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Why is the Higgs Boson tachyonic?

  1. Mar 29, 2006 #1
    Why is the (mass)^2 term of the Higgs Boson negative in the Standard
    Model Lagrangian to start with? I understand that along with the H^4
    term, it gives rise to dynamical symmetry breaking, which explains a
    bunch of stuff (and eventually gives us a physical mass term for the
    Higgs). But I find such a negative mass^2 rather unnatural, as it
    implies that if symmetry remains unbroken (at high enough
    temperatures), the Higgs Boson is an on-shell tachyon!

    How does one justify this leap of faith in introducing the -ve mass^2,
    or does it have some higher energy explanation in string theory or
    something?

    Thanks,
    Souvik
     
  2. jcsd
  3. Mar 30, 2006 #2
    Souvik wrote:

    > Why is the (mass)^2 term of the Higgs Boson negative in the Standard
    > Model Lagrangian to start with? I understand that along with the H^4
    > term, it gives rise to dynamical symmetry breaking, which explains a
    > bunch of stuff (and eventually gives us a physical mass term for the
    > Higgs). But I find such a negative mass^2 rather unnatural, as it
    > implies that if symmetry remains unbroken (at high enough
    > temperatures), the Higgs Boson is an on-shell tachyon!


    The "false vacuum", i.e., the higgs field 0, is not stable. That is most
    easy to understand in the classical limit, where the potential has a
    maximum rather than a minimum at H=0. The same holds true in the
    quantised theory, at least in the perturbative realm which is valid for
    the weak interactions since the coupling constant is small.

    Thus, there are no tachyons in the standard model. The SU(2) x U(1)
    (chiral) local gauge theory is broken down to electromagnetic U(1), and
    thus three of the Higgs fields (in the minimal standard model one has a
    Higgs doublet, i.e., four real fields) become "would-be Goldstone
    bosons", eaten up by 3 of the four gauge fields providing the
    additional degrees of freedom of massive vector bosons (massless vector
    bosons have two physical degrees of freedome, while massive ones have
    three!).
    >
    > How does one justify this leap of faith in introducing the -ve mass^2,
    > or does it have some higher energy explanation in string theory or
    > something?


    Of course, the symmetry breaking is put in by hand. We do not know of a
    more fundamental reason for it (yet?).

    A Higgs phenomenon, where we know a more fundamental reason, is the
    spontaneous breaking of electromagnetic symmetry in superconducturs:
    The ground state is a "condensate" of cooper pairs, and those are
    charged and so providing the symmetry breaking.

    --
    Hendrik van Hees Texas A&M University
    Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
    Fax: +1 979/845-1899 College Station, TX 77843-3366
    http://theory.gsi.de/~vanhees/ mailto:hees@comp.tamu.edu
     
  4. Mar 30, 2006 #3
    The Higgs boson is *not* tachyonic. The field \Phi which has a
    potential like a mexican hat is *not* the Higgs field. You get
    the Higgs field by choosing a gauge and a vacuum for your
    action, developping the action in the neighbourhood of that
    vacuum.

    Then you get massless Goldstone bosons which get `eaten' by
    the vector bosons in order to get the right number of degrees
    of freedom for a massive particle (three for spin one). The Higgs
    boson is the massive scalar that rests after after taking out the
    Goldstone bosons. The coupling of the vector bosons to the
    Higgs field is something like an effective mass term.

    As the vacuum you chose is stable, the Higgs's mass squared
    is positive. You should look this up. It's in all serious textbooks
    on QFT.

    {This is all about the `bare' Higgs mass. Of course, the real
    observable mass gets radiative corrections but in the end the
    mass squared is positive as it should.}
     
  5. Mar 30, 2006 #4
    Souvik wrote:
    > Why is the (mass)^2 term of the Higgs Boson negative in the Standard
    > Model Lagrangian to start with? I understand that along with the H^4
    > term, it gives rise to dynamical symmetry breaking, which explains a
    > bunch of stuff (and eventually gives us a physical mass term for the
    > Higgs). But I find such a negative mass^2 rather unnatural, as it
    > implies that if symmetry remains unbroken (at high enough
    > temperatures), the Higgs Boson is an on-shell tachyon!


    That's exactly the reason why m^2<0 - it forces a broken symmetry.
    See also the entry ''What about particles faster than light''
    in my theoretical physics FAQ at
    http://www.mat.univie.ac.at/~neum/physics-faq.txt


    Arnold Neumaier
     
  6. Mar 30, 2006 #5
    Arnold Neumaier wrote:
    > That's exactly the reason why m^2<0 - it forces a broken symmetry.
    > See also the entry ''What about particles faster than light''
    > in my theoretical physics FAQ at
    > http://www.mat.univie.ac.at/~neum/physics-faq.txt


    I was interested in if the Higgs Boson is expected to be tachyonic at
    high enough temperatures corresponding to an unbroken electroweak
    symmetry, and I see that you mention the possibility in your FAQ:

    "An expansion around an unstable state gives no significant
    information,
    unless one has a system that actually _is_ close such an unstable state

    (as perhaps the very early universe)."

