Particle Physics 2002/2003 Part of the “Particle and Astroparticle Physics” Master’s Curriculum VI. Discrete Symmetries and CP violation Particle Physics I troduction, history & overview (2) ncepts (5): Units (h=c=1) Relativistic kinematics Cross section, lifetime, decay width, … Symmetries (quark model, …) antum Electro Dynamics: QED (7) Spin 0 electrodynamics (Klein-Gordon) Spin ½ electrodynamics (Dirac) Experimental highlights: “g-2”, e e , … Particle Physics II IV. Quantum Chromo Dynamics: QCD (4) • Colour concept and partons • High q 2 strong interaction • Structure functions • Experimental highlights: s , e p, … V. Quantum Flavour Dynamics: QFD (6) • Low q 2 weak interaction • High q 2 weak interaction • Experimental highlights: LEP VI. Origin of matter? (6) • Strange particles • GIM (why does the charm exist?) • K 0 -K 0, oscillations, • CP violation • B 0 -B 0 oscillations • Current CP violation experiment VII. Origin of mass? (2) • Symmetry breaking • Higgs particle: in e e and in pp File on “paling”: graven/ED_MASTER/master2003.ppt
Transcript
Particle Physics 2002/2003 Part of the “Particle and Astroparticle Physics” Master’s Curriculum
VI. Discrete Symmetries and CP violation
Particle Physics I I. Introduction, history & overview (2)II. Concepts (5):
III. Quantum Electro Dynamics: QED (7)• Spin 0 electrodynamics (Klein-Gordon)• Spin ½ electrodynamics (Dirac)• Experimental highlights: “g-2”, ee, …
Particle Physics II IV. Quantum Chromo Dynamics: QCD (4)
• Colour concept and partons• High q2 strong interaction• Structure functions• Experimental highlights: s, ep, …
V. Quantum Flavour Dynamics: QFD (6)• Low q2 weak interaction• High q2 weak interaction
• Experimental highlights: LEPVI. Origin of matter? (6)
• Strange particles• GIM (why does the charm exist?)• K0-K0, oscillations,• CP violation• B0-B0 oscillations• Current CP violation experiment
VII. Origin of mass? (2)• Symmetry breaking• Higgs particle: in ee and in pp
File on “paling”: graven/ED_MASTER/master2003.ppt
Strange Particles
Isospin
“Strangeness”
( )K su 0 ( )K sd( )K su0 ( )K sd
+1-1
-1 +1
0 0p K 0 0
p K n n 0
p K K p
E threshold (GeV) 0.91 1.5 6.0
In general: more difficult (i.e. higher threshold) to make S=-1 particles then S=+1 when target is either p or n, and beam consists of + or -
because for S=+1, need to get an antiquark from somewhere…
Quite useful: can make ‘pure’ K0 or K+ sample by running below threshold
Note: long lifetim
e!
Observations:1. High production cross-
section2. Long lifetimeConclusion:
must always be produced in pairs!
Details: create a new quantum number, “strangeness“
which is conserved by the production process
hence pair production by strong force
however, the decay must violate “strangeness”
if only weak force is “strangeness violating” then it
is responsible for the decay process
hence (relatively) long lifetime…
“V particle”: particles that are producedin pairs and thus leaves a ‘v’ trial in a bubble chamber picture
Strange Particles
mK ~ 494 MeV/c2
1. No strange particles lighter than Kaons exist• Decay must violate “strangeness”
2. Strong force conserves “strangeness”• Decay is a pure weak interaction
Isospin
“Strangeness”
( )K su 0 ( )K sd( )K su0 ( )K sd
+1-1
-1 +1
0 0 0
0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
, ,
, ,
, , ,
, , ,
Hadronic decays:
K
K
K
K
0
0
,
,
,
,
leptonic decays:
e
e
K e
K e
K e e
K e e
0 0
0 0
0
0
,
,
,
,
Semi-leptonic decays:
e
e
e
e
K e
K e
K e
K e
Semi-leptonic decays:• particle and anti-particle are
distinct from one another!• “Q=S rule”
Hadronic and leptonic decays:• particle and anti-particle
behave the same
Weak decays: K- vs. -
2
0.07546
tan c
22 23
3 2 2
17.671
K
K
m mm
m m m
2
5 1 1 5 2 2
22 3 2 3
22 2 2 2 21 2 1 2
1 22 22 2
1 2 1 2
1 2
2 12 2
1 1
4 2
4
2
132
in CM frame:
P PTr p m p m
P p P p P p p
M m m m m
M m m M m mp p
M
pd d
M
M
M
,MP
1,m1p
2 2,mp
1.3334
K
K
22 23
3 2 2
K
K
m mm
m m m
Universality of weak interactions
4
4 2
0 4 2
cos 0
sin 1
purely leptonic
semi-leptonic,
semi-leptonic,
e
e C
e C
e g
n pe g S
pe g S
, ,e LL L
e u
d
Weak doublets:
cos sinc cL L
uu
d sd
cos Cg g sin Cg
K0 decays: A problem?
