# Homework Help: Fermion Current Commutators in 2 dimensions

1. May 14, 2012

### EoinBrennan

1. The problem statement, all variables and given/known data

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:
$[J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)]$.

2. Relevant equations

Canonical equal time commutation relations:

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

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

3. 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?

1. The problem statement, all variables and given/known data

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 $\epsilon \rightarrow 0$.

2. Relevant equations

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

3. 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.

2. Mar 29, 2015

### Maybe_Memorie

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

3. Mar 31, 2015

Bump