Elementary Particle Physics
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Transcript Elementary Particle Physics
Lecture 7
●
Parity
●
Charge conjugation
●
G-parity
●
CP
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Already done a lot to understand the basic particles of nature
Lepton
universality
Strong,
weak,em ?
Isospin
singlets
175000
Isospin
multiplets
small
small
small
Neutrino
oscillations/
mass
Quark
composition
Decay
modes
Symmetry and QM demands much of
what is observed!
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Transforming under parity
Convert the "handedness" of a particle.
Eg right-handed particle left-handed particle.
Wu's experiment which showed parity violation
in weak processes since left-handed antineutrinos
don't exist!
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Ways of thinking about parity
Transformation: r x, y, z r x, y, z
Further complementary ways to think about parity invariance.
(a) Parity transformation is equivalent to making a reflection
and then a rotation. Since nature is invariant to a rotation (angular
momentum conservation), is the mirror image of a process/particle equally
possible.
+
(b)
v
Not possible.
ms=-1/2
ms=1/2
Possible.
real
”virtual”
ˆ
H ; Hˆ x, y, z Hˆ x, y, z ; x, y, z , t x, y, z , t
t
Is x, y, z , t x, y, z , t
2
2
? Is the Hamiltonian invariant ?
Look for a conserved quantity (this lecture).
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Intrinsic Parity
Consider closed system of particles.
Parity transformation is: ri ri ' ri ; ri position vector of a particle at the origin.
Consider single particle (easily generalised to many particles)
Parity operator: Pˆ r , t Pa r , t (7.01) r , t "mirror image" of particle
Pa phase factor, a particle, eg e , u etc.
i p r Et
i p r Et
Eigen function of momentum: r , t e
Pˆ r , t P e
(7.02)
P
P
For a particle at rest p 0, P r , t e iEt
a
ˆ iEt P e iEt
Pˆ P r , t Pe
a
(7.03)
Pˆ 2 P r , t Pa2 P r , t Pa 1 (7.04)
eigen state of Pˆ with eigen value Pa .
A particle has an intrinsic parity Pa .
This is an intrinsic property of a particle, like baryon number or lepton number.
Showed this for particles which can be at rest. General - photons can also have an
intrinsic parity (to come)
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Intrinsic Parity
Let's give our particle a some definite orbital angular momentum: nlm is a parity eigenstate.
nlm r Rnl r Yl m , (7.05)
Y00 ,
r r
1
3
3
; Y10 ,
cos ; Y11 , sin ei (7.06)
4
8
4
r r ; ; (7.07)
3
3
cos
cos
4
4
Y00 ,
1
4
Y11 , -
3
3
3
i
i
sin e ei
sin ei
sin e
8
8
8
; Y10 ,
Yl m , 1 Yl m , (7.08)
l
l
Pˆ nlm r Pa nlm r Pa 1 r nlm r
Parity=Pa 1
l
(7.09).
Generalise to two particles c, d rotating in centre-of-mass frame:
Pc Pd 1 , l =relative orbital angular momentum. (7.10)
l
For a Hamiltonian invariant under parity: Pˆ , Hˆ 0 (7.11)
We have a conserved quantity with which we can study parity violation (weak)/invariance (strong and em).
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Calculating parity
Parity is a multiplicative quantum number. It is, like spin, an intrinsic property of a particle.
i.e. System of particles A, B with orbital angular momentum l :
PABC PA PB 1 (7.12)
l
1
fermions: Pfermion Pantifermion 1 (7.13)
2
1, Pantifermion 1 (7.14)
From Dirac equation: spinConvention: Pfermion
Ground state mesons (no orbital angular momentum):
Pmes -1 (7.15)
Check and measure with data.
Particle
Pseudoscalar mesons (s=0)
,K,h,D,B
Vector mesons (s=1)
K*,w,
Parity
(P)
-1
-1
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Parity of the photon
Decay of atomic energy level with single photon emission.
Selection rules (dipole): l 1, S 0 (7.16)
Eg electron in excited H -atom.
Recall (7.09):
l=+-1
Pˆ nlm r Pa 1 nlm r
l
Parity before photon emission=Pa 1
Parity after photon emission=Pa 1
Pa 1 Pa 1
l
l 1
l 1
l
Atomic energy levels
P
P 1 1 (7.17)
1
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Parity of the electron
1
fermions: Pfermion Pantifermion 1 (7.13)
2
e e form positronium (short-lived state - forthcoming lecture)
Parapositronium L 0, S 0
From Dirac equation: For spin
e e
Parity conservation: LHS: Pe Pe
: RHS: P 1
2
Pe Pe 1
l
l
positronium
(7.18)
Measurements of l confirm (7.13).
