Introduction to Particle Physics I

!   Risto Orava

!   Spring 2015

symmetries

SYMMETRIES •  symmetry principles •  parity – space inversion •  charge conjugation (C) •  charge conjugation parity (CP) •  charge conjugation parity time (CPT) •  isospin •  G-parity, SU(3)flavour and SU(3)colour

global and local gauge symmetries will be discussed later during the course, for reference see Halzen&Martin Chapter 2

symmetries and non-observables •  permutation symmetry: Bose-Einstein and Fermi-Dirac Statistics

•  continuous space-time symmetries: translation, rotation, acceleration,…

•  discrete symmetries: space inversion, time inversion, charge inversion

•  unitary symmetries: gauge invariances: U1(charge), SU2(isospin), SU3(color)

⇒  If a quantity is fundamentally non-observable it is related to an exact symmetry

⇒  If a quantity could - in principle - be observed by an improved measurement, the symmetry is said to be broken

NOETHER THEOREM:

SYMMETRY CONSERVATION LAW

symmetries and non-observables

r1!"

r2!"

d!"

an example: potential energy V between two particles:

absolute position is a non-observable: the interaction is independent of the choice of 0rigo

symmetry: V is invariant under arbitrary space translations:

r2!"→ r2!"+ d!"

r1!"→ r1!"+ d!"

V =V r1!"− r2!"

( )ddtp1!"!+ p2!"!

( ) = F1!"!+ F2!"!

= − ∇1! "!+∇2

! "!( )V = 0

then: over-all momentum is conserved:

0 0’

particle 2

particle 1

rotations of an equilateral triangle – a finite symmetry operation:

•  rotate clockwise by 120o •  rotate clockwise by 240o

•  leave it alone. 3D rotations of a sphere described by three angles => an infinite number of symmetry operations, since the angles could have any values. in the symmetry operations above : (1) a symmetry operation followed by another symmetry operation is itself a symmetry operation. (2) no operation signifies a symmetry operation as well

symmetries

a

c b

a symmetry is an operation which leaves the system invariant

symmetries... (3) for each symmetry operation, there is a symmetry operation which undoes it, so that the net result is doing nothing. (4) order of symmetry operations! The above properties of symmetry operations define a mathematical object, called a group - group theory is the mathematical study of symmetries and fundamental for all physics. How about the order of symmetry op’s: is symmetry operation no. 1 followed by symmetry operation no. 2 the same as doing no. 2 followed by no. 1? This is sometimes true, as in the case of the triangle rotation group, and sometimes not, as in the case of the sphere rotation group. Groups for which it is true are called Abelian groups, others are non-Abelian groups. Note: The group structure of the gauge symmetry of our theories is Abelian for forces in which the force carriers do not carry charge (EM force) and non-Abelian for those that do (strong and weak interactions).

symmetries... The interesting groups in physics are often groups of nxn matrices. A unitary matrix has U-1 = Ũ*. A special unitary matrix SU(n) is a unitary matrix with determinant = 1. An orthogonal matrix is a real unitary matrix, so that O-1 = Õ. The rotation group of the sphere is described by SO(3). The quantum descriptions of angular momentum are given by SU(2), the Pauli spin matrices. exp(iαjσj) SU(2) and SU(3) are the most important groups in particle physics. One often talks about a representation of a group if for every element a, there is matrix for which ab = c ∈ Ma ⇒ MaMb = Mc. A representation is faithful if each element of the group is represented by a different matrix. The lowest dimension faithful representation is called the fundamental representation. Why are symmetries important? Because each symmetry implies a conservation law, and conservation laws provide a simple way of conveying a great deal of useful information.

Example: Orthogonal group in 3 dimensions O(3), linear group in 3 dimensions restricted to those transformations which leave x2+y2+z.2

invariant. This invariance condition imposes 6 conditions on the 9 parameters. Thus the 3-parameter group.

symmetries... translation symmetry in time ⇒ conservation of energy (E) translation symmetry in space ⇒ conservation of momentum (p) symmetry under rotation ⇒ conservation of angular

momentum (L) in quantum mechanics:

⇒

[ ] tAAHidtAd ∂∂+= /ˆˆ,ˆ//ˆ !

0/ˆ/ =∂∂= tHdtEd

[ ] 0/,ˆ// =∂∂−= xiHidtpd !!

if H does not depend on t

if H does not depend on x

translation symmetry

consider a particle in position For a single free particle, the Hamiltonian is:

There is no dependence on the position of the particle - only on the derivative, thus H(x) = H(x’). The Hamiltonian is invariant under translation. the translation operator D is defined by

Expanding this to 1st order in (small) dx, gives

51

xxxx !!!!δ+=→ '

2222)( mmpxH +∇−=+=!

)()( xxxD !!!δ+Ψ=Ψ

)()()( xxxxx !!!!!Ψ∇⋅+Ψ=+Ψ δδ

assume: properties of an isolated physical system remain the same in a translation by an arbitrary amount dx

translation symmetry... Therefore, The conservation law for linear momentum follows by first applying D to the wavefunction Φ(x) = H(x)Ψ(x), giving for the left hand side And and, therefore, If an observable commutes with the Hamiltonian, it means it does not change with time: It is a conserved quantity. So this is the conservation law for momentum. The result can be generalised to a many particle system, and the same chain of reasoning appears often in physics,linking conservation laws to symmetries.

Ψ⋅+=Ψ )1()( pxxD!!

δ

)()()()()()()()( xDxHxxxHxxxxHxxxD !!!!!!!!!!!!Ψ=+Ψ=+Ψ+=+Φ=Φ δδδδ

[ ] 0, =HD

p,H[ ] = 0

parity violation Up to 1956 physicists were convinced that the laws of nature were left-right symmetric. Strange?

