Congregating bosons/fermions
Dave Snead
occupy the same quantum state, but bosons congregate in identical states.
Now composite particles with an even number of fermions are bosons, and if
these bosons in identical states congregate, aren't some of their identical
state fermions congregating also? Doesn't this lead to multiple fermions
occupying the same quantum state?
Thanks in advance,
Dave Snead
John Baez
Dave Snead <dsn...@charter.net> wrote:
We've discussed this extensively already here on s.p.r., so
you should probably use Google to look for articles containing
the phrase "composite boson". Briefly, if you have a composite
particle made of fermions, this particle cannot act exactly like
a fundamental boson, since the Pauli exclusion principle still
applies. The fun thing is to see how particles of this sort
can nonetheless do a decent job of *approximately* acting like
fundamental bosons, at least at low densities. In our previous
discussions we figured out roughly how this works.
Joe Marshall
"John Baez" <ba...@galaxy.ucr.edu> wrote in message news:aeugs8$p10$1...@glue.ucr.edu...
>
> Briefly, if you have a composite
> particle made of fermions, this particle cannot act exactly like
> a fundamental boson, since the Pauli exclusion principle still
> applies. The fun thing is to see how particles of this sort
> can nonetheless do a decent job of *approximately* acting like
> fundamental bosons, at least at low densities.
So there seems to be a mechanism by which you can distinguish
`composite' bosons from `true' bosons. Has anybody ever tried
doing a physical test what are believed to be `true' bosons?
Uncle Al
BCS theory of superconductivity. Cooper pairs are pairs of electrons
with conjugate momenta coupled by lattice phonons. Cooper pairs form
a degenerate Bose fluid, which is why superconductivity works - to
ruin the state you must elevate all of the electrons not just scatter
a pair here and there. Type II supercons sort of split the
difference.
He-3 is fermonic. It can be condensed into superfluid. Alkali metal
atoms are fermions. They are the working material for Bose-Einstein
condensates in vacuum.
It isn't a fermion if it is identically paired to make an integral
spin composite.
--
Uncle Al
http://www.mazepath.com/uncleal/eotvos.htm
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
John Baez
Joe Marshall <prunes...@attbi.com> wrote:
>"John Baez" <ba...@galaxy.ucr.edu> wrote in message
>news:aeugs8$p10$1...@glue.ucr.edu...
>> Briefly, if you have a composite
>> particle made of fermions, this particle cannot act exactly like
>> a fundamental boson, since the Pauli exclusion principle still
>> applies. The fun thing is to see how particles of this sort
>> can nonetheless do a decent job of *approximately* acting like
>> fundamental bosons, at least at low densities.
>So there seems to be a mechanism by which you can distinguish
>`composite' bosons from `true' bosons.
Yes: compress the hell out of a Bose-Einstein condensate; if the
bosons are made of constituent fermions you'll notice it when
the density gets high enough - it won't act like a Bose-Einstein
condensate anymore. Its pressure will become higher.
However, there is an infinitely cheaper way to see if a boson is
composite: namely, try to smash it into pieces!
BUT:
Don't forget this crucial point: the more tightly bound a composite
boson is, the more it acts like a 'true' boson - you need to
make Bose-Einstein condensates of very high density or smash the
suckers at very high energies to reveal them for the sham they are.
This is just a special case of a very general fact: we can never
be 100% sure our 'fundamental' particles won't turn out to be
composite... until we get a Theory of Everything that is Certainly True.
>Has anybody ever tried
>doing a physical test what are believed to be `true' bosons?
High-energy tests of Standard Model can be used to put upper limits
on the deviation of the photon from being a 'point particle', i.e. a
fundamental gauge boson. I don't know how good the current limits
are, but yeah, people do think about this stuff.
The W and Z and gluons are probably harder to get really clean
data on - and of course nobody has even *seen* the Higgs. Those
are all the supposedly fundamental bosons in the Standard Model.
We've talked a bunch here about "preon models" where these
bosons aren't fundamental - but so far, there's no experimental
evidence for these models.
Of course, once people thought the pion was fundamental!
Oz
John Baez writes
>Don't forget this crucial point: the more tightly bound a composite
>boson is, the more it acts like a 'true' boson - you need to
>make Bose-Einstein condensates of very high density or smash the
>suckers at very high energies to reveal them for the sham they are.
But isn't this self-limiting?
If the composite particle is very tightly bound then it looks very small
but then you would then expect it to contain a lot of binding energy and
be rather massive.
So protons are rather massive, although admittedly I am told that the
quarks are (internally) lightly bound.
But what about an electron (or worse a neutrino)?
It's light and naively might not have enough mass to have enough
internal energy to bind really tiny things together in a very small
space.
Hmmm. I do, though, seem to be predicting that massive quarks are also
composite (assuming there is evidence that they have mass). Pity we
can't make a quark out of a bound pair of gluons, if even just a
transitory state. Doubtless that would fall foul of the SM.
I'll go back to lurking. I must have been infected with some viral
infection due to the proximity of an enthusiastic wizard. Probably not
fatal though.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
John Baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>John Baez writes
>>Don't forget this crucial point: the more tightly bound a composite
>>boson is, the more it acts like a 'true' boson - you need to
>>make Bose-Einstein condensates of very high density or smash the
>>suckers at very high energies to reveal them for the sham they are.
>If the composite particle is very tightly bound then it looks very small
>but then you would then expect it to contain a lot of binding energy and
>be rather massive.
Actually, a bound state is *lighter* than than the sum of the
masses of its constituents. That's why it takes energy to pull it apart!
This is how fusion and fission bombs work: they convert bound
states of protons and neutrons to more tightly bound states,
releasing energy...
... which wizards can use to fry people who ask novice questions on s.p.r.!
>I'll go back to lurking.
Yes, you'd better take cover now. :-)
Hendrik van Hees
> However, there is an infinitely cheaper way to see if a boson is
> composite: namely, try to smash it into pieces!
Well, this recipe sometimes does not work. If you try to smash, e.g, a pion
into pieces it comes out a hell of particles but you'll never break it up
into a pair of a free quark and a free anti-quark. That's roughly what we
call confinement of the strong interactions.
> Of course, once people thought the pion was fundamental!
I think it is always a question of energy wether one thinks about particles
to be elementary or not. On a low-energy scale a pion can be described
quite well with elementary quantum fields. All the effective hadronic
theories use this fact. Only if we look with higher energies we see the
compositeness of the pion.
--
Hendrik van Hees Fakult‰t f¸r Physik
Phone: +49 521/106-6221 Universit‰t Bielefeld
Fax: +49 521/106-2961 Universit‰tsstrafle
http://theory.gsi.de/~vanhees/ D-33615 Bielefeld
Squark
> High-energy tests of Standard Model can be used to put upper limits
> on the deviation of the photon from being a 'point particle', i.e. a
> fundamental gauge boson. I don't know how good the current limits
> are, but yeah, people do think about this stuff.
>
> The W and Z and gluons are probably harder to get really clean
> data on - and of course nobody has even *seen* the Higgs. Those
> are all the supposedly fundamental bosons in the Standard Model.
Hey, and what's with the gluon? Of course, there's confinement and all,
but if you take it as making the gluon no longer a true fundumental
particle, then you should list the glueballs in. Otherwise, what are
their fundumental constituents? So there certainly are more fundumental
bosons in old SM, no? Btw, I have never got quite to the bottom of what
should be called a "fundumental particle" in an interacting QFT.
