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Lecture 7: Symmetries II
•
Charge Conjugation
•
Time Reversal
•
CPT Theorem
•
Baryon & Lepton Number
•
Strangeness
•
Applying Conservation Laws
Useful Sections in Martin & Shaw:
Section 4.6, Section 2.2
C-Parity (charge conjugation)
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 out
that the strong force is as well. But, again, not the weak force
(otherwise there would be left-handed anti-neutrinos that couple in
a similar way to left-handed neutrinos... there aren’t !)
For particles with distinct anti-particles (x = e, p, , n, ...)
C ∣ x, > = ∣ x,  > where ∣
a state characterized
x,  > by
 a particle x and
a wave function 
For particles which no not have distinct anti-particles (y = , , ...)
C∣
where Cy is a ''phase factor" of 1
y,  > = Cy ∣ y,  (like
> for parity) used to determine
C-conservation in interactions
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
>
Example:
Experimentally, 0   but never
0  
How is this reconciled with the concept of C-parity ??
C∣
C

> = C ∣

>
using a similar argument
as for parity, C = 1
C ∣  > = CC(-1)L∣  > C ∣  > = CCC (-1)L ∣ 
But 0 spin is zero, so L=0
= C2 ∣  >
= ∣  >
Ah! So we can never get
0  
if C-parity is conserved !!
= C2 C∣  >
= C∣  >
= ∣  >
Time Reversal
In analogy with parity, we could try
However, note that if we start with

H (x,t) = i ℏ (x,t)
t

Now apply t  t H (x,t) = i ℏ t(x,t)
t  t
But we want the Schrodinger equation to be invariant!
Wigner,
1931
This can be patched up by taking the complex conjugate
T [H (x,t)] = T i ℏ

(x,t) ]
t

*(x,t)
t
Both (x,t) and *(x,-t) satisfy the same equation.
H *(x,t) = i ℏ
Thus we define
T S(x,t) = S*(x, t)
x,tx,t
T Tx,tTx,t
*{ Tx,t}*{ Tx,t}
≠{ Tx,t}{ Tx,t}
Note that if
''anti-linear"
So the operator is not Hermitian and the eigenvalues are not real!
 *  *
 n*G n)  gn  n* n  gn
 (G n)* n  gn* n* n  gn*
Hermitian 
so
 gn gn*
so gn must be real
So, unlike other symmetries, time-reversal does not give rise to
real conserved quantities (i.e. no conservation laws per se)
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 :
Note that
P p = p
TPp=p
and
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, sa) + b(pb, sb)  a(pa, sa) + b(pb, sb)
apply
T P  a(pa, sa) + b(pb, sb)  a(pa, sa) + b(pb, sb)
the spin-averaged
rates of these should
thus be the same
under T P symmetry
''Principle of Detailed Balance" (confirmed in various EM & strong interactions)
Charge
Conjugation
-X
+X
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
+X
-X
Switch
coordinate
definitions
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
+X -X
Switch
coordinate
definitions
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
-X
+X
Switch
coordinate
definitions
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
-X +X
Switch
coordinate
definitions
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
-X
Time
Reversal
+X
Switch
coordinate
definitions
Charge
Conjugation
-X
+X
-X
Parity
+X
Flip
orientation
in time
-X
Time
Reversal
+X
Switch
coordinate
definitions
-X
+X
Run movie
backwards
in time
CPT
CPT Theorem
(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 !!
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
( Note that if, for example, CP is violated, then T must be violated )
Baryon and Lepton Number
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 NO Experimental
The same has been assumed to be true for Leptons.
evidence 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" :



e  
e 
}
}
}
HOT OFF THE PRESS!
These can be violated by
''neutrino oscillations"
Le L L
}L
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
Symmetry Summary:
Transformation
Symmetry Type
translation
rotation
time
(Lorentz
global, continuous
global, continuous
global, continuous
global, continuous
(''space-time")
rotation in
''isospin-space"
global, continuous
(''internal")
electromagnetic
scalar/vector potential


parity
charge conjugation
time reversal
Conserved Quantity
local, continuous, gauge
linear momentum
angular momentum
energy
CM velocity)
isospin
additive
charge
(global, continuous, gauge)
(global, continuous, gauge)
baryon number
lepton number
global, discreet
global, discreet
global, discreet
''P-value"
''C-value"
none!
multiplicative
In Addition: Local, Lorentz-Invariant, Quantum Field Theories  CPT
Strangeness






K0



p
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 for
a strong decay:
tS ~ 1 fm / c = (1015m) / (3x108m/s) = 1023s
but the observed decays took 1010s
(practically forever!)
It can be shown that this is about
what you’d expect for a weak decay:
First consider:
n  p + e + e
from Fermi Golden Rule: 1/t   (density of final states)
dNe ~ pe2 dpe
dN ~ p2 dp
 dN = dNe dN ~ pe2 dpe p2 dp
since the distributions for
e and e are unconstrained
in a 3-body decay
and, assuming the proton gets basically no kinetic energy,
the energy, E , available for the e & e is determined.
If E  Ee + E
 p ≃ E  Ee
then, for a given value of Ee ,
Thus, d = dN/dE
dp ≃ dE
= dN/dp ~ pe2 (E Ee)2 dpe
E
and, in the limit Ee ≫ me ,
 ~  Ee2 (E Ee)2 dEe
0

  E5
For -decay, E ≃ mn  mp ~ 1.3 MeV
and the neutron lifetime is ~ 1000 s
In case of the strange particles: m = 1116 MeV
mK = 498 MeV
so take an
average value
of E~750 MeV
Thus, we’d expect a weak decay timescale of order
tW = 1000 s (1.3/750)5 = 1011s
( 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.
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
strong
Interaction:
   + p
weak
charge:
1 + 1 = 0 + 0
1 + 1 = 0 + 0
1 = 1 + 0
1 = 1 + 1
lepton
number:
0+0=0+0
0+0=0+0
0=0+0
0=0+0
baryon
number:
0+1=1+0
0+1=1+0
1=0+1
1=0+1
strangeness:
0 + 0 = 1 + 0
0+0=1+1
Isospin (I3) :
1 + 1/2 = 0 + 0
1 + 1/2 = 0  1/2

Example:

Identify the neutral particle
X
    + X
Lepton number: 0
= 0 + ?
Baryon number: 0
= 0 + ?
 L=0
 B=0
= 0 + ?
 J=0
Spin:
0
Candidates: 
But, mX < m  m
(X)
X = 
X   + 
Lepton #: ? = 0 + 0
Baryon #: ? = 0 + 0
Spin:
? = 0 + 0
L=0
B=0
J=0
(as before) Candidates: 
But,
&
mX > m + m
 < 1018 s
(c < 0.3 nm)
+p  X + 
Strangeness: 0 + 0 = ? + 
X = K

(X)
p
Y   + p
Lepton #: ? = 0 + 0
Baryon #: ? = 0 + 1
Spin:
? = 0 + 1/2
strong
interaction
 S = +1

Y
(X)
X = Kor K



X
L=0
B=1
 J = 1/2
Candidates: n 
But, mY > m + mp
(Yn)
   + 
(Y)
S=0,1 (weak decay) (Y)
Y=