LHC finds new Chi-b (3P) particle

[Greg Bernhardt]

The Large Hadron Collider (LHC) on the Franco-Swiss border has made its first clear observation of a new particle since opening in 2009.

It is called Chi_b (3P) and will help scientists understand better the forces that hold matter together.

http://www.bbc.co.uk/news/science-environment-16301908

 

[lpetrich]

That article links to [1112.5154] Observation of a new chi_b state in radiative transitions to Upsilon(1S) and Upsilon(2S) at ATLAS

It's a new bottomonium state, a bottom-antibottom quark bound state. Wikipedia summarizes the PDG listings in Quarkonium (quark-antiquark bound state), List of mesons

Quarkonium states can easily be classified by orbital angular momentum L, total spin S, total angular momentum J, and radial quantum number n.

The known ones for bottomonium are:

eta-b: L=0, S=0, J=0, n=1
upsilon: L=0, S=1, J=1, n=1,2,3,4,?,?
chi-b: L=1, S=1, J=0,1,2, n=1,2,3 (1,2 known previously, 3 just discovered)
upsilon: L=2, S=1, J=2, n=1

Spin-orbit splitting makes three separate chi-b states for each n value, one for each value of J. From (largest)-(smallest) mass, this splitting is 53 MeV for n=1 and 13 MeV for n=2. The n=3 discoverers report only an average mass; it may be difficult for them to resolve this state's spin-orbit splitting.

Strictly speaking, the splitting is caused not only by the familiar L.S effect, but also by the tensor force: (r.S)^2 - (1/3)(r^2)(S^2). One can untangle these two effects if one has states for all three J values.The hard part of quarkonium modeling is the radial potential functions, but this new state's mass agrees with some model predictions to within the model's accuracy for other chi-b states. One can use lattice and perturbative QCD here, but even there, it'll be important to see how well one can do, as well as to estimate alpha-s, the QCD coupling constant at quarkonium energies.

With the LHC's success in discovering a bottomonium state, could the LHC teams discover more states? The bottomonium table has some rather glaring gaps, like for S=0 states.

 

[Vanadium 50]

The splitting between the 0++, 1++ and 2++ states is hyperfine, not fine. It's due to the quark's chromomagnetic moment, which goes as 1/m. That's why the chi_c splittings are of order 45 MeV, but the chi_b splitting is only 15.

Quarkonium states come in two bunches, you have the triplet Psis, Chis and Upsilons, which are CP-even, and singlet etas and h's, which are CP-odd. States in the same bunch can decay to each other by E1 transitions, but can only cross a bunch via a weaker M1 transition. At e+e- machines, only the first bunch is directly accessible (and at hadron machines the first bunch is certainly easier to work with), which is why the second bunch is less thoroughly investigated.

The remaining first bunch family includes three F-wave states. These have never been seen anywhere.

 

[lpetrich]

Actually, that splitting is caused by both spin-orbit (fine structure) and spin-spin (hyperfine) tensor effects. Unlike the case with chemical elements, spin-orbit and spin-spin effects are similar in size, because both particles have the same mass.

That's also true of positronium: Spectroscopy of excited state positronium -- Hyperfine Interactions, Volume 89, Number 1 - SpringerLink

Including [1109.4476] Observation of the h_b states and Beyond, the bottomonium list becomes:

eta-b(n¹S₀): L=0, S=0, J=0, n=1
upsilon(n³S₁): L=0, S=1, J=1, n=1,2,3,4,?,?
h-b(n¹P₁): L=1, S=0, J=1, n=1,2
chi-b(n³P_1,2,3): L=1, S=1, J=0,1,2, n=1,2,3 (1,2 known previously, 3 just discovered)
upsilon(1³D₂): L=2, S=1, J=2, n=1

The new one's mass: 10.539 +/- 0.004 (stat.) +/- 0.008 (syst.) GeV

I once calculated the sizes of the spin-orbit and spin-spin splittings of charmonium and bottomonium states. For the chi-q states, I used

{J=L-1 value, J=L value, J=L+1 value}
The spin-orbit effect I found as a multiple of <L.S> = (J(J+1) - L(L+1) - S(S+1))/2
{-(L+1), -1, L}
The tensor spin-spin effect I found as a multiple of <(3(n.S)² - (S.S))/2>
{-(L+1)/(2(2L-1)), 1/2, -1/((2(2L+3))}

Charmonium:

1S - L=0, n=1: Scalar spin-spin: 116.6 MeV
2S - L=0, n=2: Scalar spin-spin: 49.1 MeV
1P - L=1, n=1: Scalar spin-spin: -0.6 MeV
1³P - L=1, S=1, n=1: Spin-orbit: 35.0 MeV, Tensor spin-spin: 40.6 MeV

Bottomonium:

1S - L=0, n=1: Scalar spin-spin: 71.4 MeV
1P - L=1, n=1: Scalar spin-spin: 1.6 MeV
2P - L=1, n=2: Scalar spin-spin: 0.5 MeV
1³P - L=1, S=1, n=1: Spin-orbit: 13.7 MeV, Tensor spin-spin: 13.1 MeV
2³P - L=1, S=1, n=2: Spin-orbit: 9.3 MeV, Tensor spin-spin: 9.1 MeV

The scalar spin-spin term is sizable only for L = 0, much as for the inverse-square case.

