Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
5  
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. –  José Figueroa-O'Farrill Jan 1 '12 at 6:05
    
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. –  Zohar Ko Jan 1 '12 at 19:08
    
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. –  user6818 Jan 2 '12 at 21:42
    
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. –  Zohar Ko Jan 3 '12 at 12:55
    
@Zohar Can you give a reference for such an analysis about the supertrace of the mass-matrix of a theory? –  user6818 Jan 7 '12 at 0:52
show 1 more comment

1 Answer 1

There are lots of questions here! I think I can answer at least some...

  • First of all, you are aware that the fields in $W$ and $K$ are superfields? These contain the entire supermultiplet, so they must be complex in general. This is a short entry but it links to others: http://en.wikipedia.org/wiki/Superfield
  • As mentioned by Jose in his comment, the superpotential is just a holomorphic function. Remember that the superpotential is in some sense not physical; it only shows up in the Lagrangian - the real physics of the theory - via things like $\frac{\partial W}{\partial \phi_m}^* \frac{\partial W}{\partial \phi_n}$ in your example. It is these that must be real, not $W$ itself, and they will be. I'm not sure why you say there would be terms like $\phi^*\phi^*$... Everything is mod-squared in that $V$, so it must be real.
  • I agree with Zohar that the minus sign is probably a "typo" on Seiberg's part. My first thought when I skimmed your question was that there was $SU(2)$ contraction with an $\epsilon_{\alpha\beta}$, because one gets terms that look like that all the time, but after reading it, I don't think that's the case here.
  • To understand the supertrace in the effective potential, see http://arxiv.org/pdf/hep-ph/0111209v2.pdf by S. Martin. Eq. 1.2 is exactly what Seiberg wrote, but with the supertrace written out explicitly with the $(-1)^{2s_n}(2s+1)$ prefactor. To understand why the one-loop effective potential looks like $Str(M^4 log\frac{M^2}{M_{cutoff}^2})$, consider a couple things.
  • First, the thing inside the supertrace looks just like something you'd get doing a one-loop integral for some field, except that here $M$ is a matrix. Second, the prefactor takes care of what you get due to the multiplicity of the particles and the minus sign associated with fermion loops.

But let's say we don't believe this expression and calculate everything explicitly. We would diagonalize the mass matrix, thus finding the mass eigenvalues $m_i$ and then we'd put each of them one at a time into a loop diagram and calculate basically $\Sigma_i m_i^4 log\frac{m_i^2}{M_{cutoff}^2}$.

But notice that if we diagonalize $M$ to, say, $M'$, then $M'$ has the $m_i$ along it's diagonal. Then clearly

$\Sigma_i log\frac{m_i^2}{M_{cutoff}^2}=Tr(log\frac{M'^2}{M_{cutoff}^2})$.

Furthermore,

$Tr(log\frac{M'^2}{M_{cutoff}^2})= log(Det\frac{M'^2}{M_{cutoff}^2})$.

This follows because $M'$ is diagonal, so that obviously $Det(M'^2)=\Pi_i m_i^2$ and thus

$log(Det(M'^2))=log(\Pi_i m_i^2)=\Sigma_i log (m_i^2)=Tr(log(M'^2))$.

Now, if we diagonalized with some matrix $U$, then

$Tr(log\frac{M'^2}{M_{cutoff}^2})= log(Det\frac{M'^2}{M_{cutoff}^2}) = log(Det\frac{(UMU^\dagger)^2}{M_{cutoff}^2})$,

and since $U$ is unitary, $Det(UMU^\dagger)=Det(M)$. Thus,

$Tr(log\frac{M'^2}{M_{cutoff}^2})= log(Det\frac{M^2}{M_{cutoff}^2})=Tr(log\frac{M^2}{M_{cutoff}^2})$.

But from above

$Tr(log\frac{M'^2}{M_{cutoff}^2})=\Sigma_i log\frac{m_i^2}{M_{cutoff}^2}$.

Thus

$Tr(log\frac{M^2}{M_{cutoff}^2})=\Sigma_i log\frac{m_i^2}{M_{cutoff}^2}$.

The point of all this is to show that you can save yourself a lot of time in computing the effective potential by using the expression on the left rather than calculating each term individually.


  • You ask where fermion masses will come from. They follow because the fields in $W$ and $K$ are superfields.
  • Zohar already mentioned the mass matrices coinciding if $f=0$. I will just add that you know if SUSY is unbroken the superpartners have the same mass. Looking at that prefactor from Martin's eq. (1.2) that I quoted above should make clear how then their contributions to the supertrace cancel. The scalars are spin zero while the fermions are spin 1/2 and have a relative minus sign. There are twice as many scalars but the fermions get a factor of 2 in the prefactor of the supertrace. Thus they cancel.
  • "Integrating out" is not the same as $m\rightarrow 0$, though in this case I think that still gave you the right idea. Generally when people talk about integrating out, it means they integrate out high-momentum modes, giving some effective theory valid up to $M_{cutoff}$, above which things are integrated out, and/or replacing particular Feynman diagrams with effective vertices, such as when people talk about a gluon-gluon-Higgs effective coupling, which really takes place through a loop involving other particles, particularly the (quite heavy) top quark. The fields that have masses above the cutoff will no longer appear as fields in the effective theory but will instead give effective operators in the Lagrangian or effective vertices in the Feynman rules. http://en.wikipedia.org/wiki/Effective_field_theory might be helpful.
  • It's hard, bordering on impossible, to guess about the significance of the $\phi_2=-\frac{hX\phi_1}{m}$ without more context. Such a value obviously makes the last term in the $V$ you've written disappear, but why that is a big deal, I can't say. I have a feeling he was trying to say something about when nonzero vevs would break SUSY but I really have no context here.
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.