Lecture 6: Symmetries I
•
Symmetry & Unifying Electricity/Magnetism
•
Space-Time Symmetries
•
Gauge Invariance in Electromagnetism
•
Noether’s Theorem
•
Isospin
•
Parity
Useful Sections in Martin & Shaw:
Sections 5.3, 6.1, Append C.1, C.2
When something is changed,
something else is unchanged
Lab Frame
v
+q
 +
 +
 +
F
I
(pure magnetic)
B
 +
In Frame of
Test Charge
+q
F
(pure electrostatic)

+
+
Symmetry:
The effect of a force
looks the same when
viewed from reference
frames boosted in the
perpendicular direction

Electricity & Magnetism
are identically the same
force, just viewed from
different reference frames
 +
+

UNIFICATION !!
 +
(thanks to Lorentz invariance)
Lorentz Lorentz
expanded contracted
Lab Frame
v
+q
 +
 +
 +
F
I
(pure magnetic)
B
 +
In Frame of
Test Charge
+q
F
(pure electrostatic)

+
+
 +
+
 +
Lorentz Lorentz
expanded contracted
Symmetry of
Maxwell’s Equations

Relativity !!
Space-Time Symmetries
m1
F
m2
x1
x2
|
|
If the force does not change when translated to a different point in space, then
force felt at x2 : F = m2 x2
recoil felt at x1
subtracting:
¨
: F = m ¨
x
m ¨
x + m x¨ = 0
1 1
2
˙
2
1 1
˙
d/dt [ m2 x2 + m1 x1 ] = 0
m2 v2 + m1 v1 = constant
Translational Invariance  Conservation of Linear Momentum
Consider a system with total energy
E=
1
m
2
x2
assume this basic
description also holds
at other times
+V
dE
dV dx
= mxx +
dt
dx dt
dV
= mxx +
x
dx
but dV/dx = F = mx

(Newton’s 2nd law)
dE
= m x x m x x = 0
dt
 E = constant
Time Invariance  Conservation of Energy
Gauge Invariance in Electromagnetism:
''local"
symmetry

A  A + ∇(x,t)
    (x,t)
t
 x,t A

 ∇[
+
∇(x,t)
t
t

E = ∇ A
t

= ∇ A
=E
t
B=∇A
 ∇  [A + ∇(x,t)]
=∇A =B
Gauge Invariance  Conservation of Charge
(Wigner, 1949)
To see this, assume charge were not conserved
So a charge could
be created here by
inputing energy E
And destroyed here,
with the output of
some energy E
PoP !
PoP ! q
x1
E = q(x1)
x2
E = q(x2)
Thus we will have created an overall energy E  E = q { (x2)  (x1) }
So, to preserve energy conservation, if  is allowed to
vary as a function of position, charge must be conserved
Noether’s Theorem
Continuous Symmetries  Conserved ''Currents"
(Emmy Noether, 1917)
Gauge symmetry from another angle...
Take the gauge transformation of a wavefunction to be
  eiq
where  is an arbitrary ''phase-shift" as a function of space and time
Say we want the Schrodinger equation to
be invariant under such a transformation
clearly we’re in trouble !

∂
i
2
 2m
∇
∂t
Consider the time-derivative for a simple plane wave:  = Aei(px-Et)
  Aei(pxEt+q)
/t  = i ( E + q /t ) 
Note that if we now introduce
an electric field, the energy level
gets shifted by q
here’s the problem!
/t  = i ( E + q + q /t ) 
But we can transform    /t, thus cancelling the offending term!
(a similar argument holds for the spatial derivative and the vector potential)
Gauge invariance REQUIRES Electromagnetism !!
Another example...
Special Relativity:
Invariance with respect to reference frames
moving at constant velocity  global symmetry
Generalize to allow velocity to vary arbitrarily at different points
in space and time (i.e. acceleration)  local gauge symmetry
Require an interaction
to make this work
GRAVITY!
All known forces in nature are
consequences of an underlying
gauge symmetry !!
or perhaps
Gauge symmetries are found
to result from all the known
forces in nature !!
Pragmatism:
Symmetries (and asymmetries) in nature are
often clear and can thus be useful in leading to
dynamical descriptions of fundamental processes
True even for ''approximate" symmetries !
Isospin
Note that mp = 938.3 MeV
mn = 939.6 MeV
Both are found in the nucleus, apparently held to each other by pions
n
)
n
)
p
p
proton & neutron appear to be swapping identities  ''exchange force"
This means, to conserve charge, pions must come in 3 types: q = -1, 0, +1
So there appears to an ''approximate" symmetry here
Noether’s theorem says something must be conserved...
Call this ''Isospin" in analogy with normal spin,
so the neutron is just a ''flipped" version of the proton
I=1/2 System 
(just 2 states)
p
n
I3 : 1/2 -1/2
Some way pions can be produced
Impose isospin
conservation
p + p  p + n + +
I3(+) = +1
 p + p + 
 p + p +  +  
 I=1 for the pions
I3() = 0
I3() = -1
(similar arguments for other particle systems)
So we can think of these particle ''states" as the
result of a (continuous) ''rotation" in isospin-space
Example:
What are the possible values of the isotopic spin
and it’s z-component for the following systems
of particles:
a) + + p
b)  + p
p
a)
p : I = 1/2, I3 = +1/2
n
+
+ : I = 1, I3 = +1