A “gedanken” experiment: Consider two perfectly mirror symmetric cars:

“L” and “R” are fully symmetric, Each nut, bolt, molecule etc. However the engine is a black box

Person “L” gets in, starts, ….. 60 km/h

Person “R” gets in, starts, ….. What happens?

What happens in case the ignition mechanism uses, say, Co60 β decay?

“L” “R”

gas pedal driver

gas pedal driver

parity – space inversion The parity operator, , changes

If Ĥ is invariant under P , particles should be eigenstates of P, and P will be conserved (as a multiplicative quantum number). This is true in classical physics. The laws of physics do not depend on whether you are looking at a mirror.

(vectors)

(axial vectors)

^ ^

P rr !!−→

P of eigenstatean is )r( if ,)()()(ˆ !!!!ΨΨ=−Ψ=Ψ rPrrP

)r(P)r()r(-P)r(PP 2 !!!!Ψ=Ψ=Ψ=Ψ

11ˆPPP 22 ±=⇒=⇒≡= PPI

pdtrdPpPrrP !!

!!!−==−= )(ˆ)(ˆ;)(ˆ

σσ!!!!!!

=⇒=×= )(ˆ)(ˆ)(ˆ PLprPLP

parity – space inversion... The parity transformation is given by a complete reversal of the space components. A system is invariant under the parity transformation,i.e. if the Hamiltonian remains unchanged under this transformation:

This is true for systems which undergo EM or SI interactions, but is violated by WI. The parity operator is defined by

where a labels the particle which is described by the wavefunction Ψ.

3,2,1),,('),( =−=→= ixxxxxx ioio µµ

)()( ii xHxH −=

),(),(ˆ xtPxtP a!!

−Ψ=Ψ

parity – space inversion... Since applying the transformation twice gets you back to the beginning, Pa

2 = 1 and so Pa = ±1. Consider a free particle wavefunction, i.e. a solution of the Klein-Gordon equation,

then

Thus if the particle is in its rest frame, i.e. p=0, it is an eigenstate of parity. Pa is called the ”intrinsic parity” of particle a, or usually just ”parity” of particle a. Particles with definite orbital angular momentum are also eigenstates of parity, with The factor depending upon comes from the space symmetry of the spherical harmonics familiar from atomic physics.

( )[ ]Etxpixipxp −⋅=⋅−=Ψ!!

exp)exp()( µ

),(),(),(ˆ)(ˆ ipaa

ipp xtPxtPxtPxP i Ψ=−Ψ=Ψ≡Ψ µ

l

laPP )1(−= l

vector reflections

mirror: parity transformation

cross product

vector

parity – space inversion...

Thus under P positions change, but spins do not. Note, however, that helicities do change:

Particles will have an intrinsic parity. This may seem strange for a fundamental particle, but not for a composite one such as an atom or a meson. Since leptons and quarks – and fermions in general – are conserved, their intrinsic parity is arbitrary. However, relativistic QM (the Dirac equation) tells us that the intrinsic parity of an antifermion is opposite of the fermion. By convention, we take the parity of fermions to be +1 and the parity of antifermions to be –1.

λλσσ

λ −=⋅

≡ )(ˆ Ppp!!!!

helicity All particles have spin and three momentum For fermions , two possible arrangements

p!

f

p!

f

right handed

left handed

p!

spin vector same as momentum vector

spin vector opposite momentum vector

21

=s

!p

parity – space inversion parity operations are denoted P -reflects the wave function through the origin:

if the system is to remain invariant then we require:

•  Then, if Ψ represents a parity state: which implies

),(),(ˆ xtxtP !!−Ψ=Ψ

22 ),(),( xtxt !!−Ψ=Ψ

),(),( xtxt !!−Ψ±=Ψ

),(),(ˆ xtxtP !!−Ψ±=Ψ

parityeven is ),(),(ˆ xtxtP !!−Ψ+=Ψ

parity odd is ),(),(ˆ xtxtP !!−Ψ−=Ψ

parity of particles

intrinsic parity – fermions –  consider electrons and positrons represented by

a wave function Ψ.

–  Dirac equation is satisfied by a wave function representing both electrons and positrons ⇒ related and it can be shown

–  Strong and EM reactions always produce e+e- pairs.

–  Arbitrarily have to set one =1 and the other = -1.

( ) ( )txtx e ,P,P !! −Ψ=Ψ ±

1PP −=−+ ee

parity of fermions assign positive parity state to particles, negative to antiparticles:

make same assumption about quarks to be consistent:

1PPP

1PPP

−===

===

+++

−−−

τµ

τµ

e

e

1PPPPPP1PPPPPP−======

======

tbcsud

tbcsud

What do we expect for the parity of mesons?

For a central potential,

Since spins do not change under P,

We therefore predict that the p meson should have P = -1 since l=0. This quantity can be measured and indeed was measured in 1951.

parity of mesons

^

rinsicspinspacemeson PPP int)(ˆˆ ΨΨ=Ψ

),()( ϕθχ mlspace Yr!=Ψ

),()1(),(),(ˆ ϕθϕπθπϕθ ml

lml

ml YYYP −=+−=

1int )1()1()1()1( +−=−−=−= llrinsic

l PP

parity of mesons... Panofsky et al., observed the following reaction produced by a low energy π-meson on a deuterium:

The parity of each side must be equal – assuming that parity is conserved, i.e.

deuterium has: The π- has low momentum, and

nnd →−π

spacedd PPPP −− =ππ

)conventionby ,1(1)1( ),(0 and 1 +===−=↑↑== npl

d PPPnpls

-- PP i.e. 1)1(0d πππ

π ==−=⇒=×Δ= −

−d

lspacedPprl !!

parity of mesons... The 2 neutrons must have odd under interchange, because the two neutrons are identical fermions (Pauli Exclusion Principle):