Fundumental field is easier, but sometimes there are different "field
presentations" for the same theory, isn't it so? Take bosonisation for
instance. Oh, well...
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and use "com" rather than
"exe")
Squark
> John Baez writes
>
> >Don't forget this crucial point: the more tightly bound a composite
> >boson is, the more it acts like a 'true' boson - you need to
> >make Bose-Einstein condensates of very high density or smash the
> >suckers at very high energies to reveal them for the sham they are.
>
> But isn't this self-limiting?
>
> If the composite particle is very tightly bound then it looks very small
> but then you would then expect it to contain a lot of binding energy and
> be rather massive.
It's a nice argument, only problem: binding energy is _negative_ :-) It's
a potential _well_ you have hard time escaping from, not a potential
_hill_. The harmonic osciallator's potential energy is -kx^2/2, the
potential energy between a q charge and a -q charge with r spacing is
-q^2 / [r^2 (4 pi epsilon0)], the potential energy between two gravitating
masses m_1 and m_2 is -G m_1 m_2 / r^2 etc.
So, it's in principle possible for a light particle to be composed of
massive ones, and there was even a theory about all particles composed from
the Lambda-hyperon and - the electron I think? - predating the quark model.
Anyway, someone else on this group must remember it better than I.
[Btw, there was another post I sent a whiel ago to this threat, so if this
one gets first I would ask the moderators (almost said the
motherators :-) ) to check out what's going on out there]
Joe Marshall
"John Baez" <ba...@galaxy.ucr.edu> wrote in message news:aft4g0$581$1...@glue.ucr.edu...
> In article <Hd7R8.141941$6m5.1...@rwcrnsc51.ops.asp.att.net>,
> Joe Marshall <prunes...@attbi.com> wrote:
>
> >"John Baez" <ba...@galaxy.ucr.edu> wrote in message
> >news:aeugs8$p10$1...@glue.ucr.edu...
>
> >> Briefly, if you have a composite
> >> particle made of fermions, this particle cannot act exactly like
> >> a fundamental boson, since the Pauli exclusion principle still
> >> applies. The fun thing is to see how particles of this sort
> >> can nonetheless do a decent job of *approximately* acting like
> >> fundamental bosons, at least at low densities.
>
> >So there seems to be a mechanism by which you can distinguish
> >`composite' bosons from `true' bosons.
>
> Yes: compress the hell out of a Bose-Einstein condensate; if the
> bosons are made of constituent fermions you'll notice it when
> the density gets high enough - it won't act like a Bose-Einstein
> condensate anymore. Its pressure will become higher.
>
> However, there is an infinitely cheaper way to see if a boson is
> composite: namely, try to smash it into pieces!
>
So if I were to try to smash a photon with something large and heavy,
say, an atom of lead, and it came apart into an electron and positron,
I could conclude that photons aren't `real' bosons?
Oz
John Baez writes
>In article <ag59rd$am$1...@inky.its.caltech.edu>,
>Oz <O...@upthorpe.demon.co.uk> wrote:
>
>>John Baez writes
>
>>>Don't forget this crucial point: the more tightly bound a composite
>>>boson is, the more it acts like a 'true' boson - you need to
>>>make Bose-Einstein condensates of very high density or smash the
>>>suckers at very high energies to reveal them for the sham they are.
>
>>If the composite particle is very tightly bound then it looks very small
>>but then you would then expect it to contain a lot of binding energy and
>>be rather massive.
>
>Actually, a bound state is *lighter* than than the sum of the
>masses of its constituents. That's why it takes energy to pull it apart!
Ooops, that's true.
I can't say I've ever really considered this subatomically.
<The sky opens and an ethereal voice booms down>
"You are encouraged by the management to engage brain before replying"
<Oz peers inquisitively up at unusual sight>
<There is a flash>
<Oz is left semi-naked as the remains of his carbonised clothes flutter
to the ground.>
<the sky closes unhurriedly>
Um. (cough> That's a viewpoint worth pursuing for a bit.
OK, let's start from the basics to make sure I haven't missed a trick.
<Oz pauses, realises this is a forlorn hope, and then continues
fearlessly>
Let's just consider what are considered elementary particles for a
start. I will ignore gravity (rather weak anyway).
Electron: produces an electric field. Suggestions that one could
plausibly attribute the mass to it's associated electric field.
Neutrino: Very light. Does it 'produce' a weak force field?
Given it's low mass, and the very high mass of W,Z, I imagine any such
field would be horribly short range and horribly weak. A very small and
weak field would imply a low mass (a somewhat recursive argument).
Quark: produces a colour and a charge field. Presumably is 'light' when
in a nucleus (and strongly bound) but becomes increasingly massive if
isolated. 'Free' mass greater than a quark-antiquark pair, so never seen
free.
Photon: Hmmm. From my position of unfathomable ignorance this seems the
most attractive to consider composite (very unexpected). An electron and
a positron brought very close together would presumably become lighter
and the external electric field become much smaller. It even has spin 1
too. Of course you would have to dump some energy during the merging
process, perhaps to a conveniently close atom. Even better if you could
assign some residual energy to something akin to 'orbital speed' then
you might expect any residual energy to be related to some frequency or
other....
However I am confident this scenario has long since been examined and
found wanting. Probably for a very obvious reason that will result in
temporary conflagration.
W's & Z: carry the weak force. Very massive.
Gluons: Carry colour force. Massive??? Not seen free anyway.
Hmmmm. That's an odd combination of particles when you think about it.
Mind you if I knew more about it then it might not be so strange.
Really it would be much nicer if everything correlated with the strength
of the relevant force.
So the masses went:
Neutrino -> electron -> quark, which they probably do.
and
W,Z -> photon -> gluon, which they don't.
On the other hand let's look at the known free particles:
Neutrino -> electron -> pion (I think: quark-antiq pair)
W,Z -> photon -> gluino (gluon-antig pair)
Hmm, the spins don;t add up, do they?
Anyway, this is chemistry.....
A wise person would not post this.
I think I'll eat my lunch in the bunker.
John Baez
Squark <fii...@yahoo.com> wrote, mod spelling corrections:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<aft4g0$581$1...@glue.ucr.edu>...
>> High-energy tests of Standard Model can be used to put upper limits
>> on the deviation of the photon from being a 'point particle', i.e. a
>> fundamental gauge boson. I don't know how good the current limits
>> are, but yeah, people do think about this stuff.
>>
>> The W and Z and gluons are probably harder to get really clean
>> data on - and of course nobody has even *seen* the Higgs. Those
>> are all the supposedly fundamental bosons in the Standard Model.
>Hey, and what's with the gluon? Of course, there's confinement and all,
>but if you take it as making the gluon no longer a true fundamental
>particle, then you should list the glueballs in.
If you include glueballs you really should include mesons,
especially because observed glueball candidates like the f_0(1500)
and f_J(1710) are quite possibly "hybrids" - quantum superpositions of
glueballs and mesons. But more importantly, I don't really see why you
want to list glueballs as fundamental particles in the first place!
>Btw, I have never got quite to the bottom af what
>should be called a "fundamental particle" in an interacting QFT.
