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Kähler potential vs full effective potential

+ 15 like - 0 dislike

In evaluating the vacuum structure of quantum field theories you need to find the minima of the effective potential including perturbative and nonperturbative corrections where possible.

In supersymmetric theories, you often see the claim that the Kähler potential is the suitable quantity of interest (as the superpotential does not receive quantum corrections). For simplicity, let's consider just the case of a single chiral superfield: $\Phi(x,\theta)=\phi(x)+\theta^\alpha\psi_\alpha(x) + \theta^2 f(x)$ and its complex conjugate. The low-energy action functional that includes the Kähler and superpotential is $$ S[\bar\Phi,\Phi] = \int\!\!\!\mathrm{d}^8z\;K(\bar\Phi,\Phi) + \int\!\!\!\mathrm{d}^6z\;W(\Phi) + \int\!\!\!\mathrm{d}^6\bar{z}\;\bar{W}(\bar\Phi) $$ Keeping only the scalar fields and no spacetime derivatives, the components are $$\begin{align} S[\bar\Phi,\Phi]\big|_{\text{eff.pot.}} = &\int\!\!\!\mathrm{d}^4x\Big(\bar{f}f\,\frac{\partial^2K(\bar\phi,\phi)}{\partial\phi\partial{\bar\phi}} + f\,W'(\phi) + \bar{f}\, W(\phi)\Big) \\ \xrightarrow{f\to f(\phi)} -\!&\int\!\!\!\mathrm{d}^4x\Big(\frac{\partial^2K(\bar\phi,\phi)}{\partial\phi\partial{\bar\phi}}\Big)^{-1}|W'(\phi)|^2 =: -\!\int\!\!\!\mathrm{d}^4x \ V(\bar\phi,\phi) \end{align}$$ where in the second line we solve the (simple) equations of motion for the auxiliary field. The vacua are then the minuma of the effective potential $V(\bar\phi,\phi)$.

However, if you read the old (up to mid 80s) literature on supersymmetry they calculate the effective potential using all of the scalars in the theory, i.e. the Coleman-Weinberg type effective potential using the background/external fields $\Phi(x,\theta)=\phi(x) + \theta^2 f(x)$. This leads to an effective potential $U(\bar\phi,\phi,\bar{f},f)$ which is more than quadratic in the auxiliary fields, so clearly not equivalent to calculating just the Kähler potential. The equivalent superfield object is the Kähler potential + auxiliary fields' potential, as defined in "Supersymmetric effective potential: Superfield approach" (or here). It can be written as $$ S[\bar\Phi,\Phi] = \int\!\!\!\mathrm{d}^8z\;\big(K(\bar\Phi,\Phi) + F(\bar\Phi,\Phi,D^2\Phi,\bar{D}^2\bar{\Phi})\big) + \int\!\!\!\mathrm{d}^6z\;W(\Phi) + \int\!\!\!\mathrm{d}^6\bar{z}\;\bar{W}(\bar\Phi) $$ where $F(\bar\Phi,\Phi,D^2\Phi,\bar{D}^2\bar{\Phi})$ is at least cubic in $D^2\Phi,\bar{D}^2\bar{\Phi}$. The projection to low-energy scalar components of the above gives the effective potential $U(\bar\phi,\phi,\bar{f},f)$ that is in general non-polynomial in the auxiliary fields and so clearly harder to calculate and work with than the quadratic result given above.

So my question is: when did this shift to calculating only the Kähler potential happen and is there a good reason you can ignore the corrections of higher order in the auxiliary fields?

This post has been migrated from (A51.SE)
asked Oct 21, 2011 in Theoretical Physics by Simon (325 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

2 Answers

+ 11 like - 0 dislike

Indeed, your question has nothing to do with the distinction between 1PI and Wilsonian. The answer is that the terms which contain nontrivial dependence on $D^2\Phi$ are to be dropped if the breaking of supersymmetry is small compared to the natural ("supersymmetric") mass scale in the problem. You can see this by noting that the effective potential has to be of the form $f^2 F(f/M^2)$ where $f$ is the SUSY breaking scale and $M$ is some supersymmetric scale (which can the VEV of some modulus, too). Another way to see this is that terms with more powers of $D^2\Phi$ have a higher engineering dimension and thus have to be divided by some SUSY scale, so their effect disappears as f/M^2->0.

