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Where is Higgs boson?

  1. Apr 2, 2008 #1
    How come that higgs boson hasn't been found yet? According to wikipedia,The non-observation of clear signals leads to an experimental lower bound for the Standard Model Higgs boson mass of 114.4 GeV at 95% confidence level.

    Since the Ferimlab Tevatron has 10 times more power for the particle collisions, shouldn't be rational to confirm Higgs boson's existence experimentally?

    I've read that LHC startup is at about 10TeV. Does this mean that LHC MUST observe Higgs boson (if it really exists of course!)?
  2. jcsd
  3. Apr 2, 2008 #2


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    The lower bound on the Higgs boson comes from LEP-II not seeing it. As to why the Tevatron hasn't seen it yet: it's not a very easy signal to work out. Lots of backgrounds. The Tevatron just might not have enough luminosity to produce enough of them to overcome background events.

    If the LHC doesn't see a Higgs, then it isn't there! It will have plenty of luminosity and energy reach. We'll just have to wait and see...
  4. Apr 2, 2008 #3
    In addition to the luminosity that Blenchman indicated, there is also the fact that this does not refer to fundamental collisions center of mass energy. (Anti)Protons are made up of more fundamental quarks and gluons, and those are really the colliding particles.
  5. Apr 2, 2008 #4
    If this is a question which interests you you may want to follow Dorigo's blog, "a quantum diaries survivor". Dorigo works at the tevatron and remains optimistic that they may be able to actually find the Higgs there. He keeps posting updates on the baby steps the Tevatron is taking toward that goal.

    Here he has an article from last week in which he shows some data consistent with the idea the Higgs is *at* 115-130 GeV. Unfortunately the data as he explains is very far from conclusive and may well just be background noise-- in fact, the "signal" in his plot is stronger at 160 GeV, where the Higgs has already been excluded! (So I'm not qualfied to interpret these plots, but it sounds like if we take this data seriously we're supposed to believe that the strong signal at 160GeV was background noise, but the weaker signal at 115GeV was real? Hmm...) Anyway, this at least does seem to give reason to be optimistic that maybe if the Tevatron keeps gathering data they'll eventually be able to form some kind of conclusions on the Higgs.

    But there are no guarantees, as people have noted above it's not the case that just because your beam runs at 1 TeV and the particle exists at 115 GeV, that you'll be able to discern that 115 GeV particle in your data.
    Last edited: Apr 2, 2008
  6. Apr 2, 2008 #5
    Thank you all for you answers!

    How can we be so sure that LHC should definitely observe the Higgs boson, that is it will have sufficient luminosity to do so? Is there any theoretical prediction that sets an upper bound for the energy or the luminosity required to find Higgs? If so, is Tevatron capable to approach this upper bound?

    As for the quarks and gluons, what you really mean is that the strong interactions between them happen at lower energies, which are not sufficient to produce Higgs? If so, is there any theoretical description that informs us what is the amount of energy that is "lost" in the process?
  7. Apr 2, 2008 #6


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    You have to calculate the cross section for higgs production, and the decay rates, compare to the cross section for background events in the Standard Model, and compute the luminosity needed to overcome those backgrounds. I do not know the numbers, but my understanding is that the Tevatron is hard-pressed. MAYBE they can do it, but it's not very promising. The LHC is a MUCH larger machine, and assuming that it performs up to expectations, it will definitely find the Higgs if it is there, unless there is something hiding it. There are "doomsday"-like scenarios where the Higgs might not be found at the LHC, but such cases are pretty contrived; although if we don't see the Higgs, we will have to look closely at them.

    You have to take the Parton Distribution Functions into account. This is done in most Monte Carlo programs like PYTHIA, for example.
  8. Apr 24, 2008 #7
    First, unitarity requires the Higgs boson to have mass less than 1 TeV. If you want, I can try to give some details in a week or two when I should have more time. So the LHC will have plenty of energy to see this, and tons of work went into making sure it will have enough luminosity either to detect or rule out the Higgs once and for all.

    Second, here is a simplified way to think about what's going on with the quarks and gluons. Ignoring gluons and virtual particles, by colliding a proton with an antiproton, as at the Tevatron, we are really colliding three quarks (in the proton) with three anti-quarks (in the antiproton). If the Higgs boson is produced, it will be from one of the quarks interacting with one of the antiquarks. But if we assume the quark has one-third of the proton's total energy, and the antiquark has one third of the antiproton's total energy, then the quark-antiquark energy is really around ~650 GeV, not the total ~2000 GeV that the proton and antiproton have.

    The Parton Distribution Functions blechman mentions describe the probabilities that quarks or gluons (collectively, "partons") will have a certain fraction of the proton's total energy. Even in the contrived example I gave above, you see that there is less energy to work with than you might expect, and taking into account the gluons, sea quarks, and PDFs makes things even more difficult.
  9. Apr 24, 2008 #8
    I'd be curious to read these details whenever you have time. I am particularly curious how the 1 TeV limit follows from "unitarity"? Thanks.
  10. Apr 24, 2008 #9
    In the meantime, you may check the references given, for instance, in the particle data group review of particle on Higgs bosons searches
  11. Apr 24, 2008 #10
    The time-consuming part of it will be trying to convert things into non- (or less-)technical language. If you're feeling up to the technical stuff, you can check out section 2.6 of the Higgs Hunter's Guide and references therein -- it's partially available through Google Books (books.google.com), so you can get a taste by searching there.

