Continuing Studies in Particle Physics: Charmonium Then and Now R. Cahn LBNL Nov. 10, 2005, Nov. 17,...

Continuing Studies in Particle Physics:Charmonium Then and Now

R. Cahn

LBNLNov. 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.

Nov. 10, 2005 Charmonium Then and Now 2

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:€

1S0,3S1

€

2S +1LJ

€

C = (−1)L +S , P = (−1)L +1

€

Γ(1S0) =4α 2

M 2R(0)

2; Γ(3S1) =

16

9π(π 2 − 9)

α 3

M 2R(0)

2

€

R(0)2

= 4π ψ (0)2 Wave function at origin

squared = density


November Revolution (1974 version)

€

σ =2J +1

(2S1 +1)(2S2 +1)

4π

k 2

ΓinΓout /4

(E − E0)2 + Γtot2 /4

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 ?€

→ e+e−

€

cc

€

3S1

Why is resonance so asymmetrical?


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?

€

m(J /ψ ) = 3.091GeV

m(ψ (2S)) = 3.686 GeV


G parity and hadronic decays

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

€

G = Ce−iπI y

€

G(π ) = −1

€

η → 3π

1.5x10-4 1.5x10-2

4 x10-3 3.4x10-2

4 x10-3 2.9x10-2€

(π +π −)

€

2(π +π −)

€

3(π +π −) €

(π +π −)π 0

€

2(π +π −)π 0

€

3(π +π −)π 0

12. Discuss the branching fractions above.


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.

€

ηc (0−+),ψ (1−−), χ c 0(0++), χ c1(1++), χ c2(2++)

14. Which one of these is problematic? Why?

The measured widths in MeV are, respectively,

34(BaBar), 0.09, 10, 0.9, 2.1.


€

(3770)

15. Why is the (3770) so much broader than the (3686)?

16. What is the likely assignment for (3770) for (S,L,J)?


3D1 - 3S1 Mixing

• The spin-dependent potential for charmonium is

• Where

17. Which interaction can mix 3D1 and 3S1?

€

V0 +3

2m2

1

r

dV0

drL • S +

1

12m2S12

1

r

dV0

dr−

d2V0

dr2

⎛

⎝ ⎜

⎞

⎠ ⎟+

σ 1 • σ 2

6m2∇ 2V0

€

S12 =3σ 1 • rσ 2 • r −σ 1 • σ 2r

2

r2


Estimated mixing

18. If the (3770) were pure 3D1, 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?

€

Γ(ψ (3770) → e+e−) = 0.033 keV

€

Γ(ψ (3770) → e+e−) = 0.26 keV

Γ(ψ (3686) → e+e−) = 2.1 keV

€

|ψ (3770) >= cosφ | 3D1 > +sinφ | 3S1 >


€

(2S) → J /ψ ππ

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)


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 JP =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?

€

B− → K−X(3872);X → J /ψπ +π −

€

cc€

Γ < 2.3 MeV

€

cc


X(3872) Contradictions

Belle hep-ex/0505037

€

X → J /ψ γ

€

X → J /ψ πππ

25. How do these results conflict with the charmonium picture?


Bound State?

€

D0 D*0

€

m(D+) =1869.4 ± 0.5 MeV

m(D0) =1864.6 ± 0.5 MeV

m(D*+) = 2010.0 ± 0.5 MeV

m(D*0) = 2006.7 ± 0.5 MeV

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 JPC?

€

D0 D*0


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.

€

(s)dx

x


Observation of Structure at 4.26 GeV


€

dσ ISR

dx= W (s, x)σ e +e−(s(1− x))

x =1−M 2

s

W (s, x) = β (x β −1 −1+ x /2)

β =2α

π2ln

s

me

−1 ⎡

⎣ ⎢

⎤

⎦ ⎥= 0.087

σ res =12π 2

MΓoutBFinδ(s(1− x) − M 2)

(3686) Y(4260)

Nevts 12,000 125

0.108 0.157

Γ(ee) 2.1 keV ?

BF( +-) 0.317 ?

Determine for Y(4260): BF( +-) x Γ(ee)


Limit on Γ(ee)

€

4πα 2

3sOne 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(Yee). What is then the lower limit on Γ(Y +)? Compare to Γ(’ +).


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,

Γin,out are the partial widths of the resonance into the initial and final states. Γtot 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 .

€

σ =2J +1

(2S1 +1)(2S2 +1)

4π

k 2

ΓinΓout /4

(E − E0)2 + Γtot2 /4


2. Integral of Breit Wigner

€

dE∫ σ = dE∫ 2J +1

(2S1 +1)(2S2 +1)

4π

k 2

ΓinΓout /4

(E − E0) + Γtot2 /4

=2J +1

(2S1 +1)(2S2 +1)

4π

4k 2ΓinΓout

2π

Γtot

If J=1, S=1/2

€

dE∫ σ =6π 2

s

ΓinΓout

Γtot


3.Width of the

€

€

10 nb GeV =6π 2

(3.09GeV )2Γee 0.389 mb GeV 2

4.3 keV = Γee

72 keV = Γtot

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


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 QCDPotentialAt 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).