Yeah, this is a very subtle issue which only German algebraic
quantum field theorists have the patience for. I was just using
the term "fundamental particle" the way working-class physicists
do, namely a field that shows up in the Lagrangian - not worrying
about the fact that the same field theory can sometimes be described
using different Lagrangians!
If you want a concept with a simple precise definition, it's
"eigenstate of the Hamiltonian". Then quarks, gluons, glueballs
and neutrons are out, while photons, protons, hydrogen atoms, water
molecules and buckyballs are in!
(Unless the proton turns out to be unstable....)
Steve R Blattnig
On Tue, 9 Jul 2002, Squark wrote:
> It's a nice argument, only problem: binding energy is _negative_ :-) It's
> a potential _well_ you have hard time escaping from, not a potential
> _hill_. The harmonic osciallator's potential energy is -kx^2/2, the
> potential energy between a q charge and a -q charge with r spacing is
> -q^2 / [r^2 (4 pi epsilon0)], the potential energy between two gravitating
> masses m_1 and m_2 is -G m_1 m_2 / r^2 etc.
> So, it's in principle possible for a light particle to be composed of
> massive ones, and there was even a theory about all particles composed from
> the Lambda-hyperon and - the electron I think? - predating the quark model.
> Anyway, someone else on this group must remember it better than I.
>
How does this relate to the case of the proton? if one considers the
protobn to be made up of three quarks, most of the mass comes from the
gluons and sea quarks. The valance quarks only have a mass of a few MeV
according to the particle data group. So it would seem that the binding
energy in this case is actually making the proton much more massive.
Am I missing something obvious?
Steve
Squark
Hendrik van Hees <he...@physik.uni-bielefeld.de> wrote in message news:<afua72$h584k$4...@fu-berlin.de>...
> I think it is always a question of energy wether one thinks about particles
> to be elementary or not.
This is an intriguing remark. Do you think then, that there is an infinite
ladder of "inner particle structure"? How do you imagine a theory which such
a ladder? It certainly can't be an ordinary QFT?
Aaron Bergman
>
>Btw, I have never got quite to the bottom of what
>should be called a "fundumental particle" in an interacting QFT.
Don't bother. The whole particle concept becomes quite worthless
in strongly interacting field theories. MNSHO is that the idea of
a particle only makes sense in the infinitesimal neighborhood of
a time-independent free theory (which may be effective).
>Fundumental field is easier, but sometimes there are different "field
>presentations" for the same theory, isn't it so? Take bosonisation for
>instance. Oh, well...
Sure, and there are plenty of other dualities out there, too. I
don't see why fundamental anything is really a useful concept.
Defining just what a quantum field theory is is a rather
difficult concept. Unless you're conformal, of course.
What I'd like is to define a quantum field theory as some "space
of fields" with a "measure" and then define things via the path
integral.
Removing the quotation marks is left as an exercise to the
reader.
Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
Squark
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<aghb8d$klr$1...@glue.ucr.edu>...
> In article <939044f.02070...@posting.google.com>,
> Squark <fii...@yahoo.com> wrote, mod spelling corrections:
> >but if you take it as making the gluon no longer a true fundamental
> >particle, then you should list the glueballs in.
>
> If you include glueballs you really should include mesons,
> especially because observed glueball candidates like the f_0(1500)
> and f_J(1710) are quite possibly "hybrids" - quantum superpositions of
> glueballs and mesons. But more importantly, I don't really see why you
> want to list glueballs as fundamental particles in the first place!
I was just pointing out your list is incomplete: you should either throw
the gluons or the glueballs in, depending on your definition of a
"fundumental particle".
> I was just using the term "fundamental particle" the way working-class
> physicists do, namely a field that shows up in the Lagrangian...
Then the gluon is definitely in!
> If you want a concept with a simple precise definition, it's
> "eigenstate of the Hamiltonian". Then quarks, gluons, glueballs
> and neutrons are out, while photons, protons, hydrogen atoms, water
> molecules and buckyballs are in!
Hmm, but something containing gluons has to be in! Take any state with
gluons and decompose it into Hamiltonian eigenstate. Actually, glueballs
are such eigenstates in "pure gluodynamics", say (aren't they?), and in
the real world, you've got to have a mixture of gluons and something
else which is. Of course, the proton contains some, but something else
should too - somehow, otherwise there would be too little of those energy
eigenstates compared to the amount of degrees of freedom, ain't so?
Hmm, coming to think about it, it's not clear how what I'm saying comes
to terms the whole zoo of unstable particles out there. Maybe to complete
the picture one must throw in Hamiltonian eigenstates which are some
exicted "soups" in which those particles form as fast as decay???
Best regards,
A Somewhat Confused Squark
John Baez
Joe Marshall <prunes...@attbi.com> wrote:
>"John Baez" <ba...@galaxy.ucr.edu> wrote in message
>news:aft4g0$581$1...@glue.ucr.edu...
>> However, there is an infinitely cheaper way to see if a boson is
>> composite: namely, try to smash it into pieces!
>So if I were to try to smash a photon with something large and heavy,
>say, an atom of lead, and it came apart into an electron and positron,
>I could conclude that photons aren't `real' bosons?
No, I was speaking at the pop-physics level of
precision, without all the necessary qualifications.
Phrases like "smash it to pieces" are supposed to
indicate that I'm feeling relaxed. :-)
The moral of your example is:
One has to be careful to distinguish scattering events that
are predicted by the Standard Model, in which the photon
is a fundamental particle, from deviations that could
be due to compositeness - and from deviations with
other explanations! Symptoms of compositeness don't
even have to involve seeing a particle "fall apart"
into "pieces" - witness the flurry of interest in quark
substructure when people saw deviations from what they
thought QCD predicted in jet results at the Tevatron:
http://ppewww.ph.gla.ac.uk/preprints/1999/08/dis99_WG3/node11.html
Confinement is another example of how compositeness
can be tricky: you can't "smash open" a pion and see
its pieces short of heating it to the temperature where
confinement no longer holds (about 1 trillion Kelvin).
In technicolor models, the "fundamental" particles of the
Standard Model are made of more fundamental "preons",
confined by a gauge field called "technicolor".
Saco
Very interesting question. Here is the answer I came up yesterday.
First consider hypothetical example of non-interacting bosons being
able to condensate in the same place. Composite bosons should consist
of charged fermions so the hole system would be chargeless. Since
charged particles can not occupy the same place, this solves the
problem of real space condensation.
Now, in impulse space, nothing prohibits composite boson to
condensate. Pauli principle still works, all fermions in composite
bosons might have different impulses in such a way that all composite
bosons will have the same impulse.
Regards,
Saco
"Everyone goes through changes
Looking to find the truth
Don't look at me for answers
Don't ask me
I don't know" Ozzy Osbourne
Paul Reilly
> Actually, a bound state is *lighter* than than the sum of the
> masses of its constituents. That's why it takes energy to pull it apart!
>
Not necessarily. Its *energy* is less than the sum of the *energies* of
its constituents, but consider a proton (or indeed any hadron!!!):
sum of (real) quark masses =~ 20 MeV/c^2 + gluon masses = 0
<< Proton mass =~ 1835 MeV/c^2
The quarks are confined to a small area and have a ton of zero-point
kinetic energy, plus colour and electromagnetic field energy (virtual quarks,
gluons, and photons is another way to look at it). These virtual and real
particles have total average momentum zero in the proton's center of mass
frame, so all their energy (in that frame) contributes to the mass
(m = sqrt ( E^2 - p^2) with c=1) but the sum of the masses is certainly
less than the mass of the composite particle...