The π-d has j=1 (l=0+ s=1) ⇒ nn must have j=1. Possible combinations for the case j=1:

l 0 1 1 2 (l=1 and s=1) is the only combination with s 1 0 1 1 l+s = even

spinspacetotal ΨΨ=Ψ

spacel

space Ψ−→Ψ )1(

spins

spin Ψ−→Ψ +1)1( )(2

1 :0 ↑↓−↑↓=s )(2

1 :1 ↑↓+↑↓=s

even bemust )1()1( 1 slsl +⇒−=− ++

−− ==−=−=⇒=⇒ππPPPl d

lnn 1)1(1

Note: (-1)s+1 -factor enters because the triplet spins state (s=1) is symmetrical and the singlet spin state (s=0) is antisymmetrical with regards the need to exchange the positions of the two identical neutrons!

parity of mesons •  mesons are quark antiquark pairs:

•  for l=0 (spins anti-parallel) then P = -1. –  look at charged pion interactions with

nucleons

–  complex relationship and not straightforward take as given for now. requires more QM than 3rd students have.

( ) ( ) 111PPP +ʹ′ −=−= llqqM

parity of baryons •  Baryons contain three quarks:

•  Dealing with ground state Baryons in most cases, hence l=0

( ) ( )

( ) ( ) 1)1)(1(11PPPP

1)1(1)1(1PPPP

312312312321

312312321

0

−→−−=−=−=

+=−→−=−−=

+++

+

llllllqqqB

llllqqqB

q1

q2

q3

l12 l3

parity of photons

•  photons: –  if applied to electric field, reverses field

lines direction – hence the photon wave function must be

odd under the parity transformation – photon also has spin 1 parity

conservation not straight forward. – neutral pion decays, problematic just

under P transformations

),(),(ˆ xtxtP !"−Ψ=Ψ

parity of photons… •  parity of the photon can be derived from the effect of the parity operator on classical electromagnetic fields •  in relativistic QM, the photon is represented by its •  4-vector potential (MKS units), therefore,

Therefore, , the photon has negative parity.

),/( AcA!

ϕµ =

lyequivalentor ),/(ˆˆ AcPAP!

ϕµ =

rrqrErEP ˆ

41E since )()(ˆ

2oπε

=−=!!!!

EEEPEt/B-rBrBP!!!!!!!!!

×∇=−×−∇=×∇×∇=∂∂+= )()()(ˆ and since )()(ˆ

ABBAPBA!!!!!!

−=×−∇=×−∇=⇒×∇= )(ˆ

γγ −=P

parity of photons... An argument analogous to that give for translational invariance can be used to show that if the Hamiltonian is invariant under the parity transformation, parity is conserved, i.e.

Measurements of the reaction: , where the have known angular momentum 0, shows that the photons have odd orbital angular momentum l. The parity of the final state. is given by so we can deduce that as , the electron and positron have opposite parity.(In fact, it can be shown from the Dirac equation that this is true for all fermions.) The sign of parity is defined, by convention, to be positive for particles and negative for antiparticles.

[ ] [ ]PHHP ˆ,,ˆ =

γγ→−+ee −+ + ee

lP )1(2 −γ 1)1(2 −=−=−+l

ee PPP γ

parity of hadrons... The parities of hadrons follow from the product of the parities of the quarks multiplied by the parity of their angular momentum wavefunction. The parity of the photon can be deduced from the classical limit. Here, the photon is due to an oscillating electromagnetic field, and this field obeys the equation:

where ρ is the charge density. Under parity transformation, this becomes

since ∇ changes sign. Pγ is the intrinsic parity of the photon. So to keep the equation invariant under this transformation, we must have Pγ = -1.

oioio xxxxE ερ /),(),( =⋅∇

!

oioio xxxxEP εργ /),(),( −=−⋅∇−

!

photon helicity •  photon has spin 1, therefore JZ =-1, 0, +1

p!

γ JZ= 1

p!

γ JZ= -1

p!

γ JZ=0

p! p! p!

parity... it was assumed that the laws of physics are symmetric under parity, i.e. that the laws of physics cannot tell left from right... in mid 50’s a problem arose with the K+ -meson: two different decay modes were seen: Θ+ → π+πο, and τ+ → π+π+π- they seemed to have the same mass and lifetime, but were tought to be different particles because they had different parities.

the most likely case for the τ is all lpp = 0, since Q = mK – 3mπ = 494 MeV – 3 (140) MeV = 74 MeV. Then Pτ = (Pπ)3 = -1. - experimental analysis of these decays confirmed the result

Pθ = (−1)lππ Pπ2 = (−1)lππ

sθ ≡ Jθ ≡ 0 ⇒ lππ = 0 since sθ = 0 ⇒ Pθ = +1

parity... In 1956 T.D. Lee and F. Yang: whilst there was strong evidence that the strong and EM interactions were invariant under parity, there were no tests for the weak interactions. Madame C.S. Wu in 1956: parity strongly violated. - she polarized Co60 so that the spins were mostly pointing to a single direction, and then observed that most of the electrons are emitted in the direction of the spin. Although this does not seem very striking, it is a violation of symmetry under parity since the spin does not change direction and the momentum of the electron does.

S pe

pe

S

P

evidence for parity violation

Co60

50% e-

50% e-

50% e-

50% e-

MIR

RO

R W

OR

LD

REA

L W

OR

LD

Co60

Co60

Co60 Co60

Co60

e-

Co60

magnetic field

magnetic field e-

e-

e-

parity... It turned out that parity is maximally violated by the weak interactions – WI only acts on fermions with left-handed helicity in the limit v → c. Note, that in general, helicity is not Lorentz invariant - if a right-handed particle is viewed by an observer with a higher velocity, it looks left-handed. However, ν is either massless or close enough to be massless that we cannot tell the difference. Then, helicity is Lorentz invariant for the neutrino. All ν’s are left-handed and all anti-ν ’s are right-handed. (Any right-handed neutrino would not interact with matter.) How do we know this? The helicity of the neutrino can be deduced from the π+ decay:

π+ → µ+ ν Since J = 0 S = 0 helicity of µ+ must be left-handed (= -1). The helicity of µ was measured from its decays in 1957.

