (x + \epsilon) ( i \delta^{3} (x - y - 2 \epsilon)) \psi_{L} (y - \epsilon) - \psi^{\dagger}_{L} (y + \epsilon) ( i \delta^{3} (y - x + 2 \epsilon)) \psi_{L} (x - \epsilon) \\ = i \delta^{3} (x - y - 2 \epsilon) \psi^{\dagger}_{L} (x + \epsilon) \psi_{L} (y - \epsilon) - i \delta^{3} (y - x + 2 \epsilon) \psi^{\dagger}_{L} (y + \epsilon) \psi_{L} (x - \epsilon)

Is this all correct?

Homework Statement

I am then asked to evaluate \langle 0 \vert [J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] \vert 0 \rangle in the massless case, and the limit as \epsilon \rightarrow 0.

Homework Equations

I am given that \langle 0 \vert \psi^{\dagger}_{L} (t,x) \psi_{L} (t,y) \vert 0 \rangle = \frac{1}{x - y}.

The Attempt at a Solution

So \langle 0 \vert [J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] \vert 0 \rangle = \langle 0 \vert i \delta^{3} (x - y - 2 \epsilon) \psi^{\dagger}_{L} (x + \epsilon) \psi_{L} (y - \epsilon) - i \delta^{3} (y - x + 2 \epsilon) \psi^{\dagger}_{L} (y + \epsilon) \psi_{L} (x - \epsilon) \vert 0 \rangle \\ = \langle 0 \vert i \delta^{3} (x - y - 2 \epsilon) \psi^{\dagger}_{L} (x + \epsilon) \psi_{L} (y - \epsilon) \vert \rangle - \langle 0 \vert i \delta^{3} (y - x + 2 \epsilon) \psi^{\dagger}_{L} (y + \epsilon) \psi_{L} (x - \epsilon) \vert 0 \rangle

I'm not quite sure how operators like \langle 0 \vert act on the \delta terms.

But it seems like the answer will be:

\langle 0 \vert [J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] \vert 0 \rangle = i \delta^{3} (x - y - 2 \epsilon) \frac{1}{x - y -2 \epsilon} - i \delta^{3} (y - x + 2 \epsilon) \frac{1}{y - x + 2 \epsilon}

As \epsilon \rightarrow 0 we get: \langle 0 \vert [J^{0}_{0} (t,x), J^{0}_{0} (t, y)] \vert 0 \rangle = i \delta^{3} (x - y) \frac{1}{x - y} - i \delta^{3} (y - x) \frac{1}{y - x}.

Is this = 0?

Any feedback would be greatly appreciated. I have very little support from my lecturer and I'm feeling a bit overwhelmed by Quantum Field Theory.

 

[Maybe_Memorie]

Bumping this as it is the exact question I posted yesterday.

 

[Maybe_Memorie]

Bump