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  Can i use the y representation of the chiral covariant derivatives on a superfield that contains both y and y?

+ 2 like - 0 dislike
990 views

Imagine I want to compute this

D2Dα(ΦΦ)

where the D-s are super-covariant derivatives and Φ is a chiral superfield. Following the notation of this review http://arxiv.org/abs/hep-ph/9709356 on supersymmetry, the chiral covariant derivatives are (page 33)

Dα=θαi(σμθ)αμDα=θα+i(θˉσμ)αμ

D˙α=θαi(ˉσμθ)˙αμD˙α=θ˙α+i(θσμ)˙αμ

let's now consider the following coordinate change

xμyμ=xμ+iθˉσμθ


  
the review claims (page 34) that in these y,θ,θ coordinates the chiral covariant derivative are

Dα=θα2i(σμθ)αyμDα=θα+2i(θˉσμ)αyμ

D˙α=θαD˙α=θ˙α

we can as well go to coordinates

xμˉyμ=xμiθˉσμθ

where in page 35 it is claimed that the chiral covariant derivatives take the form

Dα=θαDα=θα

D˙α=θα2i(ˉσμθ)˙αˉyμD˙α=θ˙α+2i(θσμ)˙αˉyμ

so far so fine. It is obvious that for some computations the choice of the clever coordinates will make computations considerably less cumbersome. This is where my problems arise. The computation I want to do contains a chiral superfield, which can be expressed most compactly using the y,θ coordinates, whilst the antichiral superfield is better written using ˉy,θ. My question is, is itlegitimate to use the chiral covariant derivatives is the ˉy,ˉθ form, even though what i wanna take the derivative of contains all yˉy,θ,θ? even if the answer is no, what i the wisest way to perform my computation?

asked Jul 17, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

1 Answer

+ 0 like - 0 dislike

I still am unsure about wether it is legitimate to use one representation of the super-covariant derivatives on something that is y and y dependent (even though I suspect that the answeris no). Nonetheless, I have found that an efficient way of making the computation not so cumbersome is writing the chiral superfield in y coordinates, where it is very simple, and writing the antichiral superfield also in y coordinates. This one will be longer. The reason for this choice is that this makes the last two super covariant derivatives trivial since they only involve a derivative over the Grasmann variable θ

Thus, we expand both chiral and antichiral superfields in y coordinates, we multiply them, we take the first super-covariat derivative, and we keep only terms with θ2, since only these will survive the last two derivatives.

answered Jul 18, 2015 by Dmitry hand me the Kalashnikov (735 points) [ revision history ]

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