S S µ+ ν

π+

charged pion decays

•  pion has spin 0 - spins of decay particles must sum to zero.

41027.1

%100−+

++

×≈→

≈→

BR

BR

ee ν

νµπ µ

π+ νµ µ+

revisit parity: bosons •  parity effects different particles/fields depending on

their type: –  spin-1 particles:

•  (Polar) Vectors

• Axial Vectors • 

–  spin-0 particles: • Scalars no dimension and transform to

themselves PS = +1 • Pseudoscalars

–  spin-2 particles: • Tensors transform with ∴ PT = +1

)1( )()()(ˆ :)( −=−=−= VPxVxVxVPxV !!!

prL momentumangular e.g.

1)1()1( :)(')(!!!

!!

×=

+=−×−=× APxVxV

1)1( :)(")(')( 3 −=−=×⋅ psPxVxVxV !!!

W-boson •  the W boson:

–  only sees left-handed particles

–  only sees right-handed anti-particles –  hence if right-handed neutrinos exist, they

would not interact with W’s and therefore not weakly

•  massive particles:

–  not pure helicity states. –  right-handed = +1, left-handed –1

–  left-handed massive particle has helicity: -v/c

symmetries – C, CP and CPT A 2-element symmetry group is given by the ”charge-conjugation” operator C which changes a particle to an antiparticle, leaving its position, momentum, and spin unchanged, i.e.

Unlike parity, only particles which are their own antiparticles can be eigenstates of C. The photon has C = -1 since both E & B fields change sign when the electron changes sign. This implies: neutral, flavourless mesons are taken as quark-antiquark pairs. for a bound state of two quarks of the same flavour, the wave function would have to be antisymmetric on interchange of the two quarks for a system consisting of a quark and an antiquark, no such restriction exists

1 with ˆ ,ˆ 2 ±=−== CaCaCIC

µµ ϕϕ AAcAcCAC −=−−== ),/(),/(ˆˆ !!

symmetries – C for the over-all state function to be antisymmetric on interchange

This is easy to see experimentally, since

B(πo → γγγ)/B(πo → γγ) < 3.1 × 10-8.

This also shows that C is conserved by the EM interactions. It is also conserved by the strong interactions. However, it is clearly not conserved by the weak interactions since does not exist.

echspinspace argΨΨΨ=Ψ

)(ˆargarg qqCC echech ↔Ψ=Ψ

spins

spinspacel

space Ψ−→ΨΨ−→Ψ +1)1( ,)1(

))1( with this(compare )1( 1++ −=−=⇒ lsl PC

1 has +=⇒ Coπ

llC νν =ˆ

charge conjugation •  charge conjugation (C) swaps a particle for

its antiparticle.

–  if a set of laws obey this operation then if they predict the behaviour of a particle they predict the behaviour of an antiparticle.

•  time reversal (T) represents reversing the process. i.e. one can get back to the start by reversing the process

charge conjugation •  swaps particles by antiparticles:

– apply to mesons

–  therefore π± is not a charge conjugation eigenstate

– apply two subsequent charge conjugation operations - get the original particle back

+−+ ≠→ πππC

00C ππ ±→

charge conjugation - photon

•  by swapping the electric charge that produces a photon, get a reversed electric field

•  implies that photon is anti-symmetric under charge changes

•  for the pion:

γγ ±→C

1C −=γ

( )2

0

1CCC

2

0 −==

→

γγπ

γπ

symmetries - CP consider a combination of two symmetry operations: CP takes a left-handed ν into a right-handed ν -thus it seems that the weak interactions could conserve CP. The parity operator reverses 3-momentum , but leaves angular momentum unchanged. The helicity of a particle is defined as aprojection of the spin on the momentum: The helicity changes sign when operated on by the parity operator. Weak interactions couple differently to helicity = -1 particles than to helicity = +1 particles (we’ll discuss this much more in the coming weeks!) This means that WI’s are not, in general, invariant under swapping helicityor swapping particles for their anti-particles. C and P are not good symmetries of WI’s. Let us consider next the case of neutral Ko –mesons.

-

pp!!!!

σσ

λ⋅

=

!p!r × !p

particles and helicity

•  fermions usually come in left and right-handed varieties: fL and fR.

•  how are they effected by the C and P operations?

C

Lf RfP

Lf Rf

CP

neutrinos

•  the neutrinos only come in one helicity – a particle is left handed – an antiparticle is right handed

C

Lν RνP

Lν Rν

CP

CP conservation

•  in general, the weak interaction does not conserve parity or charge conjugation

•  it does, however, almost satisfy CP conservation

–  this is a C and P operation one after the other

–  take product of C and P to calculate

•  example neutrinos

C

Lν RνP

Lν Rν

CP

CP transformations: Nickel

neutral kaon mixing

•  because the d and s quarks are not weak eigenstates, a mixing between the two neutral kaons can take place

•  this box diagram demonstrates the process by which a kaon transforms into an anti-kaon

d

s

s

d

u

u

WW0K 0K

neutral kaon mixing is an state and is a state -these are produced by strong interactions. -If there were only strong and EM interactions, this would be the end of the story, since they carry different flavours and flavour is conserved by these interactions. However, the weak interactions do not conserve flavour, and the Ko can change into an anti-Ko, and vice versa, by the second-order weak interaction (see Figure). If weak interactions conserve CP, would expect the weakly decaying states to be eigenstates of CP:

