Why is the lightest Higgs not a free parameter in SUSY? (SE)convinced me to write a new, not too long text about the way to construct minimally supersymmetric Lagrangians for \(d=4\) supersymmetric quantum field theories.

Just to be sure, the poster above asked why the Higgs mass seems to be freely adjustable in the Standard Model but there are various constraints on the Higgs mass in the Minimal Supersymmetric Standard Model – for example, the lighter Higgs mass can't be too much heavier than the Z-boson.

I answered that a heavy Higgs boson in the Standard Model also leads to some trouble such as instabilities but the bulk of my answer – which you can read if you click at the SE link above – was dedicated to an explanation why the Higgs masses can't be arbitrarily scaled in supersymmetric theories.

A short answer is that the quartic coupling \(\lambda\) in the \(\lambda h^4\) quartic (fourth-order) self-interaction of the Higgs field – which is an increasing function of the Higgs mass, assuming a fixed given vacuum expectation value (vev) – is no longer arbitrary in the MSSM. Instead, it is given by various combinations of \(g^2\) and \(g^{\prime 2}\) gauge couplings for the \(SU(2)\times U(1)\) electroweak gauge group. This follows from some insights from "101 SUSY model building". In this text, I would like to sketch how the \(\NNN=1\) supersymmetric Lagrangians in \(d=4\) may be constructed in some more detail.

This text may be viewed as a continuation of various previous SUSY texts on TRF, especially Supersymmetry: transformations of superspace. All the material I describe here may be found in SUSY textbooks such as the book by Michael Dine.

Fine. First of all, you must realize that supersymmetric quantum field theories are "just a subset" of quantum field theories. Unlike string theory, they're not "generalizing" quantum field theories in any sense. On the contrary, they are restricting quantum field theories, they are choosing a subset of them that exhibits the new nice symmetry, supersymmetry. So you may write all the superfields "in components" and you obtain an ordinary quantum field theory with \(j=0\) scalar fields, \(j=1/2\) spinor fermionic fields, and \(j=1\) gauge fields.

Now, I will be mostly focusing on renormalizable quantum field theories in \(d=4\). Roughly speaking, they're theories whose coupling constants are classically dimensionless or they have the units of a positive power of mass. Couplings with units of a positive power of length (i.e. negative power of mass) – those that you would be forced to place in front of very complicated high-mass-dimension terms such as \(h^{10}\) – would produce "nonrenormalizable theories", theories in which multi-loop Feynman diagrams would be increasingly more divergent because high powers of the loop energy, \(p^{k}\), would naturally arise to cancel the powers of these "naughty coupling constants" that have the units as negative powers of energy.

We also reduce our attention to fields with spin at most \(j=1\). The fields with \(j=3/2\) already require a local supersymmetry to get rid of some unphysical polarizations – so they're inevitably theories of supergravity if they're consistent at all. Fields with \(j=2\) must be linked to gravity in one way or another and fields with \(j\gt 2\) don't admit interactions consistent with the crucial new gauge symmetries at all.

And I will assume at most two-derivative terms in the Lagrangian. These conditions are fair but they reduce the room for possible quantum field theories dramatically.

**Players**

First, which fields can we use to play the game? When it comes to their spin (or, more generally, representations under the Lorentz or super-Poincaré algebra), we may only use two possible types of fields:

- Chiral superfields, unifying \(j=0\) and \(j=1/2\) excitations
- Vector superfields, unifying \(j=1\) and \(j=1/2\) excitations

Using the superspace which is very helpful for \(\NNN=1\) theories in \(d=4\), we may rewrite a chiral superfield (a superfield is a field that depends on normal bosonic spacetime coordinates as well as the new, Grassmann or fermionic coordinates \(\theta^\alpha\) and/or \(\bar\theta^{\dot\alpha}\)) in terms of ordinary fields (that only depend on the bosonic coordinates \(x^\mu\)) as:\[

\Phi(x,\theta)= \phi(x)+\sqrt{2}\theta\psi(x)+ \theta^2 F(x).

