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Homework Help: Commutator of Charges of the charged Weak Currents

  1. Dec 15, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm having a bit of trouble evaluating the following commutator

    $$ \left[T^{+},T^{-}\right] $$

    where [itex]T^{+}=\int_{M}d^{3}x\:\bar{\nu}_{L}\gamma^{0}e_{L}=\int_{M}d^{3}x\:\nu_{L}^{\dagger}e_{L} [/itex]


    [itex] T^{-}=\int_{M}d^{3}x\:\bar{e}_{L}\gamma^{0}\nu_{L}=\int_{M}d^{3}x\:e_{L}^{\dagger}\nu_{L}[/itex]

    a step necessary to prove that the [itex]\text{SU}(2)_L[/itex] is a part of the GWS Electroweak gauge symmetry group. Discussing with my professor I've been confused as to how I should proceed with this. Initially, I wrote:


    according to Paschos (Electroweak Theory) where my something of an unfamiliarity with QFT leaves me wondering exactly how I inserted that Dirac delta, but I see it's necessary.

    2. Relevant equations

    3. The attempt at a solution
    I thought of using the fermionic algebra, the anticommutator [itex]\left\{ a(y),a^{\dagger}(x)\right\} =1 \Rightarrow a(y)a^{\dagger}(x)=\left\{ a(y),a^{\dagger}(x)\right\} -a^{\dagger}(x)a(y)[/itex] along with the Dirac delta should let me reach [itex]\left[T^{+},T^{-}\right] = \frac{1}{2}\int_{M}d^{3}x\:\left(\nu_{L}^{\dagger}\nu_{L}-e_{L}^{\dagger}e_{L}\right)[/itex]

    and therefore


    but my professor says that this is wrong. I also tried [itex]\left[T^{+},T^{-}\right]=\left[\bar{\chi}_{L}\gamma^{0}\tau^{+}\chi_{L},\bar{\chi}_{L}\gamma^{0}\tau^{-}\chi_{L}\right][/itex] but I got weirded out by the [itex]u\bar{u}[/itex] factors and lucked out.

    Could someone construct the proof or guide me addressing some of these issues?
  2. jcsd
  3. Dec 17, 2015 #2


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    Science Advisor

    Where did you get the delta function from? And you don’t need it because the anticommutation relations produce the delta function for you. Just use the following identity, which relates commutators with anticommutators
    [AB,CD] =& A \big\{ C , B \big\} D - C \big\{ A , D \big\} B \\
    & + \big\{ A , C \big\} DB - AC \big\{ B , D \big\} .
    Now, take [itex]A = \nu^{\dagger}(x)[/itex], [itex]B = e(x)[/itex], [itex]C = e^{\dagger}(y)[/itex] and [itex]D = \nu (y)[/itex]. Then, the only non-zero anticommutators are
    \big\{ \nu^{\dagger}(x) , \nu (y) \big\} = \big\{ e^{\dagger}(x) , e(y) \big\} = \delta^{3}(x-y) .
    So, you get
    [T_{+},T_{-}] &= \int d^{3}x d^{3}y \left( \nu^{\dagger}(x)\nu(y) \delta (x-y) - e^{\dagger}(y) e(x) \delta(x-y) \right) \\
    &= \int d^{3}x \left( \nu^{\dagger}(x)\nu(x) - e^{\dagger}(x) e(x) \right) \\
    &= 2 \int d^{3}x \left( \nu^{\dagger}(x) , e^{\dagger}(x) \right) \left( \frac{\tau^{3}}{2} \right) \begin{pmatrix} \nu(x) \\ e(x) \end{pmatrix} \\
    &= 2 T^{3} .
  4. Dec 18, 2015 #3
    I wrote the delta function just plain following Paschos's book, assuming I should be doing it according to some QFT rule. Didn't really make much use of it and pretty much ignored it, but getting the x=y properties along the way.

    Now that's more like it samalkhaiat, that's along the lines of what I thought it would be. I missed a factor of 2 in my own work, which I will check again, but I'm all in all glad not to see it get much more complicated. Thanks!
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