Algebra of the generator of supersymmetry transformations?

[alialice]

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.
Can you help me please?
Thank you very much!

 

[LBloom]

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.

 

[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?
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

 

[haushofer]

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.

 

[alialice]

Thank you!
So $\left\{ \bar{\theta}_{\alpha}, \frac{\partial}{\partial \bar{\theta}^{\beta}} \right\}=1$? Why?

 

[LBloom]

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.

 

[haushofer]

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.

 

[alialice]

Thank you for the help!

 

[alialice]

Which is the relation between theta and bar theta?
Or better, what is the result of
$\left\{ \theta_{\beta}, \frac{\partial}{\partial \bar{\theta}^{\alpha} }\right\}$ ?

 

[haushofer]

Apply to a test function and use the grassman properties.

 

[haushofer]

Did you manage to find it out?

 

[alialice]

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)!