6. Spectra without spin

Coulomb Harmonic oscillator

QCD

1s 1s

1s

2s 2s1p

1p

1p2s

1d


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.

€

ω(nx + ny + nz +3

2)


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.


7. Spectra with spin

ηc(1S0)

ηc(1S0)

(3S1)

(3S1)

(3D1)

c0(3P0)

c1(3P1)hc(1P1)

c2(3P2)


8. Radiative Transitions

ηc(1S0)

ηc(1S0)

(3S1)

(3S1)

(3D1)

c0(3P0)

c1(3P1)hc(1P1)

c2(3P2)

E1

M1

Suppressed M1

Obscure and not shown:

€

h1c → χ c1γ (M1)

h1c → χ c0γ (M1)


9. Hadronic Transitions to

ηc(1S0)

ηc(1S0)

(3S1)

(3S1)

(3D1)

c0(3P0)

c1(3P1)hc(1P1)

c2(3P2)

‚η0

0

0

Also

€

ηc '→ χ c 0π0

χ c0,c2 →η cπ0

0

00


Branching Fractions from RPP


hep-ex/0508037 (CLEO)


10. G parity

• C has the effect of complex conjugation

• If +=(1+i2)/√2, then -=(1-i2)/√2.

0= 3

• e-iIy is a rotation by about the y axis, so 1-1, 2 2, 3-3

• G(+)=- +, etc.


11. Decays of the η• 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 η, which is obviously

electromagnetic.• Because isospin is not an exact symmetry, we can think of the

0 and η as each being mixtures of I=0 and I=1. Then the decay η3 proceeds through this mixing.

η


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.


13. Charmonium decay widths

• The decay of the ηc 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

€

Γ∝ s

2

M 4| R /(0) |2

€

Γ∝ s

3

M 2| R(0) |2€

Γ∝ s

2

M 2| R(0) |2


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 η and let the two photons have polarization vectors 1 and 2 and momenta k1 and k2=-k1.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.€

η • (ε1 ×ε2)€

€

η • ((k1 − k2) × (ε1 ×ε2))


• 15. (3770) is above threshold for the decay to D+D-

• 16. Only 3D1 and 3S1 are possible and the (3770) is too close to (3686) to be another 3S1, so 3D1 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 3D1 to 3S1.

• 18. The radial wave function varies as r2 near the origin, so

€

Γ(e+e−)∝α 2 | R' '(0) |2

M 6


• 19.Let the amplitudes for s-wave and d-wave be As and Ad be normalized so , in keV. We introduce a mixing angle so

• Also

€

Γ=| A |2

€

| cosφ AS + sinφ AD |2= 2.1

| −sinφ AS + cosφ AD |2= 0.26

€

| AD |2= 0.033

€

AS2 + AD

2 = 2.1+ 0.26

cos2φ(AS2 − AD

2 ) + 2sin2φ AS AD = 2.1− 0.26

cos2φcos0.235 + sin2φ sin0.235 = cos(0.677)

ϕ = −13o, 26o


• 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.

€

(2S) → J /ψ ππ


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. 3D2, 1P1

• 24.

€

DD

€

J /ψππ

€

DD

€

3D2 → χ c1,c 2γ; 1P1 → η cγ


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/ .

€

X → J /ψ γ


• The actual eigenstates would be C-eigenstates.

• The simplest state would be an s-wave giving C=1, J=1, P=+.

€

D0 D*0

€

±D 0D*0

€

D0 D*0

Bound State?


ISR

€

( s − q,r q )2 = M 2

s − 2 sq + q2 −r q 2 = M 2

s 1−q

s /2

⎛

⎝ ⎜

⎞

⎠ ⎟= s(1− x) = M 2

x =1−M 2

s=1−

3.092

10.582= 0.91

For dimensional reasons, we need me. To get convergence change x-1 to x -1 . The scale for the radiation must have an in it.

€

01 dx β∫ x β −1 =1 (nice!)

β =2α

π2ln

s

me

−1 ⎛

⎝ ⎜

⎞

⎠ ⎟


€

Γ(Y → J /ψπ +π −) × BF(Y → e+e−)

€

#events∝W (s, x)Γ(→ J /ψπ +π −)BF(→ e+e−)

Mε

3.686 4.26

Events 12,000 125

x 0.879 0.838

W(s,x)/ 0.564 0.594

M 3.686 4.26

0.108 0.157

Γout BFin 666 eV 5.2 eV

€

666 ×4.26

3.69×

0.108

0.157×

0.564

0.594×

12

0.125= 5.2


Limit on Γ(ee)

The peak cross section for the BW:

€

4π

k 2

3

4BFinBFout ≅

12π

sBFin

Peak in units of R:

€

9

α 2BRin

Suppose this is less than 0.5

€

BFin < 3×10−6, Γ(Y → J /ψπ +π −) >1.7MeV

For comparison

€

Γ(ψ '→ J /ψπ +π −) = 0.089MeV

Continuing Studies in Particle Physics: Charmonium Then and Now R. Cahn LBNL Nov. 10, 2005, Nov. 17,...

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Transcript of Continuing Studies in Particle Physics: Charmonium Then and Now R. Cahn LBNL Nov. 10, 2005, Nov. 17,...