Tuesday, July 13, 2010

What a light Higgs would mean for particle physics

Recent rumors about the associated production of Higgs boson with bottom quarks, despite their being denied by the Fermilab P.R. department, make it more likely than not that the Higgs boson is lighter than 135 GeV, possibly close to the 115 GeV figure suggested by CERN's previous LEP collider right before it was stopped by Luciano Maiani in 2000.

If this value is correct, what could we say about the inner workings of the Universe and the future of particle physics? Before I get to this hot question, let me discuss a more elementary one:

Why the Higgs has to exist?

The electromagnetic and weak interactions have been described by the electroweak theory for more than 40 years. The theory implies that the weak interactions - such as the beta decay - are mediated by new particles, the W+, W-, and Z_0 bosons.

These particles generalize the photon - and they extend the electromagnetic U(1)_{electromag} gauge symmetry to SU(2)_{isospin} x U(1)_{hypercharge} where the electromagnetic U(1) generator is written as "Y/2 + T_3", or the sum of the hypercharge U(1) generator and the z-component of the isospin.

The theory worked so well that its authors, Glashow, Salam, and Weinberg - I was tempted to write Green, Schwarz, Witten whose initials are also GSW - received their Nobel prize in 1979, four years before the W and Z bosons they predicted were even observed!

But don't get me wrong, the W,Z bosons were detected at CERN in 1983 which produced another, experimental Nobel prize for van der Meer and Rubbia in 1984.

From a theoretical, or top-down perspective, it's clear why we need something such as the Standard Model to explain the weak interactions. The dependence of the beta decay on polarizations implies that the interaction is of the vector-pseudovector (V-A) type.

Renormalizability implies that you need spin-one intermediate particles and a pair of cubic interactions to generate the four-fermion interactions first studied by Fermi in the 1930s. Consistent spin-one fields have to be gauge fields. And the corresponding gauge symmetry has to be broken by the Higgs sector for the W,Z bosons - and all the fermions - to acquire their masses that differ from their isospin partners. Gauge symmetries are pretty - and they remain equally pretty when they're spontaneously broken which must be done to match the observed data.

Don Garbutt: The Higgs Field

The simplest way to break the electroweak symmetry is to add a scalar field that is electrically neutral (under "Q"), so that electromagnetism remains an unbroken long-range force (the photon remains massless because it doesn't interact with the Higgs), but carries nonzero "Y" and "T_3" charges so that W,Z bosons become massive and the weak nuclear force becomes a short-range force.

The elementary Higgs field, added to the SU(2) x U(1) theory by Salam and especially Weinberg, is something that you need but you're not necessarily excessively proud about, at least according to Glashow who called the Higgs boson "Weinberg toilet" for this very reason.

However, you may find gauge symmetries too formal and abstract an idea. Instead, you may want to follow the bottom-up approach described in the book on the left side (written by a QFT teacher of mine in Prague) and start with the particles we can observe. There must exist well-defined formulae for their scattering.

The total probability of different processes can never exceed 100%. If you are satisfied with the tree-level approximations for the formulae, you can nicely add the new particles and new interactions and show - in a step-by-step fashion - that they're inevitable to keep many kinds of scattering probabilities below 100%. You never encounter a "gauge symmetry" if you construct the Standard Model in this way; however, the final theory you obtain by this requirement of "tree unitarity" is fully equivalent to the Standard Model derived from the "pretty" principles of gauge symmetries.

The last particle you have to add in this bottom-up algorithm is the Higgs boson. It has to be there for the W-W scattering to remain below 100% of probability up to a TeV. Why is it so? It's because the Standard Model without a Higgs boson would predict that the collisions of two longitudinally polarized W bosons would exceed 100% at energies of order 1 TeV.

Many people have claimed that they can describe the electroweak interactions without any Higgs. But you should always ask them what is their actual formula for the (longitudinal) W-W scattering cross section, why it agrees with the established formulae at low energies (given by the Standard Model), but also why it doesn't produce probabilities above 100% at high energies because this is what the "Standard Model minus the Higgs boson" automatically implies.

This question of yours would be the right litmus test and it is directly linked to the main reason why people think that the Higgs must actually exist; the Higgsless "thinkers" who don't solve this problem are just denying it (or are unaware of it) and their ideas can't be interesting.

These probabilities of the W-W scattering would be given by the exchange of virtual W bosons, Z bosons, and photons. However, probabilities should never exceed 100%. The only way how to stop the unwelcome growth of the probability near 1 TeV is to add additional diagrams where the W bosons exchange a new scalar particle, the Higgs boson. The leading power-law divergences are cancelled and the probability stays below 100% for much longer.

So what if the Higgs is light?

