1. Particle Physics II – CP violation (also known as “Physics of Anti-matter”) Lecture 4
N. Tuning
Niels Tuning (1)

2. Plan Wed 9 Feb: Anti-matter + SM
Mon 13 Feb: CKM matrix + Unitarity Triangle
Wed 15 Feb: Mixing + Master eqs. + B0J/ψKs
Wed 15 Mar: CP violation in B(s) decays (I)
Mon 20 Mar: CP violation in B(s) and K decays (II)
Wed 22 Mar: Exam
Final Mark:
if (mark > 5.5) mark = max(exam, 0.8*exam + 0.2*homework)
else mark = exam
In parallel: Lectures on Flavour Physics by prof.dr. R. Fleischer
Tuesday + Thrusday
Niels Tuning (2)

4. Recap uI W dI u W d,s,b Diagonalize Yukawa matrix Yij Mass terms
Quarks rotate
Off diagonal terms in charged current couplings u
d,s,b W
Replace the derivative with the covariant derivative introducing the force carriers: gluons, weak vector bosons and photon.
The constants g_s, g and g’ are the corresponding coupling constants of the gauge particles.
Niels Tuning (4) 4

5. CKM-matrix: where are the phases?
Possibility 1: simply 3 ‘rotations’, and put phase on smallest:
Possibility 2: parameterize according to magnitude, in O(λ): u
d,s,b W
Niels Tuning (5)

6. This was theory, now comes experiment
We already saw how the moduli |Vij| are determined
Now we will work towards the measurement of the imaginary part
Parameter: η
Equivalent: angles α, β, γ .
To measure this, we need the formalism of neutral meson oscillations…
Niels Tuning (6)

7. Neutral Meson Oscillations (1)
Start with Schrodinger equation:
Find eigenvalue:
Solve eigenstates:
Eigenstates have diagonal Hamiltonian: mass eigenstates!
(2-component state in
P0 and P0 subspace)
Niels Tuning (7)

8. Neutral Meson Oscillations (2)
Two mass eigenstates
Time evolution:
Probability for |P0>  |P0> !
Express in M=mH+mL and Δm=mH-mL  Δm dependence

9. Meson Decays Formalism of meson oscillations: Subsequent: decay P0 f
Interference
P0 f
P0P0 f
Interference
(‘direct’) Decay

10. Classification of CP Violating effects
CP violation in decay
CP violation in mixing
CP violation in interference
Niels Tuning (10)

11. Now: Im( λf) CP violation in decay CP violation in mixing
CP violation in interference
We will investigate λf for various final states f

12. CP violation: type 3
Niels Tuning (12)

13. Classification of CP Violating effects - Nr. 3:
Consider f=f :
If one amplitude dominates the decay, then Af = Af
CP violation in interference
Niels Tuning (13)

14. |λf|=1 Final state f : J/ΨKs Interference between B0→fCP and B0→B0→fCP
For example: B0→J/ΨKs and B0→B0→ J/ΨKs
Lets’ s simplify …
For B0 we have:
Since fCP =fCP we have:
The amplitudes |A( B0→J/ψKs)| and |A(B0→J/ψKs)| are equal:
|λf|=1 B0 d B0
Niels Tuning (14)

15. Relax: B0J/ΨKs simplifies…
ΔΓ=0
Niels Tuning (15)

16. λf for B0 ® J/yK0S c J/y K0 B0 b s d b d s d
|Vcs Vcd|^2 mc^2 = (1*lambda)^2 mc^2 vs. |Vts Vtd|^2 mt^2 = (lambda^2*lambda^3)^2 mt^2  suppressed by lambda^8, enhanced by (mt/mc)^2 << 1 s d
Niels Tuning (16)

17. Time-dependent CP asymmetry
λf for B0 ® J/yK0S
Time-dependent CP asymmetry
Theoretically clean way to measure b
Clean experimental signature
Branching fraction: O(10-4)
“Large” compared to other CP modes!
|Vcs Vcd|^2 mc^2 = (1*lambda)^2 mc^2 vs. |Vts Vtd|^2 mt^2 = (lambda^2*lambda^3)^2 mt^2  suppressed by lambda^8, enhanced by (mt/mc)^2 << 1
Niels Tuning (17)

