## Algebra of the generator of supersymmetry transformations?

We consider a superfield $\Phi$$\left(x^{\mu}, \theta_{\alpha}\right)$.
For a small variation $\delta \Phi$ = $\bar{\epsilon} Q \Phi$
where the supercharge $Q_{\alpha}$ is given by:
$Q_{\alpha}$=$\frac{\partial}{\partial \bar{\theta}^{\alpha}}$-$\left(\gamma^{\mu} \theta \right) _{\alpha} \partial _{\mu}$
They satisfy the algebra:
$\left\{ Q_{\alpha}, Q_{\beta} \right\}$ = -2$\left( \gamma^{\mu} C \right)_{\alpha \beta} \partial_{\mu}$
where C is the charge coniugation matrix.
How can I demonstrate this? The exercise is to calculate explicitely the anticommutator.
Thank you very much!

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 Do you know the rules for taking the derivatives of Grassmann variables? I think all you need is that the anticommutator of θ and its respective derivative is 1.
 I'm not sure of knowing it exactly... can you control my assumptions please? The anticommutator of two theta is zero: $\left\{ \theta_{\alpha}, \theta_{\beta} \right\}=0$ The anticommutator of the derivatives is also zero: $\left\{ \frac{\partial}{\partial \bar{\theta}^{\alpha}}, \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}=0$ You say that the anticommutator $\left\{ \theta_{\alpha}, \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}$ is equal to 1? And what is the result of an anticommutator like $\left\{ \left( \gamma^{\mu} \theta \right)_{\alpha} \partial_{\mu} , \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}$ = ? $\partial_{\mu}$ commute or anticommute with a theta? Thank you

Recognitions:

## Algebra of the generator of supersymmetry transformations?

 Quote by alialice I'm not sure of knowing it exactly... can you control my assumptions please? The anticommutator of two theta is zero: $\left\{ \theta_{\alpha}, \theta_{\beta} \right\}=0$ The anticommutator of the derivatives is also zero: $\left\{ \frac{\partial}{\partial \bar{\theta}^{\alpha}}, \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}=0$ You say that the anticommutator $\left\{ \theta_{\alpha}, \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}$ is equal to 1?
No, because there you differentiate wrt theta-bar. Not wrt theta.

 $\partial_{\mu}$ commute or anticommute with a theta? Thank you
They commutate; $\partial_{\mu}$ is a derivative wrt a bosonic coordinate.

 Thank you! So $\left\{ \bar{\theta}_{\alpha}, \frac{\partial}{\partial \bar{\theta}^{\beta}} \right\}=1$? Why?
 The same anticommutator relation also holds for the unbarred thetas. As to why that relation holds, I don't know if I can give an intuitive reason. In general the property is just axiomized to get desired properties. Any good introduction SUSY text will probably give a good motivation.
 Recognitions: Science Advisor See e.g. eq.(11.47) of the book "Supersymmetry demystified", which is a very nice first exposure to SUSY: $$\{ \partial_a, \theta^b \}f = \partial_a (\theta^b f) + \theta^b \partial_a f = \partial_a \theta^b f - \theta^b \partial_a f + \theta^b \partial_a f = \delta_a^b f$$ where the minus-sign comes from the fact that you work with Grassmannian variables.
 Thank you for the help!
 Which is the relation between theta and bar theta? Or better, what is the result of $\left\{ \theta_{\beta}, \frac{\partial}{\partial \bar{\theta}^{\alpha} }\right\}$ ?
 Recognitions: Science Advisor Apply to a test function and use the grassman properties.
 Recognitions: Science Advisor Did you manage to find it out?
 Yes I managed, thanks, but I did't apply the anticommutator to a test function, I wrote theta as a bar theta(it appears a conjugation matrix in 4 dim or a gamma matrix in 2 dim)!