#1
Nov406, 03:28 PM

P: n/a

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 onshell 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
Nov406, 03:28 PM

P: n/a

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 onshell 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 "wouldbe 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: +1 979/8451411 Cyclotron Institute, MS3366 Fax: +1 979/8451899 College Station, TX 778433366 http://theory.gsi.de/~vanhees/ mailto:hees@comp.tamu.edu 


#3
Nov406, 03:28 PM

P: n/a

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 onshell 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/physicsfaq.txt Arnold Neumaier 


#4
Nov406, 03:28 PM

P: n/a

Why is the Higgs Boson tachyonic?
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
Nov406, 03:28 PM

P: n/a

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/physicsfaq.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) highenergy 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 


#6
Nov406, 03:28 PM

P: n/a

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/physicsfaq.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) > highenergy 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 


#7
Nov406, 03:28 PM

P: n/a

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 higgsrelated 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. 


#8
Nov406, 03:28 PM

P: n/a

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/physicsfaq.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 finitetemperature 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 standardmodel 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: +1 979/8451411 Cyclotron Institute, MS3366 Fax: +1 979/8451899 College Station, TX 778433366 http://theory.gsi.de/~vanhees/ mailto:hees@comp.tamu.edu 


#9
Nov406, 03:28 PM

P: n/a

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 


#10
Nov406, 03:28 PM

P: n/a

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 expandedaroundcondensate) 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 


#11
Nov406, 03:29 PM

P: n/a

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, antiyukawian field? 


#12
Nov406, 03:34 PM

P: n/a

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 KleinGordon 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/ 


Register to reply 
Related Discussions  
Where is Higgs boson?  High Energy, Nuclear, Particle Physics  10  
Higgs Boson  High Energy, Nuclear, Particle Physics  1  
Why is the Higgs Boson tachyonic?  General Physics  13  
Higgs Boson  High Energy, Nuclear, Particle Physics  5  
Higgs boson  Beyond the Standard Model  17 