- #1
EoinBrennan
- 12
- 0
Homework Statement
Given the current: [itex]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) [/itex] with [itex]\psi_{L} = \frac{1}{2} (1 - \gamma^{5}) \psi_{D}[/itex].
Use the canonical equal time commutation relations for fermions to compute the equal time commutator:
[itex][J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)][/itex].
Homework Equations
Canonical equal time commutation relations:
[itex]\{\psi_{a} (x), \psi^{\dagger}_{b} (y)\} = i \delta^{3} (x - y) \delta_{a b}[/itex]
[itex]\{\psi_{a} (x), \psi_{b} (y)\} = \{\psi^{\dagger}_{a} (x), \psi^{\dagger}_{b} (y)\} = 0[/itex]
The Attempt at a Solution
So [itex][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)[/itex]
From here I'm not sure what path to take.
[itex]\{ \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) [/itex]
Subbing this into the commutation relation gives
i.e. [itex][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)[/itex]
With [itex]\psi^{\dagger}_{L} (x + \epsilon) \psi^{\dagger}_{L} (y + \epsilon) = \frac{1}{2} \{ \psi^{\dagger}_{L} (x + \epsilon), \psi^{\dagger}_{L} (y + \epsilon) \} = 0[/itex], etc.
So now I have
[itex][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)[/itex]
Is this all correct?
Homework Statement
I am then asked to evaluate [itex]\langle 0 \vert [J^{\epsilon}_{0} (t,x), J^{\epsilon}_{0} (t, y)] \vert 0 \rangle[/itex] in the massless case, and the limit as [itex]\epsilon \rightarrow 0[/itex].
Homework Equations
I am given that [itex]\langle 0 \vert \psi^{\dagger}_{L} (t,x) \psi_{L} (t,y) \vert 0 \rangle = \frac{1}{x - y}[/itex].
The Attempt at a Solution
So [itex]\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[/itex]
I'm not quite sure how operators like [itex]\langle 0 \vert[/itex] act on the [itex]\delta[/itex] terms.
But it seems like the answer will be:
[itex]\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}[/itex]
As [itex]\epsilon \rightarrow 0[/itex] we get: [itex]\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}[/itex].
Is this [itex]= 0[/itex]?
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.