cos sinc cL L
uu
d sd
0.6351 0.0018Br K
Z0 boson will now couple to uu and d’d’ ...
cos sinc c
sin c cos sinC Cs s d
;
cos sinwith
sin cos
C
c cC
c c
d dV
s s
V
0 9(7.25 0.16)10LBr K
This generates a “FCNC”, (Flavour Changing Neutral Current)… need to do more.
cos sin sin cos 0!c c c c
Or, to put it the other way around:The absence of FCNC requires V to be unitary
† *TC C C CV V V V 1
the reason there are no FCNC is that VC is unitary:
d d s s dd ss
K0 decays: enter the charmTo (almost) cancel this diagram, lets introduceanother up-type quark, and have it interact through a W with the orthogonal combination of (d,s)
cos sinc cL L
cc
s ds
This new ’c’ quark causes an additional diagramthat (almost) cancels the one above…
If mumc, then the cancellation would be complete! This is called “GIM suppression” leads to a prediction of the charm mass of 1.5—2 GeV, prior to the discovery of J/
,
,
e L L
L L
e
u c
d s
C
d dV
s s
In general: the weak eigenstates are not the mass eigenstates!
If all quarks were the same mass, this could not happen as we could take any linear combination of quarks as the mass eigenstates…
And as long as V is unitary, there will be no FCNC!note: can (and will) extend this to 3 families laterQ: what happens if VC=1 (i.e. C=0)?
A: the s quark (and thus all S0 particles) would be stable!!!
Q: how many independent parameters does V have when there are 2 generations (i.e. is C all there is?). How about 3 generations?A: 2*22-22-(2*2-1)=1; 2*32-32-(2*3-1)=4
Intermezzo: Discovery of the J/
By studying the decay of strange particles, the existence of the charm and its properties(eg. mass, weak couplings) were predicted before its discovery – but nobody believed it!
Bro
okh
aven:
JS
LAC
:
Sam Ting and Burt Richter got the1976 Nobel prize for their discovery
Back to K0 decays…
Phys. Rev. 97, 1387 (1955)
0K
0K
0K0K
0K
0K
Known:1. K0 can decay to +- Hypothesized:1. K0 has a distinct anti-particle K0
Claims:1. K0 (K0) is a “particle mixture” with two distinct lifetimes2. Each lifetime has its own set of decay modes3. No more than 50% of K0 (K0) will decay to +-
K0 and CP symmetryKnown decay:
Assuming CP symmetry, this should be possible as well:
0K
0K
0 0K K Assuming the reversereaction is allowed, particlecan “mix” into anti-particle,and vice-versa…
How does this system evolve in time? (ignore decays for the time being)
0 0 ( )( ) ( ) ( )
( )
a tt a t K b t K
b t
i Ht
0
0K
K
MH
M
,KK
K
MH M
M
Mixing causes tiny off-diagonal element:
0 0
0 0
1,
21
,2
K
K
K K m M
K K m M
0
0
,
,
K
K
K m M
K m M
With completely different eigenstates!
K0 decay and CP: K1 and K2
Phys Rev 103,1901 (1956)
Only the CP even state (K1) can decay into 2 pions(which are CP even)The odd K2 state will decay into 3 particles instead (,, e,…).
