# Presentation: 087962

## Continuing Studies in Particle Physics:Charmonium Then and Now

- R. Cahn
- LBNL
- Nov. 10, 2005

- Reference: “Lectures on the New Particles,”         J. D. Jackson, SLAC Summer Institute on Particle Physics, 1976.

## Positronium

- e+e- bound state discovered by Martin Deutsch, 1951
- Two forms: para, ortho -
- [remember:             ]
- For fermion-antifermion:
- Para decays to two photons, ortho to three:

- Wave function at origin squared = density

- Charmonium Then and Now

- Nov. 10, 2005

## November Revolution (1974 version)

- 1.What do the symbols in the Breit Wigner formula mean?
- 2.What is the area under a resonance curve?
- The area under the  cross section is 10 nb GeV.
- The branching fraction for                 is about 6%.
- 3.What is the width of the  ?
- 4.If this is a          state, how do we know it is a     ?

- Why is resonance so asymmetrical?

- Charmonium Then and Now

- Nov. 10, 2005

## Charmonium Spectrum

- 5. Theoretically, we expect the quark-antiquark potential to be a sum of a Coulombic piece and a linear piece.  Why?
- 6. Ignoring spin dependent forces, guess what the charmonium spectrum might look like.  It might help to think about the spectrum for the 3-d harmonic oscillator. Include two s-wave levels and one p-wave level.  Use
- 7. Put in the splittings due to spin. Label each state both in spectroscopic notation and as JPC.
- 8. What electromagnetic transitions are allowed?  Label them as E1, M1 etc.
- 9.What hadronic transitions might occur from higher charmonium levels to the ?  What constraints come from isospin?

- Charmonium Then and Now

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## G parity and hadronic  decays

- Symmetry of strong interactions:
- 10. Show that:
- 11. What can you say about the decay                 ?

|  | 1.5x10-4 |  | 1.5x10-2 |
| --- | --- | --- | --- |
|  | 4 x10-3 |  | 3.4x10-2 |
|  | 4 x10-3 |  | 2.9x10-2 |

- 12. Discuss the  branching fractions above.

- Charmonium Then and Now

- Nov. 10, 2005

## Other hadronic charmonium decays

- 13. Guess the form of the equation for the hadronic width of the states below in terms of s, the radial wave function R, and the mass M of the state.

- 14. Which one of these is problematic? Why?

- The measured widths in MeV are, respectively,
- 34(BaBar),   0.09,  10,  0.9,   2.1.

- Charmonium Then and Now

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## Discussion of Questions

- 1.
-        J is the spin of the resonance, 2S1,2+1 are the number of spin states for the incident particles, k is the cm momentum,
-      in,out are the partial widths of the resonance into the initial and final states. tot is the total width, E0 is the cm energy at the resonance.  Useful conversions are 0.197 GeV fm = 1,       0.389 mb GeV2=1 .

- Charmonium Then and Now

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## 2. Integral of Breit Wigner

- If J=1, S=1/2

- Charmonium Then and Now

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## 3.Width of the 

- Actually integral is of hadronic events, so better estimate is       72 keV/0.88=82 keV. PDG says  91 keV.

- Charmonium Then and Now

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## 4. Assignment of 

- We know JPC=1-- and P=(-1)L+1, C=(-1)L+S, so L is even and S is odd.  Thus S=1, and L=0,2... The lowest state should have L=0, so the  is 3S1. The resonance is asymmetrical because of initial state radiation.

- 5. Static QCDPotential

- At short distances, QCD is weak and dominated by single (massless) gluon exchange, so it is analogous to a Coulombic potential.  At long distances, QCD is confining and a linear potential gives confinement (and is confirmed by lattice calculations).

- Charmonium Then and Now

- Nov. 10, 2005

## 6. Spectra without spin

- Coulomb

- Harmonic oscillator

- QCD

- 1s

- 1s

- 1s

- 2s

- 2s

- 1p

- 1p

- 1p

- 2s

- 1d

- Charmonium Then and Now

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## 3-d Harmonic Oscillator

- The 3-d harmonic oscillator is just the sum of oscillators in the x, y, and z directions.
- Each level can be specified by three integers: nx,ny,nz.
- The energy of the level is
- The lowest level (0,0,0) has degeneracy 1, but the next is three-fold degenerate: (0,0,1),(0,1,0),(1,0,0).
- Because our potential is spherically symmetric it must be possible to classify the states in terms of angular momentum. Thus these three must be p-wave.
- The next level is six-fold degenerate:(1,1,0),...(2,0,0).  This level is the sum of degenerate s-wave and d-wave states.