I'm certain Dr. Baez understands this better than I do, but I have the same
pleasure a fourth grader has here in pointing out when the teacher has made
a mis-statement :)
Paul
ksh95
Uncle Al <Uncl...@hate.spam.net> wrote in message
> with conjugate momenta coupled by lattice phonons. Cooper pairs form
> a degenerate Bose fluid, which is why superconductivity works - to
> ruin the state you must elevate all of the electrons not just scatter
> a pair here and there. Type II supercons sort of split the
> difference.
>
> He-3 is fermonic. It can be condensed into superfluid. Alkali metal
> atoms are fermions. They are the working material for Bose-Einstein
> condensates in vacuum.
>
> It isn't a fermion if it is identically paired to make an integral
> spin composite.
Yea, but all of these are still quasi-bosons like Baez mentioned
earlier. Commutators don't lie.
vittorio
Hi,
Aaron Bergman wrote:
>
> In article <939044f.02070...@posting.google.com>, Squark wrote:
.........cut...........
> What I'd like is to define a quantum field theory as some "space
> of fields" with a "measure" and then define things via the path
> integral.
I put a question:
in your approach, where is the hilbert space of states?
Are you always able(in all possible cases) to rebuild it from the
expectetion values you
find using path integral?
vittorio
Gordon D. Pusch
You are missing the zero-point kinetic energy of the valence quarks.
Since they are confined in a region an order of magnitude smaller than
their Compton wavelengths, the uncertainty in their momenta will have
them rattling around inside the proton at ultrarelativistic velocities.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Aaron Bergman
>In technicolor models, the "fundamental" particles of the
>Standard Model are made of more fundamental "preons",
>confined by a gauge field called "technicolor".
I think you're confusing preon models and technicolor.
Technicolor is (roughly) a composite Higgs. There's no reason for any other
particle to be composite.
Saco
> Hendrik van Hees <he...@physik.uni-bielefeld.de> wrote in message news:<afua72$h584k$4...@fu-berlin.de>...
> > I think it is always a question of energy wether one thinks about particles
> > to be elementary or not.
>
> This is an intriguing remark. Do you think then, that there is an infinite
> ladder of "inner particle structure"? How do you imagine a theory which such
> a ladder? It certainly can't be an ordinary QFT?
Recently I came across a very interesting article
http://xxx.lanl.gov/abs/hep-th/0207081
If one would take Tchebyscheff polynomial, used there, of higher and
higher degree, this theory can generate more and more elementary
particles.
But this won't give infinite ladder of "inner particle structure".
Infinite ladder of "inner particle structure" is possible within
ordinary QFT, if one ends up in a macroscopic chaotic phase, which is
possible in non-perturbatively non-renormalizable QFT. By this I mean
that in the lattice QFT, one might end up in a phase, where order
parameter of sublattice varies chaotically. When taking a continuum
limit in this phase, one can identify arbitrary 4 sublattice as a
components of the staggered fermion, hence we'll have a theory with
infinite ladder of "inner particle structure".
This just says that, when doing lattice QCD simulations, one has to
keep track of the sublattice order parameters as well.
Gordon D. Pusch
> Alkali metal atoms are fermions.
Check again. All the alkali-metal BEC experiments that I am familiar with
used isotopes with _odd_ atomic weights, such as Rb-87, Li-7, or Na-23.
Odd atomic weight plus odd atomic number means they have an odd number
of protons, the SAME odd number of electrons, and an _EVEN_ number of
neutrons. Twice an odd number is even, and an even number plus an
even number is even. Therefore, the alkali metal atoms used have
all been _bosons_, not fermions.
Aaron Bergman
In article <3D2E92DB...@physics.it>, vittorio wrote:
>Aaron Bergman wrote:
>
>> What I'd like is to define a quantum field theory as some "space
>> of fields" with a "measure" and then define things via the path
>> integral.
>
>I put a question:
>in your approach, where is the hilbert space of states?
It's not an approach; it's a dream.
>Are you always able(in all possible cases) to rebuild it from the
>expectetion values you
>find using path integral?
I don't particularly see the need for a Hilbert space.
A.J. Tolland
On Mon, 15 Jul 2002, Aaron Bergman wrote:
>
> In article vittorio wrote:
> >Aaron Bergman wrote:
> >
> >> What I'd like is to define a quantum field theory as some "space
> >> of fields" with a "measure" and then define things via the path
> >> integral.
>
> >Are you always able(in all possible cases) to rebuild it from the
> >expectetion values you
> >find using path integral?
>
> I don't particularly see the need for a Hilbert space.
Well, you may not need it, but if your observables form a
C*-algebra and your path integral gives a way of computing a linear
functional on this algebra, you can use the GNS construction to recover a
Hilbert space.
--A.J.
Squark
sa...@moon.yerphi.am (Saco) wrote in message news:<85160b48.02071...@posting.google.com>...
> Infinite ladder of "inner particle structure" is possible within
> ordinary QFT, if one ends up in a macroscopic chaotic phase, which is
> possible in non-perturbatively non-renormalizable QFT. By this I mean
> that in the lattice QFT, one might end up in a phase, where order
> parameter of sublattice varies chaotically. When taking a continuum
> limit in this phase, one can identify arbitrary 4 sublattice as a
> components of the staggered fermion, hence we'll have a theory with
> infinite ladder of "inner particle structure".
What does the phrase "one can identify ... as a components of the
staggered fermion" mean I don't quite understand. You mean the action
behaves as if it is such? How do you define "chaotic variation of the
order parameter" I also can imagine only vaguely.
Saco
"Dave Snead" <dsn...@charter.net> wrote in message news:<uh3t2bk...@corp.supernews.com>...
> I've got an elementary particle physics question. No two fermions ever
> occupy the same quantum state, but bosons congregate in identical states.
> Now composite particles with an even number of fermions are bosons, and if
> these bosons in identical states congregate, aren't some of their identical
> state fermions congregating also? Doesn't this lead to multiple fermions
> occupying the same quantum state?
Very interesting question. Here is the answer I came up yesterday.
First consider hypothetical example of non-interacting bosons being
able to condensate in the same place. Composite bosons should consist
of charged fermions so the hole system would be chargeless. Since
charged particles can not occupy the same place, this solves the
problem of real space condensation
Saco
fii...@yahoo.com (Squark) wrote in message news:<939044f.02071...@posting.google.com>...
> sa...@moon.yerphi.am (Saco) wrote in message news:<85160b48.02071...@posting.google.com>...
> > Infinite ladder of "inner particle structure" is possible within
> > ordinary QFT, if one ends up in a macroscopic chaotic phase, which is
> > possible in non-perturbatively non-renormalizable QFT. By this I mean
> > that in the lattice QFT, one might end up in a phase, where order
> > parameter of sublattice varies chaotically. When taking a continuum
> > limit in this phase, one can identify arbitrary 4 sublattice as a
> > components of the staggered fermion, hence we'll have a theory with
> > infinite ladder of "inner particle structure".
>
> What does the phrase "one can identify ... as a components of the
> staggered fermion" mean I don't quite understand. You mean the action
> behaves as if it is such? How do you define "chaotic variation of the
> order parameter" I also can imagine only vaguely.