There is a huge difference between K0 and K0 in phasespace (~600x!). So the CP even state will decay much faster
0 0
12
K KK
0 0
22
K KK
K1 and K2 are their own antiparticle, but one is CPeven, the other CP odd:
CP: +1 -1
2 vs. 3 particle phase
space
More on time evolution1 1 1 1 1
2 2 2 2 2
1
2
( ) , with ,2
( ) , with .2
i t
i t
iK t e K M
iK t e K M
20 0 0;I K K t g t I
20 0 0;I K K t g t I
1 21 1 2 2( ) i t i tT t e K K e K K
0 0 0 0
0 0 0 0
1 2
( )
( ) ,
( )2
i t i t
g t K K K K
g t K K K K
e eg t
1 22 1
2 cos4
cosh cos2 2
t t t
t
g t e e e m t
e tm t
Tag K0 and K0 decay by semileptonic decay(remember the S=Q rule?)
K1 decays
K2 decays
Testing CP conservation Easy to create a pure K2 beam:
just “wait” until the K1 component has decayed…If CP conserved, should not see the decay to 2 pions in this K2 beamThis is exactely what was tested by Cronin & Fitch in 1957…
Main background: +- 0
K2+-
Effect is tiny: about 2/1000
… and for this experiment they got the Nobel price in 1980…
Interference2 1
2
1 2
2
,1
.1
L
S
K KK
K KK
KL and KS are no longer orthogonal:
0 0
0 0
,
.
S
L
K p K q K
K p K q K
2
2
2 1
2 1
1 ,
1 .
p
q
, , ,
( )
2
S Li t i tS S L L
S L S L S L
T t e K K e K K
im
2 2
2
1 1S LK K
0 0 0 0
0 0 0 0
( )
( )
g t K K K K
p qg t K K K K
q p
0 00 01if then oscillation rate q p K K K K
CPLEAR, Phys.Rep. 374(2003) 165-270
T violation in mixingt=0 t
( )g t
( )qg t
p
0K
0K
0K
( )g t
( )pg t
q
0K
0K
0K
ee
ee
ee
ee
36.6 1.6 10
0.9967 0.0008 1
TA t
q p
4
4
14
1e
ee
TeI t q p
AI t
I t It
qt p
CP
Note: 1. This measurement allows one to make
an ABSOLUTE distinction between matter and anti-matter
2. Don’t need to know the specific value of decay amplitudes; only need:
2 22 20 00 00; 0e K e K e K e K
(2x)2 ways to decay…t=0 t Amplitude
( ) ( )q
g t A g t Ap
( )g t
A
A
( )qg t
p
0K
0K
0K
( )g t
A
A
( )pg t
q
0K
0K
0K
CP
( ) ( )p
g t A g t Aq
( ) ( )A g t g t
1( ) ( )A g t g t
Aq
p A
0
0
A K
A K
3 ways to break CP
A A
1q
p
0
CP violation in decay
CP violation in mixing
CP violation in the interference between mixing and decay
* 12 sin
4
sinh sin2 2
S Lt t t
t
g t g t e e ie m t
e ti m t
1 22 1
2 cos4
cosh cos2 2
t t t
t
g t e e e m t
e tm t
20
20 1
K A g t g t
K A g t g t
3 Ways to break CP
2 220
20 2 2
2
1 12
K g t g t g t g t
K g t g t g t g t
1 2
1 2
220
20 2
2
1 1 21 1 1 cos 2 1 sin
4 1 1 1
1 11 1 21 cos 2 1 sin
4 1 1 1
t t t
t t t
K e e e m t m t
K e e e m t m t
2 220
20 2 2
2
12
K g t g t g t g t
K g t g t g t g t
2 220
0 2 2
2 2
2
1 2
K g t g t g t g t
K g t g t g t g t
2 220
0 2 2
2
2
2
1
2
K g t g t
g t g t g t g t
K g t g t
g t g t g t g t
1 2
1 2
1 2
0
2
0
12 cos
41
2 cos4
1 2 sin
21
4
S L
t t t
t t t
t t t
t
K e e e m t
e e e m t
e e e m t
K e e
1 2
2
2
2 cos
1 1 2 cos
4
1 2 sin
2S L
t t
t t t
t t t
e m t
e e e m t
e e e m t
1 2
1 2
2 20
0
2 2
11 2 1 cos
41
2 sin2
1 1 11 2 1 cos
4
S L
t t t
t t t
t t t
K e e e m t
e e e m t
K e e e m t
2
1 2 sin
2S Lt t te e e m t
1 2
1 2
2 20
0 2 2
2
11 2 1 cos
41
2 sin2
1 11 2 1 cos
4
S L
t t t
t t t
t t t
K e e e m t
e e e m t
K e e e m t
1 2 sin
2S Lt t te e e m t
1 2
1 2
2 2
0
2
2 2
0
2 2
1 2 1 21
4 2 1 cos 2 sin
1 2 1 21 1
4 2 1 cos 2 sin
t t
t
t t
t
e eK
e m t m t
e eK
e m t m t
1 2
1 2
2
2
202
2
2
22
0
2 2
2
1 2
1 211 2
4 1 cos2
1 2 2 sin
1 2
1 21 1 21
4 1 cos2
1 2 2 sin
t t
t
t t
t
e e
Km t
em t
e e
Km t
em t
The Final Result…
0
122
0 like like interfer
2 cos
ence term
S LS L
tt t
SL
K K
Ke e mt e
KK KS L
If =0: only “KS” like decays
If 0: not only “KL” like decays, down by ||
2, but alsointerference contribution, down
by ||
The interference term has a sign difference for K0 and K0bar!