18. Remember: C and P eigenvalue
C and P are good symmetries (not involving the weak interaction)
Can associate a conserved value with them (Noether Theorem)
Each hadron has a conserved P and C quantum number
What are the values of the quantum numbers
Evaluate the eigenvalue of the P and C operators on each hadron P|y> = p|y>
What values of C and P are possible for hadrons?
Symmetry operation squared gives unity so eigenvalue squared must be 1
Possible C and P values are +1 and -1.
Meaning of P quantum number
If P=1 then P|y> = +1|y> (wave function symmetric in space) if P=-1 then P|y> = -1 |y> (wave function anti-symmetric in space)
Niels Tuning (18)

19. Remember: P eigenvalues for hadrons
The p+ meson
Quark and anti-quark composite: intrinsic P = (1)*(-1) = -1
Orbital ground state  no extra term
P(p+)=-1
The neutron
Three quark composite: intrinsic P = (1)*(1)*(1) = 1
P(n) = +1
The K1(1270)
Quark anti-quark composite: intrinsic P = (1)*(-1) = -1
Orbital excitation with L=1  extra term (-1)1
P(K1) = +1
Meaning: P|p+> = -1|p+>
Niels Tuning (19)

20. Intermezzo: CP eigenvalue
Remember:
P2 = 1 (x  -x  x)
C2 = 1 (ψ  ψ  ψ )
 CP2 =1
CP | f > =  | f >
Knowing this we can evaluate the effect of CP on the K0
CP|K0> = -1|K0>
CP| K0> = -1|K0 >
K0 is not CP eigenstate, but flavour eigenstate (sd) !
Mass eigenstates:
|KS> = p| K0> +q|K0>
|KL> = p| K0> - q|K0>
|Ks> (CP=+1) → p p (CP= (-1)(-1)(-1)l= =+1)
|KL> (CP=-1) → p p p (CP = (-1)(-1)(-1)(-1)l=0 = -1)
( S(K)=0  L(ππ)=0 )
Niels Tuning (20)

21. CP eigenvalue of final state J/yK0S
CP |J/y> = +1 |J/y>
CP |K0S> = +1 |K0S>
CP |J/yK0S> = (-1)l |J/yK0S>
( S(J/y)=1 )
( S(B)=0  L(J/yK0S)=1 )
Relative minus-sign between
state and CP-conjugated state:
|Vcs Vcd|^2 mc^2 = (1*lambda)^2 mc^2 vs. |Vts Vtd|^2 mt^2 = (lambda^2*lambda^3)^2 mt^2  suppressed by lambda^8, enhanced by (mt/mc)^2 << 1
Niels Tuning (21)

22. Time dependent CP violation
If final state is CP eigenstate then 2 amplitudes (w/o mixing):
B0f and B0B0f
B0 - B0 oscillation is periodic in time  CP violation time dependent
Amplitude 1
Amplitude 2 +
e-iφ
(i=eiπ/2 δ=90o)
(φ=2β)
Niels Tuning (22)

23. Time dependent CP violation
If final state is CP eigenstate then 2 amplitudes (w/o mixing):
B0f and B0B0f
B0 -B0 oscillation is periodic in time  CP violation time dependent 1 +
e-iφ
(i=eiπ/2 δ=90o)
(φ=2β)
Niels Tuning (23)

24. Sum of 2 amplitudes: sensitivity to phase
What do we know about the relative phases of the diagrams?
B0  f
B0  B0  f
f(strong)=f
Decays are identical
f(strong)=f
K0 mixing exactly cancels Vcs
f(weak)=0
f(weak)=2b
f(mixing)=p/2
There is a phase difference of i between the B0 andB0
Niels Tuning (24)

25. Sum of 2 amplitudes: sensitivity to phase
Now also look at CP-conjugate process
Investigate situation at time t, such that |A1| = |A2| :
Directly observable result (essentially just from counting) measure CKM phase b directly!
N(B0  f)  |A|2  (1-cosf)2+sin2f
= 1 -2cosf+cos2f+sin2f
= 2-2cos(p/2-2b)
 1-sin(2b) + =
Γ(Bf)=
sin(f)
p/2+2b CP
1-cos(f) + =
sin(f)
Γ(Bf)=
p/2-2b
N(B0  f)  (1+cosf)2+sin2f
= 2+2cos(p/2-2b)
 1+sin(2b)
1+cos(f)
Niels Tuning (25)

