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Some questions on a version of the O'Raifeartaigh model

+ 6 like - 0 dislike
39 views

This form is taken from a talk by Seiberg to which I was listening to,

Take the Kahler potential ($K$) and the supersymmetric potential ($W$) as,

$K = \vert X\vert ^2 + \vert \phi _1 \vert ^2 + \vert \phi_2\vert ^2 $

$W = fX + m\phi_1 \phi_2 + \frac{h}{2}X\phi_1 ^2 $

  • This notation looks a bit confusing to me. Are the fields $X$, $\phi_1$ and $\phi_2$ real or complex? The form of $K$ seems to suggest that they are complex - since I would be inclined to read $\vert \psi \vert ^2$ as $\psi ^* \psi$ - but then the form of $W$ looks misleading - it seems that $W$ could be complex. Is that okay?

Now he looks at the potential $V$ defined as $V = \frac{\partial ^2 K}{\partial \psi_m \partial \psi_n} \left ( \frac {\partial W}{\partial \psi_m} \right )^* \frac {\partial W}{\partial \psi_n}$

(..where $\psi_n$ and $\psi_m$ sums over all fields in the theory..)

For this case this will give, $V = \vert \frac{h}{2}\phi_1^2 + f\vert ^2 + \vert m\phi_1 \vert ^2 + \vert hX\phi_1 + m\phi_2 \vert ^2 $

  • Though for the last term Seiberg seemed to have a "-" sign as $\vert hX\phi_1 - m\phi_2 \vert ^2 $ - which I could not understand.

I think the first point he was making is that it is clear by looking at the above expression for $V$ that it can't go to $0$ anywhere and hence supersymmetry is not broken at any value of the fields.

  • I would like to hear of some discussion as to why this particular function $V$ is important for the analysis - after all this is one among several terms that will appear in the Lagrangian with this Kahler potential and the supersymmetry potential.

  • He seemed to say that if *``$\phi_1$ and $\phi_2$ are integrated out then in terms of the massless field $X$ the potential is just $f^2$"* - I would be glad if someone can elaborate the calculation that he is referring to - I would naively think that in the limit of $h$ and $m$ going to $0$ the potential is looking like just $f^2$.

  • With reference to the above case when the potential is just $f^2$ he seemed to be referring to the case when $\phi_2 = -\frac{hX\phi_1}{m}$. I could not get the significance of this. The equations of motion from this $V$ are clearly much more complicated.

  • He said that one can work out the spectrum of the field theory by "diagonalizing the small fluctuations" - what did he mean? Was he meaning to drop all terms cubic or higher in the fields $\phi_1, \phi_2, X$ ? In this what would the "mass matrix" be defined as?

The confusion arises because of the initial doubt about whether the fields are real or complex. It seems that $V$ will have terms like $\phi^*\phi^*$ and $\phi \phi$ and also a constant term $f^2$ - these features are confusing me as to what diagonalizing will mean.

Normally with complex fields say $\psi_i$ the "mass-matrix" would be defined the $M$ in the terms $\psi_i ^* M_{ij}\psi_j$ But here I can't see that structure!

  • The point he wanted to make is that once the mass-matrix is diagonalized it will have the same number of bosonic and fermionic masses and also the super-trace of its square will be $0$ - I can't see from where will fermionic masses come here!

  • If the mass-matrix is $M$ then he seemed to claim - almost magically out of the top of his hat! - that the 1-loop effective action is $\frac{1}{64\pi^2} STr \left ( M^4 log \frac{M^2}{M_{cut_off}^2} \right ) $ - he seemed to be saying that it follows from something else and he didn't need to do any loop calculation for that!

I would be glad if someone can help with these.

This post has been migrated from (A51.SE)
asked Dec 30, 2011 in Theoretical Physics by user6818 (955 points) [ no revision ]
A brief comment on your first question: the superpotential is a holomorphic function, hence complex unless (locally) constant. It appears in the lagrangian density as the real part of its integral over "half the superspace". Chiral superfields are therefore complex: indeed they define holomorphic coordinates on a Kähler manifold.

This post has been migrated from (A51.SE)
Indeed, all the fields are complex. The rest is more or less standard in QFT, except for the minus sign error, which you got right.

This post has been migrated from (A51.SE)
Jose - Thanks for the explanation. @Zohar Ko I can understand that in principle this is standard QFT but I can't exactly do what he seems to say should be done. Like in what sense is the potential just $f^2$ and how is the mass-matrix (with fermionic masses!) going to emerge here. I have done this kind of calculation elsewhere but here things don't seem to fit in. And the last expression for the 1-loop effective action looks as mysterious. It would be great if you can sketch out the calculation.

This post has been migrated from (A51.SE)
The mass matrix of the bosons and fermions coincides if $f=0$. Hence, in the super-trace over M^4\log M^2 there would be perfect cancelation if $f=0$. The answer is nonzero if $f\neq0$ and it scales like f^2.

This post has been migrated from (A51.SE)
@Zohar Can you give a reference for such an analysis about the supertrace of the mass-matrix of a theory?

This post has been migrated from (A51.SE)
I believe that this is what Mr. O'R did in his original paper... The foundations of this approach were of course developed by Coleman-Weinberg, which is a very nice paper.

This post has been migrated from (A51.SE)

1 Answer

+ 0 like - 0 dislike

I found Stephen P. Martin's review very useful for this analysis.

This post has been migrated from (A51.SE)
answered Jan 27, 2012 by Tali (0 points) [ no revision ]
@Tail, thanks for answering, but please provide more information (e.g. link to the review you refer to, and at least a sentence what is there).

This post has been migrated from (A51.SE)
@Piotr Migdal I guess the review being referred to is this one. http://arxiv.org/pdf/hep-ph/9709356v6.pdf But I don't see this STr there which I have quoted. It would be great if Tali can point out where this is there in that large review.

This post has been migrated from (A51.SE)

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