# SuSy Algebra

by ChrisVer
Tags: algebra, susy
 P: 779 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.....
 P: 779 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 $θ\bar{θ}$ and use a chain rule derivative???)...But I am not sure if the dotted derivatives "see" the undotted spinors and the opposite...
 P: 779 SuSy Algebra So in fact, you did a chain rule in the 1st two terms- first the derivative acting on the spinor $\theta$ and then acting on "something" that will appear next... correct? I understand that since $\partial_{a} (\theta^{b} \theta^{c})= \delta_{a}^{b} \theta^{c} - \theta^{b} \delta_{a}^{c}$ there also appears the minus