\] That's very simple. This superfield is called "chiral" – the adjective is linked to "hand" in Greek and in physics, it always represents (left-right-asymmetric) things that distinguish the left from the right (hand but not only hand) – because the superfield only depends on \(\theta^\alpha\) which are left-handed spinors but it doesn't depend on the complex conjugate \(\bar\theta^{\dot \alpha}\) which are the right-handed spinors.

If I screwed the usual conventions for which indices are left and which are right, then I apologize but the error doesn't matter too much because the complex (or Hermitian) conjugate field to a chiral superfield always has to exist as well and it depends on the opposite \(\theta\)'s, and has the opposite handedness.

In the formula for \(\Phi\), the first term is a complex scalar scalar field \(\phi(x)\). The second term, with the \(\sqrt{2}\) conventional normalization, contains a Weyl spinor field \(\psi_\alpha(x)\) which must be contracted with the coordinates \(\theta^\alpha\) to get rid of the spinor index \(\alpha=1,2\). The final term is proportional to \(\theta^2\) which is the product of the two components of the two-component spinor \(\theta^\alpha\) – I won't write it with the values of indices because it would reveal that my notation is ambiguous as the superscripts may denote indices as well as powers. Note that this \(\theta^2\) is Lorentz-invariant. The dynamical part of the term is \(F(x)\) which is again a bosonic field but an auxiliary one. It's an F-term in the field.

The mass dimensions of the component fields \(\phi,\psi,F\) are \({\rm mass}\), \({\rm mass}^{3/2}\), and \({\rm mass}^2\), respectively. This is clear from the fact that \(\theta^\alpha\) has the dimension of \({\rm length}^{1/2}\). This simple scaling already tells you that \(|F|^2\) is a pretty nice term that may appear in the ordinary Lagrangian. I wrote the absolute value because the action has to be real but \(F\) is inevitably complex, much like \(\phi\) and \(\psi_\alpha\): chiral things have to be complex because they may be used as variables in holomorphic functions such as the superpotential.

Now, the other type of the player is the vector superfield, unifying a Yang-Mills gauge field with a Majorana \(j=1/2\) spinor. Indeed, the information in a \(d=4\) Majorana spinor is the same as in the \(d=4\) Weyl spinor but here I choose the "Majorana" terminology because the gauge fields are naturally "real", not complex and holomorphic, so this should hold for the spinor term in it, too. And "real spinors" are called Majorana spinors.

A vector superfield isn't chiral so it depends both on \(\theta^\alpha\) and \(\bar\theta^{\dot \alpha}\):\[

\eq{

V &= i\chi -i\chi^\dagger-\theta\sigma^\mu \theta^* A_\mu +\\

&+ i\theta^2 \bar\theta \bar\lambda -i\bar\theta^2\theta \lambda + \frac 12\theta^2\bar\theta^2 D.

}

\] I should have used \(\bar\theta\) before as well (update: fixed retroactively) but now I introduced the bars for the \(\theta\)'s with the dotted indices. Note that the normal "gauge field" term with the vector index is multiplying the \(\theta\bar\theta\) structure exactly because the product of the two spinors of opposite chirality transforms as a vector (we may have inserted the 4D Pauli matrices \(\sigma^\mu\) in between the two spinors). The field \(A_\mu\) is the normal gauge field, with the units of mass. Again, you see that the last term – multiplied by all the four theta components – namely the D-term has the units of \({\rm mass}^2\) again, so \(D^2\) may appear and will appear in the ordinary Lagrangian.

We have already noticed but let me repeat it again. Even though the main bosonic field in the vector superfield \(A_\mu\) has a Lorentz vector index, we could actually write this field as one of the terms in a superfield that is a scalar \(V(x,\theta,\bar\theta)\) without any vector indices! The different components of the vector are encoded in the dependence on the theta's.

