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  A Question about Supercovariant Derivative

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8097 views

I posted this question here, but received no answers.

Let Φ(x,θ,ˉθ) be a complex superfield. Let K(Φ,ˉΦ) be a function of Φ and ˉΦ. How to prove the following identity?

d4xd2θd2ˉθK(Φ,ˉΦ)=116d4xD2ˉD2K(Φ,ˉΦ)|θ=ˉθ=0,

where Dα=αi(σμ)α˙βˉθ˙βμ, and ˉD˙α=ˉ˙αi(ˉσμ)˙αβθβμ 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,θ,ˉθ) be a function on superspace. One can always expand it in terms of θ,ˉθ as f(x,θ,ˉθ)=c1(x)+θc2(x)+ˉθc3(x)+θ2ˉθ2c4(x).

By definition, the integral over d4θ=d2θd2ˉθ is f(x,θ,ˉθ)d4θ=c4(x).

By a direct computation, f(x,θ,ˉθ)d4θ=116[θαθαˉθ˙αˉθ˙αf(x,θ,ˉθ)]|θ=ˉθ=0.

Introducing a simple coordinate change, yμ=xμ+iθ˙ασμα˙αˉθ˙α, we can rewrite the fermionic derivatives as: Dα=θα+2iσμα˙αˉθ˙αyμ,ˉD˙α=ˉθ˙α.

This transformation is usually employed when dealing with chiral and anti-chiral superfields, but here it simplifies the computation since ˉD2=ˉD˙αˉD˙α trivially equals (/ˉθ˙α)(/ˉθ˙α), while DαDαf(x,θ,ˉθ)=(θα+2iσμα˙αˉθ˙αyμ)2f(x(y),θ,ˉθ)=θαθαf(x(y),θ,ˉθ)+O(θ,ˉθ),

 where O(θ,ˉθ) are first-order terms in the fermionic variables, which vanishes when we apply [...]|θ=ˉθ=0. So we have derived that: 

116[D2ˉD2f(x,θ,ˉθ)]|θ=ˉθ=0=f(x,θ,ˉθ)d4θ.

Your particular result follows once you expand the superfield functional K[Ψ,ˉΨ] 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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