# Anyone has done calculations with the Wess-Zumino model?

1. Apr 24, 2009

### nrqed

This is driving me crazy. I am calculating the order g^2 (two loops) vacuum energy in the Wess-Zumino model, and I don't get a cancellation. The tadpole diagrams cancel out but the others don't. I have checked that I get the correct results for the one-loop corrections to the scalar propagators, but I can't find *any* reference that provides some details on that calculation (I have about 15 books on SUSY, including a collection of reprints and of physics reports).
This is the kind of calculation that is probably assigned as a homework in courses in SUSY, so someone here must have done it at some point, or at least coul dpoint me to a reference providing some details (by that I mean the actual expressions for at least some of the diagrams involved). I know what diagrams are supposed to cancel each other, but I can't make it work.

I will provide details of my calculation is anyone asks.

Thanks,

Patrick

2. May 14, 2009

### nrqed

No one?

3. May 14, 2009

### librab103

Can you please attach you calculations and some of us will take alook at it for you.

4. May 26, 2009

### nrqed

Fair enough. Although I doubt anyone will find the mistake if they haven't done the problem before.

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

$$L_1 = - \frac{1}{2} g^2 (A^2 + B^2)^2$$

$$L_2 = - M g (A^3 + A B^2)$$

$$L_3 = - g A \overline{\Psi} \Psi$$

$$L_4 = - ig B \overline{\Psi} \gamma_5 \Psi$$

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

$$i L_1 - \frac{1}{2} ( L_2^2 + L_3^2 + L_4^2 + 2 L_2 L_3 + 2 L_2 L_4 + 2 L_3 L_4 )$$

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

$$I \equiv i g^2 \int D(z-z) D(w-w)$$

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??

Thanks!

Last edited: May 26, 2009
5. Jun 12, 2009

### nrqed

Has anyone done this calculation? Sorry for being annoying with this but I would really like to understand what I am doing wrong

6. Jun 18, 2009

### blechman

Hey Pat.
I've been away for a while so I've just seen your post. Is there any reason in particular that you're doing this in component diagrams? There is only one supergraph!

I've never done the calculation before myself. It's not immediately obvious that the supergraph vanishes (holomorphy allows it), so it must just vanish by brute force. Hmm... I don't like the sound of that.

I'll think about it. I'm in the middle of something else right now, but I'll try to come up with a good answer if you didn't figure it out first. Let me know.

7. Jun 18, 2009

### blechman

You know.... I'm being kinda stupid in my last post. The 2-loop supergraph that contributes to the vacuum renormalizes the Kahler potential by a (quadratically divergent) constant. In rigid supersymmetry this automatically vanishes when you do the Grassmann integrals, but in supergravity, this generally does not vanish and is actually a serious problem in supergravity-mediated susy breaking theories - it more or less irreparably damgaged the consistency of the "no-scale" models that were popular in the 80s and early 90s; see Bagger, Poppitz, Randall, hep-ph/9505244, especially Appendix D.

Of course, this doesn't really answer your question - what are the explicit cancellations in component diagrams... Are you sure you didn't screw up symmetry factors? That's usually where I mess up when doing these kinda calculations...

8. Jun 18, 2009

### nrqed

Hey!!

Thanks for your feedback. I hope that some of that time was for vacations.

I want to get to supergraph techniques but I wanted to be able to do calculations in different ways to really ensure that I understood the cancellations. So I wanted to do some calculations with the component fields, with our without the auxiliary field eliminated, before doing supergraphs. But I am stuck right at the start.

The symmetry factors are of course what bother me the most. But I have checked many many times all of them. This is why I provided my values in my post, so that someone could tell me if any of these factors are incorrect.

The problem is that here are also log divergent diagrams and they don't cancel out either.

I have looked at a paper by Zumino where he draw the diagrams that should cancel out. He just says that these diagrams identically cancel out (no details are provided). And yet, when I use his rules (these are rules before F is eliminated), I don't get zero.

I will write the details of the log divergent diagrams later tonight.

Very interesting. I really want to learn about supergraph techniques and supergravity.Thansk for the reference.

I wish I could be as stupid as you :tongue2:

9. Jun 18, 2009

### blechman

You are right to do so! But it is amusing that you decide to start with a two-loop problem! A MUCH easier calculation is to prove that the superpotential (M and g) does not get renormalized at one loop. The supergraph approach is trivial (there are no diagrams!), and the explicit calculation in components is also straightforward.

Well, I cannot promise you I can go through your calculation in detail. I CAN tell you that the answer should be zero (Zumino is no fool!) Maybe I'll try to do the problem over the weekend, if I have time.

Not the best reference to learn supergraphs from! But a very nice paper with very important results (and I have a good relationship with the authors ). I "learned" supergraph techniques from Peter West's textbook (take note of the quotations!). That's a good place to start.

No you don't. Trust me!!

10. Jun 18, 2009

### Haelfix

Can you link the paper by Zumino? I also dont see any obvious tricks or cancellations in the component form other than brute force, but there probably is one. Its a bit of an ongoing problem, we get too lazy and rely too much on clever tricks/cool formalisms like supergraphs and consequently lose calculational muscle and then problems like this can show up.