Pair production of e , e makes it impossible to ever determine e or e absolutely.
Have to define Pe 1, Pe 1
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Intrinsic parity
Particle
Parity (P)
Electron, muon, tau
+1
-1
+1
-1
-1
-1
-1
+1
Positron,antimuon,antitau
Quark
Antiquark
Photon
Pseudoscalar mesons (s=0,l=0) ,K,h,D,B..
Vector mesons (s=1,l=0)
K*,w,
Ground state baryons (l=0) p,n,S..
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Question
●
Before Wu’s experiment the t and particles
were observed with the same spin, mass,
charge. They were thought to be different
particles because they decayed into states with
different parities
P 1
P ?
t
●
0
0
0
Calculate the parity of the pion system from the
second decay. Assume no orbital angular
momentum in the final pion system.
●
State which particle t and is.
●
Which of the decays violates parity ?
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Charge conjugation: C
Charge conjugation Cˆ converts each particle to its antiparticle, to .
particle which is its own antiparticle, eg 0 , , 0 ..
Cˆ | C |
ˆ ˆ | |
CC
C 1 (7.19)
a particle which is not its own antiparticle, eg , K ,
ˆ ˆ | a C C | a C C 1 (7.20)
Cˆ | a Ca | a ; Cˆ | a Ca | a CC
a a
a a
C =C -parity is a useful quantum number for particles which are their own antiparticles
and are eigenstates of Cˆ , eg 0 , , . Can also be extended into G-parity (later).
Ca ,Ca are arbitary phase factors with no physical significance.
Cˆ changes the sign of all "internal" quantum numbers:
charge, lepton number, baryon number, strangeness, charm,bottomness, I 3 .
Mass, energy, momentum, spin unchanged.
Cˆ , Hˆ 0 (7.21) when charge conjugation symmetry is respected: em and
strong forces.
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Question
Show, with the example of a neutrino, that
charge conjugation is not a symmetry of the
weak force.
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C-parity (C)
• Same game as for parity! We’ve found a symmetry
of the em and strong forces, but not of the weak.
• Find a quantity conserved in strong and em
processes.
• Most particles are not eigenstates of Cˆ
• Particles which are eigenstates are their own
particles, eg 0,,0
• Can also construct eigenstates using particle
0
,
antiparticle pairs, eg
• Particles or multiparticle states have eigen value
known as an intrinsic C-parity quantum number, eg
has C1
• C is a multiplicative quantum number like parity.
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Calculating C
Eigenvalue of Cˆ for pion-pion pair ( )
given by Cˆ | ; l (1)l | ; l (7.22)
1
Eigenvalue of Cˆ for spin- fermion-antifermion pair (ff )
2
given by Cˆ | ff ; l , s (1)l s | ff ; l , s (7.23)
l orbital angular momentum, s total spin angular momentum
for combined ff state.
C for the lowest mass hadron states (l=0 )
0 uu dd ff (spin 0) C 1
0 uu dd ff
(spin 1) C 1
Unless otherwise stated, we will be dealing in this lecture
with systems of particles for which l=0.
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Question
Using the decay calculate C 0 .
0
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C and P
Particle
P
C
Pseudoscalar
mesons (s=0)
,K,h,D,B
-1
+1
Vector mesons
-1
-1
-1
+1
-1
N/A
Charged
Leptons
1
N/A
Charged antileptons
-1
N/A
C value only applicable
for particles which are
their own anti-particles
(s=1) K*,w,
Photon
Ground state
baryons
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Formalism
A particle is characterised by the form JPC, eg 1-J=total angular momentum, P=parity(+ = +1,- = -1),
C=charge conjugation number (+ = +1, - = -1)
In certain situations C is not a useful quantum
number – most particles are not eigenstates of C:
JP is used.
Eg : 0 , 0 : 0 , K : 0 , :1.
P=1
C=1
”even” parity, P=-1
”even” C-parity, C=-1
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”odd” parity
”odd” C-parity
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G-parity
The strong force is invariant under isospin and respects charge conjugation.
Can we combine these symmetries to get another useful symmetry ?
Combine Cˆ charge conjugation with Rˆ rotation around " y " (or 2) axis in isospin space.
For pions Rˆ 0 0 ; Rˆ
(7.24)
Same algebra of angular momentum as for isospin.
3
3
3
0
0
0
Spherical harmonic Y1 ,
cos , Y1 ,
cos
cos Y1 , (7.25)
4
4
4
3
3
3
i
i
1
i
1
Y11 ,
sin e , Y1 ,
sin e
sin e =Y1 , (7.26)
8
8
8
3
3
3
i
1
i
1
i
1
Y1 ,
sin e , Y1 ,
sin e
sin e =Y1 , (7.27)
8
8
8
For neutral pion Cˆ 0 0 (7.28)
ˆ ˆ (7.29) Gˆ 0 G 0 0
Define Gˆ CR
G 1. (7.30)
Eigenvalue of Gˆ G-parity or G - conserved in strong interactions.
