Lecture 7: Symmetries II• Charge Conjugation • Time Reversal• CPT Theorem• Baryon & Lepton Number• Strangeness• Applying Conservation Laws

Section 4.6, Section 2.2

Useful Sections in Martin & Shaw:

Changes particle to anti-particle (without affecting linear or angular momentum)

Electromagnetism is obviously symmetric with respect to C-parity(flip the signs of all charges and who would know?). It turns outthat the strong force is as well.

For particles with distinct anti-particles (x = e, p, , n, ...)

C ∣ x, > = ∣ x, > where ∣ x, > a state characterizedby a particle x anda wave function

For particles which no not have distinct anti-particles (y = , , ...)

C ∣ y, > = Cy ∣ y, >

where Cy is a ''phase factor" of 1

(like for parity) used to determine C-conservation in interactions

C-Parity (charge conjugation)

But, again, not the weak force(otherwise there would be left-handed anti-neutrinos that couple ina similar way to left-handed neutrinos... there aren’t !)

For multi-particle systems

C ∣ x, x > = ∣ x, x > = | x, x >

depending on whether the system is symmetric or antisymmetric under the operation

Example: consider a pair in a state of definite orbital angular momentum L

C ∣ , L > = (-1)L ∣ , L >

since interchanging and reverses their relative position vector in the spatial wave function

Experimentally, 0 but never 0

C ∣ > = C ∣ >

C ∣ > = CC(-1)L∣ >

= C2 ∣ >

= ∣ >

C

using a similar argument

as for parity, C = 1

C ∣ > = CCC (-1)L ∣ >

= C2 C ∣ >

= ∣ >

= C ∣ >

Ah! So we can never get 0 if C-parity is conserved !!

Example:How is this reconciled with the concept of C-parity ??

But 0 spin is zero, so L=0

Time ReversalIn analogy with parity, we could try t t

Now apply t t H (x,t) = i ℏ (x,t) t

But we want the Schrodinger equation to be invariant!

Both (x,t) and *(x,-t) satisfy the same equation.

Thus we define T S(x,t) = S*(x, t)

H (x,t) = i ℏ (x,t) t

However, note that if we start with

H *(x,t) = i ℏ *(x,t) t

This can be patched up by taking the complex conjugate

T [H (x,t)] = T i ℏ (x,t) ] t

Wigner, 1931

Note that if x,t x,t T Tx,t Tx,t *{ Tx,t } *{ Tx,t }

≠ { Tx,t } { Tx,t } ''anti-linear"

So the operator is not Hermitian and the eigenvalues are not real!

So, unlike other symmetries, time-reversal does not give rise toreal conserved quantities (i.e. no conservation laws per se)

Hermitian * *

so n* G

n) g

n

n*

n g

n

(G n)*

n g

n*

n*

n g

n*

gn g

n* so g

n must be real

In practice this is difficult to test (how do you ''reverse time" in an experiment?)

It’s more useful to consider the combination T P :

P L = P (x m dx/dt) = x m dx/dt T P L = (x m dx/dt) = L

(assume this holds for spin)

a(pa, s

a) + b(p

b, s

b) a(p

a, s

a) + b(p

b, s

b)

a(pa, s

a) + b(p

b, s

b) a(p

a, s

a) + b(p

b, s

b)

apply T P

the spin-averagedrates of these shouldthus be the sameunder T P symmetry

''Principle of Detailed Balance" (confirmed in various EM & strong interactions)

Note that P p = pT P p = p

and

-X +X

ChargeConjugation

ChargeConjugation

-X +X -X +X

Flip orientation in time

Parity

ChargeConjugation Parity

-X +X -X +X +X -X

Flip orientation in time

Switch coordinatedefinitions

ChargeConjugation Parity

-X +X -X +X +X -X

Flip orientation in time

Switch coordinatedefinitions

ChargeConjugation Parity

-X +X -X +X

Flip orientation in time

Switch coordinatedefinitions

-X+X

ChargeConjugation Parity

-X +X -X +X -X +X

Flip orientation in time

Switch coordinatedefinitions

ChargeConjugation Parity

-X +X -X +X -X +X

Flip orientation in time

Switch coordinatedefinitions

TimeReversal

ChargeConjugation Parity

TimeReversal

-X +X -X +X -X +X -X +X

Flip orientation in time

Switch coordinatedefinitions

Run movie backwards in time

CPT

1) Integer spin particles obey Bose-Einstein statistics (bosons)

2) 1/2 - spin particles obey Fermi-Dirac statistics (fermions)

3) Particles and antiparticles must have identical masses & lifetimes

4) All internal quantum numbers of antiparticles are opposite to those of the corresponding particles

(independently discovered by Pauli, Luders and Bell and Schwinger)

States that if a quantum field theory is invariant under Lorentz transformation, then C P T is an exact symmetry !!