    For instance, in less-technical language (copied from Wikipedia), "unitarity means that the sum of probabilities of all possible outcome of any event is always 1" -- en.wikipedia.org/wiki/Unitarity_(physics)
  12. May 20, 2008 #11
    So, yeah, I'm afraid I'm not able to give a good layperson's explanation for the unitarity bound on the Standard Model higgs mass. The basic idea is that the probabilities of certain interactions (specifically, vector boson scattering) depend on the higgs mass, and if the higgs mass is too large, those probabilities end up greater than 100%. We interpret such a nonsensical result as either setting an upper limit on the higgs mass in the framework of the Standard Model, or indicating the scale at which some new physics (such as supersymmetry or technicolor) must come into play to make the Standard Model an inappropriate framework at that scale.

    I don't consider that satisfying because I have no non-technical way of explaining why the problems arise. I'll briefly sketch out the outline of the technical derivation, but I wouldn't expect anyone other than folks who've done some graduate quantum field theory to be able to follow it, no matter how much detail I give.

    We start off analyzing vector boson scattering ([tex]WW \to WW[/tex], [tex]W^+W^- \to ZZ[/tex], etc) using the goldstone boson equivalence theorem. This roughly states that at high energies (much greater than the masses of the [tex]W[/tex] and [tex]Z[/tex]), the longitudinal components of the [tex]W[/tex] and [tex]Z[/tex] behave like massless scalar goldstone bosons. This is an aspect of the common saying that the [tex]W[/tex] and [tex]Z[/tex] each "eat" a goldstone boson degree of freedom to become massive through spontaneous symmetry breaking. This leads to the result (which also follows more generally from considering polarization factors) that at high energies, the amplitude for longitudinal gauge boson scattering dominates the amplitude for transverse gauge boson scattering.

    For example, at high energy the amplitude for longitudinal [tex]W^+_LW^-_L \to W^+_LW^-_L[/tex] scattering is
    [tex]\mathcal A = -\frac{m_h^2}{v^2}\left(\frac{s}{s - m_h^2} + \frac{t}{t - m_h^2}\right),[/tex]
    [tex]v = \left(\frac{G_F \sqrt 2}{(\hbar c)^3}\right)^{-1/2} = 246 {\rm\ GeV}[/tex]
    is a constant known accurately from precision electroweak experiments, and [tex]m_h[/tex] is the mass of the higgs.

    To connect to unitarity, we break that amplitude up into a partial-wave decomposition,
    [tex]\mathcal A = 16\pi\sum_{l = 0}^{\infty}(2l + 1)P_l(\cos\theta)a_l.[/tex]
    The optical theorem, which must hold if unitarity is not violated, gives bounds on the coefficients [tex]a_l[/tex], in particular
    [tex]|\Re(a_l)| \le \frac{1}{2}[/tex],
    which follows from
    [tex]\Im(a_l) = |a_l|^2 = \Re(a_l)^2 + \Im(a_l)^2 \Rightarrow \Re(a_l)^2 = \Im(a_l)(1 - \Im(a_l)) \le \frac{1}{4}.[/tex]

    We can find an expression for [tex]a_0[/tex] by integrating over
    [tex]\int_{-1}^1 d\cos\theta \to \int_{-s}^0 dt[/tex]
    [tex]a_0 = \frac{1}{16\pi s}\int_{-s}^0\mathcal A dt = -\frac{m_h^2}{16\pi v^2}\left[2 + \frac{m_h^2}{s - m_h^2} - \frac{m_h^2}{s}\log\left(1 + \frac{s}{m_h^2}\right)\right][/tex]
    The high-energy limit [tex]m_h^2 \ll s[/tex] gives
    [tex]\frac{m_h^2}{8\pi v^2} \le \frac{1}{2} \Rightarrow m_h \le 870 {\rm\ GeV},[/tex]
    while if we try to get rid of the higgs, which we can do by taking its mass to infinity, [tex]m_h \to \infty[/tex], [tex]m_h^2 \gg s[/tex], we find
    [tex]\frac{s}{32\pi v^2} \le \frac{1}{2} \Rightarrow \sqrt s \sim 1.8 {\rm\ TeV}[/tex]
    as the maximum scale at which new physics must appear to preserve unitarity (conservation of probability).

    As a final note, if I recall correctly, you can get a slightly lower bound of
    [tex]m_h \le 710 {\rm\ GeV}[/tex]
    by performing a similar calculation for [tex]W^+_LW^-_L \to Z_LZ_L[/tex] scattering, which is more complicated since it's inelastic and has identical particles in the final state.
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