# Will the Higgs mix with glueballs?

1. Dec 3, 2008

### franoisbelfor

Both the Higgs boson and glueballs are expected to be 0++ particles: spin 0, positive C and P parity. Thus the two particles will probably mix - or not? Will this have any effect on either of the two?

François

2. Dec 3, 2008

### malawi_glenn

mix in what sense?

3. Dec 3, 2008

### JustinLevy

Isn't the Higgs supposed to be its own anti-particle? Which I would think means it is not appropriate to assign a parity quantum number to the particle. Similarly, can't a photon "have" +/- parity, as it too is its own anti-particle?

4. Dec 3, 2008

### malawi_glenn

A photon has instrinic parity -1

5. Dec 3, 2008

### JustinLevy

Consider positronium in a singlet state, which decays via electromagnetic interactions.

e+ e- ==> gamma gamma

according to your statement, the parity is
-1 * +1 == -1 * -1

Are you telling me electric interactions violate parity?
If not, then I am severely misunderstanding something here. Please do help.

Last edited: Dec 3, 2008
6. Dec 3, 2008

### malawi_glenn

are you suggesting then that parity does not have any parity at all, what will happen with parity conservation then? ;-)

Parity of positronium is: -(-1)^L where L is the relative angular momentum between e+e-

So FIRST of all you must consider what state you are mention ;-) I assume the S-state (L=0).

Parity of that state is -1.

BUT photons have instrinsic angular momentum aswell (spin), so there can be relative angular momentum between the emitted photons. Hence the photons will be polarized, and one can measure that and confirm that the 2photons came from a $$^1S_0$$-state of positronium.

See: Perkins, Introduction to High energy physics, p. 89-90

7. Dec 3, 2008

### hamster143

I think you need a degree 2 vertex coupling between particles to have "mixing". E.g. neutrinos mix because there's a vertex that couples $$\nu_e$$ with $$\nu_{\mu}$$. Higgs would not couple to a single glueball.

Or would it?

Last edited: Dec 3, 2008
8. Dec 3, 2008

### JustinLevy

I'm saying that the lagrangian is symmetric to parity, but that doesn't mean every particle must be an eigen state of that symmetry. What is the C parity number of the electron for instance? For the photon and higgs, I would think they are not eigen states of either parity operator.

My expectations come down to this:
An antiparticle should have opposite quantum numbers of the particle.

If that is wrong, that is where my mistake is. If that is wrong, please tell me what an anti-particle is. I'm sorry this is getting so remedial, but I do appreciate your time helping explain things to me.

I don't have that book, but a gradstudent friend of mine does. I'll ask to borrow it.

Yes, I meant the ground state. If I didn't say it was excited, I thought that was implied. Sorry about that.

I don't understand your answer though, because for angular momentum to be conserved, the photon pair must have total angular momentum zero. Doesn't this require J=L+S=0, and if the photons are spin polarized relative to each other to have opposite spins then total S=0 for this state and therefore L=0 for this state. So shouldn't the parity still be -1 * -1 if we call the parity of the photon -1 ?

Searching google I found some lecture notes that say the photon can't be assigned an intrinsic parity since it has no rest state (different from my reasoning), but convention gives a photon from an electric dipole interaction a +1, and from a magnetic interaction a -1.
http://www.physics.ohio-state.edu/~kass/P780_L6_sp03.ppt#2
If you need to already know what is going on to assign a parity value to it, why assign a value at all?

9. Dec 3, 2008

### malawi_glenn

It depends on what that lecturer means by "Strictly speaking, we can not assign a parity to the photon since it is never at rest. "

Look at page 94 in Martins book "particle physics" for instance. It is the vector potential which corresponds to the photon wavefunction in QM (Also see Mandl Quantum Field theory ch 1)

You will have the same problem with pi0 -> gamma gamma decay.

the pages I was referring to was in 3rd ed but you can perhaps find something of vaule in here. Page 68-69 covers parity of pi0 from its 2gamma decay, if you'll find it illuminating.

I have to think more about angular momentum conservation.

10. Dec 3, 2008

### JustinLevy

Thanks for all the helpful suggestions.