-

-

oooo KKPCKKPC −=−= ˆˆ and ˆˆ

1) (2

1)(2

1ˆˆ KKKKKPC oooo ≡−+=−

CP 12

( Ko + Ko ) = − 12

( Ko + Ko ) ≡ K2

sdK 0 K 0 sd

neutral kaon mixing... K1 should decay to states with CP = +1, while K2 should decay to states with CP = -1. For example: parity of a two-pion state is +1*, C of 2πo is clearly +1, since C of each πo is +1. For π+π- ,the same type of argument for bosons as earlier used for the fermions works: by Bose-Einstein statistics, the total wave function:

must be even under interchange. Ψspace is even since L= 0, Ψspin is even since all S = 0 ⇒ is even, i.e. CP(2π) = +1

⇒ K1 → 2π and K2 → 3π... There is much more energy available for the 2π decay than for the 3π decay, and the K1 will therefore have a shorter lifetime than the K2. Experimentally: τK1 = 9 10-11 s and τK2 = 5 10-8 s. Within its lifetime, K1

travels a few centimeters, while K2 travels several meters.

* see the discussion on τ-Θ puzzle

echspinspace argΨΨΨ=Ψ

neutral kaon mixing... This is a striking prediction; if Ko’s were produced at a target, 2π decays would be observed close to the target, and 3π decays further away. This was predicted by Gell-Mann and Pais in 1955 and observed by Lederman in 1956. The neutral kaons are thus produced by the strong interactions as Ko

and Ko, and they decay as K1 and K2. This system forms a beautiful quantum mechanical laboratory (c.fg. linearly and circulary polarized light or a particle with spin in the x- and z-directions). The system provides a way to search for CP violation in weak interactions. If K2 → 2π, CP symmetry is violated. It was later discovered - by Cronin and Fitch in 1964 – that CP is in fact violated. The effect was small, 0.3% of all K2 decays. The result showed, for the first time, that physics laws for matter and antimatter were not the same.

-

neutral kaon mixing... Two of the K2 decay modes are:

K2 → π+e- νe these two states are CP reflections of each other. Since CP is not conserved, K2 → π-e+ νe the latter has a slightly higher rate than the former, thus giving an absolute difference between matter and antimatter.

The long-lived Ko, now called KL is mostly K2 with a small admixture of K1:

)(11

122KKKL ε

ε+

+=

CP conservation

The combined operation- CP - is a close to being a good symmetry of WI’s. For particles in their rest frame, the CP operator has the effect of simply transforming them to their antiparticles at rest. However, CP violation has also been observed in some particle decays - as discussed earlier and so far no significant deviations have been onserved. Improved measurements are still required for finding out whether: CP violation is really consistent with the SM, and can it be used to explain why there seems to be so many more particles than anti- particles? The BaBar and Belle experiments were built to find answers to these questions, presently the LHCb experiment has the greates potential of making discoveries in this sector.

time reversal symmetry, T, says that - everything else being equal - the rate for A+B → C+D is the same as that for C+D → A+B. it does not lead to any conserved quantum numbers combination of relativity and QM requires CPT to be a good symmetry and there is no basic reason not to have T-violating effects.. one prediction of CPT symmetry is that every antiparticle has the same mass and lifetime as its partner particle. The most sensitive test is the Ko – anti-Ko mass difference, in which the agreement is fine to 2 parts in 1018. CP violation can only occur in the weak interactions if there is a complex phase in the mixing angle matrix. This only happens for three or more families.*

time reversal invariance

-

* The proof of this can be found in Section 12.13 of Halzen & Martin.

CPT invariance Local Field theories always respect:

•  Lorentz Invariance •  symmetry under CPT operation (an electron = a positron travelling back in time) => Consequence: mass of particle = mass of anti-particle:

•  Question 1: The mass difference between KL and KS: Δm = 3.5 x 10-6 eV => CPT violation?

•  Question 2: How come the lifetime of KS = 0.089 ns while the lifetime of the KL = 51.7 ns?

•  Question 3: BaBar measures decay rate B-> J/ψ KS and Bbar-> J/ψ KS. Clearly not the same: how can it be?

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

† 1

1

M p p H p p CPT CPT H CPT CPT p

p CPT H CPT p p H p M p

∗−

∗ ∗−

= =

= = =

=> Similarly the total decay-rate of a particle is equal to that of the anti-particle

Answer 3: Partial decay rate ≠ total decay rate! However, the sum over all partial rates (>200 or so) is the same for B and Bbar.

Answer 1 + 2: A KL ≠ an anti-KS particle!

(Lüders, Pauli, Schwinger)

(anti-unitarity)

Isospin •  neutron and proton seem to interact - through

strong nuclear force – in a similar way; the only difference seems to be due to the proton charge

•  treat the neutron and the proton as identical particles with different spins directions; the spin degree of freedom is called the Isospin: I. I3 is its projection

•  proton and neutron make an Isospin doublet, I=½:

21,

21, 33 −=+= InIp

Isospin... neutron was discovered in 1932: (1) neutron mass is almost identical to the proton mass, Δm≈10-3. (2) nuclear isotopes ⇒ strong interactions identical for the proton and neutron? ⇒ Werner Heisenberg suggested in 1932 that proton and neutron could be consideredas different ”spin” states of a single particle.

a 1 0 N = p = n = b 0 1

The ”isotopic spin” - ”isospin” - is not be confused with any sort of angular momentum - it is called "spin", only because its algebra is the same – in group theoretical language, p and n form a 2-dimensional representation of SU(2).

Isospin... Werner Heisenberg suggested that strong interactions were invariant under rotations of isospin ⇒ isospin is conserved in strong interactions (as angular momentum is conserved in situations in which there is rotational invariance). However: isospin is not a fundamental symmetry of the strong interactions, but an approximate and ”accidental” symmetry, which derives from a more basic symmetry. First, we have to transfer isospin from the nucleons to the quarks: u and d quarks have isospin ½ and all other quarks have isospin 0. Isospin symmetry derives from the fundamental symmetry that all quarks carry - the colour charge. Therefore, the strong forces cannot tell one flavour from another. The only difference is the mass of the quarks. The mass of the quarks is an ill-defined concept since all quarks are confined in hadrons. There are two different types of quark masses which physicists use.