First, we should appreciate that the Higgs would be the first known and detected elementary scalar particle in the Universe. During the years, various physicists speculated that Nature wouldn't allow spinless particles. These champions of preon theories etc. were always wrong but they would be proved wrong.

With the new particle, one should look at the Higgs potential, also known as the Mexican hat potential, Landau buttocks potential, or the wine bottle potential.

Note that the Higgs potential has a stationary point at "h=0" ("h" is called "phi" above). However, it's a maximum, not a minimum as you would expect, so Nature actually doesn't stay there. It's an unstable point and Nature wants to save energy. So She spontaneously rolls down to the bottom - in any direction. She does so at every point of space.

Consequently, even though all the laws of physics are symmetric under the SU(2) x U(1) rotations - and that includes the rotationally symmetry Higgs potential - the observed phenomena won't be symmetric because the vacuum has to choose a random direction in order to minimize the energy. And this random choice will break the pre-existing symmetries between the different directions.

This breaking mechanism, called the Higgs mechanism, belongs to the more general group of "spontaneous symmetry breaking" mechanisms in which the fundamental laws are symmetric but the observed phenomena won't be because the properties of the vacuum break the symmetry.

How does the Higgs mass affect the potential?

Well, we know that if there's a single Higgs field that generates the W,Z masses etc., the vacuum expectation value has to be "phi = 247 GeV" or so. That's the horizontal distance from the "phi = 0" maximum of the wine bottle potential in the middle to any point on the "circle of minima" at the bottom of the wine bottle potential where the Higgs field actually wants to sit.

We know what the vacuum expectation value (vev) approximately is because the W,Z boson masses are determined by this vev and the gauge couplings of SU(2) x U(1) that are known as well - because they can be calculated from the W/Z boson mass ratio and from the fine-structure constant (the strength of electromagnetic interactions).

However, what hasn't been known so far is the overall normalization of the potential.

The wine bottle graph above may be written down as
-A phi2 + B phi4,
with some appropriate coefficients "A,B". Note that there is a minus sign before (positive) "A" (usually written as "mu^2/2") because "phi=0" is a minimum, not a maximum, of the potential.

But at the end, the potential has to be bounded from below because the vacuum would otherwise be completely unstable and it would "escape to infinity" on this "phi space". So "B" (usually and conventionally written as "lambda/4") has to be positive. Consequently, there exists some minima determined by "A/B".

Because the position of the minimum is at "246 GeV", assuming the conventional normalization of the "phi" kinetic terms - the terms in the action with the derivatives - the only freedom we have is to rescale both "A" and "B". Moreover, the squared mass of the Higgs boson is determined by the second derivative of the potential expanded around the minimum - any point on the circle.

So the lighter the Higgs boson is, the shallowed the wine bottle potential becomes. For a light Higgs boson, both "A" and "B" are pretty low.

Renormalization running of the potential

However, quantum mechanics allows you to create virtual particle pairs for a while - and do analogous effects. These effects actually imply that the potential above etc. is just an approximation. Moreover, the approximation is associated with an energy scale. If you change the energy scale, i.e. if you go to shorter or longer distances, physics changes a bit.

At least, the coefficients change.

For example, the fine-structure constant that determines the strength of the electric attraction in some natural dimensionless units is 1/137.036 or so at very low energies. However, at higher energies, the denominator is smaller than 137.036.

Analogously, the coupling constants "A", "B" slightly depend on the energy scale, too. Such a dependence is usually very slow, logarithmic in Nature: fast, power-law dependence is usually captured by the classical approximation well. But quantum mechanics loves to add subdominant, power-law "running" of the constants. You may imagine that the constants such as "A,B" in the Lagrangian love to depend on the energy in this way:
A(E) = A(E0) + C log(E/E0)
This formula is morally correct but you must use more accurate ones if you really care about the numbers. If you measure "A" at some energy scale "E_0" and you increase your energy to "E" which is 2.718 times higher, "A" changes by a constant called "C" in the formula above.

In which way do the couplings change if you go to higher energies?

I have already said that the electromagnetic interaction is always getting stronger. The strong interactions between the quarks - the Yang-Mills interactions - have the unusual ability to grow weaker at higher energies i.e. shorter distances. This "antiscreening" comes from the "magnetic" interactions of the gluons for themselves and it is the cause of the "asymptotic freedom": quarks inside the protons behave as nearly free particles.

The asymptotic freedom was the main property of the new theory of quark forces, Quantum Chromodynamics, that also allowed the Nobel committee to give the Nobel prize to Gross, Wilczek, and Politzer.