26. 2 amplitudes 2 phases Remember!
Necessary ingredients for CP violation:
Two (interfering) amplitudes
Phase difference between amplitudes
one CP conserving phase (‘strong’ phase)
one CP violating phase (‘weak’ phase)
2 amplitudes
2 phases
Niels Tuning (26)

27. Remember!
2 amplitudes
2 phases
Niels Tuning (27)

28. eiφ e-iφ What do we measure? Γ( B f) = |A1+A2ei(φ+δ)|2
We measure quark couplings
There is a complex phase in couplings!
Visible when there are 2 amplitudes:
matter
anti-matter
Γ( B f) = |A1+A2ei(φ+δ)|2
Γ( B f) = |A1+A2ei(-φ+δ)|2
Proof of existence of complex numbers…
Niels Tuning (28)

29. B0→J/ψKs B-system - Time-dependent CP asymmetry BaBar (2002)
Niels Tuning (29)
BaBar (2002)

30. Consistency with other measurements in (r,h) plane
4-fold ambiguity because we measure sin(2b), not b
Prices measurement of sin(2β) agrees perfectly with other measurements and CKM model assumptions The CKM model of CP violation experimentally confirmed with high precision! 2 1
without sin(2b) h 3 4 r
Method as in Höcker et al, Eur.Phys.J.C21: ,2001
Niels Tuning (30)

31. βs: Bs0 ® J/yφ : Bs0 analogue of B0 ® J/yK0S
Replace spectator quark d  s
Niels Tuning (31)

32. βs: Bs0 ® J/yφ : Bs0 analogue of B0 ® J/yK0S
Niels Tuning (32)

33. Remember: The “Bs-triangle”: βs
Replace d by s:
Niels Tuning (33)

34. βs: Bs0 ® J/yφ : Bs0 analogue of B0 ® J/yK0S
Differences: B0
B0s
CKM
Vtd
Vts ΔΓ ~0
~0.1
Final state (spin)
K0 : s=0
φ: s=1
Final state (K)
K0 mixing -
Niels Tuning (34)

35. βs: Bs0 ® J/yφ A║ A┴ A0 B0 B0s CKM Vtd Vts ΔΓ ~0 ~0.1
Vts large, oscilations fast,
need good vertex detector A║ A0 A┴
l=2
l=1
l=0
3 amplitudes B0
B0s
CKM
Vtd
Vts ΔΓ ~0
~0.1
Final state (spin)
K0 : s=0
φ: s=1
Final state (K)
K0 mixing -
Niels Tuning (35)

36. “Recent” excitement (5 March 2008)
Niels Tuning (36)

37. Bs  J/ψФ : Bs equivalent of B J/ψKs !
The mixing phase (Vtd): φd=2β
B0  f
B0  B0  f
Wolfenstein parametrization to O(λ5):
Niels Tuning (37)

38. Bs  J/ψФ : Bs equivalent of B J/ψKs !
The mixing phase (Vts): φs=-2βs
B0  f
B0  B0  f
Vts - s Ф Bs Bs Ф s s s
Vts
Wolfenstein parametrization to O(λ5):
Niels Tuning (38)

39. Bs  J/ψФ : Bs equivalent of B J/ψKs !
The mixing phase (Vts): φs=-2βs
B0  f
B0  B0  f
Vts - s Ф Bs Bs Ф s s s
Vts
Niels Tuning (39)

40. Measure γ: B0s  DsK-/+ : both λf and λf2 +
Γ(Bf)= 2 +
Γ(B f )=
NB: In addition B s  DsK-/+ : both λ f and λf
Niels Tuning (40)

41. Measure γ: Bs  DsK-/+ --- first one f: Ds+K-
This time | Af||Af|, so |λ|1 !
In fact, not only magnitude, but also phase difference:
Niels Tuning (41)