Isospin... (1) The ”constituent” mass: the mass which a quark appears to have in a hadron in terms of properties (such as magnetic moments). For u and d quarks, it is ~1/3 of the mass of the nucleon. (2) The ”current” mass: the ”bare” mass, which one would insert into the fundamental theory, i.e. the difference between the current and the constituent masses is due to the effects of the strong interactions. Theory suggest that the u and d current or bare masses are very small: mu ≈ 4 MeV and md ≈ 7.5 MeV. Even though the u and d masses may be different by a factor of two, they are both so light that it does not matter. They are both effectively massless, with a mass difference of the order of 1% of the effective mass generated by the strong interactions. This is why one can consider isospin symmetry as an accidental one. ⇒ isospin symmetry is an excellent symmetry of SI. Since EM effects are also at the same level of 1%, it is usually impossible to separate small isospin violations from EM effects. Note: EM does not conserve isospin since the charge of the u quark is different from the charge of the d quark.

Isospin... What are the consequences of isospin symmetry? First, the hadrons form isospin multiplets. Consider the pseudoscalar mesons: π mesons form an isospin 1 triplet: Note the minus sign which is needed in order to get antiparticles to transform properly under SU(2). Here the state is taken to be a u. In addition, we would expect there an isosinglet state which could be identified as the η. In fact, there is also the state, which could mix with the η. We do not understand why, but the mixing does take place.

1,121;

21

21;

21

33 −=====−=−=+ IIIIduπ

0,121,

21

21,

21

21,

21

21,

21

21

21

=⎥⎦

⎤⎢⎣

⎡−−−=−= dduuoπ

1,121;

21

21;

21

33 −=−==−====− IIIIduπ

21,

21

[ ]dduu +21

state 0 ssI =

Isospin... Isospin operators obey the same commutation relations as the angular momentum operators ⇒ rules for addition of isospin vectors. The 2-nucleon system can now be in a simultaneous eigenstate of I2 with eigenvalue i(i+1) and of I3 with eigenvalue l. Since the isospin of each nucleon is 1/2 by the usual rules of the addition of angular momentum, the total isospin is either i=1 (a triplet state with l=-1,0 or +1) or i=0 (a singlet state with l=0). The corresponding eigenfunctions can be abtained with the help of usual Glebsch-Gordan coefficients (for usage of GG coefficients see Particle Data Booklet):

2. and 1 nucleons of isospins ),2()1( III +=

2211213, ,,);,(,21

21λλλλλλ

λλ

iiiCi ii∑=

{ }

{ })1()2()2()1(210,0

)2()1(1,1

)1()2()2()1(210,1

)2()1(1,1

npnp

nn

npnp

pp

−=

=−

+=

=

triplet eigenvectors

singlet eigenvector

Isospin... The η and η’ are, approximately:

The vector mesons are:

The K and K* mesons form I = 1/2 doublets:

-

[ ]ssdduu 261

−+=η

[ ]ssdduu ++=31'η

[ ] uddduuduo ,21,,, −=−+ ρρρ

[ ] ss ,2

1=+= φω dduuo

21,

21

21,

21,,

21,

21

21,

21,,

−==

−==

+

−

susdKK

usdsKK

o

o

Isospin - Baryons The baryon wave functions are more complicated: three quarks, q1 q2 q3 The first two can form an I=0 state or an I=1 state assuming that they are u and d.

When the third quark is added to the I=0 state, it will give I=1/2; if added to the I=1 state, it will give either I=1/2 or I=3/2, depending on the symmetry. If the third quark has I=0, such as the s quark, then we will have I=0 or I=1. The proton and neutron have I=1/2; Σ+, Σο, Σ- have I=1, while Λ has I=0. The difference between Σο and Λ is now clear. Both are uds states, but the ud diquark is I=1 in Σο and I=0 in Λ; Ξ- and Ξo form an I=1/2 state, since they contain only one u or d quark.

[ ]122121 :0 qqqqI −=

[ ]122121 :1 qqqqI +=

Isospin - Baryons Why there is no partner of the proton with Q=2 or Q=-1. The proton wave function has minus sign in it, either from the I=0 diquark or from combining the third quark with the I=1 diquark. If all quarks are the same, these wave functions vanish. The baryon decuplet has the simplest wave function with all plus signs. The Δ's have I=3/2. Isospin conservation is an important tool in predicting relative reaction rates and decay branching fractions. The first and classic application of isospin was pp scattering by Fermi in 1951.

23,

23

21,

211,1 ==+ pπ

21,

21

32

21,

23

31

21,

211,1 −−−=−=− pπ

Isospin - Baryons For the production of the Δ++ and Δo, only the amplitude is relevant

This is not the whole story. The Δ++ always decays into π+p, but the Δo

can decay into either π-p or πon.

23

3)/()( =Δ→Δ→ −+++ opp ππσ

21,

210,1

32

21,

211,1

31

21,

23

−+−=−=Δo

31)(/)( =→Δ→Δ − allBpB oo π

933)(/)( =×=→Δ→→Δ→ −−++++ pppp o ππσππσ

isospin and weak decays even though isospin is a conserved quantity only of SI, it is useful in limiting the effect of the SI in weak decays. a typical case is the decay of D+ and Do -mesons.

