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Fermion Current Commutators in 2 dimensions

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    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].

    2. Relevant 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]

    3. 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?

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

    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].

    2. Relevant 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].

    3. 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.
     
  2. jcsd
  3. Mar 29, 2015 #2
    Bumping this as it is the exact question I posted yesterday.
     
  4. Mar 31, 2015 #3
    Bump
     
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