Absent that, there are software packages that could probably tell you which diagrams are different than your results..

11. Jun 19, 2009

### nrqed

Thanks for the suggestion, I will look into that
Thanks. Of course, I know that Zumino is right, I am just puzzled by what I must be doing wrong. I will post a detail run through of how I get all my symmetry factors, so it should be easy to spot a mistake (if the mistake is in the calculation of the symmetry factors).

I understand that you may not have time to check this! I will post a much more detailed post later today and you could check some part of it.

Great, thanks for the suggestions.

12. Jun 19, 2009

### nrqed

The paper is "Supersymmetry and the vacuum" by Zumino, Nuclear Physics B89 (1975) 535-546.

The Feynman rules he uses are given in

"Broken supergauge symmetry and renormalization", Iliopoulos and Zumino, Nuc. Phys. B76 (1974), 310-332.

I have hard copies, I haven't yet checked if they are available online (probbaly at the CERN archives, but I am not familiar how to search articles there).

13. Jun 30, 2009

### nrqed

I will explain how I got all my symmetry factors, in case someone can spot a mistake. I will also show my results for the sunset diagrams in a different post.
This comes from the interaction $$A^4/2$$. If we pick any A, there are three ways to connect it with the remaining three A. The two A left over must be contracted together. So there are three possible contrations, giving a factor of 3/2 (I won't bother about the signs in this post, just the symmetry factor)

This comes from $$2 A^2 B^2 /2$$ an dthe symmetry factor is obviously 1 (the two A must be contracted together and the two B must also be contracted together)

Oops, I had meant to write two B loops in my post! The term is B^4/2 and the symmetry factor is obviously the same as for A^4/2

we have $$A^3(x) A^3(y)/2$$. Any of the three A(x) can be connectd with any of the thee A(y), giving a factor of 9. In order to get a dumbell diagram, the two remaining A(x) must be contracted together and the two remaining A(y) must also be contracted together. So an overall factor of 9/2
This comes from $$A^3(x) A(y) B^2(y)$$. The B must be contracted together, There are three possibilitie sto contract the A(y) with any of the A(x), so an overall factor fo 3.
This comes from $$A(x) B^2(x) A(y) B^2(y) /2$$ . The symmetry factor for the dumbell diagram is clearly 1.

This comes from $$A(x) \overline{\Psi}(x) \Psi(x) A(y) \overline{\Psi}(y) \Psi(y) /2$$ The symmetry factor is 1. We get the product of two traces,

$$Tr(\gamma^\mu p_\mu + m ) Tr(\gamma^\mu k_\mu + m) = 16 m^2$$

Times the factor fo 1/2, we get an overall coefficient of 8.

we have $$A^3(x) A(y) \overline{\Psi}(y) \Psi(y)$$
The symmetry factor if 3 (three ways to contract A(y) with one of the A(x)). We get a factor fo 4M from the trace, so an overall factor of 12.

This comes from the interaction $$A(x) B^2(x) A(y) \overline{\Psi}(y) \Psi(y)$$
This time the symmetry factor is obviously 1 and we stll get a factor fo 4M from the trace so an overall factor fo 4.
So that's how I got my sum.

Pat

14. Jul 20, 2009

### nrqed

Has anyone spotted my mistake(s)?? (see the previosu post, where I give the details of
my calculation)

15. Jul 22, 2009

### blechman

Hey Patrick.
I admit I'm not being very careful (just scribbling down some quick notes on a scrap piece of paper) but are you sure of your formulas? I do things in terms of COMPLEX scalars and WEYL fermions to make the symmetry factors a bit easier to work out. The dumbbell diagrams look to be proportional to

$$\int d^4x d^4y [m D(0)]^2 D(x-y) \neq I$$

while the double-lobe diagram is proportional to I. When I do the calculation, I get:

(dumbbell of fermion + boson) = -(dumbbell of two bosons)

while the dumbbell of two fermions cancels the double-lobe integral after some gymnastics with the integrands.

I was doing things quickly and not very carefully, but it seems to work...

Did you try my suggestion of proving the nonrenormalization of g? One loop problem - MUCH easier!

16. Jul 22, 2009

### blechman

OK, I went through and double checked my calculations a bit more carefully, and I now stand by my previous post!

Phew!

17. Jul 22, 2009

### nrqed

Hi Andrew,

Thanks a lot for looking at this!

I don't quite see the difference, so let me be more explicit. It seems to me that a the Fourier transform to momentum space of a dumbell diagram is of the form (dropping the factors of i)

$$\int d^4 x d^4y \int d^4k \frac{1}{k^2-m^2} \int d^4p \frac{e^{-i (x-y)}}{p^2-m^2} \int d^4q \frac{1}{q^2-m^2}$$

Carrying out the x integral I get a $$\delta^4(p)$$ and the remaining y integration gives an overall volume factor that appears in all the vacuum diagrams. Discarding this volume factor, I then ger

$$- \frac{1}{m^2} \int d^k d^4q \frac{1}{q^2-m^2} \frac{1}{k^2-m^2}$$

which is what I defined as my integral I (times -1/m^2).