In some physical scenarios these corrections could be important, but since in dynamical models having rigorous control over the physics usually entails having SUSY-breaking as a small effect, in most of the literature these terms are dropped.

This post has been migrated from (A51.SE)
answered Oct 28, 2011 by Zohar Ko (650 points) [ no revision ]
Thanks Zohar, that all sounds fairly reasonable! Do you know the reference for where this type of reasoning was first applied? Or a canonical reference for this?

This post has been migrated from (A51.SE)
This reasoning is probably as old as the realization that SUSY can be broken. Instead of referring you to arcane computations from three decades ago, I'd recommend you take a look at appendix A.5. of hep-th/0602239. They study the effective action due to some massive fields which are integrated out and show that to leading order (i.e. $|f|^2$) the effective potential can be neatly packaged in a K\"ahler potential. They also emphasize the existence of higher order corrections (higher order in $f$), which are not captured by the effective K\"ahler potential.

This post has been migrated from (A51.SE)
If you are also interested in retracing the footsteps of this computation, you should take a look at hep-th/9605149...

This post has been migrated from (A51.SE)
Welcome Zohar, and thanks for your contribution. I hope with time this becomes a really useful place for our community, so I am glad every time I see a familiar face...

This post has been migrated from (A51.SE)
Zohar: The fact that the leading order eff pot in $|f|^2$ can be packaged into a Kähler potential seems kind of trivial. But I agree with you and ISS that for small susy breaking it is probably enough. I think that, when combined with calculational difficulties, this motivates the common practice of ignoring the full effective potential.

This post has been migrated from (A51.SE)
Whether or not the corrections in $f^2$ should be ignored is model dependent and the mere fact that they may be harder to calculate does not by itself justify ignoring them (and by they way, at one-loop nothing is difficult, not even getting these corrections right)

This post has been migrated from (A51.SE)
+ 5 like - 0 dislike

There are two types of effective actions, the one particle irreducible (1PI) (Coleman-Weinberg) and the Wilsonian.

The variables in the 1PI are the vacuum condensations of the fields, i.e., it is "classical". In principle it is computed by performing the path integral with sources and then replacing the sources by the vacuum condensates through a Legendre transform. This action includes all the quantum corrections of the theory and its potential term determines its vacua. This action needs not be local. In prcatice, this action can be computed only approximately by loop expansion, and its expansion suffers from IR divergences in the case of massless fields.

The second type of effective action is the Wilsonian effective action where the modes of energy beyond some given scale are integrated. The basic variables here are the low energy modes of the fields. This action is quantum mechanical in the sense that it does not include the radiative corrections of the low energy modes and still the path integral on them must be performed. This action is local and does not suffer from IR divergences, for these reasons it is used in supersymmetry breaking computations. Please see the following review by Tanedo (and references therein) describing the distinction between the two types of effective actions in the context of supersymmetry.

Now, regarding to the computation in the first paragraph of the question, if the "tree-level" Kahler potential is used, it is just a computation of the tree scalar potential.

This post has been migrated from (A51.SE)
answered Oct 22, 2011 by David Bar Moshe (3,545 points) [ no revision ]
Thanks David, but I'm not sure how the 1PI vs Wilsonian effective actions addresses my question. It might, but it's not obvious... Even loop calculations now seem to focus purely on the Kähler potential and claim that [it's the SUSY analogue of the Coleman-Weinberg effective potential](http://arxiv.org/abs/hep-th/9605149) or that ["Kähler potential, the superpotential and the gauge kinetic function" encode all info about the low-energy effective action](http://inspirehep.net/record/696662). (By a weird coincidence I downloaded Flip's "Seibergology" review last night - haven't read it all yet!)

This post has been migrated from (A51.SE)

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