A(KS)
(t)
A(KL)
K0(t=0)
If CP were conserved, KL wouldn’t decay
to , and there would be no interference…
A(KS)
(t)
-A(KL)
K0(t=0)
CP
0
0
,
.
S L
S L
K K K
K K K
CPLEAR Detector@CERNUse the strangeness conservation of the strong interactions to perform Tagged K0 and K0 production:
0
0
0
0
K
K
K K Kp pK K K
At t=0, events with a K+ are a pure K0bar sampleK- are a pure K0 sample
CPLEAR
Results: CP in InterferenceCPLEAR, PLB 1999
(K0-K0)/(K0+K0)
decaytime / S
Approx equal KSand KL rate:Maximal interference!
Mainly KSdecays
Mainly KLdecays
K0barK0
Note: rates are normalized to each other in the range (,)
Interference maximal:
2ln 1
2ln 12
L S
S S L
I K I K
t
Details…
CP violation: when?
P f
Pbar
Introduce A, Abar into the pictureCP violation seems to occurs in interferenceWhat kinds of interference can we have?•Mixing•Decay•Mixing vs. Decay
Basic Equations: Neutral Meson Mixing0 0
1 1 2 2( ) ( ) ( ) ( ) ( ) ...t a t P b t P c t f c t f If• At t=0, only a(t) and b(t) are non-zero• We are only interested in a(t) and b(t), and not ci(t)• t is large compared to the strong-interaction scaleThen one can make an approximation (Wigner-Weisskopf) which considerably simplifies things:
In general, want to know the time evolution of:
00 0 0 0 0ˆ ˆˆ ˆ fi m E t
f
di b t P V P a t P V P b t P V f c t edt
st em
di t H H V tdt
em st
em st
em st
( )ˆ
ˆ( )
( )i i
i H H t
i H H t
i H H t
a t a t e
b t b t e
c t c t e
0 01 1 2 2
ˆ( ) ( ) ( ) ( ) ( ) ...ˆ ˆ ˆ ˆt a t P b t P c t f c t f
ˆ ˆd
i t V t tdt
0 00 0 ˆˆ ˆ ˆg g g fi E m t i E m t i E E t
g ff
di c t g V P a t e g V P b t e g V f c t edt
0
0 0 2
00
ˆˆ ˆlimfi E m t
ff
ec t f V P a t f V P b t V
m E i
O
00 0 0 0 0ˆˆ ˆ ˆ fi m E t
f
di a t P V P a t P V P b t P V f c t edt
Basic Equations: Neutral Meson Mixing
11 12
21 22
a t a t a tdi
b t b t b tdt
Λ
is not Hermitian: otherwise mesons would only oscillate, and never decay… instead:
2
i Λ M Γ
,
.
M M†
†
1( ),
2
( ).i
M Λ Λ
Γ Λ Λ
†
†
00 0 0 0 0ˆ ˆˆ ˆ fi m E t
f
di b t P V P a t P V P b t P V f c t edt
0
0 0 2
00
ˆˆ ˆlimfi E m t
ff
ec t f V P a t f V P b t V
m E i
O
00 0 0 0 0ˆˆ ˆ ˆ fi m E t
f
di a t P V P a t P V P b t P V f c t edt
0
1 1lim i x
x i x
P
00
ij ij ff
i V f f V jM m i V j
m E
P
02ij ffi V f f V j m E
Virtual Intermediate States
Real Intermediate States
Basic Equations: Neutral Meson Mixing
is not Hermitian: otherwise mesons would only oscillate, and never decay… instead:
2
i Λ M Γ
,
.