The Higgs "B" interaction - in front of the quartic term (one with the fourth power) - is somewhat analogous to the fine-structure constant. This interaction is also growing stronger at higher energies, due to the Higgs self-interaction. Note that the God particle is giving mass to other particles but in some sense, He is giving mass to Himself, too - although this mass, coming from the self-interactions also known as its own potential energy, seems a bit mundane and irreligious. ;-)

So the Higgs quartic self-interaction given by the "B" term is getting stronger at higher energies. Eventually, it could reach the "Landau pole" and the theory would become defunct. Well, the perturbative expansions would surely break down but there are good reasons to think that the theory, without being completed to a more robust one, would become inconsistent even at the non-perturbative level.

The quartic coupling "B" can't really exceed a threshold of order one for any energy scale. A value above this threshold would make the power expansions in "B" diverge. This fact means that the Higgs can't be too heavy for the Standard Model to work. It must be lighter than 800-1000 GeV or so.

(Supersymmetry and high-precision measurements of particle physics are more stringent: both of them independently imply that the Higgs is almost certainly lighter than 200 GeV.)

Fine. So I just explained that the Higgs mass - and the "A,B" coefficients - can't be too high. But that may become an irrelevant academic question if the Higgs is indeed light. If the Higgs is as light as 115 GeV or so, we should focus on the behavior of the Higgs potential for low values of "A,B".

If "A,B" are low, it is much easier to change their sign by the running - by adding "C times logarithms", using the toy model above. So in some sense, the physics is even more sensitive to the quantum effects.

It turns out that while the quartic interaction "B" wants to grow stronger because of the Higgs self-interactions (namely because of Feynman diagrams with a Higgs loop, and two pairs of legs attached at two points), there are also terms that prefer to weaken "B" at higher energies.

These terms are analogous to the "antiscreening" of the gluons that led to the asymptotic freedom.

In particular, the interactions of the Higgs scalars with the fermions - leptons and quarks - have this opposite, weakening effect. Because the top quark is much heavier than all other quarks and leptons, it is really the top quark that has the strongest interaction with the Higgs boson. And that's why the top quark is the most important player in this discussion, too.

Because the top quark mass is about 173 GeV, it is actually heavier than the Higgs if the rumors are correct. It is not hard to believe - and one can check - that the loop interactions involving the heavier top quark are actually more important than the self-interactions of the lighter Higgs boson. And the top quark is making the Higgs self-interaction weaker at higher energies.

So the coupling "B" is going down if you increase the energy scale "E".

It turns out that for the Higgs mass around 115 GeV, the coupling "B" actually goes negative at energies of order 10^{6} GeV. Nature must be able to tell particles at these high energies what to do when they collide. Be sure that such particles exist. In fact, cosmic rays detected on the Earth routinely exceed 10^{10} GeV.

So the laws must tell them something to do. And if the "B" coefficient were negative, the quartic potential would actually be unbounded from below, much like the function "-phi^4". The vacuum would be totally unstable. In fact, some higher-order quantum effects could stabilize the potential at vastly higher values of "phi" than the electroweak ones but it's reasonable to think that such dramatic changes of physics are incompatible with the observations.

What will Mother Nature tell these 10^{10} GeV cosmic rays when they collide?

It must tell them that things are fine. Things have to be fine, indeed. The coefficient "B" cannot grow negative. The problem is that the Standard Model - the full Standard Model, including all the known elementary particles plus the 115 GeV Higgs boson - does predict that "B" goes negative.

Armageddon! What can we do about it?

You don't have to do anything. You must rely on Nature that she solves Her problems. After all, we observe that she has solved them because the world hasn't disappeared within a fraction of a second. So there must be new effects that our discussion has neglected. And these effects prevent "B" from getting negative at 10^{6} GeV.

In other words, there must be new physics below one million gigaelectronvolts!

What this physics can be? Is the Higgs boson composite? Well, it seems extremely unlikely if the Higgs boson is light. Why is it so? The technicolor and other "composite" theories of the Higgs boson need to produce the fermion masses much like other theories of the Higgs.

The new structures needed for the fermion masses actually predict new pseudo-Goldstone bosons that are predicted to be lighter than the Higgs itself. So if the Higgs is light, and if - as it seems - there are no similarly light new pseudo-Goldstone bosons, the "composite" Higgs theories, at least the known ones (with some possible exception of highly contrived ones), are dead.

Now, assuming that the Higgs boson remains an elementary point-like particle at 10^{6} GeV, there must exist other new physical phenomena that keep "B" positive in the old-fashioned way, by positive contributions to the potential. These new particles must interact with the Higgs - at least indirectly.