Do and D+ have different isospin structures in their final states, and can have different lifetimes. In fact, they differ by a factor of 2.5.

c

u

s

u

d u W-: ΔI=1

ΔI3=1

c

d d s

d

u

W+

21,

21−

21,

21

21,

23

21,

21

23,

23

- -

-

- -

-

D0 (cu)

D+(cd )

isospin and weak decays consider the semileptonic decays: there is only one isospin final state in each case, so the partial width for these two modes must be the same. This gives a relationship between the semileptonic fractions and the lifetimes:

Do

D+

W+:ΔI=0

W+

c u

c

d d

s

u s

e+

νe

νe

e+

21,

21−

21,

21

∑ ΓΓ=Γ=ΓΓ=i

itot totiitot /B , ,/!τ

)(/)()(/)()()()()()()( ooslsl

otot

osltotsl

oslsl DDBDDBDDBDDBDD ττ =⇒Γ=Γ⇒Γ=Γ +++++

- -

- -

isospin and weak decays...

experimental results: Bsl(D+) = (17.2 ± 1.9)% Bsl(Do) = (6.75 ± 0.29)% τ(D+) = (1.057 ± 0.015)ps τ(Do) = (0.415 ± 0.004)ps Bsl(D+)/τ(D+) = (0.163 ± 0.018)ps-1

Bsl(Do)/τ(Do) = (0.163 ± 0.007)ps-1

universal conserved quantities

•  Fundamental Physics Conservation Laws –  Energy Conservation –  Conservation of Momentum –  Conservation of Angular Momentum

•  Universal Particle Physics Conservation Laws –  Baryon Number:

• 1/3 for quarks, -1/3 antiquarks –  Lepton Number

• Separate Number for each family of leptons. –  Charge Conservation

symmetries – G-parity Charge Conjugation symmetry is limited to neutral states. Isospin relates charged and neutral states for hadrons with the u and d quarks. How about a combination of the two symmetry operations? G-parity is defined as:

where R2 represents a rotation about the isospin z-axis. R2(180o) on a π+ flips it to a π-, then C flips it back to a π+. The only question is: What is the spin after these operations? To determine this we consider the I3 = 0 member of a multiplet (the only one which will be returned into itself by R2(180o)). This is the same as the effect on Yl

0(θ,ϕ) of a rotation about the y-axis in ordinary space: θ → π - θ, ϕ → π - ϕ. The Yl

0(θ,ϕ) are of the form al coslθ + al-2 cosl-2θ ... and

G ≡ CR2 (180o )

R2 (180o )Yl

o(θ,φ) = (−1)I C Ψ I3=0

symmetries – G-parity... There is an arbitrary phase for the I3 = 0 members of the multiplet, but we choose it so that all members of a multiplet have the same G-parity

Thus,an N-pion state has G = (-1)N. This is quite powerful, since it implies that a meson will decay strongly into either an even or odd number of pions, depending on its G-parity. (Note that G-parity contains no new information that we did not already have with C and isospin – it just combines these two in a useful manner.)

Note also that G-parity is useful in case the states contain u and d quarks,only. It does not work with K-mesons, since

ππ CG I ˆ)1(ˆ −=

G K − = CR2 K− =C Ko = Ko

Why not to extend these ideas to the s quark, as well? From the Σ*-Δ mass difference, the s quark should be about 150 MeV heavier than the u and d quarks. This is still a minor difference compared to the proton mass. Murray Gell-Mann followed these lines of thought in 1962. Instead of having two quarks invariant under rotations of SU(2), he took three quarks to be invariant under rotations of SU(3). Just as SU(2) is the rotation group generated by the three Pauli spin matrices, SU(3) is generated by eight 3x3 matrices. Therefore the name ”Eightfold Way”. The problem with SU(3) is that it is not a very good symmetry because of the large s-quark mass. Isospin is good to the 1% level, SU(3) is sometimes good only to the 50% level... After charm was discovered in 1976, people tried to extend flavour symmetry to SU(4), but it was useless as a symmetry, and quickly fell out of favour.

symmetries – SU(3)

symmetries – SU(3)... The most important use of SU(3) is in the construction of multiplets, which are representations of the group. For example, the SU(2) representations are familiar to us and represent the different symmetries. Spin ½ x Spin ½ → Spin 0 + Spin 1:

In SU(2), particles and antiparticles are in the same representation. In SU(3), they are not

3 = 3 = u d

s d u

s -

- -

-

2⊗ 2 =1⊗ 3

symmetries – SU(3)... A graphical way to construct the representations, say is to place the center of a on each point of b and apply the following rules:

(1) a triangle has uniform density (2) otherwise, one node on the boundary, two at the first inner layer, etc.

To construct the mesons,

= +

⇒  The nonet of mesons forms a SU(3) octet + SU(3) singlet.

dd d u

s dd

dd

ds us

du

sd su

du

ds us

su sd

du du dd dd dd

a⊗ b

3⊗ 3

3⊗ 3= 8⊗1

The singlet will be completely symmetric:

1/√3 (uu + dd + ss) This is very close to the η (a pseudoscalar), but is in fact a mixture of ω and φ (vector mesons). For the baryons, i.e. the baryons form a decuplet, two octets, and a singlet. The decuplet is completely symmetric in flavour. The singlet is completely antisymmetyric in flavour, i.e.

1/√6 (uds + dsu + sud – dus – usd – sdu). The octets have a mixed symmetry.

symmetries – SU(3)...

3⊗ 3= 6⊗ 3,3⊗ 3⊗ 3= (6⊗ 3)⊗ 3=10+8+8+1

symmetries – SU(3)... •  construct the proton wave function. •  colour (1/√6 εijk) and space (all l=0) wave functions are trivial, just

write down the flavour and spin wave functions. •  need to have spin ½ for the proton: take two quarks in a spin 0 state,

then it is automatically spin ½ . For proton spin up:

The flavour wave function must have the same symmetry for the product to be symmetric.