Now, what about the gauge symmetry? For \(U(1)\) fields, we liked to write the ordinary gauge transformation as\[

A_\mu\to A_\mu + \partial_\mu \lambda

\] or so. But this is obsolete because \(A_\mu\) is just one component in the expansion of a superfield. We want to construct supersymmetric theories so we must use the whole superfield \(V\) with the component \(A_\mu\) and all the other components, too. Consequently, \(\lambda\) must be promoted to a superfield, too. If you study how it can work so that you reproduce pretty much the same dynamics, you will realize that \(\lambda\) must be promoted to a chiral superfield and the gauge transformation acting on a \(U(1)\) vector superfield is\[

V \to V+i\Lambda -i\Lambda^\dagger

\] where \(\Lambda\) generalizes \(\lambda\) and is a chiral superfield. We added the complex conjugate term as well because \(V\), while not a chiral field, was real (or Hermitian, as an operator). Note that if we decided \(\Lambda\) to be a non-chiral, vector-like superfield, it would make all the degrees of freedom in \(V\) unphysical. That would be too much of a good thing, too much of a gauge symmetry.

You must have observed that our supersymmetric version of the gauge transformation only transforms \(V\) by a multiple of \(\Lambda\) rather than its spacetime derivatives. But it's OK because the old gauge field \(A_\mu\) is the \(\theta\bar\theta\)-proportional term in \(V\) and if you rewrite the gauge transformations in components, you will see that \(A_\mu\) de facto picks the derivative of \(\Lambda\).

For vector superfields, it's also useful to define the "gauge-invariant field strength" superfield generalizing the non-supersymmetric \(F_{\mu\nu}\) as\[

W_\alpha = -\frac{1}{4} \bar D^2 D_\alpha V

\] where the \(D\) objects are the superderivatives (dimensionally, they're "square roots" of normal bosonic derivatives). This "gauge-invariant field strength" superfield has a spinor index, unlike all previous superfields we have discussed (which were scalars so far), but unlike \(V\), it is moreover chiral and "fermionic". So the leading component without \(\theta\)'s is \(-i\lambda_\alpha\), a multiple of the gaugino spinor field, \(\theta_\alpha\) multiplies \(D\), the D-term, plus some multiple of \(F_{\mu\nu}\), and the \(\theta^2\) monomial multiples some first spacetime derivatives of \(\lambda^{*\alpha}\). I don't want to go into these things because much of the expansion of superfields into components involves a messy algebra and we have certainly gotten to that point already. ;-)

However, the \(U(1)\) gauge transformation of a charged chiral superfield is simple:\[

\Phi\to e^{-iq\Lambda} \Phi.

\] It's almost the same as it has been in non-supersymmetric theories. Moreover, those exponential formulae may be rather easily generalized from \(U(1)\) to non-Abelian gauge groups.

**Gauge-invariant kinetic terms**

Great. We want some actions that are renormalizable, gauge-invariant, and healthy. How do we construct the kinetic terms that produce e.g. the familiar Klein-Gordon term \(\partial_\mu \phi\cdot \partial^\mu\phi\) when expanded into components? Well, the superspace formula is actually easier than in non-supersymmetric theories once again. Instead of the spatial derivatives with \(\mu\)-indices carefully contracted, the kinetic term for the chiral superfield is simply\[

\LL_{\rm kin} = \int \dd^4 \theta \sum_i \Phi_i^\dagger \Phi_i.

\] There ain't no explicit derivatives here. This term is integrated over the whole four-dimensional fermionic part of the superspace. And that's the reason why the two spatial derivatives are produced at the end, out of the auxiliary components. You need to work hard to understand why all these things work but at the end, it's just some hard algebra. However, the term above – which I already summed over all the chiral superfields labeled by the index \(i\) – isn't gauge-invariant assuming that these fields carry charges. How do we make it gauge-invariant?

