1. Continuing Studies in Particle Physics: Charmonium Then and Now R. Cahn LBNL Nov. 10, 2005, Nov. 17, 2005 Jan. 18, 2006, Jan. 27, 2006 Reference: “Lectures on the New Particles,” J. D. Jackson, SLAC Summer Institute on Particle Physics, 1976.

2. Nov. 10, 2005Charmonium Then and Now2 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

3. Nov. 10, 2005Charmonium Then and Now3 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?

4. Nov. 10, 2005Charmonium Then and Now4 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 J PC. 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?

5. Nov. 10, 2005Charmonium Then and Now5 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.

6. Nov. 10, 2005Charmonium Then and Now6 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.

7. Nov. 10, 2005Charmonium Then and Now7 15. Why is the  (3770) so much broader than the  (3686)? 16. What is the likely assignment for  (3770) for (S,L,J)?

8. Nov. 10, 2005Charmonium Then and Now8 3 D 1 - 3 S 1 Mixing The spin-dependent potential for charmonium is Where 17. Which interaction can mix 3 D 1 and 3 S 1 ?

9. Nov. 10, 2005Charmonium Then and Now9 Estimated mixing 18. If the  (3770) were pure 3 D 1, how would its partial width to e+e- depend on the radial wave function? 19. A theoretical estimate indicates that if there were no mixing, we would expect. The measured rates are If what might  be?

10. Nov. 10, 2005Charmonium Then and Now10 20. Let L be the orbital angular momentum of the . What values are allowed? Let K be the orbital angular momentum of the J/  -(  ) system. What values are allowed? Does the mass spectrum suggest anything? BES: PRD 62,032002 (2000)

11. Nov. 10, 2005Charmonium Then and Now11 X(3872) Belle saw 21. Belle said. Why is this narrowness expected given the observed decay channel? 22. This indicates that it was a state with J P =0 --, 1 +-, 2 --,… Why? 23. Leading candidates were d-wave and p-wave states. Give the spectroscopic notation for such possible states. 24. What E1 transitions would be expected?

12. Nov. 10, 2005Charmonium Then and Now12 X(3872) Contradictions Belle hep-ex/0505037 25. How do these results conflict with the charmonium picture?

13. Nov. 10, 2005Charmonium Then and Now13 Bound State? 26. If X(3872) is a bound state would C be a good quantum number? Would isospin be a good quantum number? What would you expect for J PC ?

14. Nov. 10, 2005Charmonium Then and Now14 Initial State Radiation 27. In e+e- annihilation in the cm, what fraction of the electron’s energy must be radiated when running at the  (4s) so that a J/  can be resonantly produced? 28. If x is the fraction of the electron’s energy that is radiated away, we might guess that the formula for the radiative cross section is something like where  is some function of the cm energy squared, s. What else must it depend on? What sets the scale for  ? If we integrate over x to get the “total spectrum,” it will diverge. Suggest a simple fix for this.

15. Nov. 10, 2005Charmonium Then and Now15 Observation of Structure at 4.26 GeV

16. Nov. 10, 2005Charmonium Then and Now16  (3686) Y(4260) N evts 12,000125  0.1080.157  (ee) 2.1 keV? BF(   +  - ) 0.317? Determine for Y(4260): BF(   +  - ) x  (ee)

17. Nov. 10, 2005Charmonium Then and Now17 Limit on  (ee) One unit of R= What is the peak cross section for an e+e- resonance in units of R? Use this to get a rough upper limit on the BF(Y  ee). What is then the lower limit on  (Y   +  )? Compare to  (  ’   +  ).

18. Nov. 10, 2005Charmonium Then and Now18 Discussion of Questions 1. J is the spin of the resonance, 2S 1,2 +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, E 0 is the cm energy at the resonance. Useful conversions are 0.197 GeV fm = 1, 0.389 mb GeV 2 =1.

19. Nov. 10, 2005Charmonium Then and Now19 2. Integral of Breit Wigner If J=1, S=1/2

20. Nov. 10, 2005Charmonium Then and Now20 3.Width of the  Actually integral is of hadronic events, so better estimate is 72 keV/0.88=82 keV. PDG says 91 keV.

21. Nov. 10, 2005Charmonium Then and Now21 4. Assignment of  We know J PC =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 3 S 1. 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).

22. Nov. 10, 2005Charmonium Then and Now22 6. Spectra without spin Coulomb Harmonic oscillator QCD 1s 2s 1p 2s 1d

23. Nov. 10, 2005Charmonium Then and Now23 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: n x,n y,n z. 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.