(which gives I= ½ )

↑↓↑−↑↓=Ψ )(21

spin

uduudflavour )(21

−=Ψ

)nspermutatio cyclic2(181

+↑↑↓−↑↓↑−↓↑↑=

↓↑↑+↓↑↑↑↑↓↑↑↓

↑↓↑+↑↓↑−↓↑↑−↓↑↑

↑↑↓+↑↑↓−↑↓↑−↑↓↑∝ΨΨ

duuduuduu

duuuu-ddu-uuudududuuududuu

uududuuududuflavourspin

symmetries – SU(3)...

symmetries – SU(3)... Before the quark model, magnetic moments of the baryons were a major puzzle. For a point spin ½ particle:

µ = q/mc S → magnitude µ = q!/2mc. We define µN = e !/2mNc, and get experimentally:

µp = 2.79 mN and µn = -1.91 mN These values, and the other magnetic moments of the baryons are (approximately) readily explained by the quark model: The quark magnetic moments will be: µu = (2/3)e!/2muc µd = -(1/3)e!/2mdc µs = -(1/3)e!/2msc

symmetries – SU(3)... The proton magnetic moment is now:

µp = < ψp µ ψp >

= (1/18) [4(2µu – µd) + µd + µd] × 3 ⇒ µp = (4/3) µu – (1/3) µd If we take mu = md = mN/3, then

µu = 2 µN and µd = - µN ⇒ µp = 3 µN compared to 2.79 – good to 7.5%. ⇒ µn = (4/3) µd– (1/3) µu = -2 µn compared to –1.91 – good to 4.5%. The usual procedure is to use µp to fix mu and md and µΛ to fix ms, then predict all the others.

symmetries – SU(3)... This is how the SU(3) is structured – helps to understand the nature of colour. The three colours form a fundamental representation of SU(3). Note: This SU(3)colour has nothing to do with SU(3)flavour. Red, green, and blue Anti-red, anti-green, are a 3 of colour. and anti-blue form a 3 of colour. For mesons, qq, ( this singlet is completely symmetric = (1/√3)δij) For baryons, qqq, (this singlet is completely anti- symmetric = 1/ √6 εijk)

HADRONS ARE COLOUR SINGLETS ≡ COLOURLESS.

-

3⊗ 3 = 8⊗1

3⊗ 3⊗ 3=10⊗8⊗8⊗1

quark conserved quantities

•  Q represents the charge, B the Baryon number, I Isospin, J spin, P parity, I3 Isospin Projection, S Strangeness, Bottomness, T Topness.

•  All quarks have zero lepton number. Anti-quarks have the opposite sign for all of the above numbers.

Quark Mass (MeV) Q B I(JP) I3 S C B~ T u (up) 1-5 +2/3 1/3 ½( ½+) ½ 0 0 0 0

d (down) 3-9 -1/3 1/3 ½( ½+) -½ 0 0 0 0 s (strange) 75-170 -1/3 1/3 0( ½+) 0 -1 0 0 0 c (charm) 1.15-1.35 GeV +2/3 1/3 0( ½+) 0 0 1 0 0 b (bottom) 4.0-4.4 GeV -1/3 1/3 0( ½+) 0 0 0 -1 0

t (top) 174.3±5.1 GeV +2/3 1/3 0( ½+) 0 0 0 0 1

lepton conserved quantities

•  Leptons have no isospin. Antiparticles have opposite charge and lepton number.

•  Isospin related to up and down quark flavours.

Lepton Mass (MeV) τ (second) Q L JP Le Lµ Lτ Decay BR e (electron) 0.510 stable -1 1 ½+ 1 0 0 νe (neutrino) ~0 stable 0 1 ½+ 1 0 0

µ (muon) 105.6 2.2 × 10-6 -1 1 ½+ 0 1 0

µνν ee−→ γνν µee −→

98.6% 1.4%

νµ ~0 stable 0 1 ½+ 0 1 0

τ (tau) 1.78 GeV 3.0 × 10-13

-1 1 ½+ 0 0 1 τµννµ −→ µνν ee−→

→hadrons

18% 18% 64%

ντ ~0 stable 0 1 ½+ 0 0 1

particle properties

0 +1 0 0 -1 +1 0 0 ±1 0 -1 Q

+1 0 0 -½ ½ 0 No 497.7 K0 -1 0 0 -½ ½ 0 No 493.6 K-

+1 0 0 +½ ½ 0 No 493.6 K+

1 1 0

0 0 0 0 0 B

0 0 -½ ½ ½ No 939.6 n 0 0 ½ ½ ½ Yes 938.2 p -1 0 +½ ½ 0 No 497.7 K0

0 0 0 0 0 No 548.8 η

0 0 1 1 L

0 0 1 0 No 134.9 π0

0 ±1 1 0 No 139.6 π±

0 ½ Yes ~0 ν

0 ½ Yes 0.51 e S I3 I Spin Stable Mass Particle

conservation

Lτ Lµ

Le

Time Reversal

Charge Conjugation

Parity Flavour Strangeness

Lepton # Baryon #

Weak EM Strong Quantity

symmetry and non-observables Non-observables Symmetry

Transformations Conservation Laws or Selection Rules

Difference between identical particles

Permutation B.-E. or F.D. statistics

Absolute spatial position Space translation momentum

Absolute time Time translation energy

Absolute spatial direction Rotation angular momentum

Absolute velocity Lorentz transformation generators of the Lorentz group

Absolute right (or left) parity

Absolute sign of electric charge charge conjugation

Relative phase between states of different charge Q

charge

Relative phase between states of different baryon number B

baryon number

Relative phase between states of different lepton number L

lepton number

Difference between different co- herent mixture of p and n states

isospin

r!→−r!

e e→−

t→ t +τr!→ r!+Δ"!

iQe θψ ψ→

ψ→ eiNθψiLe θψ ψ→

p pU

n n⎛ ⎞ ⎛ ⎞

→⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r!→ "r!