If I do the lobe diagram, I get exactly the same integrand. For example, if I contract the fields in $$A^4(x)$$ and I Fourier transform the x, I simply get

$$\int d^4x \int d^4 p \int d^4q \frac{1}{p^2-m^2} \frac{1}{q^2 -m^2}$$

which again produces an overall volume factor times I.

I must be missing something obvious!

Thanks again for having a look. I am clearly missing something obvious.

I want to try it, definitely! It seemed to me that I would need to have a large number of terms to compute in the expansion of $$e^{-i \int d^4x {\cal L}_{int}}$$. I guess I should just work out the Feynman rules and word directly with Feynman diagrams. I wll have a look at it and get back with questions!

Thanks

Pat

18. Jul 23, 2009

### blechman

WARNING: There are errors in this post. See below (post #20) for an explanation.

I think you're missing a minus sign! Ignoring the symmetry factors for the moment, the figure-8 diagram is your integral. The dumbbell diagrams are my integral, which after some integration gymnastics is MINUS the figure-8 integral.

Now symmetry factors. Again, I use Weyl fermions and complex scalars, which make the symmetry factors MUCH easier (since $\phi$ and $\phi^*$ are distinguishable). The Lagrangian (same as yours in terms of these variables) is:

$$\mathcal{L}=-g(\phi\psi\psi + {\rm h.c.}) - gm\phi^*\phi(\phi+\phi^*) - g^2(\phi^*\phi)^2$$

Using "F" to denote a fermion loop and a "B" to denote a boson loop, I find three classes of diagrams:

F ==== F : $(1)(-i)^2\frac{(-g)^2}{2}\int d^4x\int d^4y [-2mD(0)]^2D(x-y)$

F ==== B : $(2)(-i)^2\frac{(-g)^2}{2}\int d^4x\int d^4y [-2mD(0)][+mD(0)]D(x-y)$ WRONG!!!

B ==== B : $(4)(-i)^2\frac{(-g)^2}{2}\int d^4x\int d^4y [+mD(0)]^2D(x-y)$

The first number are the Wick-contraction factors:

FF: the only thing you can do is switch all the arrows, and this corresponds to switching x and y, which is included in the integration and hence not a new diagram.

FB: There are two $\phi\phi^*$ contractions you can make in the boson loop.

BB: Same as FB for BOTH loops, and you can have either $\phi$ or $\phi^*$ over the bridge, but this is again the same diagram, so it doesn't count.

The $(i)^2(-g)^2/2$ comes from the Feynman rule/Dyson expansion; (-2) for fermion loops -- recall these are Weyl fermions; and factors of m in the fermion propagator since it is helicity-changing (why using Weyl fermions makes it easier!) and an m for each 3-boson vertex. Putting it all together we find:

$$I_{\rm dumbbell}=-2g^2 I_B$$

where $I_B$ is the "blechman integral" (hehe).

Finally, we have the figure 8 diagram:

$$I_8=(2)(-ig^2)\int d^4x[D(0)]^2$$

Again, there are two ways to contract the $\phi\phi^*$, and the Feynman rule is $(-ig^2)$.

So in the end, we find:

$${\rm Answer}=-2g^2\int d^4x[D(0)]^2\left\{i+\int d^4y~m^2D(x-y)\right\}$$

I leave it to you to prove that the quantity in curly brackets vanishes, which is a straightforward exercise in integration.

Last edited: Jul 23, 2009
19. Jul 23, 2009

### nrqed

Thank you so much for taking the time to write this.
I haven't found my mistake yet (seems to be that I am off by a factor of 2 in afew diagrams) but it helps me greatly to see the breakdown of the contributions, and it is very instructive to see the calculation in Weyl language.

I may have more questions, so stay tuned!

Patrick

20. Jul 23, 2009

### blechman

ERROR: I see that I made a mistake with the F==B dumbbells - in fact, you have a Wick contraction factor of FOUR (4), not 2, since you can have either a $\phi$ or a $\phi^*$ crossing the bridge, and these are NOT equivalent diagrams since one loop is a fermion and one is a boson! So indeed, it looks like the cancellation is not perfect.

But you know what? WHY should it be? In particular, I am looking at the two-loop sunset diagram coming from exchange of three scalars. To understand the divergence structure of this diagram, you start shrinking propagators (Weinberg Power Counting Theorem). Lo! When I shrink one of those propagators I get the figure-8 diagram! This suggests that you need the figure 8 diagram to cancel divergences there as well, so indeed it looks like these diagrams BETTER not cancel on their own!!

Note, by the way, that all other diagrams are safe, since if you shrink a line in the fermion sunset diagrams you get dimension 5 and 6 operators which you know by renormalization theory cannot be divergent.

So I think, in fact, that you (Patrick) were actually right all along, and that the reason the C.C. was not canceling was because you didn't do all the diagrams.

Thoughts?