First, about "chaotic variation of the order parameter". This means
that the order parameter is not uniform like in (para/ferro)magnetic
phase, it is not periodic with period two like in antiferromagnets,
but it varies chaotically throughout the space like in spin glasses.
The phrase "one can identify ... as a components of the staggered
fermion" means that staggered fermions have four sublattices where one
puts components of Dirac bispinors. If one has two fermions, then,
they might be 8 sublattices or 4 sublattices with two variable each,
which is the same in the continuum limit. Similarly one might have
infinitely many fermions with infinitely many sublattices or just one
field describing infinitely many fermions interacting in a strange
way.
John Baez
Squark <fii...@yahoo.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<aghb8d$klr$1...@glue.ucr.edu>...
>> If you include glueballs you really should include mesons,
>> especially because observed glueball candidates like the f_0(1500)
>> and f_J(1710) are quite possibly "hybrids" - quantum superpositions of
>> glueballs and mesons. But more importantly, I don't really see why you
>> want to list glueballs as fundamental particles in the first place!
>I was just pointing out your list is incomplete: you should either throw
>the gluons or the glueballs in, depending on your definition of a
>"fundumental particle".
I don't think so.
>> I was just using the term "fundamental particle" the way working-class
>> physicists do, namely a field that shows up in the Lagrangian...
>Then the gluon is definitely in!
On that we agree.
>> If you want a concept with a simple precise definition, it's
>> "eigenstate of the Hamiltonian". Then quarks, gluons, glueballs
>> and neutrons are out, while photons, protons, hydrogen atoms, water
>> molecules and buckyballs are in!
>Hmm, but something containing gluons has to be in!
Yes: protons, hydrogen atoms, water molecules and buckyballs, for
example.
>Take any state with
>gluons and decompose it into Hamiltonian eigenstates.
You have to be a bit careful, because there's no observable called
"gluon number". When I said above that protons contain gluons I
was speaking in the usual way working-class particle physicists do.
I certainly wasn't saying there's some observable called "gluon
number" which is nonzero for a proton. So, you have to decide
for yourself what counts as "a state with gluons". There's no
such thing as "1-gluon state" in the real world, so you have to
try something else. If you consider a proton to be a state with
gluons, and you decompose that into Hamiltonian eigenstates, you get...
a proton!
>Actually, glueballs
>are such eigenstates in "pure gluodynamics", say (aren't they?),
I'm not sure - you may be right. However, in the real world
are certainly just resonances, not eigenstates of the Hamiltonian.
>Hmm, coming to think about it, it's not clear how what I'm saying comes
>to terms the whole zoo of unstable particles out there.
Good - maybe you're beginning to see why I was questioning
your decision to include glueballs as fundamental particles. :-)
vittorio
Hi all,
A.J. Tolland wrote:
> On Mon, 15 Jul 2002, Aaron Bergman wrote:
>
>>In article vittorio wrote:
>>
>>>Aaron Bergman wrote:
>>I don't particularly see the need for a Hilbert space.
why not? what is quantum theory for you? if you don't have a Hilbert
space, in particular you haven't asymptotyc states and you don't have
an S-matrix but you need the S-matrix for experimental reasons.
What do experimental physicists measure at particle accelerator?
>>
>
> Well, you may not need it, but if your observables form a
> C*-algebra and your path integral gives a way of computing a linear
> functional on this algebra, you can use the GNS construction to recover a
> Hilbert space.
Well,
I am not an expert in this topic, but, when you use the algebraic
approach in
the context of QFT, you use a theorem that allows you to build a
rapresentation
of the algebra as operators acting on a suitable Hilbert space.
When you build this rapresentation you have to fix the coupling constants of
the theory, but the coupling constants is running in field theory.
Isn't this a problem? Don't we have a different rapresentation for every
choice of the energy scale?
Thus, which is the physical Hilbert space?
vittorio
> --A.J.
>
John Baez
In article <aggfkj$6gr$1...@blinky.its.caltech.edu>,
Oz <O...@upthorpe.demon.co.uk> wrote:
>Let's just consider what are considered elementary particles for a
>start. I will ignore gravity (rather weak anyway).
>
>Electron: produces an electric field. Suggestions that one could
>plausibly attribute the mass to it's associated electric field.
That suggestion was plausible enough for Lorentz to get very
interested in it, but it's worth noting that this is *not* what
the electron mass is attributed to in the Standard Model. Instead,
the electron's mass is attributed to its coupling to the Higgs field.
>Neutrino: Very light. Does it 'produce' a weak force field?
Yes, but again these days the neutrino masses are attributed
to their couplings to the Higgs.
>Quark: produces a colour and a charge field. Presumably is 'light' when
>in a nucleus (and strongly bound) but becomes increasingly massive if
>isolated. 'Free' mass greater than a quark-antiquark pair, so never seen
>free.
Because they're confined, it's tricky to define the masses of quarks,
but people still do, and by this definition the up and down are quite
light, the strange a bit heavier, the bottom a lot heavier and the top
vastly heavier still. Again, all these masses are attributed to
their couplings to the Higgs.
>Photon: Hmmm. From my position of unfathomable ignorance this seems the
>most attractive to consider composite (very unexpected).
Huh? There's an old theory due to Schrodinger that the photon
is a neutrino-antineutrino bound state (since spin-1/2 plus spin-1/2
can give spin-1) but this is awfully hard to credit these days, and
I don't know why you like it. Anyway, since we're talking about
masses: the Higgs is uncharged, so the photon doesn't couple to it
at all, and the photon is perfectly massless in the Standard Model.
This is very unlike the W and Z, which are also gauge bosons, but
which couple to the Higgs, and develop whopping masses as a result
(making the weak force weak).
>Hmmmm. That's an odd combination of particles when you think about it.
>Mind you if I knew more about it then it might not be so strange.
No, the Standard Model is actually quite strange - the more the more
you think about it. You left out the electron's big brothers by
the way: the mu and tau. More massive, since they have strong
couplings to the Higgs.
I bet you're getting downright sick of me saying "coupling to the Higgs".
And of course the punchline is that nobody has seen the bloody Higgs!
In a couple of years we should know a lot more about whether it really
exists or not... this should be very interesting.
A.J. Tolland
me wrote:
> Well, I am not an expert in this topic, but, when you use the algebraic
> approach in the context of QFT, you use a theorem that allows you to
> build a representation of the algebra as operators acting on a suitable
> Hilbert space. When you build this representation you have to fix the
> coupling constants of the theory, but the coupling constants is running
> in field theory. Isn't this a problem? Don't we have a different
> representation for every choice of the energy scale?
Nope. Coupling constants run in field theory when you are
renormalizing. I regard renormalization as an artifice of effective field
theory. The coupling constants in the path integral definition should be
the "bare" constants which define the model.
> Thus, which is the physical Hilbert space? vittorio
Again, the "bare" space. You don't really have a Hilbert space in
effective field theory, because effective field theories ignore high
energy states. Completeness should fail.
--A.J.
Hendrik van Hees
> Yes, but again these days the neutrino masses are attributed
> to their couplings to the Higgs.