- Charmonium Then and Now

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## Guess at the spectrum for Coulomb + linear

- This is very heuristic!  We simply guess that the p-wave state lies halfway between the pure Coulomb and pure harmonic oscillator cases, I.e. it is 3/4 the way from the first s-wave to the second.  This gives us 3.091 + (3/4)0.595=3.537GeV.  In fact the center of gravity of the p-wave states is about 3.525 GeV.

- Charmonium Then and Now

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## 7. Spectra with spin

- c(1S0)

- c(1S0)

- (3S1)

- (3S1)

- (3D1)

- c0(3P0)

- c1(3P1)

- hc(1P1)

- c2(3P2)

- Charmonium Then and Now

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## 8. Radiative Transitions

- c(1S0)

- c(1S0)

- (3S1)

- (3S1)

- (3D1)

- c0(3P0)

- c1(3P1)

- hc(1P1)

- c2(3P2)

**E1**

**M1**

- Suppressed M1

- Charmonium Then and Now

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## 9. Hadronic Transitions to 

- c(1S0)

- c(1S0)

- (3S1)

- (3S1)

- (3D1)

- c0(3P0)

- c1(3P1)

- hc(1P1)

- c2(3P2)

- ‚

- 0

- 0

- 0

- Also

- Charmonium Then and Now

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## 10. G parity

- C has the effect of complex conjugation
- If +=(1+i2)/√2, then -=(1-i2)/√2.
- 0= 3
- e-iIy is a rotation by  about the y axis, so 1-1, 2 2, 3-3
- G(+)=- +, etc.

- Charmonium Then and Now

- Nov. 10, 2005

## 11. Decays of the 

- Since the        is C-even and has I=0, it is G-even.
- Being G-even, it can’t decay hadronically into an odd number of pions.
- This is no surprise since we observe , which is obviously electromagnetic.
- Because isospin is not an exact symmetry, we can think of the   0 and  as each being mixtures of I=0 and I=1.  Then the decay 3 proceeds through this mixing.

- 

- Charmonium Then and Now

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## 12. Hadronic Decays of the 

- Since the        is C-odd and has I=0, it is G-odd.
- We expect it to decay to odd numbers of pions.
- The final states with even numbers of pions, however, are not all that suppressed.
- This is not surprising since the BF for e+e- is 6%, and goes through a virtual photon.
- We expect qqbar production through the same mechanism with a rate about 3(1/9 +1/9 +4/9)=2 times as big, i.e., 12%.
- These decays then contribute to the even-pion final states.

- 

- Charmonium Then and Now

- Nov. 10, 2005

## 13. Charmonium decay widths

- The decay of the c is analogous to that of parapositronium:
- The decay of the  is analogous to that of orthopositronium:
-       We need three gluons because the two-gluon decay makes    only C=+.
- For the p-wave states the wave function vanishes at the origin, so we have instead

- Charmonium Then and Now

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## 14.The problem with J=1 p-wave

- This argument fails for J=1 because of Yang’s theorem (first proved by Landau): a spin-one particle cannot decay into two photons.  To see this, work in the rest frame of the decaying particle, whose polarization vector is **** and let the two photons have polarization vectors ****1 and ****2 and momenta **k**1 and **k**2=-**k**1.The decay amplitude must be a scalar and must be even under interchange of 1 and 2, because of Bose statistics.  The amplitude must be linear in each polarization vector.  You will quickly see that this is impossible.  E.g.
-      Is odd, not even.  Why doesn’t                                                 work?
- With a little thought you can convince yourself that this applies also to the decay to two gluons, even though they are not identical particles.  Thus other mechanisms must account for the decay of the J=1 state.

- Charmonium Then and Now

- Nov. 10, 2005