total I3 = 1 + 1/2 = 3/2
thus, the only value of total Isospin we can have is also I = 3/2
b)
p : I = 1/2, I3 = +1/2
 : I = 1, I3 = 1
total I3 = 1 + 1/2 = 1/2
thus, possible values of total Isospin are: I = 1  1/2 = 1/2
or I = 1  1/2 = 3/2
Parity
P F(x) = F(-x)
 discreet symmetry (no conserved ''currents")
consider the scattering
probability of the following:
y
m2
m1
x
y
P x = x
m1
P dx/dt =  dx/dt
m2
x
y
m1
m2
So, even though x and dx/dt are each odd under parity, the scattering probability,
PS , is even (i.e. P PS = PS)
 parity is multiplicative, not additive
x
Also note that parity does not reverse the direction of spin!
+z
z
z
z
flip z
+z
flip ''velocity"
direction
+z
z
+z
(stand on your head)
parity
But, for orbital angular momentum in a system of particles,
it depends on the symmetry of the spatial wave function!!
+z
z
z
z
z
flip z
+z
flip ''velocity"
direction
+z
flip x & y
positions
+z
parity
+z
not the same !
z
(stand on your head)
Intrinsic parity of the photon from ''first-principles":
∇ E(x,t) = (x,t)/0
P (x,t) = (x,t)
P ∇ = ∇
thus, we must have P E(x,t) = E(-x,t)
for Poisson’s equation to remain invariant
But also E = ∇  A/t =  A/t
(in absence of free charges)
and since /t doesn’t change the parity  P A(x,t) = A(x,t)
But A basically corresponds to the photon wave function:
A(x,t) = N (k) exp[i(kxt)]
Thus, the intrinsic parity of the photon is 1 (or  = 1 )
However, the effective parity depends on the angular momentum
carried away by the photon from the system which produced it:
P =  (1) l
(i.e. radiation could be s-wave, p-wave etc.)
but for an isolated photon, this cannot be disentangled!!
The (547) meson has spin 0 and is observed to decay via the
electromagnetic interaction through the channels:
Example:
  0 +  0 + 0
   +  + 0
and
From this, deduce the intrinsic parity of the  and explain why
the decays:   0 + 0
and    +  are never seen
 = (
)3
L12
L3
(1) (1)
L12
but final state must have zero total angular
momentum since the initial state has spin 0
L3

 L12 = L3
Ltot = L12 + L3 = 0
 = (

1
P = P P P
L3 2
{(1) }
)3
= ()3 = ( )3 = 1
However, for 2-pion final states we would have:
but we must have L=0, so
P = ()2 = 1
P = ()2 (1)L
and is thus forbidden
Evidence for Parity Conservation
(in strong/electromagnetic interactions)
1) Polarized protons scattering off a nucleus show no
obvious asymmetry towards spin-up vs spin-down directions
2) Ground state of deuteron (np) has total angular momentum
J=1 and spin S=1. Thus, the orbital angular momentum could
take on values of l = 0 (m=1), l = 1 (m=0) or l = 2 (m = 1)
But the observed magnetic moment is consistent with a
superposition of only S and D waves (l=0, 2). This can be
reconciled if
P (p+n) = P (d)
p n = p n (1) l
By convention,
p = n  +1
& also
 so l must be even
e-  +1
 and the relative parities of the other particles then follow
(anti-fermions have the opposite parity, anti-bosons have the same parity)
Parity is a different animal from other symmetries in many respects...
It is often impossible to determine the absolute parity (assigned +1 or -1)
of many particles or classes of particles. So we essentially just assume that
basic physical processes are invariant with respect to parity and construct
theories accordingly, making arbitrary assignments of parity when necessary
until we run into trouble.
Weak interaction violates parity !!
(so, ''left" and ''right" really matter... weird!)