Choose Cˆ (7.31). Obs! Results don't depend on our choice of arbitary phase
Choose that all isomultiplet members have same G. Gˆ
(7.32)
Generally for an isomultiplet, eg , , 0 : G (-1) I C (7.33)
C C -parity for neutral member of isomultiplet, I isospin.
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Particles for which G-parity is relevant
G G parity, a useful quantum number.
Which particles are eigenstates of Gˆ ?
Particles carrying no flavour quantum numbers (S , C , B ) or baryon
number.
Gˆ G , Gˆ G
Gˆ K a K 0
Gˆ 0 b 0
Some eigenvalues (not exclusive list):
w , ,, '... C 1 ; I 0
From (7.30) G (-1)0 (1) 1
, , ... C 1 ; I 1
G (-1)1 (1) 1
(7.34)
Since G is a multiplicative conserved quantity.
states of N -pions have G 1 (7.35)
N
explains why some particles decay strongly to 2 and others to 3 .
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Some decays explained with G-parity
Eg
2 LHS: G 1,
3 LHS: G 1,
RHS: G 1 1 ok!!
2
RHS: G 1 1 forbidden!
3
w 3 LHS : G 1, RHS: G 1 1 ok!!
3
w 2 LHS : G 1, RHS: G 1 1 forbidden!!
2
,
"Explains" decays of many other particles: ',,h,h '.....
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Conserved quantities/symmetries
Quantity
Strong
Weak
Electromagnetic
Energy
Linear momentum
Angular momentum
Baryon number
Lepton number
Isospin
-
-
Flavour (S,C,B)
-
Charges (em, strong
and weak forces)
Parity (P)
-
C-parity (C)
-
G-parity (G)
-
-
CP
-
T
CPT
Coming up
-
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CP
• C and P are not separately respected in
weak decays
• What about CP ?
( left - handed)
( left - handed!! )
X
( right - handed)
Apply C
Now apply P
Combined CP transformed decay ok!
• Original and CP-transformed decays occur
with same rate. CP symmetry is respected
in many weak processes.
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Neutral kaons
We define a neutral K0 by its quark content (sd), mass (498
MeV), spin (0), isospin (I=1/2,I3=-1/2) - a normal particle !
Consider production of K 0 :
eg . p n K 0 K p p
Strong decays forbidden
(no flavour violation!)
It decays weakly with time:
Non-exponential decay time!
K 0 seems to comprise two other
particles (K1 and K 2 ) with definite lifetimes t 1 ,t 2
t 1 1010 s, t 2 5 108 s
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Strategy
Test the hypothesis that CP is a good symmetry of the weak
force.
Try to form CP eigenstates from K0 and K0 and check they decay
in CP-conserving ways (recall and C conservation.)
Pˆ | K 0 | K 0
Cˆ | K 0 | K 0
Pˆ | K 0 | K 0 (7.36)
Cˆ | K 0 | K 0 (7.37)
ˆ ˆ | K 0 | K 0
CP
ˆ ˆ | K 0 | K 0 (7.38)
CP
ˆ ˆ:
Normalised eigenstates of CP
1
1
0
0
0
0
0
| K
|
K
|
K
|
K
|
K
|
K
(7.39)
2
2
2
ˆ ˆ | K 0 | K 0
ˆ ˆ | K 0 | K 0 (7.40)
CP
CP
1
1
2
2
0
1
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1
| K 0 | K 0
2
1
| K 0
(| K10 | K 20 )
2
| K10
1
| K 0 | K 0 (7.39)
2
1
| K 0
(| K10 | K 20 ) (7.41)
2
| K 20
Hypothesise that the CP eigenstates are the two neutral states
observed in the K 0 decay.
If CP is conserved then K10 decays into CP 1 states , 0 0 and
K 20 decays into CP 1 states
K10
and
0 , 0 0 0 .
K 20 .