In non-supersymmetric theories, we would have to replace the partial derivatives \(\partial_\mu\) in the kinetic terms by the covariant derivatives \(D_\mu=\partial_\mu -ieA_\mu\). What about the supersymmetric theories in the superfield formalism? Well, we just insert \(e^V\) in between \(\Phi_i^\dagger\) and \(\Phi_i\):\[

\LL_{\rm kin} = \int \dd^4 \theta \sum_i \Phi_i^\dagger e^V \Phi_i.

\] Because many things simplify in the superspace, this simple insertion does the job. The exponential of the vector superfield \(V\) simply does the right "quasi gauge transformation" of the chiral superfield that guarantees that the result is gauge-invariant. Well, you may check the gauge invariance of this term. We have said how \(V\) transformed under a gauge transformation – we added \(\Lambda\) and \(\Lambda^\dagger\) to it. But when exponentiated, \(\exp(i\Lambda)\) and \(\exp(-i\Lambda^\dagger)\) – sorry if I switched the signs – exactly undo the same exponential factors that, as we have said, appear in the gauge transformation rules for the chiral superfields \(\Phi_i\) and \(\Phi_i^\dagger\). So the result is gauge-invariant.

We have said some things about the "non-chiral" part of the Lagrangian, one integrated both over \(\theta\) and \(\bar\theta\). But there's an interesting "chiral" part of the Lagrangian, the so-called superpotential \(W\). It only depends on the chiral superfields and it must depend holomorphically (or physicists could often say "analytically", ignoring the fact that for mathematicians, the latter adjective is somewhat more constraining). The superpotential term in the action is\[

\LL_W = \int \dd^2\theta\, W(\Phi_i) +\text{c.c.}.

\] The c.c. (complex conjugate) term to the first one depends on the opposite \(\theta\)'s and the fields \(\Phi^\dagger_i\), of course. The first interacting \(d=4\) supersymmetric theory that was studied was the so-called Wess-Zumino model. It had one chiral superfield and a quadratic-cubic superpotential, \(a\Phi^2+b\Phi^3\). That's the most general form that produces renormalizable interactions.

The funny thing is that if you carefully derive the equations of motion not only for the "old-fashioned crucial" components of the superfields but also for the auxiliary terms \(D_i,F_i\), you will realize that\[

F_i^* = -\pfrac {W}{\Phi_i}

\] is an equation of motion that "eliminates" the auxiliary term \(F_i\). However, if you study the superpotential part of the Lagrangian which matters – if you integrate over two theta's, what's left is the "highest-order" F-term-like component – you will realize that the superpotential Lagrangian may be rewritten into components – into the purely non-supersymmetric QFT language – as\[

\LL_W = \pfrac{W}{\Phi_i} F_i + \frac{\partial^2 W}{\partial \Phi_i\Phi_j} \psi_i\psi_j

\] which is capable of producing some potential terms for the scalars and some mass-like terms for the fermions. If you insert our equation of motion for \(F_i\), you will realize that the superpotential actually induces – in the non-supersymmetric language – a normal potential (this is not a vector superfield, just a clash of notation)\[

V = |F_i|^2 = \abs{ \pfrac{W}{\Phi_i} }^2.

\] So if you want to determine the "normal" potential, you differentiate the superpotential with respect to individual chiral superfields and square the absolute value of the result. That's why at most cubic superpotentials produce quartic ordinary potentials: the derivative of a cubic function is a quadratic one and the square of the latter is a quartic function.

Well, similar algebra applies to the vector superfields as well. The normal potential will also have contributions from \(D^2\), the squared auxiliary terms in the vector superfields, much like it had the \(|F|^2\) terms from the chiral superfields:\[

V =\sum_i \abs{F_i}^2 +\sum_a \frac{1}{2 g_a^2} (D^a)^2.

\] We have used some convention about whether or not the gauge coupling \(g_a\) is included in the normalization of the vector superfields; you know similar choices that you do in non-supersymmetric gauge theories. The equation of motion for the D-terms of the vector superfields tells you\[

D^a = \sum_i (g^a \phi_i^* T^a \phi_i).