42. Measure γ: Bs  DsK-/+ B0s  Ds-K+ has phase difference ( - ):
Need B0s  Ds+K- to disentangle  and :
Niels Tuning (42)

43. Next CP violation in decay CP violation in mixing
CP violation in interference

44. γ B - → D0 K- (Time integrated) Bs0 → DsK+- (Time dependent)
( )
B - → D0 K- (Time integrated)
Bs0 → DsK+- (Time dependent)
Necessary ingredients for CP violation:
Two (interfering) amplitudes
Phase difference between amplitudes
one CP conserving phase (‘strong’ phase)
one CP violating phase (‘weak’ phase)

45. γ (GLW) D0K+ fCPK+ B+ GLW: CP eigenstate: D0→ K+K-(ππ) bc favoured
Vub*
Vcb*
( )
B- →D0K-
Relative phase: γ B+
D0K+
fCPK+
bc favoured
bu suppressed
GLW:
CP eigenstate: D0→ K+K-(ππ)

46. γ (GLW) D0K+ fCPK+ B+ GLW: CP eigenstate: D0→ K+K-(ππ) bc favoured
Vub*
Vcb*
( )
B- →D0K-
Relative phase: γ B+
D0K+
fCPK+
bc favoured
bu suppressed
GLW:
CP eigenstate: D0→ K+K-(ππ)

47. γ (GLW) D0K+ fCPK+ B+ GLW: CP eigenstate: D0→ K+K-(ππ) bc favoured
bu suppressed
GLW:
CP eigenstate: D0→ K+K-(ππ)

48. γ (ADS) D0K+ B+ ADS: K-π+K+ B or D Cabibbo favoured: D0→ K+π-
Vub*
Vcb*
( )
B- →D0K-
Relative phase: γ B+
D0K+
bc favoured
bu suppressed
K-π+K+
Cabibbo supp.
Cabibbo allowed.
ADS:
B or D Cabibbo favoured: D0→ K+π-

49. γ (ADS) D0K+ B+ ADS: K-π+K+ B or D Cabibbo favoured: D0→ K+π-
Vub*
Vcb*
( )
B- →D0K-
Relative phase: γ B+
D0K+
bc favoured
bu suppressed
K-π+K+
Cabibbo supp.
Cabibbo allowed.
ADS:
B or D Cabibbo favoured: D0→ K+π-
“Difference in suppression”
“Average suppression”

50. γ (ADS) D0K+ B+ ADS: K-π+K+ B or D Cabibbo favoured: D0→ K+π-
Suppressed mode for the B- is relatively more suppressed than for the B+… B+
D0K+
bc favoured
K-π+K+
Cabibbo supp.
Cabibbo allowed.
bu suppressed
ADS:
B or D Cabibbo favoured: D0→ K+π-
BR~ 2 x 10-7

51. CP violation in Decay? (also known as: “direct CPV”)
First observation of Direct CPV in B decays (2004):
BABAR
hep-ex/
Phys.Rev.Lett.93:131801,2004
4.2s
BABAR
HFAG:
ACP = ± 0.012
Niels Tuning (51)

52. CP violation in Decay? (also known as: “direct CPV”)
First observation of Direct CPV in B decays at LHC (2011):
LHCb
LHCb-CONF
LHCb
Niels Tuning (52)

53. 2 amplitudes 2 phases Remember!
Necessary ingredients for CP violation:
Two (interfering) amplitudes
Phase difference between amplitudes
one CP conserving phase (‘strong’ phase)
one CP violating phase (‘weak’ phase)
2 amplitudes
2 phases
Niels Tuning (53)

54. Direct CP violation: Γ( B0 f) ≠ Γ(B0f )
CP violation if Γ( B0 f) ≠ Γ(B0f )
But: need 2 amplitudes  interference
Amplitude 1
Amplitude 2 +
Only different if both δ and γ are ≠0 !
 Γ( B0 f) ≠ Γ( B0f )
Niels Tuning (54)

55. Hint for new physics? B0Kπ and BKπ0
Redo the experiment with B instead of B0…
d or u spectator quark: what’s the difference ??
B0Kπ
Average
3.6s ?
B+Kπ
Average
Niels Tuning (55)

