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  A Question about Supercovariant Derivative

+ 2 like - 0 dislike

I posted this question here, but received no answers.

Let $\Phi(x,\theta,\bar{\theta})$ be a complex superfield. Let $K(\Phi,\bar{\Phi})$ be a function of $\Phi$ and $\bar{\Phi}$. How to prove the following identity?

$$\int d^{4}x\int d^{2}\theta d^{2}\bar{\theta}K(\Phi,\bar{\Phi})=\frac{1}{16}\int d^{4}x D^{2}\bar{D}^{2}K(\Phi,\bar{\Phi})|_{\theta=\bar{\theta}=0},$$

where $D_{\alpha}=\partial_{\alpha}-i(\sigma^{\mu})_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\partial_{\mu}$, and $\bar{D}_{\dot{\alpha}}=\bar{\partial}_{\dot{\alpha}}-i(\bar{\sigma}^{\mu})_{\dot{\alpha}\beta}\theta^{\beta}\partial_{\mu}$ are the supercovariant derivatives. 

asked Mar 28, 2019 in Theoretical Physics by Libertarian Feudalist Bot (270 points) [ no revision ]

1 Answer

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It is just a direct application of the definitions, but you can simplify the computation of the squared supercovariant derivatives a little bit by introducing a coordinate change in the superspace, as I will describe below.

First recall the definitions. Let \(f(x,\theta,\bar\theta)\) be a function on superspace. One can always expand it in terms of \(\theta,\bar\theta\) as \[f(x,\theta,\bar\theta)=c_1(x)+\theta c_2(x)+\bar\theta c_3(x)+\theta^2 \bar{\theta}^2 c_4(x) .\]

By definition, the integral over \(d^4\theta=d^2\theta d^2\bar\theta\) is \[\int f(x,\theta,\bar\theta) d^4\theta = c_4(x).\]

By a direct computation, \[\int f(x,\theta,\bar\theta) d^4\theta = \frac{1}{16}\Bigg[\frac{\partial}{\partial \theta_\alpha}\frac{\partial}{\partial \theta^\alpha}\frac{\partial}{\partial {\bar\theta}^\dot\alpha}\frac{\partial}{\partial {\bar\theta}_\dot\alpha} f(x,\theta,\bar\theta) \Bigg] \Bigg|_{\theta=\bar\theta=0}.\]

Introducing a simple coordinate change, \(y^\mu=x^\mu+i \theta^\dot\alpha \sigma^{\mu}_{\alpha \dot\alpha}\bar\theta^\dot\alpha\), we can rewrite the fermionic derivatives as: \[D_\alpha=\frac{\partial}{\partial \theta^\alpha} + 2i \sigma^{\mu}_{\alpha \dot\alpha} \bar\theta^{\dot\alpha} \frac{\partial}{\partial y^\mu}, \bar{D}_\dot\alpha = -\frac{\partial}{\partial \bar\theta^{\dot\alpha}}.\]

This transformation is usually employed when dealing with chiral and anti-chiral superfields, but here it simplifies the computation since \(\bar{D}^2=\bar{D}^\dot\alpha \bar{D}_\dot\alpha\) trivially equals \((\partial / \partial \bar\theta_\dot\alpha) (\partial / \partial \bar\theta^\dot\alpha)\), while \[D_\alpha D^\alpha f(x,\theta,\bar\theta)=\Bigg(\frac{\partial}{\partial \theta^\alpha} + 2i \sigma^{\mu}_{\alpha \dot\alpha} \bar\theta^{\dot\alpha} \frac{\partial}{\partial y^\mu} \Bigg)^2 f(x(y),\theta,\bar\theta) = \frac{\partial}{\partial \theta^\alpha} \frac{\partial}{\partial \theta_\alpha} f(x(y),\theta,\bar\theta)+\mathcal O (\theta,\bar\theta), \] where \(\mathcal O (\theta,\bar\theta)\) are first-order terms in the fermionic variables, which vanishes when we apply \([ ... ]\big|_{\theta=\bar\theta=0}\). So we have derived that: 

\[\frac{1}{16} \big[ D^2\bar{D}^2 f(x,\theta,\bar\theta) \big] \big|_{\theta=\bar\theta=0}=\int f(x,\theta,\bar\theta)d^4\theta.\]

Your particular result follows once you expand the superfield functional \(K[\Psi,\bar\Psi]\) in term of the superfields and apply the above identity.

answered Mar 29, 2019 by Igor Mol (550 points) [ no revision ]

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