\] So these D-terms are bilinear in the bosons from the chiral superfields, with charges (or matrices of the generators) playing the role of the coefficients. And the \((D^a)^2\) term in the "normal potential" therefore produces quartic (fourth-order) terms in the scalars.

*That's the way and the only way how the quartic self-interaction for the Higgs fields are generated in the Minimal Supersymmetric Standard Model.*

Indeed, there can't be any cubic term \(h^3\) in the superpotential \(W\) because \(W\) has to be gauge-invariant but you surely can't produce a gauge-invariant singlet as the third power of a charged doublet field. So all the quartic terms in the "normal potential" \(V\) have to arise from the D-terms, from the vector potentials. And that's why the quartic coupling constant(s) \(\lambda\) is (are) linked to various combinations of \(g^2\) for various factors in the gauge group. And that's why supersymmetry predicts inequalities for the Higgs masses, e.g. that at the tree level, the lightest Higgs boson has to be lighter than the Z-boson. (This inequality gets loosened already if you include one-loop corrections, especially from the top quark loops, and Higgs boson masses up to \(130\GeV\) would be compatible with the MSSM as a result.)

**The most general Lagrangian**

Using the pieces we have already encountered, the most general Lagrangian with at most two derivatives may be written as\[

\eq{

\LL &= \int \dd^4\theta\, K(\Phi_i,\Phi^\dagger_i)+\\

&+ \int\dd^2 \theta\, W(\Phi_i) + \text{c.c.}+\\

&+ \int \dd^2\theta \,f_a(\Phi) (W_\alpha^{(a)})^2+\text{c.c.}

}

\] The first term is non-chiral and depends on the so-called Kähler potential which is a non-holomorphic function of the chiral superfields, i.e. a function of them and their complex conjugates. For the normal free scalars, you need \(K\) composed of \(\abs{\Phi_i}^2\) terms. For them to be gauge-invariant, you have to insert the \(e^V\) objects in between.

The functions determining the terms on the remaining two lines are holomorphic fields of the chiral superfields. That's also why we have to add the complex conjugate terms by hand; the action has to remain real. It's the superpotential \(W\) and the gauge coupling function \(f\) – separate for each factor of the gauge group – which has the form\[

f(g^2) = \frac{8\pi^2}{g^2}+ia+{\rm const.}

\] The imaginary part \(ia\) automatically and inevitably produces a term "counting the instantons" proportional to \(F\wedge F\). The real part is dominated by the \(1/g^2\)-like term but the function may be shifted by loop corrections once you start to calculate them.

At any rate, if you only want to define a classical theory that will produce a renormalizable quantum theory, the form of the supersymmetric theory is extremely constrained. All the potential-like interactions must be encoded in the at most cubic, gauge-invariant superpotential \(W\); it's the most adjustable information about the theory that knows about the Yukawa couplings and similar properties of "pure matter". The Kähler potential must be de facto quadratic and is uniquely determined. The gauge coupling functions reduce to the constant gauge couplings (but they also know about the "axionic" imaginary parts).

*So the structure is pretty much determined once you choose your gauge group; invent your collection of chiral superfields and their charges or representations; and choose their holomorphic function \(W\).*

It may be useful to mention that the functions \(K, W, f^a\) determine the dynamics even if you consider local i.e. gauged supersymmetry – i.e. if you study theories of supergravity. However, supergravity isn't renormalizable, anyway, so it isn't justified to demand that you will get a renormalizable theory. For this reason, these functions are pretty much unrestricted in supergravity theories. Of course, \(W\) and \(f^a\) must still be holomorphic functions. But the functions are often non-polynomial and the Kähler potential may actually define the Kähler potential of a curved Kähler manifold.

I guess that most readers who managed to penetrate up to this point agree that supersymmetric theories in \(d=4\) are elegant, determined just by a few choices, but they still give you all the dynamics you need to describe the real world: kinetic terms for scalars, fermions, and gauge fields; gauge couplings for charged fields and Yang-Mills gauge fields themselves; Yukawa couplings; cubic and quartic couplings for scalars.