Not necessarily, since there is still the possibility that neutrinos are
Majorana particles, and Majorana mass terms are allowed to be put in by
hand without spoiling any "goodies" of the SM. For an overview see
http://de.arxiv.org/abs/hep-ph/0202058
--
Hendrik van Hees Fakultät für Physik
Phone: +49 521/106-6221 Universität Bielefeld
Fax: +49 521/106-2961 Universitätsstraße
http://theory.gsi.de/~vanhees/ D-33615 Bielefeld
vittorio
Hi A.Tolland,
thanks for the answer.
You wrote:"...effective field theories ignore high energy states.
Completeness
should fail."
I'm not sure that I understood correctly your proposition.
Do you mean:
1) asymptotyc states of the interacting theory do not span the hilbert
space?
or
2) the theory is not well defined in teh hi energy limit and thus we need
another theory such that QFT is its low energy limit(perhaps you're
thinking at
string theory?)?
I think you mean 2). Don't you?
Could you suggest me something to read?
Thanks in advance
vittorio
Aaron Bergman
In article <3D36F148...@physics.it>, vittorio wrote:
>
>A.J. Tolland wrote:
>> On Mon, 15 Jul 2002, Aaron Bergman wrote:
>>
>>>In article vittorio wrote:
>>>
>>>>Aaron Bergman wrote:
>
>>>I don't particularly see the need for a Hilbert space.
>
>why not?
Exactly because of what you say below.
>what is quantum theory for you?
The path integral, whatever it means.
>if you don't have a Hilbert
>space, in particular you haven't asymptotyc states and you don't have
>an S-matrix but you need the S-matrix for experimental reasons.
Why should there be asymptotic states in the real world? The
whole idea of scattering essentially means that you can have a
free theory at infinity, throw things in to interact, and then
have another free theory in the far future. It's nice when this
is true -- we've gone quite far with the idea -- but I don't see
why it should be universal. The S-matrix should be, IMO, a
derived concept when applicable, not a fundamental concept.
>What do experimental physicists measure at particle accelerator?
Scattering, of course.
>> Well, you may not need it, but if your observables form a
>> C*-algebra and your path integral gives a way of computing a linear
>> functional on this algebra, you can use the GNS construction to recover a
>> Hilbert space.
I'll get to this at some point, I hope. I'm not sure I see the
need for all the things that give rise to the GNS construction,
for example.
Chris Jacobs
"John Baez" <ba...@galaxy.ucr.edu> schreef in bericht
news:ah6r2j$3e6$1...@glue.ucr.edu...
>
> In article <aggfkj$6gr$1...@blinky.its.caltech.edu>,
> Oz <O...@upthorpe.demon.co.uk> wrote:
>
> >Let's just consider what are considered elementary particles for a
> >start. I will ignore gravity (rather weak anyway).
> >
> >Electron: produces an electric field. Suggestions that one could
> >plausibly attribute the mass to it's associated electric field.
>
> That suggestion was plausible enough for Lorentz to get very
> interested in it, but it's worth noting that this is *not* what
> the electron mass is attributed to in the Standard Model. Instead,
> the electron's mass is attributed to its coupling to the Higgs field.
See: Lectures on Physics II chapter 28.
Do you mean that without the Higgs the electromagnetic field would
have no momentum?
John Baez
Aaron Bergman <aber...@princeton.edu> wrote:
>In article <agh9tf$k82$1...@glue.ucr.edu>, John Baez wrote:
>>In technicolor models, the "fundamental" particles of the
>>Standard Model are made of more fundamental "preons",
>>confined by a gauge field called "technicolor".
>I think you're confusing preon models and technicolor.
>Technicolor is (roughly) a composite Higgs. There's no reason for any other
>particle to be composite.
Okay... I guess I was just using "technicolor" as a flashy
name for any gauge field that confines more fundamental
particles into the fundamental particles in the Standard Model... sorry.
(Note clever pun: "technicolor", "flashy" - get it?)
Squark
> that the order parameter is not uniform like in (para/ferro)magnetic
> phase, it is not periodic with period two like in antiferromagnets,
> but it varies chaotically throughout the space like in spin glasses.
Okay, and how does it lead to the next conclusion?
> The phrase "one can identify ... as a components of the staggered
> fermion" means that staggered fermions have four sublattices where one
> puts components of Dirac bispinors. If one has two fermions, then,
> they might be 8 sublattices or 4 sublattices with two variable each,
> which is the same in the continuum limit. Similarly one might have
> infinitely many fermions with infinitely many sublattices or just one
> field describing infinitely many fermions interacting in a strange
> way.
I think I got the feeling of what you mean: instead of a field with
an infinity of components at each lattice site, you've got the flavours
distributed (as a staggered fermion) and get and infinity of flavour that
way. However, I still don't see how this situation arises... What do you
define as a staggered fermion? Any collection of field observables on a
sublattice which transforms correctly in the continuum limit? Isn't it too
general a definition? And don't you get an infinite ladde "up" this way,
not "down"? (that is, you build more and more composite particles from the
fundumental ones rather than having no fundumental, but any particle being
decomposable).
Saco
> > fermion" means that staggered fermions have four sublattices where one
> > puts components of Dirac bispinors. If one has two fermions, then,
> > they might be 8 sublattices or 4 sublattices with two variable each,
> > which is the same in the continuum limit. Similarly one might have
> > infinitely many fermions with infinitely many sublattices or just one
> > field describing infinitely many fermions interacting in a strange
> > way.
>
> I think I got the feeling of what you mean: instead of a field with
> an infinity of components at each lattice site, you've got the flavours
> distributed (as a staggered fermion) and get and infinity of flavour that
> way. However, I still don't see how this situation arises... What do you
> define as a staggered fermion? Any collection of field observables on a
> sublattice which transforms correctly in the continuum limit? Isn't it too
> general a definition? And don't you get an infinite ladde "up" this way,
> not "down"? (that is, you build more and more composite particles from the
> fundumental ones rather than having no fundumental, but any particle being
> decomposable).
In general, particle contents of the theory will depend on particular
interaction under study. Currently we assume that there are fixed
number of elementary particles in QFT. But, in principle, it is
possible to have a theory which has different number of elementary
particles depending on energy/coupling constants.
Staggered fermions were introduced to avoid fermion doubling problem
and to keep chiral symmetry by putting components of Dirac bispinors
on four different sublattices. Now consider a scalar field with some
nonlinear interaction which for some values of the coupling constants
has a phase with four sublattices. Then, one can find a fermionic
field which is macroscopically identical to the scalar field in that
phase. By macroscopically identical I mean that the expectation values
of the Dirac bispinor components and the expectation values of scalar
field are identical on all four sublattices.
Similarly, one can have a scalar field which, for instance, at low
temperatures behaves as a single uniform field, then, at high
temperatures it turns into bounded state of infinitely many fermions.
All these are speculations but, at least, it gives a mechanism for
particle creation in lattice QFT.
Mark
>>if you don't have a Hilbert
>>space, in particular you haven't asymptotyc states and you don't have
>>an S-matrix but you need the S-matrix for experimental reasons.
>Why should there be asymptotic states in the real world? The
>whole idea of scattering essentially means that you can have a
>free theory at infinity, throw things in to interact, and then
>have another free theory in the far future.
The actual idea of scattering is that you have a state at the
(-) spacetime boundary of a FINITE spacetime region (namely,
that which encompasses comprises the preparation-detection events),
and a state at the (+) spacetime boundary.
The region is V, its boundary is (V+) - (V-), V+ and V- are
finite spacelike surfaces, and V is compact.