    Then you say something I do not follow:

    "But in that case there are no relevant excitations (tachyons), since
    the whole process (inflation) of motion towards a more stable state
    proceeds so rapidly that excitations do not form and everything can be
    analyzed semiclassically."

    What if there's no inflation and we're just doing a (futuristic)
    high-energy experiment probing the symmetry unbreaking scale? Would we
    expect a tachyonic Higgs Boson to whizz by, given no new physics kicks
    in by then?

    -Souvik
     
  7. Mar 31, 2006 #6
    Souvik wrote:

    > Arnold Neumaier wrote:
    >
    >>That's exactly the reason why m^2<0 - it forces a broken symmetry.
    >>See also the entry ''What about particles faster than light''
    >>in my theoretical physics FAQ at
    >> http://www.mat.univie.ac.at/~neum/physics-faq.txt

    >
    > I was interested in if the Higgs Boson is expected to be tachyonic at
    > high enough temperatures corresponding to an unbroken electroweak
    > symmetry, and I see that you mention the possibility in your FAQ:
    >
    > "An expansion around an unstable state gives no significant
    > information,
    > unless one has a system that actually _is_ close such an unstable state
    > (as perhaps the very early universe)."


    Note that I say 'close to' and not 'in'.


    > Then you say something I do not follow:
    >
    > "But in that case there are no relevant excitations (tachyons), since
    > the whole process (inflation) of motion towards a more stable state
    > proceeds so rapidly that excitations do not form and everything can be
    > analyzed semiclassically."


    Actually this is not well understood. People do simulations and explore
    possibilities, but there is nothing definite. In particular, inflation
    is always treated semiclassically, as far as I have seen. (But I don't
    know all the literature and might be wrong in this.)

    In my opinion, tachyons are completely unphysical objects, for reasons
    of dynamical consistency. They appear only through formal perturbations
    which don't make real sense.


    > What if there's no inflation and we're just doing a (futuristic)
    > high-energy experiment probing the symmetry unbreaking scale?


    I don't think there can be a completely unbroken phase.
    If there were one, it would be unstable under the slightest perturbation
    (which must exist since any model is only an approximation), and
    immediately develop inflationary motion.


    Arnold Neumaier
     
  8. Mar 31, 2006 #7

    Oz

    User Avatar

    Souvik <souvik1982@gmail.com> writes
    >"But in that case there are no relevant excitations (tachyons), since
    >the whole process (inflation) of motion towards a more stable state
    >proceeds so rapidly that excitations do not form and everything can be
    >analyzed semiclassically."


    Presumably, in the very early universe, if tachionic higgs-related
    particles did commonly exist (as a gas) in large number, then the
    universe would rapidly tend to significant homogeneity over a very large
    volume.

    Its fortunate we don't see such a homogenous early universe ....

    --
    Oz
    This post is worth absolutely nothing and is probably fallacious.
     
  9. Mar 31, 2006 #8
    Souvik wrote:

    > Arnold Neumaier wrote:
    >> That's exactly the reason why m^2<0 - it forces a broken symmetry.
    >> See also the entry ''What about particles faster than light''
    >> in my theoretical physics FAQ at
    >> http://www.mat.univie.ac.at/~neum/physics-faq.txt

    >
    > I was interested in if the Higgs Boson is expected to be tachyonic at
    > high enough temperatures corresponding to an unbroken electroweak
    > symmetry, and I see that you mention the possibility in your FAQ:
    >
    > "An expansion around an unstable state gives no significant
    > information,
    > unless one has a system that actually _is_ close such an unstable
    > state
    >
    > (as perhaps the very early universe)."


    I cannot answer what Arnold means in his FAQ. For me that sounds rather
    strange.

    The answer to your first question is quite easy. Of course at finite
    temperature, you do not have tachyons either. A way to understand the
    issue is to calculate the effective potential of the Higgs field in
    finite-temperature quantum field theory.

    The parameters, appearing in the standard model (couplings, masses,
    etc.), are renormalized in the vacuum to fit the observed data about
    particles. As we know, this fit is very successful, since no
    statistically significant deviation from standard-model behaviour has
    been observed so far (take asside the neutrino masses and oscillations
    which can be incorporated into the standard model without principle
    difficulties).

    Then, at finite temperature, there are no further divergences, but you
    have finite contributions to the masses and couplings which depend on
    temperature (and chemical potentials of conserved charges). At high
    enough temperatures, a zero Higgs field becomes a stable configuration,
    and you have a phase transition from the phase where gauge symmetry is
    broken (Higgs phase) to a phase where it is not.

    --
    Hendrik van Hees Texas A&M University
    Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
    Fax: +1 979/845-1899 College Station, TX 77843-3366
    http://theory.gsi.de/~vanhees/ mailto:hees@comp.tamu.edu
     
  10. Mar 31, 2006 #9
    Darth Sidious wrote:
    > The Higgs boson is *not* tachyonic. The field \Phi which has a
    > potential like a mexican hat is *not* the Higgs field. You get
    > the Higgs field by choosing a gauge and a vacuum for your
    > action, developping the action in the neighbourhood of that
    > vacuum.