Two neutral kaons states had been observed decaying
weakly: t 1 1010 s, t 2 5 108 s
The 2 decay will happen faster because of greater
phase space
- associate K10 ,t 1:CP 1 and K 20 ,t 2 : CP 1
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(7.42)
27
Neutral Kaons and Strangeness Oscillations
1
(| K 0 | K 0 )
2
1
0
| K
(| K10 | K 20 )
2
1
(| K 0 | K 0 ) (7.39)
2
1
0
| K
(| K10 | K 20 ) (7.41)
2
| K10
| K 20
Consider in the kaon rest frame and allow a decay
t
t
i
i
i
(
m
)
t
i
(
m
)t
1
2
1
1
im1t
2t1
im2
2t1
2t1
2t 2
o
0
0
0
K (t )
K1 e e
K2
K1 e
K 20
e e
e
2
2
(7.43)
Consider a particle produced at t=0 as a K0 . Amplitude that it is still a K0
at a later time t:
i
i
i
(
m
)
t
i
(
m
)t
1
2
1
2t1
2t 2
o
o
0
0
0
K K (t )
K1 K 2 ) e
K1 e
K 20
2
1
1
i
(
m
)
t
i
(
m
)t
1
2
1
2t1
2t 2
o
o
K K (t ) e
e
(7.45)
2
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(7.44)
28
Probability that it is still a K0 at a later time t:
K o K o (t )
2
K o K o (t )
i ( m2
) t i ( m1
)t
i ( m2
)t
1 i ( m1 2t1 )t
2t 2
2t1
2t 2
e
e
e
e
4
i
2
K o K o (t )
Similarly:
K o K o (t )
2
2
i
i
t
t
i
i
i
i
i
(
m
m
)
t
i
(
m
m
)t
2
1
1
2
1 t1
t2
2t1 2t 2
2t1 2t 2
e e e
e
4
2
K o K o (t )
i
t
t
t 1 1
( )
1 t1
t2
2 t1 t 2
i ( m2 m1 ) t
i ( m1 m2 ) t
e e e
e
e
4
t
t
t 1 1
( )
1 t1
t2
2 t1 t 2
e e 2e
cos (m2 m1)t (7.46)
4
t
t
t 1 1
( )
1 t1
t2
2 t1 t 2
e e 2e
cos (m2 m1)t (7.47)
4
Strangeness oscillation!!
Beam of particles, initially pure K 0 .
After time t (in particles rest frame)
study intensity of the beam and
composition of K 0 , K 0 .
Eg compare rates of K 0 p uds
(not possible as strong reaction for K 0 , why??)
and K 0 p K 0 , K 0 p K 0 .
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Kaon oscillations
Intensity
(Niebergall et al., 1974)
K
Oscillations observed.
0
From many experiment: m2 m1 3.5 106 eV (7.44)
(Obs!
m2 m1
mK
1014 -tiny, m2 m1 from other experiments. )
0.25
K
0
2
0
4
6
8
Time in K0 rest frame (x10-10 s)
50% of the beam | K10 (short lived)
has died away leaving the
1
| K 20
| K 0 | K 0 with apparently
2
equal contributions of | K 0 ,| K 0 .
The | K 20 decays very slowly (t 5 108 ).
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Question
To measure the oscillations of a beam of neutral
kaons of energy 10 GeV how large should an
experiment be ?
(Niebergall et al., 1974)
Intensity
●
Oscillation pattern visible over time
K
0
scale tosc
109 s in kaon rest frame.
E 10
Speed v c ;
20 .
m 0.5
Distance travelled in the lab
ctosc 20 3 108 10 9 6m.
Pretty small compared with
neutrino oscillations!
K
0
2
0
4
6
8
Time in K0 rest frame (x10-10 s)
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Some interpretation and comparisons
K 0 oscillation
Neutrino oscillation (2-component approximation)
(1) Particles are produced in a flavour
(1) Particles are produced in a beam in a flavour
eigenstate, eg .
eigenstate by (s ).
(2) As they pass through space the proportion of flavour
, t oscillates ( t ).
(2) As they pass through space the proportion of
of flavour s , s oscillates (K 0 K 0 ).
(3) This is a simple quantum mechanical effect.
The is not a mass eigenstate, rather a mixture
(3) This is a simple quantum mechanical effect.
The K 0 is not a mass eigenstate, rather a mixture
0
1
of two mass eigenstates: 1 , 2 .
0
2
of two mass and CP eigen states K , K .
(4) The beam intensity stays constant since the
(4) The beam intensity drops since the K 0 decays.
neutrinos do not decay (nothing to decay down
to perhaps!).
(5) Oscillations seen over distance scales 1000 km.
(6) The t transformation can't be
(5) Oscillations seen over distance scales
m.
(6) The K 0 K 0 transformation can be understood by
the Feynman diagram of a weak process in which
strangeness is violated (next lecture).
understood by any Feynman diagram since
lepton number violation "can't" happen at a
vertex.
K0
K0
|S|=2
Strangeness violated
t
L L t 1
Lepton number violated!
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Summary
●
Discrete symmetries
Parity (P)
Charge conjugation (C-parity)
G-parity
Fundamental symmetries of nature constrain the
behaviour of particles
CP
●
●
●
Neutral kaons
Strangeness oscillations
Next lecture – CP violation
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