Commutation relations in SUSY

In summary: We can now use the property of anti-commutativity with derivatives to simplify this further:$$\{\frac{\partial}{\partial \theta^\alpha}, \partial_\mu\} = \{\partial_\mu, \frac{\partial}{\partial \theta^\alpha}\} = \partial_\mu \{\partial_\nu, \frac{\partial}{\partial \theta^\alpha}\} = -\partial_\mu \delta^\nu_\alpha = -\
  • #1
I hope I put this in the correct section of this forum, I apologize if I didn't.

Homework Statement


It is well known that the generators

Q_\alpha = \frac{\partial}{\partial \theta^\alpha} - i \sigma^\mu_{\alpha \dot \beta} \bar{\theta}^\dot{\beta} \partial_\mu


\bar{Q}_\dot{\alpha} = -\frac{\partial}{\partial \bar{\theta}^\dot{\alpha}} + i \theta^\beta\sigma^\mu_{\beta \dot \alpha} \partial_\mu

$$ \theta^\alpha, \bar{\theta}^\dot{\beta} $$
are Grassmann variables, obey the anti-commutation relations

\{Q_\alpha, \bar{Q}_\dot{\alpha}\} = 2i \sigma^\mu_{\alpha \dot \alpha} \partial_\mu
\{Q_\alpha, Q_\beta\} = \{\bar{Q}_\dot{\alpha}, \bar{Q}_\dot{\beta}\} = 0

I am asked to explicitly verify those anti-commutation relations, say for example
$$ \{Q_\alpha, Q_\beta\} = 0 $$

Homework Equations

see above

The Attempt at a Solution

However, I'm unable to reproduce that result. I might get as far as follows, by simply expanding the anti-commutator, provided I did not make a mistake (I've never had to deal with Grassmann variables before, so that is a real possibility).

\{Q_\alpha, Q_\beta\} = \{\frac{\partial}{\partial \theta^\alpha}, \frac{\partial}{\partial \theta^\alpha}\}
- i \sigma^\mu_{\beta \dot \beta} \bar{\theta}^{\dot{\beta}} \{\frac{\partial}{\partial \theta^\alpha}, \partial_\mu\}
- i \sigma^\mu_{\alpha \dot \beta} \bar{\theta}^{\dot{\beta}} \{ \partial_\mu, \frac{\partial}{\partial \theta^\beta} \}
- \sigma^\mu_{\alpha \dot \beta} \sigma^\nu_{\beta \dot \gamma} \{ \bar{\theta}^{\dot{\beta}} \partial_\mu, \bar{\theta}^{\dot{\gamma}} \partial_\nu \}

Also, I think the last term should vanish, due to the anti-commutativity of the $$\bar{\theta}$$
Is this correct so far?
Unfortunately, I'm unable to make further progress.
Any help would be greatly appreciated. Thanks in advance.
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  • #2

Thank you for your post. I understand that you are having trouble verifying the anti-commutation relations for the given generators. I am a scientist and I would be happy to help you with this problem.

Firstly, I would like to confirm that your expansion of the anti-commutator is correct so far. The last term should indeed vanish due to the anti-commutativity of the Grassmann variables.

To continue, we can use the following properties of Grassmann variables:

1. Anti-commutativity: $$\{\theta^\alpha, \theta^\beta\} = \{\bar{\theta}^\alpha, \bar{\theta}^\beta\} = 0$$
2. Commutativity with ordinary variables: $$\{\theta^\alpha, x^\mu\} = \{\bar{\theta}^\alpha, x^\mu\} = 0$$
3. Anti-commutativity with derivatives: $$\{\theta^\alpha, \partial_\mu\} = \{\bar{\theta}^\alpha, \partial_\mu\} = 0$$

Using these properties, we can simplify the first two terms in your expansion:

\{\frac{\partial}{\partial \theta^\alpha}, \frac{\partial}{\partial \theta^\beta}\} = \{\bar{\theta}^{\dot{\alpha}} \partial_\mu, \bar{\theta}^{\dot{\beta}} \partial_\nu\} = \{ \partial_\mu, \partial_\nu\} \bar{\theta}^{\dot{\alpha}}\bar{\theta}^{\dot{\beta}} = 0

\{\frac{\partial}{\partial \theta^\alpha}, \partial_\mu\} = \{\bar{\theta}^{\dot{\alpha}} \partial_\mu, \partial_\nu\} = \partial_\mu \{\bar{\theta}^{\dot{\alpha}}, \partial_\nu\} = \partial_\mu \delta^{\dot{\alpha}}_\nu = 0

Therefore, we are left with the following expression:

\{Q_\alpha, Q_\beta\} = -i \sigma^\mu_{\beta \dot \beta} \bar{\theta}^{\dot{\beta}} \{\frac{\partial}{\partial \theta^\alpha}, \

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