    What's in a name? At high enough temperatures, when the vacuum is no
    longer necessarily at the base of the Mexican hat, do excitations of
    the \Phi field show up as tachyons?

    -Souvik
     
  11. Apr 1, 2006 #10
    Hendrik van Hees wrote:
    > At high
    > enough temperatures, a zero Higgs field becomes a stable configuration,
    > and you have a phase transition from the phase where gauge symmetry is
    > broken (Higgs phase) to a phase where it is not.


    I am not sure what you mean by a zero Higgs field. Do you mean to say
    the vacuum expectation value of the Higgs field is zero (and one may
    expand around the origin of the "Mexican Hat")? Yes, that unbreaks
    electroweak symmetry, but doesn't solve the fact that the bare
    Lagrangian (which will be important in the dynamics now) is tachyonic!

    How does the -m^2, the bare (not expanded-around-condensate) mass
    behave under renormalisation? If it increases to a positive value with
    increasing energy, it could mean that our tachyonic form of the
    Lagrangian is an approximation to some legitimate physics at high
    energies. Has anyone does this calculation? Or is it obviously a wrong
    idea?

    Thanks,
    Souvik
     
  12. Apr 8, 2006 #11
    Souvik wrote:
    > Darth Sidious wrote:
    > > The Higgs boson is *not* tachyonic. The field \Phi which has a
    > > potential like a mexican hat is *not* the Higgs field. You get
    > > the Higgs field by choosing a gauge and a vacuum for your
    > > action, developping the action in the neighbourhood of that
    > > vacuum.

    >
    > What's in a name? At high enough temperatures, when the vacuum is no
    > longer necessarily at the base of the Mexican hat, do excitations of
    > the \Phi field show up as tachyons?
    >


    Hmm and now I think another related question: Given that the top quark
    coupling to this field is strong (exactly 0.99 hc), can two top quarks
    interact using this, er, anti-yukawian field?
     
  13. May 24, 2006 #12
    In article <e0h13s$j0u$1@emma.aioe.org>,
    Hendrik van Hees <hees@comp.tamu.edu> wrote:

    >At high
    >enough temperatures, a zero Higgs field becomes a stable configuration,
    >and you have a phase transition from the phase where gauge symmetry is
    >broken (Higgs phase) to a phase where it is not.


    Of course you mean a Higgs field with zero *expectation value*
    is a stable configuration - there will be enormous random thermal
    fluctuations.

    The rest of this stuff is for the nonexperts, not folks like
    Hendrik - except for the last 2 paragraphs.

    To understand what's going on here, just imagine a ball sitting in a
    deep dish that has a little bump in the middle. Imagine the dish
    has rotational symmetry.

    If you don't shake the dish at all (zero temperature), it's unstable
    for the ball to balance right on top of the bump. Instead, it will
    randomly sit at one of the circle of places at the bottom of the dish.

    If you jiggle the dish a lot (high temperature), the ball bounces all
    over. Its *average* position will be centered right at the bump, for
    reasons of symmetry - but it's still a bit less likely to be there
    than somewhere nearer the bottom of the dish.

    The Higgs field acts like this! At zero temperature, it randomly
    picks out a nonzero value that minimizes its potential energy.
    At high temperatures, it wiggles all over, but its *average* value
    is zero.

    I'm ignoring quantum effects here, to keep things simple.

    Okay, here are those last 2 paragraphs:

    As for the Higgs boson being "tachyonic", this is just a highly
    unilluminating way to describe the fact that the potential energy
    of the Higgs has a little bump centered at zero. When its value is
    near zero, one can approximate the Higgs by a free tachyonic field -
    but the Higgs field never stays near this value, so it's usually
    a bad approximation. In other words, saying the Higgs is "tachyonic"
    is just a way of saying "zero is not a stable equilibrium".

    Moreover, the whole theory of tachyons is misunderstood by lots of
    people. You can't actually transmit information faster than light
    with tachyons, because the propagation velocity of the tachyonic
    Klein-Gordon equation

    (BOX - m^2) phi = 0

    is still 1 (the speed of light). So, don't get the idea of Higgs
    bosons sending signals faster than light.

    ---------------------------------------------------------------------
    Puzzle 24: In what part of the world is it most likely for a woman to
    claim her child's father is a dolphin?

    If you get stuck, try my puzzles page at:

    http://math.ucr.edu/home/baez/puzzles/
     
  14. May 24, 2006 #13
    :surprised
     
  15. May 24, 2006 #14

    CarlB

    User Avatar
    Science Advisor
    Homework Helper

    The presence of radiative corrections suggests that the bare masses should have a simpler structure than the measured masses. However, it is known that it is the measured masses that are simple. For the charged leptons, see:
    http://arxiv.org/abs/hep-ph/0505220

    For the neutrinos, see:
    http://www.arxiv.org/abs/hep-ph/0605074
    http://brannenworks.com/MASSES2.pdf

    Carl
     
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