A lesson suggested above is more general. New fermions typically push "B" to grow weaker at higher energies; new bosons want to make it stronger. So we actually want new bosons if our goal is to stabilize "B" and keep it positive at higher energies. I say it's our goal but it was really a homework exercise for the divine engineers that Mother Nature had to hire when She was designing the technical aspects of our Universe. ;-)

What are the new bosons that keep the Higgs potential stable up to higher energies?

Well, the first specific question you may want to ask is what is their spin. Elementary bosons in quantum field theories can only have spins 0,1,2. Moreover, there can only be one spin-two particle and it's the graviton. Its (gravitational) interactions are too weak to matter for the Higgs stabilization at 10^{6} GeV.

Also, spin-one particles have to be gauge bosons. But the Higgs is pretty unlikely to be charged with respect to another gauge symmetry. As a homework exercise, try to collect positive and negative evidence for/against such new gauge fields under which the Higgs is charged.

If you agree it's hard or unnatural, you may agree that the new bosons that help to stabilize the Higgs are spin-zero fields. They're spinless much like the Higgs itself. Why do they stay as light as 10^{6} GeV - or much less? That's another "hierarchy problem".

Much like for the Higgs itself, it can be seen that supersymmetry is the only good reason why a spinless bosonic particle may want to stay relatively light - or at least much lighter than the Planck scale.

Supersymmetry links spin-zero bosons to their spin-1/2 fermionic superpartners that may be massless or light because of the group theory: Weyl fermions have to be massless. Also, for the Higgs boson itself, SUSY may be interpreted as a mechanism that cancels the loop corrections to the Higgs' own mass - the cancellation is between the bosons and fermions (their superpartners) that interact with the Higgs.

This is how squarks (škvarky) look like in Czechia. Almost like cracklings.

The particular problem at 10^{6} GeV when the Higgs would go unstable in the absence of new physics is solved in supersymmetric theories, mostly by positive loop contributions to "B" coming from the stop squark, the superpartner of the top quark. The stop stops the decrease of "B"; this quark quarks the value of "B" up. I hope you understand what it means if someone quarks you up (MP3).

More generally, supersymmetry pretty much universally implies that the Higgs should be lighter than 130 GeV and it actually prefers values closer to 100 GeV, some of which have already been ruled out. At any rate, a light Higgs boson is much more natural in supersymmetric theories than the non-supersymmetric ones.


To summarize, a light Higgs boson pretty much proves that there has to be new physics beyond the Standard Model well below the Planck scale. The new particles should be bosons and the only natural reasons why the Higgs, or the new bosons, should stay light is supersymmetry.

The best fits based on this assumption imply that the squarks and slepton masses - and more generally, other superpartner masses - should be close to 500 GeV or so and they should be observable by the LHC, possibly by the end of 2011.

Large "tan(beta)"

The text above contains some hopefully persuasive arguments why supersymmetry would be likely with a light Higgs. Later on Tuesday, I learned additional details about the rumor. They spoke about associated production of Higgs bosons with bottom quarks.

The probability of this process grows like "tan^2(beta)" which is the ratio of the two Higgs doublets' vacuum expectation values in the minimal supersymmetric standard model where it's therefore natural to look for this signal. Consequently, the effect has a big chance to be sharply observed if we're lucky to live in a supersymmetric world with a high value of "tan(beta)".

That's exactly what the experimenters would de facto claim according to this detailed rumor.

"Tan(beta)" is widely believed to be constrained to stay in the interval between 2 and 50, in order for the couplings to stay perturbative up to the GUT or Planck scale. High values are natural to suppress extreme CP-violation in gauge-mediated supersymmetry-breaking models. They're (around 50) are also nice to get the top-bottom-tau mass hierarchy from equal Yukawa couplings, as expected from SO(10) GUTs or higher.

However, models exist with a "tan(beta)" well above 50. The most recently famous one is uplifted supersymmetry that was also claimed to explain the recent D0 observations of a new source of CP-violation (which is paradoxical because high "tan(beta)" is normally good to suppress unwanted CP-violation - here it's better to produce possibly observe CP-violation!).

These are just signals that may go away but it's clear that at a certain point in time, the amount of knowledge will be extensive enough for statistical arguments based on a few 3-sigma signals of this kind to become somewhat rational. Are we there yet?

Anniversary: the unique visitor #5,000,000 (five million) came from the University of Barcelona, a Mac-Firefox user. Click to zoom in.

Because of the timing, the last 50+50 visits before the special one included many, mostly European places - such as CERN, Oxford, Cambridge U.K., but also Wolfram.

The first 100 visitors of the 6th million included people from University of East Anglia (ClimateGate), Max Planck Institute, Durham University, Weizmann Institute, Credit Suisse, and many others.

The spectrum of visitors is very different during the U.S. daytime.

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