( Note that if, for example, CP is violated, then T must be violated )

CPT Theorem

It is an empirical observation that the number of baryons (fermions with masses the proton mass) minus the number of antibaryons is conserved in all reaction thus far observed. Thus we define the ''baryon number"B (# baryons) - (# antibaryons) as a conserved quantity

The same has been assumed to be true for Leptons.

e

e

Le L L

L}} } }

HOT OFF THE PRESS!These can be violated by ''neutrino oscillations"

Baryon and Lepton Number

NO Experimentalevidence that this is the case!!!

However, there is also a form of the rule that seems to operate which relates to individual lepton ''families" or ''generations" :

Among other things, conservation of B and L means that protons and electrons don’t decay (so matter is stable) and baryons don’t mix with leptons.

GUT models therefore predict these laws to break down at some point

Transformation Symmetry Type Conserved Quantity

translation global, continuous linear momentum rotation global, continuous angular momentumtime global, continuous energy(Lorentz global, continuous CM velocity)

rotation in ''isospin-space" global, continuous isospin

electromagnetic scalar/vector potential local, continuous, gauge charge

(global, continuous, gauge) baryon number (global, continuous, gauge) lepton number

parity global, discreet ''P-value"charge conjugation global, discreet ''C-value"time reversal global, discreet none!

(''space-time")

(''internal")additive

multiplicative

Symmetry Summary:

In Addition: Local, Lorentz-Invariant, Quantum Field Theories CPT

p

K0

Strangeness

Originally found in cosmic ray cloud chambers in 1947,Then in bubble chambers at the Brookhaven Cosmotron in 1953

The new particles were always produced in pairs (''associated production"),suggesting a new conserved quantity ''strangeness"

so define S 1 and SK = +1

The cross-section for production indicates a strong interaction,however the decay timescale is much longer than expected fora strong decay:

tS ~ 1 fm / c = (1015m) / (3x108m/s) = 1023s

but the observed decays took 1010s

It can be shown that this is about what you’d expect for a weak decay:

(practically forever!)

First consider: n p + e + e

from Fermi Golden Rule: 1/t (density of final states)

dNe ~ p

e2 dp

e

dN = dNe dN ~ p

e2 dp

e p

2 dp

since the distributions for e and

e are unconstrained

in a 3-body decay

If E Ee + E

then, for a given value of Ee , dp ≃ dE

Thus, d = dN/dE

and, in the limit Ee ≫ m

e ,

E5

and, assuming the proton gets basically no kinetic energy,the energy, E , available for the e &

e is determined.

p ≃ E Ee

~ pe2 (E

E

e)2 dp

e = dN/dp

~ Ee2 (E

E

e)2 dE

e

E

0

dN ~ p2 dp

For -decay, E ≃ mn m

p ~ 1.3 MeV

and the neutron lifetime is ~ 1000 s

In case of the strange particles: m = 1116 MeV m

K = 498 MeV

Thus, we’d expect a weak decay timescale of order

tW

= 1000 s (1.3/750)5 = 1011s ( which is certainly alot closer to what is actually observed! )

Interpretation:S is conserved by the strong interaction, which is why these particles are produced in pairs and why the individual particles cannot undergo strong decay to non-strange products.

so take an average value of E~750 MeV

However, S is not conserved by the weak interaction, which eventually does allow the and K0 to decay !

For “1st-order" Weak Interactions: S = 0, 1

Applying Conservation Laws:

Example: In the following pairs of proposed reactions, determine which ones are allowed and the relevant force at work

+ p 0 + 0 + p 0 + K0 + n + p

Interaction:

charge:

leptonnumber:

baryon number:

strangeness:

Isospin (I3) :

1 + 1 = 0 + 0

0 + 0 = 0 + 0

0 + 1 = 1 + 0

1 + 1 = 0 + 0

0 + 0 = 0 + 0

0 + 1 = 1 + 0

0 + 0 = 1 + 1

1 + 1/2 = 0 1/2

strong

1 = 1 + 0

0 = 0 + 0

1 = 0 + 1

weak

1 = 1 + 1

0 + 0 = 1 + 0

1 + 1/2 = 0 + 0

0 = 0 + 0

1 = 0 + 1

X

Example:

+ X

Lepton number: 0 = 0 + ? L = 0

Baryon number: 0 = 0 + ? B = 0

Spin: 0 = 0 + ? J = 0

Candidates:

But, mX < m m X =

(X)

Identify the neutral particle

X

Y

p

X +

Lepton #: ? = 0 + 0 L = 0Baryon #: ? = 0 + 0 B = 0Spin: ? = 0 + 0 J = 0

(as before) Candidates:

But, mX

> m + m

& < 1018 s (c < 0.3 nm)

X = K or K

Y + pLepton #: ? = 0 + 0 L = 0Baryon #: ? = 0 + 1 B = 1Spin: ? = 0 + 1/2 J = 1/2

Candidates: n

But, mY

> m + mp

(Yn)

(X)

(X)

+ (Y)

S=0,1 (weak decay) (Y)

Y =

+p X +

Strangeness: 0 + 0 = ? + S = +1

strong interaction

X = K