I am trying to do what is (almost) the simplest SUSY calculation one can think of: the calculation of the vacuum energy in the Wess-Zumino model. The result shoudl be zero but I don't get that.

Since SUSY is "beyond the standard model" physics, I decided to cross-post my question here.

Let me start by asking a simpler question and see if anyone can help.

Let me break up the Wess-Zumino interactions into four terms

[tex] L_1 = - \frac{1}{2} g^2 (A^2 + B^2)^2 [/tex]

[tex] L_2 = - M g (A^3 + A B^2) [/tex]

[tex] L_3 = - g A \overline{\Psi} \Psi[/tex]

[tex] L_4 = - ig B \overline{\Psi} \gamma_5 \Psi [/tex]

where A and B are scalar fields and Psi is a Majorana spinor, although for the terms I want to check in this post, this makes no difference, they can be treated as Dirac spinors.

I consider the contributions to the vacuum energy of order g^2 so I calculate the time

ordered expectation values of

[tex] i L_1 - \frac{1}{2} ( L_2^2 + L_3^2 + L_4^2 + 2 L_2 L_3 + 2 L_2 L_4 + 2 L_3 L_4 ) [/tex]

There are two main classes of diagrams: diagrams in which three lines connect two distinct points.

(i.e. "sunset" diagrams). These are a bit more tricky.

Simpler diagrams are those that contain two loops over independent variables (so the

diagrams contain two independent loops). These should be simple to double check.

I will only give my results for these diagrams, not the sunset diagrams for now. These diagrams should

cancel independently of the sunset diagrams (a fact confirmed by a paper of Zumino).

But I can't get it to work!

These diagrams either have the form of two circles touching each other or two circles connected by

line (forming a dumbell).

These diagrams are all proportional to the integral

[tex] I \equiv i g^2 \int D(z-z) D(w-w) [/tex]

where D is just the usual boson propagator. I will quote all my results in terms of I .

CONTRIBUTION FROM L1

Diagram with two A loops: -3I/2

Diagram with one A loop and one B loop: -I

Diagram with two A loops: -3I/2

CONTRIBUTION FROM L2^2

Dumbell diagram with two A loops: 9 I/2

Dumbell diagram with one A loop and one B loop: 3 I

Dumbell diagram with two B loops: I/2

CONTRIBUTION FROM L3^2

Dumbell diagram, with two fermion loops: 8 I

CONTRIBUTION FROM L4^2

Because there is a gamma 5, there are no dumbell type contribution

CONTRIBUTION FROM L2 L3

Contribution with one A loop and one fermion loop: - 12 I

Contribution with one B loop and one fermion loop: -4 I

CONTRIBUTION FROM L3 L4 is identically zero.

**SUM = -4 I**

Now, can anyone check any of this??