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So I'm trying to show that one choice of representation for the SuSy generators fulfills the SuSy algebra... (one of which is \left\{ Q_{a},\bar{Q_{\dot{b}}} \right\}= 2 \sigma^{\mu}_{a\dot{b}} p_{\mu})...
For
Q_{a}= \partial_{a} - i σ^{μ}_{a\dot{β}} \bar{θ^{\dot{β}}} \partial_{\mu}
\bar{Q_{\dot{b}}}= -\bar{\partial_{\dot{b}}} + i θ^{a}σ^{μ}_{a\dot{b}} \partial_{\mu}
I am not sure how I could go on with computing this anticommutation...
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})
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}
In a similar way I can write:
\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}
Now the first terms after addition cancel because the grassmann derivatives are like spinor fields, so they anticommute:
\left\{ \partial_{a},\bar{\partial_{\dot{b}}} \right\}=0
Is that a correct statement? (I think it is, but I am also asking)
The last terms give:
[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=0
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:
[(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}
=[-(σ^{ν}\bar{θ})_{a}(θσ^{μ})_{\dot{b}}- (θσ^{μ})_{\dot{b}}(σ^{ν}\bar{θ})_{a}] ∂_{ν}∂_{μ}
renaming again:
[(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
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:
(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})
But it wouldn't work out I think...I would also lose the spinoriac indices...
Then I have the middle terms:
[\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}
And here I'd like to ask, if you have any suggestion of how this can work out... I'd really appreciate it...
For
Q_{a}= \partial_{a} - i σ^{μ}_{a\dot{β}} \bar{θ^{\dot{β}}} \partial_{\mu}
\bar{Q_{\dot{b}}}= -\bar{\partial_{\dot{b}}} + i θ^{a}σ^{μ}_{a\dot{b}} \partial_{\mu}
I am not sure how I could go on with computing this anticommutation...
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})
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}
In a similar way I can write:
\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}
Now the first terms after addition cancel because the grassmann derivatives are like spinor fields, so they anticommute:
\left\{ \partial_{a},\bar{\partial_{\dot{b}}} \right\}=0
Is that a correct statement? (I think it is, but I am also asking)
The last terms give:
[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=0
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:
[(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}
=[-(σ^{ν}\bar{θ})_{a}(θσ^{μ})_{\dot{b}}- (θσ^{μ})_{\dot{b}}(σ^{ν}\bar{θ})_{a}] ∂_{ν}∂_{μ}
renaming again:
[(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
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:
(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})
But it wouldn't work out I think...I would also lose the spinoriac indices...
Then I have the middle terms:
[\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}
And here I'd like to ask, if you have any suggestion of how this can work out... I'd really appreciate it...
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