The key hypothesis is that the boundary states for V+ and V-
can be represented within the algebra of free state operators
and that the state spaces corresponding to the surfaces
V+ and V- are the same (or, more accurately; the spaces
corresponding to Cauchy surfaces C+ and C- that contain V+ and
V- and coincide outside of V+ and V-).
The transition matrix would be S(V), which would be
represented by the formal operator
T[exp(i A(V))]
A(V) being the total action (attributed to the interaction)
within V:
A(V) = integral L_I(x) dx: over x in V,
and by its definition, S(V) would satisfy the key principle:
Causality:
S(V union W) = S(V) S(W)
V separated from W by a spacelike boundary,
V on the future side, W on the past side.
This is the foundation that yields Causal formulation of quantum
field theory and -- following the additional hypothesis of
S(V) being representable by a formal power series -- Causal
Perturbation Theory; i.e., the Bogoliubov-Epstein-Glaser
formulation of scattering theory.
The interpretation of the state over V- would be the state
space of all preparation events; and over V+ the state
space of all detection events.
Steinmann in his 2000 (Axiomatic Quantum Electrodynamics, I
think the title is) discusses the flaw behind a scattering
theory based on the more usual LSW asymptotic in-out state formulation.
A.J. Tolland
> A.J. wrote:
> > Effective field theories ignore high energy states.
> > Completeness should fail.
>
> I'm not sure that I understood correctly your proposition. Do you mean:
>
> 1) asymptotyc states of the interacting theory do not span the hilbert
> space? or
>
> 2) the theory is not well defined in the high energy limit and thus we
> need another theory such that QFT is its low energy limit(perhaps you're
> thinking at string theory?)?
Actually, I think I was wrong about the completeness. There is a
Hilbert space of states. (I was thinking of an argument where you took a
sequence of states with energies approaching the cutoff scale. But you
can just throw in states with the cutoff scale, and I don't think this
causues any trouble.) The problem, rather is that this Hilbert space
isn't actually the physical state space, because it neglects higher energy
excitations.
So yes, from a physical standpoint, I meant something like 2), I
guess. The point is still that you shouldn't worry about which space is
the physical Hilbert space when doing effective QFT.
> Could you suggest me something to read?
On what?
--A.J.
John Chelen
especially for the virtual collapse of states? The probability
distributions of interference patterns might be modeled as sublattice
arrays in fermions.
sa...@moon.yerphi.am (Saco) wrote in message news:<85160b48.0207...@posting.google.com>...
>
> In general, particle contents of the theory will depend on particular
> interaction under study. Currently we assume that there are fixed
> number of elementary particles in QFT. But, in principle, it is
> possible to have a theory which has different number of elementary
> particles depending on energy/coupling constants.
>
> Staggered fermions were introduced to avoid fermion doubling problem
> and to keep chiral symmetry by putting components of Dirac bispinors
> on four different sublattices. Now consider a scalar field with some
> nonlinear interaction which for some values of the coupling constants
> has a phase with four sublattices. Then, one can find a fermionic
> field which is macroscopically identical to the scalar field in that
> phase. By macroscopically identical I mean that the expectation values
> of the Dirac bispinor components and the expectation values of scalar
> field are identical on all four sublattices.
>
> Similarly, one can have a scalar field which, for instance, at low
> temperatures behaves as a single uniform field, then, at high
> temperatures it turns into bounded state of infinitely many fermions.
>
> All these are speculations but, at least, it gives a mechanism for
> particle creation in lattice QFT.
[Moderator's note: Unnecessary quoted text deleted...
I could be wrong here, but I don't think that anything particularly
exotic is required to model interference effects in lattice QFT. -MM]
John Baez
Chris Jacobs <c.t.m....@hccnet.nl> wrote:
>"John Baez" <ba...@galaxy.ucr.edu> schreef in bericht
>news:ah6r2j$3e6$1...@glue.ucr.edu...
>> the electron's mass is attributed to its coupling to the Higgs field.
>Do you mean that without the Higgs the electromagnetic field would
>have no momentum?
No, I was talking about the electron's mass, not the momentum
of the electromagnetic field! The electron is very different
from the electromagnetic field, and mass is very different from
momentum.
The electromagnetic field would indeed have momentum without the Higgs...
... although if we somehow eliminated the Higgs from the Standard Model
while keeping everything else the same, we'd have a completely
different concept of what counts as the electromagnetic field, since
there'd be a massless photon-like particle coupling to hypercharge
instead of charge. (Ignore this last paragraph if you like, but honesty
compelled me to insert it.)
James B. Glattfelder
> ... although if we somehow eliminated the Higgs from the Standard Model
> while keeping everything else the same, [...]
Any thoughts on how to do this without destroying the
renormalizability of the theory?
John Baez
I was talking about something simple: taking the Standard
Model Lagrangian and deleting every term that contains the
Higgs field, while leaving every other term exactly the same.
If we did this the theory would still be renormalizable. The
W and Z would become massless, and I guess so would all the
quarks and leptons - though I haven't thought hard enough about
radiative corrections for these. There would be a massless
photon-like particle - a linear combination of the current
photon, W and Z - but it would couple to hypercharge instead
of charge.
I imagine you're talking about something much harder:
getting rid of the Higgs while somehow keeping the W and
Z massive. I have no particular thoughts on how to do this.
This goes to show how ambiguous it can be to talk about "changing
one thing drastically while keeping everything else the same"!
vittorio
A.J. Tolland wrote:
>>2) the theory is not well defined in the high energy limit and thus we
>>need another theory such that QFT is its low energy limit(perhaps you're
>>thinking at string theory?)?
>>Could you suggest me something to read?
>>
> On what?
>
about 2).
thanks
vittorio
Hendrik van Hees
> I was talking about something simple: taking the Standard
> Model Lagrangian and deleting every term that contains the
> Higgs field, while leaving every other term exactly the same.
> If we did this the theory would still be renormalizable. The
> W and Z would become massless, and I guess so would all the
> quarks and leptons - though I haven't thought hard enough about
> radiative corrections for these. There would be a massless
> photon-like particle - a linear combination of the current
> photon, W and Z - but it would couple to hypercharge instead
> of charge.
I am quite sure that the quarks and leptons would remain massless to any
order PT, because there are no anomalies concerning the local gauge
symmetries underlying the standard model (otherwise it would be no theory
at all). Thus due to the chiral vector structure (the famous V-A-structure)
of the electroweak sector of the standard model the particles must remain
massless in an "un-higgst" standard model gauge theory.
>
> I imagine you're talking about something much harder:
> getting rid of the Higgs while somehow keeping the W and
> Z massive. I have no particular thoughts on how to do this.
