Tuesday, 3 February 2015

supersymmetry - Can you quantize Grassmann-even superfields in the same fashion as Boson fields?


In a related Phys.SE question about supersymmetric Lagrangian L=12(S)212(P)212ˉψ/ψ, the fields S and P are said to be Grassmann-even supernumber-valued rather than real- (or complex) valued, so that supersymmetry transformations (with Grassmann-odd ε) δεS=ˉεψδεP=ˉεγ5ψδεψ=/(S+Pγ5)ε can be consistently defined.



My question is: can you still do path integral (or canonical) quantization of Grassmann-even fields (S and P) in the same fashion as real (or complex) scalar/pseudoscalar Boson fields? I am asking this because Grassmann-even supernumbers behave differently from real (complex) numbers. For example, the Grassmann-even ab (where a and b are Grassmann-odd) squares to zero (nilpotent) (ab)(ab)=(aa)(bb)=0, which is totally different from a real (complex) number.



Answer





  1. First of all, 1 complex (super)number can be viewed as 2 real (super)numbers, so it is enough to discuss real (super)numbers.




  2. A field ϕ=ϕBbody+ϕSsoul that takes values in the set R1|0 of Grassmann-even real supernumbers can be quantized in the same way as a field ϕB that takes values in the set R of real numbers.


    This is easiest to see in the path integral formalism, since an integral over a Grassmann-even real supernumber ϕR1|0 is by definition given by the corresponding integral over its body ϕBR R1|0dϕ f(ϕ) := RdϕB f(ϕB).





  3. The operator formalism can in principle be mapped to the path integral formalism.




  4. See also this related Phys.SE post.




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