"However, a boson and 'antiboson' - i.e. one of the opposite charge - have the same parity."
(pg 72 of the Perkins pook you linked to)
Weird. And it is stated without explaining why.

I have a lot ot think about. And plenty to read. Thanks.
If anyone has more insights, please do share.

11. Dec 3, 2008

### malawi_glenn

I think onw must study more QFT to understand that ;-)

12. Dec 3, 2008

### humanino

I don't think the Higgs and pure glueballs can mix, even if they had the same quantum numbers, because their respective masses are too different. The handwaving reason giving me that opinion is as following. In order for them to overlap in momenta, you need them to have very different energies, and the wavefunctions will oscillate at quite different frequencies. Or to put it another way, the phase spaces do not match.

13. Dec 4, 2008

### franoisbelfor

Ok, that makes sense. I was just wondering whether a mixing might give a Higgs mass outside the usual boundaries of between 115 and 200 GeV. But if the mixing effect is small, one can neglect this.

François

14. Dec 8, 2008

### blechman

I'm not sure I understand this argument. After all, the $\rho^0$ meson, with a mass of 770 MeV or so, can mix with the photon. A similar effect can create strong bounds on technicolor and susy models from the extra particles there. Also, in extra dimension models, the Higgs boson can mix with the radion, the 55 component of the metric. Finally, what about quark mixing - the CKM matrix is independent of the quark masses, at least in theory (that there is such an alignment in practice is actually an open coincidence!).

I think the answer to the OP's question is as follows: these effects CAN shift the mass of the Higgs, as well as the couplings of the Higgs to other particles, and AFTER these effects are taken into account is where we have to apply the bounds of electroweak precision tests and direct searches. But since we have as yet no UV-complete theory of the Higgs, so we have no idea where the Higgs mass comes from, these effects can always just be reabsorbed into the parameter we call the "Higgs boson mass" and no one would be the wiser. Same for the couplings to fermions. Gauge interactions with the Higgs are more rigid, since they're fixed by gauge invariance...

Anyway, hope that helps.

15. Dec 8, 2008

### Haelfix

Thats correct, except that you are talking about a tiny second and third order effects. The mass shift from such a process is completely trivial relative to the physics of the heavy stuff

16. Dec 8, 2008

### blechman

agreed, to a point: tiny for higgs-glueball, yes, I think you're right; and that's probably why people don't usually worry about such things in practice. However, in the more general problem (higgs-radion or technirho or "D-term SUSY breaking" models, for instance) this is (can be) a significant problem.

But in regards to the OP's question: yes. I agree with you.

17. Dec 8, 2008

### humanino

I was expecting more protest actually, so I called the argument "barely hand waving"

To comment on the rho, which I am most familiar with and we are sure is relevant to Nature, saying "770 MeV is big" is not enough. You need to compare it to the QCD scale, or the rho size, and as you know 770 MeV is actually not big (especially compared to the Higgs mass !).

In fact, it is always true in principle that the rho will mix with the photon, but the question is "when is this relevant ?" (or if you will "when can one see it ?"). At high energy, the longitudinal part of the rho will mix with the photon, but obviously not the transverse one. So you need to be in a regime where the longitudinal cross section dominates the transverse one. For instance, take electroproduction on a hadron, the question is "when will the longitudinal part of the propagator of the virtual photon fluctuating into a vector meson be dominant ?" (or even, measurable). You need both large energy and transfer compared to lambda QCD, with their ratio being either large or small. If the energy is large and the transfer is small, you will resolve the photon fluctuating into a vector meson and exchange a gluon ladder (reggeon) with the hadron. If the transfer is large and the energy is small, you will pull out quarks of both the rho and the (target) hadon with hard gluon exchanges. In any case, those processes which can be observed, described in terms of photon/vector meson mixing, and exhibit for us a proof that the photon propagator has a significant quark-antiquark mixing cover a very small part of the phase space : those description are valid in certain kinematical limits where factorisation-like theorems apply.

So my opinion until further quantitative argument is presented is that the putative scalar glueball mixing with the Higgs is in principle possible, but in practice negligible.

18. Dec 8, 2008

### blechman

I agree with you, humanino.