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:

<br /> \{ \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<br />
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)!