56. Hint for new physics? B0Kπ and BKπ0
Niels Tuning (56)

57. Hint for new physics? B0Kπ and BKπ0
T (tree)
C (color suppressed)
P (penguin)
B0→K+π-
B+→K+π0
Niels Tuning (57)

58. Next CP violation in decay CP violation in mixing
CP violation in interference

59. CP violation in Mixing? (also known as: “indirect CPV”: ε≠0 in K-system)
Look for like-sign
lepton pairs:
t=0 t
Decay
gVcb*
gVcb
Niels Tuning (59)

60. (limit on) CP violation in B0 mixing
Look for a like-sign
asymmetry:
As expected, no
asymmetry is observed…
Niels Tuning (60)

61. 2 amplitudes 2 phases Remember!
Necessary ingredients for CP violation:
Two (interfering) amplitudes
Phase difference between amplitudes
one CP conserving phase (‘strong’ phase)
one CP violating phase (‘weak’ phase)
2 amplitudes
2 phases
Niels Tuning (61)

62. CP violation in Bs0 Mixing??
D0 Coll.,
Phys.Rev.D82:032001,2010.
arXiv:
CP violation in Bs0 Mixing?? b s
“Box” diagram: ΔB=2
φsSM ~ 0.004
φsSMM ~ 0.04
Niels Tuning (62)

63. CP violation from Semi-leptonic decays
SM: P(B0s→B0s) = P(B0s←B0s)
DØ: P(B0s→B0s) ≠ P(B0s←B0s) ?
b→Xμ-ν, b→Xμ+ν
b→b → Xμ+ν, b→ b → Xμ-ν
Compare events with like-sign μμ
Two methods:
Measure asymmetry of events with 1 muon
Measure asymmetry of events with 2 muons
Switching magnet polarity helps in reducing systematics
But…:
Decays in flight, e.g. K→μ
K+/K- asymmetry

64. CP violation from Semi-leptonic decays
SM: P(B0s→B0s) = P(B0s←B0s)
DØ: P(B0s→B0s) ≠ P(B0s←B0s) ?
B0s→Ds±X0μν

65. More β…
Niels Tuning (65)

66. Time?
Niels Tuning (66)

67. Other ways of measuring sin2β
Need interference of bc transition and B0 –B0 mixing
Let’s look at other bc decays to CP eigenstates:
All these decay amplitudes have the same phase
(in the Wolfenstein parameterization)
so they (should) measure the same CP violation
Niels Tuning (67)

68. CP in interference with BφKs
Same as B0J/ψKs :
Interference between B0→fCP and B0→B0→fCP
For example: B0→J/ΨKs and B0→B0→ J/ΨKs
For example: B0→φKs and B0→B0→ φKs
Amplitude 1
Amplitude 2 +
e-iφ
Niels Tuning (68)

69. CP in interference with BφKs: what is different??
Same as B0J/ψKs :
Interference between B0→fCP and B0→B0→fCP
For example: B0→J/ΨKs and B0→B0→ J/ΨKs
For example: B0→φKs and B0→B0→ φKs
Amplitude 1
Amplitude 2 +
e-iφ
Niels Tuning (69)

70. Penguin diagrams
Nucl. Phys. B131:
Niels Tuning (70)

71. Penguins?? The original penguin: A real penguin: Our penguin:
Niels Tuning (71)

72. Funny Flying Penguin Dead Penguin Penguin T-shirt: Super Penguin:
Niels Tuning (72)

73.  The “b-s penguin” Asymmetry in SM B0J/ψKS B0φKS b sμ
“Penguin” diagram: ΔB=1
… unless there is new physics!
New particles (also heavy) can show up in loops:
Can affect the branching ratio
And can introduce additional phase and affect the asymmetry
Niels Tuning (73)

74. ? Hint for new physics?? B J/ψ Ks φ Ks B sin2β sin2βpeng g,b,…? ~~d c s φ Ks B s b d t ?
g,b,…? ~~
sin2βbccs = ± 0.03
sin2βpeng = ± 0.05
sin2β
sin2βpeng
Niels Tuning (74)
S.T’Jampens, CKM fitter, Beauty2006