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Algebra of the generator of supersymmetry transformations?

  1. Jul 7, 2012 #1
    We consider a superfield [itex]\Phi[/itex][itex]\left(x^{\mu}, \theta_{\alpha}\right)[/itex].
    For a small variation [itex]\delta \Phi [/itex] = [itex] \bar{\epsilon} Q \Phi [/itex]
    where the supercharge [itex]Q_{\alpha}[/itex] is given by:
    [itex]Q_{\alpha}[/itex]=[itex]\frac{\partial}{\partial \bar{\theta}^{\alpha}}[/itex]-[itex]\left(\gamma^{\mu} \theta \right) _{\alpha} \partial _{\mu}[/itex]
    They satisfy the algebra:
    [itex]\left\{ Q_{\alpha}, Q_{\beta} \right\}[/itex] = -2[itex]\left( \gamma^{\mu} C \right)_{\alpha \beta} \partial_{\mu} [/itex]
    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!
     
  2. jcsd
  3. Jul 7, 2012 #2
    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.
     
  4. Jul 9, 2012 #3
    I'm not sure of knowing it exactly... can you control my assumptions please?
    The anticommutator of two theta is zero: [itex]\left\{ \theta_{\alpha}, \theta_{\beta} \right\}=0[/itex]
    The anticommutator of the derivatives is also zero:
    [itex]\left\{ \frac{\partial}{\partial \bar{\theta}^{\alpha}}, \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}=0[/itex]
    You say that the anticommutator
    [itex]\left\{ \theta_{\alpha}, \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}[/itex]
    is equal to 1?
    And what is the result of an anticommutator like
    [itex]\left\{ \left( \gamma^{\mu} \theta \right)_{\alpha} \partial_{\mu} , \frac{\partial}{\partial \bar{\theta}^{\beta} }\right\}[/itex] = ?
    [itex]\partial_{\mu}[/itex] commute or anticommute with a theta?
    Thank you
     
  5. Jul 9, 2012 #4

    haushofer

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    No, because there you differentiate wrt theta-bar. Not wrt theta.


    They commutate; [itex]\partial_{\mu}[/itex] is a derivative wrt a bosonic coordinate.
     
  6. Jul 9, 2012 #5
    Thank you!
    So [itex]\left\{ \bar{\theta}_{\alpha}, \frac{\partial}{\partial \bar{\theta}^{\beta}} \right\}=1[/itex]? Why?
     
  7. Jul 9, 2012 #6
    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.
     
  8. Jul 10, 2012 #7

    haushofer

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    See e.g. eq.(11.47) of the book "Supersymmetry demystified", which is a very nice first exposure to SUSY:

    [tex]
    \{ \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
    [/tex]
    where the minus-sign comes from the fact that you work with Grassmannian variables.
     
  9. Jul 12, 2012 #8
    Thank you for the help!
     
  10. Jul 13, 2012 #9
    Which is the relation between theta and bar theta?
    Or better, what is the result of
    [itex]\left\{ \theta_{\beta}, \frac{\partial}{\partial \bar{\theta}^{\alpha} }\right\}[/itex] ?
     
  11. Jul 14, 2012 #10

    haushofer

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    Apply to a test function and use the grassman properties.
     
  12. Jul 16, 2012 #11

    haushofer

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    Did you manage to find it out?
     
  13. Jul 17, 2012 #12
    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)!
     
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