Isn't there a proof, within the Epstein-Glaser approach if I remember right,
that it is impossible to create masses of non-abelian gauge bosons without
using the Higgs mechanism? See for instance
http://de.arXiv.org/abs/hep-th/9906089
It is clear that one can create massive abelian gauge fields by using the
old Stueckelberg approach. If you like you can have a look at my PhD-thesis
on this subject (due to strange regulations of the Technical University
Darmstadt I had to write it up in German, I'm sorry) or on a recently given
talk about it, both are on my homepage:
the thesis:
http://theory.gsi.de/~vanhees/publ/doc.pdf
the talk (best viewed with acroread 5.0, it's a computer presentation):
http://theory.gsi.de/~vanhees/publ/santo-semi.pdf
--
Hendrik van Hees Fakult‰t f¸r Physik
Phone: +49 521/106-6221 Universit‰t Bielefeld
Fax: +49 521/106-2961 Universit‰tsstrafle
http://theory.gsi.de/~vanhees/ D-33615 Bielefeld
Stephen Selipsky
In article <airqop$rqr$1...@glue.ucr.edu>, ba...@galaxy.ucr.edu (John Baez)
wrote:
> We're talking about a world like in the Standard Model,
> but without a Higgs boson or QCD:
>
> In article
> <stephen.selipsky-48...@news-rich.pi.sbcglobal.net>,
> Stephen Selipsky <stephen....@iname.com> wrote:
> >If we never break SU(2) then
> >its non-Abelian coupling runs to nonperturbative values and
> >confines isocharge (but only over <scribble> 400 km).
> Wow! Really?! Weird!
Really! Of course, we should expect a weird world if we
delete fundamental architecture like QCD forces and the Higgs.
Remember SU(2) dynamics are asymptotically free like SU(3)_QCD.
The weak force fails to confine in the real world only because
spontaneous symmetry breaking decouples non-abelian loops at
momenta below M_W,Z. If we turn off EW symmetry breaking, W
and Z loops can keep contributing negatively to the beta function
while the coupling constant runs to non-perturbative values
at low momenta, just like QCD does.
To derive the scale, we need to pick physical parameters for
the no-Higgs world. Let's assume it has the real world's value
of the SU(2) weak coupling, evaluated at some momentum scale
above real M_W,Z, say q^2 = (250 GeV)^2 (I'll label 250 GeV = v,
although with no Higgs it isn't an expectation value of a field):
g^2 /(4 pi) = (e^2 / 4 pi)/sin^2(theta_w) = (1/126)/0.236 = 1/29.7
Since this alpha_2(q^2 = v^2) is 3.5 times weaker than alpha_QCD,
and SU(2) is smaller so runs slower than SU(3), the confinement
length will be exponentially longer than QCD's. Numerically we
can run the coupling to lower momenta using the solution to the
renormalization group equation's one-loop beta function (with no
particle masses or QCD condensates or bound states):
1/alpha(q^2) = 1/alpha(v^2) - b ln(q^2 / v^2)
where b = (-11/3 N + n/3 + n_s/6)/(4 pi) is negative for SU(N=2)
with n multiplets of chiral fermions and n_s = 0 scalars in the
fundamental rep. Keeping 3 quark colors for anomaly cancellation,
n = 3 generations * (1 lepton isodoublet + 3 quark colors) = 12;
we simply set the QCD coupling g_s = 0 to avoid EW symmetry
breaking through quark condensates.
This alpha_2 blows up at the "dimensional transmutation" scale
Lambda_w = v exp(1/alpha(v^2) / 2b) = 1.2e-22 GeV = (hbar c)/ 1700 km
-- my earlier scribbled 400 km used one less significant figure,
alpha_2(v^2) = 1/29. Lambda_w = 1.2e-22 GeV is the mass scale of
strong SU(2) bound states, akin to Lambda_QCD in the real world.
> >At very
> >long distances you would indeed get only one massless boson,
> >the hypercharge carrier B, while the W+, W-, W0, and left-handed
> >fermions would all be confined. Only right-handed fermions and
> >"EW-baryons", interacting via the B, would propagate beyond the
> >confinement scale.
>
> Are you saying there'd be something vaguely like "hadrons"
> built from left-hand fermions but confined by the weak force,
> about 400 kilometers in diameter? Bizarro!
Yes, exactly; only left-handed particles have SU(2) charge,
so only they would be bound. As seen above the calculation of 400
(1700) km is exponentially sensitive to bizarro world's particle
content and SU(2) interaction strength. We had b = -5/(6 pi);
if we changed to only 1 generation of quarks and leptons, b would
change to -3/(2 pi) which would reduce the confinement length to
25 micrometers-- still macroscopic.
> >Whatever "scale" means in a massless world.
> But these "weak hadrons" - pardon the oxymoron - would have
> mass, right?
Yes, around Lambda_w. Apologies for my melodramatic phrase, I
simply meant that scales are relative so any creatures able to live
in that world would have their sub-"atomic" structures determined
by Lambda_w, and would think of it as microscopic. Matter, stars,
etc. would all be proportionately bigger if they existed at all.
> What would they be like? I'm having trouble imagining this
> world, but it sounds like fun.
Maybe someone like Robert Forward will write a science fiction
treatment of light-year-sized creatures, like his "Dragon's Egg"
biochemistry of long-chain nuclei on a neutron star. But I think
it would be more fun to think about universes where the QCD coupling
is weaker than ours, thus the confinement length is longer, and we
could look inside hadrons or even use QCD forces in everyday life.
There's a long tradition of hard-science based alternate worlds
in science fiction. Asimov's "The Gods Themselves" connected our
universe to one where the weak force was stronger; good fun, in
spite of dated or garbled scientific details. John especially
would appreciate the plot reference to "spinor calculus" as abstruse,
cutting-edge, new-generation math (and used for cosmology!-)
Regards, -- Stephen
James B. Glattfelder
>
> > In article <ed230336.02080...@posting.google.com>,
> > James B. Glattfelder <j_...@gmx.net> wrote:
>
> > >ba...@galaxy.ucr.edu (John Baez) wrote in message
> > >news:<ai8rfn$pe0$1...@glue.ucr.edu>...
> Which raises the next interesting question: what about
> the origin of mass in fundamental theories of strings and
> loops?
Thomas Thiemann introduced the concept of point holonomies into the
formalism of LQG to include scalar Higgs fields in 1997
(http://xxx.lanl.gov/abs/gr-qc/9705021 or his 301-page introduction to
quantum gravity: http://xxx.lanl.gov/abs/gr-qc/0110034).
Indeed, in 2000 he started a series of articles aimed at "connecting
non-perturbative quantum general relativity with the low energy
physics of the standard model", called "Gauge Field Theory Coherent
States (GCS)" (http://xxx.lanl.gov/abs/hep-th/0005233). He announced
eight papers, the fifth entitled: "Extension to Higgs Fields and
Fermions". By the end of the year four had been written. However, the
Higgs paper is still to appear...
This year Thiemann seems to be concentrating on QFT in curved space:
"how one might obtain the semiclassical limit of ordinary matter QFT
propagating on curved spacetimes from full fledged quantum general
relativity, starting from first principles"
(http://xxx.lanl.gov/abs/gr-qc/0207030).
So he appears to have gone back to the issues perhaps best covered in
Birrell and Davies' 1982 book "Quantum Fields in Curved Space".
Already here the subtleties of merging the two concepts becomes
apparent (p. 36): "The basic generalization of the particle concept to
curved spacetime is readily accomplished. What is not so easy is the
physical interpretation of the formalism so developed. There has, in
fact, been a certain amount of controversy over the meaning - and
meaningfulness - of the particle concept when a background
gravitational field is present. In some cases [...] the concept seems
well defined, while in others [...] the notion of particle can seem
hopelessly obscure."
Again our incomplete knowledge of the nature of reality at a
fundamental scale and our formal concepts thereof seem to pose riddles
from every possible angle of attack...