# Why is the Higgs Boson tachyonic?

1. Mar 29, 2006

### Souvik

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. Mar 30, 2006

### Hendrik van Hees

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

3. Mar 30, 2006

### Darth Sidious

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.}

4. Mar 30, 2006

### Arnold Neumaier

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

5. Mar 30, 2006

### Souvik

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

6. Mar 31, 2006

### Arnold Neumaier

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

7. Mar 31, 2006

### Oz

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.

8. Mar 31, 2006

### Hendrik van Hees

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

9. Mar 31, 2006

### Souvik

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. Apr 1, 2006

### Souvik

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

11. Apr 8, 2006

### arivero@unizar.es

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?

12. May 24, 2006

### John Baez

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.

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13. May 24, 2006

### zhangjian2005

:surprised

14. May 24, 2006

### CarlB

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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