M M†
†
1( ),
2
( ).i
M Λ Λ
Γ Λ Λ
†
†
M describes oscillations, decays
2 2 * * must be positive definite
a tda t b t a t b t
b tdt
Γ Γ
00
ij ij ff
i V f f V jM m i V j
m E
P
02ij ffi V f f V j m E
Phase conventions0 0
0 0
,
.
i
i
P e P
P e P
( )12 12
( )12 12
( )21 21
( )21 21
,
,
,
.
i
i
i
i
M e M
e
M e M
e
Because of the requirement of phase independence, R has only 7 (physical) parameters
CPT invariance:T invariance:CP invariance:
Requiring CPT reduces this to X parameters (SHOW!)
11 22 11 22,M M M
11 22 11 22,M M
2 221 12 21 12,i iM e M e
2 221 12 21 12,i iM e M e
Solving the master equations
12 12
12 12
2 2
2 2
i iM Ma t a td
ib t b ti idt
M M
i t i ta t C e C e 12 12 12 122 2 2
i i iM M M
12 12 12 122 2i i
q p M M
12 12 12 122 2
i im M M M
12 12 12 1222 2
i iM M
12 12 12 1222 2
i im M M
Computing Dm and DG in K0 mixing
Okun p88?Cahn-Goldhaber, chapter 15
Or: why is kaon mixing so different from B mixingAnd why is D mixing different again???
Actually, why is the B lifetime so large? as expected, the D lifetime is much less than the K0S one
Show that mixing vanishes if all quark masses are equal
i.e. why the helium bag?Or: another way to measure dm
Enter the B meson…
Long lifetime! -> Vcb must be tiny!
Third generation -> VCKM
It mixes! -> top must be VERY heavy
And then there were 3…The CKM matrix
CKM
d d
s V s
b b
Unexpected long B lifetime! => Vcb must be small!
Eg. B+ to mu+ not observed, only limits
Mixing of neutral mesons
D0 mixing vs. Bd mixing
Mixing dominated by Vtd
Not yet observed!Experimental limit goes here
22 *s cs usm V V
22 *t td tbm V V
How do we measure mixing??
Compute Vtd from the measured dm values
Must have heavy (>100 GeV) top!As VtdVtb**2 isn’t very large (0.2**6)
2 22 2 14
26 10
6tB F B
tb td WB W
mm G fV V M S
m m
Bd mixing vs. Bs mixing
Mixing Dominated by Vtd Mixing Dominated by Vts2 6
2-12 4
-1
0.0412 ps
0.4 0.0 ps
tdd
ss ts
d
Vm
mm V
m x x
Not yet observed!Experimental limit goes here
Some other effects of O(30%)lead to the SM expectation of ~18
CP violation in mixing
Why small?Experimental results
Kaons, Bd mesons
Is P a good symmetry?
Parity violation observed in 60Co experiment in 1956.
Magneticfield
e- Parity
transformation
e-
Observed I() = 1 + (v/c) cos
with = -1
60Co 60Co
J J
More electrons emitted opposite the J direction.Parity violation!
1950 ‘60 ‘70 ‘80 ‘90 2000 2010
Parityviolation
C. Yang and T. Lee, 1956C. S. Wu, 1957
Sketch and photograph of apparatus used to study beta decay in polarized cobalt-60 nuclei. The specimen, a cerium magnesium nitrate crystal containing a thin surface layer of radioactive cobalt-60, was supported in a cerium magnesium nitrate housing within an evacuated glass vessel (lower half of photograph). An anthracene crystal about 2 cm above the cobalt-60 source served as a scintillation counter for beta-ray detection. Lucite rod (upper half of photograph) transmitted flashes from the counter to a photomultiplier (not shown). Magnet on either side of the specimen was used to cool it to approximately 0.003 K by adiabatic demagnetization. Inductance coil is part of a magnetic thermometer for determining specimen temperature.
Q: why does eg. K*0 not mix? It has the same quark content…A: it decays to K+pi-, a strong decay – it just isn’t stable enough!A2: it is a vector particle…
How do we make sure KL -> p+p- is really the same final stateAs KS->pipi ?
maybe we’re missing a particle that takes away very little momentum?NOTE: beta decay spectrum was ‘solved’ by introducing a new particle
