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Spinor notation excercise with grassman numbers

  1. Mar 26, 2012 #1
    Spinor notation exercise with grassman numbers

    I'm checking a term when squaring a vector superfield in Wess-Zumino gauge, but its really just an excercise in index/spinor notation:

    I need to square the term [itex]\left(\theta^{\alpha}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\right)V_{\mu}[/itex]

    where [itex]\theta^{\alpha}[/itex] is a Weyl spinor of grassman numbers, [itex]\bar{\theta}^{\dot{\beta}}[/itex] is a spinor in conjugate rep, [itex]\sigma^{\mu}[/itex] is a Pauli 4-vector (ie [itex]\left(1,\sigma^{i}\right)[/itex]) and [itex]V_{\mu}[/itex] is just a 4-vector.

    The answer is [itex]\frac{1}{2}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)V^{\mu}V_{\mu}[/itex] where [itex]\left(\theta\theta\right)\equiv\theta^{\alpha}\theta_{\alpha}[/itex] and [itex]\left(\bar{\theta}\bar{\theta}\right)\equiv\theta_{\dot{\beta}}\theta^{\dot{\beta}}[/itex].

    So, in detail, to square the term I want to multiply it by a similar term with upper and lower indices switched:
    [itex]\left(\theta^{\alpha}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\right)\left(\theta_{\delta}\left(\sigma_{\nu}\right)^{\delta\dot{\gamma}}\bar{\theta}_{\dot{\gamma}}\right)V^{\nu}V_{\mu}[/itex].

    First of all I switch the [itex]\bar{\theta}^{\dot{\beta}}\theta_{\delta}[/itex] for free as they commute(?)

    Then since [itex]\theta^{\alpha}\theta_{delta}=\theta^{\alpha}\theta^{\lambda}\epsilon_{\lambda\delta}=-\frac{1}{2}\left(\theta\theta\right)\epsilon^{\alpha\lambda}\epsilon^{\lambda\delta}=-\frac{1}{2}\left(\theta\theta\right)\delta^{\alpha}_{\delta}[/itex],
    and similarly [itex]\bar{\theta}^{\dot{\beta}}\bar{\theta}_{\dot{\gamma}}=\frac{1}{2}\left(\bar{\theta}\bar{\theta}\right)\delta^{\dot{\beta}}_{\dot{\gamma}}[/itex], I have

    [itex]-\frac{1}{4}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)\delta^{\alpha}_{\delta}\delta^{\dot{\beta}}_{\dot{\gamma}}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\delta\dot{\gamma}}V^{\nu}V_{\mu}[/itex].

    Or [itex]-\frac{1}{4}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}V^{\nu}V_{\mu}[/itex].

    So, does [itex]\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}=-2\delta^{\mu}_{\nu}[/itex]?

    Its possible that a minus sign comes from the commutation of the two thetas in the first step, and perhaps I should have used the conjugate pauli 4-vector [itex]\left(-1,\sigma^{i}\right)[/itex] to square the term?
     
    Last edited: Mar 26, 2012
  2. jcsd
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