24. Nov. 10, 2005Charmonium Then and Now24 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.

25. Nov. 10, 2005Charmonium Then and Now25 7. Spectra with spin c(1S0)c(1S0) c(1S0)c(1S0) (3S1)(3S1) (3S1)(3S1) (3D1)(3D1)  c0 ( 3 P 0 )  c1 ( 3 P 1 ) hc(1P1)hc(1P1)  c2 ( 3 P 2 )

26. Nov. 10, 2005Charmonium Then and Now26 8. Radiative Transitions c(1S0)c(1S0) c(1S0)c(1S0) (3S1)(3S1) (3S1)(3S1) (3D1)(3D1)  c0 ( 3 P 0 )  c1 ( 3 P 1 ) hc(1P1)hc(1P1)  c2 ( 3 P 2 ) E1 M1 Suppressed M1 Obscure and not shown:

27. Nov. 10, 2005Charmonium Then and Now27 9. Hadronic Transitions to  c(1S0)c(1S0) c(1S0)c(1S0) (3S1)(3S1) (3S1)(3S1) (3D1)(3D1)  c0 ( 3 P 0 )  c1 ( 3 P 1 ) hc(1P1)hc(1P1)  c2 ( 3 P 2 )  ‚  00 00 00 Also 00 00 00

28. Nov. 10, 2005Charmonium Then and Now28 Branching Fractions from RPP

29. Nov. 10, 2005Charmonium Then and Now29

30. Nov. 10, 2005Charmonium Then and Now30

31. Nov. 10, 2005Charmonium Then and Now31

32. Nov. 10, 2005Charmonium Then and Now32

33. Nov. 10, 2005Charmonium Then and Now33 hep-ex/0508037 (CLEO)

34. Nov. 10, 2005Charmonium Then and Now34 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.

35. Nov. 10, 2005Charmonium Then and Now35 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. 

36. Nov. 10, 2005Charmonium Then and Now36 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. 

37. Nov. 10, 2005Charmonium Then and Now37 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

38. Nov. 10, 2005Charmonium Then and Now38 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.

39. Nov. 10, 2005Charmonium Then and Now39 15.  (3770) is above threshold for the decay to D + D - 16. Only 3 D 1 and 3 S 1 are possible and the  (3770) is too close to  (3686) to be another 3 S 1, so 3 D 1 is the best guess. 17. We need to connect states whose L values differ by 2. Only the tensor force is a second rank spatial tensor. The tensor force is a  L=2 operator. It is also a  S=2 operator. This allows it to connect 3 D 1 to 3 S 1. 18. The radial wave function varies as r 2 near the origin, so

40. Nov. 10, 2005Charmonium Then and Now40 19.Let the amplitudes for s-wave and d-wave be A s and A d be normalized so, in keV. We introduce a mixing angle  so Also

41. Nov. 10, 2005Charmonium Then and Now41 20. Since J/  and  ’ are both C=-1 states, the  state must be C=+1, i.e. even angular momentum. To conserve parity, the other angular momentum must then be even, also. In addition, the two orbital angular momenta can’t differ by more than 2. Data indicate that besides the pure s-wave, there is an admixture of L(  )=2, K=0. Note that p-wave is not allowed because of C (and isospin), even though the m  distribution may suggest it.

42. Nov. 10, 2005Charmonium Then and Now42 X(3872) 21. If X decayed to it would be broad like the  (3770) and the BF to would be tiny. 22. Parity would then prevent the decay to. If X is charmonium, isospin requires that the  be I=0 and thus C=+, so X has C= -. 23. 3 D 2, 1 P 1 24.

43. Nov. 10, 2005Charmonium Then and Now43 X(3872) Contradictions We expect the  to be in an even partial wave by isospin and thus to have C=+1. Then X should have C=-1. But the decay is impossible. If the X is charmonium is has a well specified G- parity. But since J/  also has pure G-parity (-1), there can’t be decays of X to both J/   and J/  .

44. Nov. 10, 2005Charmonium Then and Now44 The actual eigenstates would be C-eigenstates. The simplest state would be an s-wave giving C=  1, J=1, P=+. Bound State?

45. Nov. 10, 2005Charmonium Then and Now45 ISR For dimensional reasons, we need m e. To get convergence change x -1 to x  -1. The scale for the radiation must have an  in it.

46. Nov. 10, 2005Charmonium Then and Now46 3.6864.26 Events12,000125 x0.8790.838 W(s,x)/  0.5640.594 M3.6864.26  0.1080.157  out BF in 666 eV5.2 eV

47. Nov. 10, 2005Charmonium Then and Now47 Limit on  (ee) The peak cross section for the BW: Peak in units of R: Suppose this is less than 0.5 For comparison