When you expand the nice superspace formulae into components, you obtain messy equations. But you shouldn't consider it as supersymmetry's fault. It's your problem that you need to rewrite formulae in a messy way to find out what's really going on. Nature doesn't have to write any messy formulae: She knows how to calculate and control Nature directly by the elegant laws and principles. ;-) Moreover, the theories you obtain by rewriting all the superfields in terms of components fields are nothing of "unprecedented messiness"; they're nothing else than the non-supersymmetric quantum field theories you used to study before you learned about supersymmetry, with some values of the couplings and other parameters (that are related to the parameters of the SUSY theories by various transformations and redefitions but that may be constrained by additional constraints implied by SUSY).

For decades, people would study \(\NNN=1\) supersymmetric theories in \(d=4\) as the only kind of supersymmetric field theories that may be relevant for the Universe around us. But as I have mentioned in several recent blog entries, the gauge fields in Nature around us could actually be parts of \(\NNN=2\) supermultiplets – manifestations of a larger supersymmetry that only holds for the gauge fields but not for the matter fields (the latter must remain in chiral superfields). That would be even more extraordinary, of course, because the gauge fields would preserve a greater fraction of the ultimate stringy beauty of Nature, beauty that has to be contaminated by various symmetry-breaking processes for Nature to get rid of Her sterility.

Thanks Lumo for this great extansion of your answer :-)

ReplyDeleteI'll devote to this text the proper amount of time it deserves tomorrow (or darn later today ...), because now it has become too late for me again ...

Good night !

This is indeed a very nice summary of how supersymmetric Lagrangians can be built :-)

ReplyDeleteSince this is very serious and not so easy stuff, I had to heavily cling to my demystified book (which contains many of the detailed calculations involving the in components expanded superfields too), meaning reconsidering certain things etc, but I finally got through :-P

Only this imiginary part of gauge coupling function catched me on the wrong foot somehow, that did not appear in Patrick Labelle's text ... :-/

Lenny Susskind explains the superspace formalism (in a slightly simplified way compared to this article) too in these nice Lectures:

https://itunes.apple.com/itunes-u/supersymmetry-grand-unification/id384233338?mt=10#ls=1

(The title of this course is some kind of a misnomer, there are no strings in it ...)

Cheers

I heard recently a remark of Nima Arkani-Hamed (the original subject was about twistor amplitudes) about a "tension" between supersymmetry and locality. A clue would be the impossibility of using a superspace for N=4 SUSY, although it it possible for N=2 and N=1 SUSY. I am not sure to have understood it correctly, because there is a difference between locality and manifest locality (and he talks about manifest unitarity and manifest locality), but could we say that too much fermionic degrees of freedom have the effect to render manifest locality (or maybe locality) impossible ?

ReplyDeleteIt's not a new or controversial statement that superspaces with too many Grassmann coordinates - too many manifest supersymmetries - become less useful and sometimes impossible to use.

ReplyDeleteThe N=4 Yang-Mills has this supersymmetry even if you don't write it in the hypothetical N=4 superspace. In N=0 or N=1 superspace, this gauge theory still has a manifest locality.

In the twistor variables, the Yangian may seem/be more fundamental and it mixes the supersymmetries with other generators etc.

Generally, it's true that formulations that make one kind of a symmetry (or locality or unitarity) manifest tend to obscure other things.

Dear Lubos,

ReplyDeleteWhat is your opinion on this news from MIT http://web.mit.edu/physics/news/spotlight/20121030_wilczek.html

It's fun. The following sentence isn't meant to be boasting but it's probably relevant: I played with a very similar model, and perhaps the same one, when I was an undergrad.

ReplyDeleteStill, it's not a full-fledged reliable theory of quantum gravity in any sense, so any conclusion may be either relevant for the real story or irrelevant and misleading.