However, the n,L energy levels have a noticeable departure from the inverse-square case.

Charmonium:

2S - 1S: 606 MeV
2S - 1P: 148 MeV

Bottomonium (S = 1 only):

2S - 1S: 563 MeV
2S - 1P: 123 MeV

3S - 2S: 332 MeV
3S - 2P: 95 MeV
2P - 1D: 99 MeV

4S - 3S: 224 MeV
4S - 3P: 40 MeV

The main potential is clearly not quite inverse square.

 

[Vanadium 50]

All those calculations are between multiplets. The splitting between the 0++, 1++ and 2++ (and 1+-) in the same multiplet is hyperfine.

The fact that the 3P is bound and the 4S is not immediately tells you the potential is not inverse square.

 

[petergreat]

This particle has a mass of only 10.5 GeV. Why has it eluded LEP and Tevatron? What makes LHC suitable for its discovery? If a future 500 GeV International Linear Collider is built, will it be capable of studying this particle?

 

[Vanadium 50]

It eluded the Tevatron probably for several reasons, but one of the important ones is luminosity. The LHC created more particles, and it created them in kinematic regions where they would be easy to reconstruct.

LEP produced way, way too few of them.

The B-factories are more interesting. They may well have dozens or even hundreds of such events on tape, and now that they know what to look for, I suspect they are looking for them now. Or at least will after the holidays.

 

[petergreat]

What's the main production channel of B-mesons at the LHC? Gluon-Gluon fusion?

 

[Vanadium 50]

It's gluon induced, but how one makes B-mesons is not relevant to this particle.

 

[lpetrich]

As I understand it, here are the strongest processes that create quarkonium states.

In electron-positron colliders, the e- and e+ combine to make a virtual photon, which in turn makes a quark-antiquark pair.

In hadron colliders, a quark that collides with another quark can radiate a virtual gluon, which in turn makes a quark-antiquark pair.

Alternately, each quark can radiate a gluon, which make a quark-antiquark pair when they run into each other.

If a quark-antiquark pair has low enough energy, it will be bound, making a quarkonium state.


The next question is why we've discovered lots of total-spin-1 charmonium and bottomonium states, but hardly any total-spin-0 ones.

One can get a clue from positronium, which can be either total-spin-1 (ortho) or total-spin-0 (para). Total-spin-0 states decay into even numbers of photons, while total-spin-1 states decay into odd numbers of photons.

This may explain why electron-positron colliders directly make mostly total-spin-1 states, which explains that discovery discrepancy.

Hadron ones are likely to make a more even mix, because both processes I've mentioned have 4 gluon vertices, giving them similar probability. The first one makes a quarkonium state from a single gluon, thus preferring total spin 1, and the second one makes a quarkonium state from two gluons, thus preferring total spin 0.

If that is correct, then it may be possible to discover more total-spin-0 charmonium and bottomonium states at the LHC.

 

[Vanadium 50]

In e+e- colliders, you are correct that you go through a virtual photon. So you can only produce states with the quantum numbers of the photon, 1^--. Those will decay via E1 transitions to other spin-triplet states, but only go by the weaker M1 to spin singlet states.

In general, the spin singlet states are broad, and no single branching fraction is large. That makes them hard to detect.

In hadron collisions, all states can be produced. However, nothing can go through a one-gluon process. A single gluon carries color, and a hadron does not, so this vertex violates color conservation. The minimum number of gluons is two. That said, there is substantial evidence that the quantum numbers of the qqbar pair formed are not the same as that of the final hadron. You could, for example, have a two gluons fuse into a ³P₂ color-octet state, which then can radiate a soft gluon, "bleaching" the quark-antiquark pair, which then end up in some other state, like a ³S₁. This appears to be a very complicated process, and no model can correctly reproduce all of the experimental phenomena.

It's unlikely that the LHC will find many of the singlet states, for the reason above. Worse, if you had a decay like h_b → 8π, how would you know in which event to start combining pions? Indeed, how did the event even get triggered on?

 

[lpetrich]

Oops, I didn't think it through about a one-gluon process making a quarkonium particle. A photon would work, though it would be relatively improbable compared to a gluon process.

Also, what makes singlets (S = 0) much broader than triplets (S = 1)? Decaying into 2 gluons vs. decaying into 3 gluons? (thinking of positronium here)

 

[Vanadium 50]

Part of it is two vs. three gluons. Part of it is just more accessible states, when you factor in all the spin-statistics.

 

[lpetrich]

A further problem. What makes the C parity of an onium state the opposite sign from what one might at first expect?

Imagine a particle x with antiparticle X. Their spin states are
Spin 0: (1/sqrt(2))*(|x+,X-> - |x-,X+>)
antisymmetric
Spin 1: |x+,X+>, (1/sqrt(2))*(|x+,X-> + |x-,X+>), |x-,X->
symmetric

Let charge conjugation C be exchanging x and X. Then C² gets the original x and X without sign changes.
Spin 0: multiply by -1
Spin 1: multiply by +1

So there's a factor of -1 sneaking in somewhere. Is C for elementary fermions really something like x -> X, X -> -x? That's a negative sign on one of them. That would make C² equal to -1 instead of +1.