SuSy Algebra

  1. ChrisVer

    ChrisVer 2,403
    Gold Member

    So I'm trying to show that one choice of representation for the SuSy generators fulfills the SuSy algebra.... (one of which is [itex]\left\{ Q_{a},\bar{Q_{\dot{b}}} \right\}= 2 \sigma^{\mu}_{a\dot{b}} p_{\mu}[/itex])...
    [itex] Q_{a}= \partial_{a} - i σ^{μ}_{a\dot{β}} \bar{θ^{\dot{β}}} \partial_{\mu}[/itex]

    [itex] \bar{Q_{\dot{b}}}= -\bar{\partial_{\dot{b}}} + i θ^{a}σ^{μ}_{a\dot{b}} \partial_{\mu}[/itex]

    I am not sure how I could go on with computing this anticommutation....

    [itex] Q_{a}\bar{Q_{\dot{b}}}= (\partial_{a} - i (σ^{μ}\bar{θ})_{a} \partial_{\mu})(-\bar{\partial_{\dot{b}}} + i (θσ^{μ})_{\dot{b}} \partial_{\mu})[/itex]

    [itex] Q_{a}\bar{Q_{\dot{b}}}= -\partial_{a}\bar{\partial_{\dot{b}}}+\partial_{a} (θσ^{μ})_{\dot{b}} p_{\mu}+(σ^{μ}\bar{θ})_{a}\bar{\partial_{\dot{b}}} p_{\mu}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}} ∂_{μ}∂_{ν}[/itex]

    In a similar way I can write:
    [itex] \bar{Q_{\dot{b}}}Q_{a}= -\bar{\partial_{\dot{b}}}\partial_{a}+\bar{\partial_{\dot{b}}}(σ^{μ}\bar{θ})_{a} p_{\mu}+(θσ^{μ})_{\dot{b}}\partial_{a}p_{\mu}+ (θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a} ∂_{μ}∂_{ν}[/itex]

    Now the first terms after addition cancel because the grassmann derivatives are like spinor fields, so they anticommute:
    [itex] \left\{ \partial_{a},\bar{\partial_{\dot{b}}} \right\}=0[/itex]
    Is that a correct statement? (I think it is, but I am also asking)

    The last terms give:
    [itex] [(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=0[/itex]
    I am saying it's equal to zero, because I'm again seeing the parenthesis (...) as spinors, so they anticomutte... and because the partial derivatives are symmetric under the interchange μ to ν, and the [...] is antisymmetric it's going to give zero:

    [itex] [(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=[(θσ^{μ})_{\dot{b}}(σ^{ν}\bar{θ})_{a}+(σ^{ν}\bar{θ})_{a} (θσ^{μ})_{\dot{b}}] ∂_{ν}∂_{μ}[/itex]
    [itex]=[-(σ^{ν}\bar{θ})_{a}(θσ^{μ})_{\dot{b}}- (θσ^{μ})_{\dot{b}}(σ^{ν}\bar{θ})_{a}] ∂_{ν}∂_{μ} [/itex]
    renaming again:

    [itex] [(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=-[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν} =0[/itex]

    So far I am certain I'm on a correct way (because I don't want as a result the double derivatives-is it in spinor or lorentz spaces)... However I'm not certain about my reasonings.... For example, in the last one, I also thought of writing:
    [itex](θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}=(θσ^{ν}\bar{1})(1σ^{μ}\bar{θ}) \propto n^{\mu\nu} (θ1)(\bar{1}\bar{θ})[/itex]
    But it wouldn't work out I think...I would also lose the spinoriac indices...

    Then I have the middle terms:

    [itex] [\partial_{a} (θσ^{μ})_{\dot{b}} +(σ^{μ}\bar{θ})_{a}\bar{\partial_{\dot{b}}}+ \bar{\partial_{\dot{b}}} (σ^{μ} \bar{θ})_{a} +(θσ^{μ})_{\dot{b}}\partial_{a}]p_{\mu}[/itex]
    And here I'd like to ask, if you have any suggestion of how this can work out.... I'd really appreciate it.....
    Last edited: Apr 16, 2014
  2. jcsd
  3. ChrisVer

    ChrisVer 2,403
    Gold Member

    Does for example, in the 1st term in the last expression, the derivative act on θ? (to give a delta Kroenicker).....
    But then I'm not sure about the derivatives appearing on the left...
    Also I could try acting with that thing on some spinors (for example a [itex]θ\bar{θ}[/itex] and use a chain rule derivative???)...But I am not sure if the dotted derivatives "see" the undotted spinors and the opposite...
    Last edited: Apr 16, 2014
  4. You are right about the first and last term, they vanish as per requirement of anticommutation and symmetry of partial derivative respectively. But you are making mistakes with middle terms, you should write them in the anticommutator form like one of the term is,
    ##i\left\{∂_a,θ^cσ^v_{cḃ}∂_v\right\}=i(∂_aθ^c)σ^v_{cḃ}∂_v-(iθ^cσ^v_{cḃ}∂_v)∂_a+(iθ^cσ^v_{cḃ}∂_v)∂_a=i(∂_aθ^c)σ^v_{cḃ}∂_v=iδ^c_aσ^v_{cḃ}∂_v##, because as you pass the derivative w.r.t. grassmann variable through a grassmann variable, there is a sign change and hence the second term. Similar term arises from differentiation w.r.t. ##{\dot{b}}##,which combined with above give desired result.
    1 person likes this.
  5. ChrisVer

    ChrisVer 2,403
    Gold Member

    So in fact, you did a chain rule in the 1st two terms- first the derivative acting on the spinor [itex]\theta[/itex] and then acting on "something" that will appear next... correct?
    I understand that since
    [itex]\partial_{a} (\theta^{b} \theta^{c})= \delta_{a}^{b} \theta^{c} - \theta^{b} \delta_{a}^{c}[/itex]
    there also appears the minus
  6. Yes.
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