- #1
Onamor
- 76
- 0
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 exercise in index/spinor notation:
I need to square the term \left(\theta^{\alpha}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\right)V_{\mu}
where \theta^{\alpha} is a Weyl spinor of grassman numbers, \bar{\theta}^{\dot{\beta}} is a spinor in conjugate rep, \sigma^{\mu} is a Pauli 4-vector (ie \left(1,\sigma^{i}\right)) and V_{\mu} is just a 4-vector.
The answer is \frac{1}{2}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)V^{\mu}V_{\mu} where \left(\theta\theta\right)\equiv\theta^{\alpha}\theta_{\alpha} and \left(\bar{\theta}\bar{\theta}\right)\equiv\theta_{\dot{\beta}}\theta^{\dot{\beta}}.
So, in detail, to square the term I want to multiply it by a similar term with upper and lower indices switched:
\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}.
First of all I switch the \bar{\theta}^{\dot{\beta}}\theta_{\delta} for free as they commute(?)
Then since \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},
and similarly \bar{\theta}^{\dot{\beta}}\bar{\theta}_{\dot{\gamma}}=\frac{1}{2}\left(\bar{\theta}\bar{\theta}\right)\delta^{\dot{\beta}}_{\dot{\gamma}}, I have
-\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}.
Or -\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}.
So, does \left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}=-2\delta^{\mu}_{\nu}?
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 \left(-1,\sigma^{i}\right) to square the term?
I'm checking a term when squaring a vector superfield in Wess-Zumino gauge, but its really just an exercise in index/spinor notation:
I need to square the term \left(\theta^{\alpha}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\right)V_{\mu}
where \theta^{\alpha} is a Weyl spinor of grassman numbers, \bar{\theta}^{\dot{\beta}} is a spinor in conjugate rep, \sigma^{\mu} is a Pauli 4-vector (ie \left(1,\sigma^{i}\right)) and V_{\mu} is just a 4-vector.
The answer is \frac{1}{2}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)V^{\mu}V_{\mu} where \left(\theta\theta\right)\equiv\theta^{\alpha}\theta_{\alpha} and \left(\bar{\theta}\bar{\theta}\right)\equiv\theta_{\dot{\beta}}\theta^{\dot{\beta}}.
So, in detail, to square the term I want to multiply it by a similar term with upper and lower indices switched:
\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}.
First of all I switch the \bar{\theta}^{\dot{\beta}}\theta_{\delta} for free as they commute(?)
Then since \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},
and similarly \bar{\theta}^{\dot{\beta}}\bar{\theta}_{\dot{\gamma}}=\frac{1}{2}\left(\bar{\theta}\bar{\theta}\right)\delta^{\dot{\beta}}_{\dot{\gamma}}, I have
-\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}.
Or -\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}.
So, does \left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}=-2\delta^{\mu}_{\nu}?
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 \left(-1,\sigma^{i}\right) to square the term?
Last edited: