Download Lect25_Parity

A Review of Some INVARIANCE PRINCIPLES
Hermitian Operators: R† = R
Unitary Operators: R† = R-1
•can be diagonalized
•have real eigenvalues
•define the physical
observables
•represent transformations of basis (vectors/functions)
•can transform operators & wavefunctions by similarity
transformations that leave the Lagrangian (and therefore the
equations of motion) invariant. (include rotations/translations)
AA = RAR† and     =R so that all computed
matrix elements:
< |A| > = < |A | >
= < |R†(RAR†)R| > = < |A| >
Furthermore:
If the Hamiltonian is unchanged by the similarity transformation i.e.
if RHR† =H then RH = HR  [H,R] = 0 and the eigenvalues
are constants of the equations of motion, i.e., we have conserved quantities
If Both Hermitian and Unitary:
R†R = R-1R = R2 = 1
R has eigenvalue of 1
this also implies:
det(R)det(R) = det(I) = 1
det(R) = 1
We’ve focussed a lot on the special det(R)=+1 cases:
simple
SO(n) rotations
SU(n)
• These are “continuous symmetries”: seamlessly connected
to the identity matrix I by infinitesimal transformations
• expressible by exponential sums of infinitesimal transformations
• such operators can be used to define a CURRENT
that obeys the continuity equation
• which we can identify with an associated CHARGE
What about the R = - class of transformations?
Recall
Rz =
cosq sinq
-sinq cosq
0
0
then compare that to
0
0
1
-cosq sinq
+sinq cosq
0
0
0
0
1
or
cosq sinq 0
-sinq cosq 0
0
0 -1
or
y
-1 0 0
0 +1 0
0 0 +1
x
z
y
+1 0 0
0 -1 0
0 0 +1
x
z
PARITY TRANSFORMATIONS ALL are equivalent
to a reflection
y
(axis inversion)
plus a rotation
x'
y
x
z
x
z
y
y
x
x
z
z
y'
y
z'
x'
z
y'
x
-1 0 0
0 -1 0
0 0 -1
The PARITY OPERATOR on
3-dim space vectors
every point is carried through the origin
to the diametrically opposite location
y
x
z
Wave functions MAY or MAY NOT have a well-defined parity
(even or odd functions…or NEITHER)
 = cos x
P  = cos(- x) = cos x = +
P
= +1
 = sin x
P = sin(- x) = - sin x = -
P
= -1
but the more general
 = cos x + sin x
P  = cos x - sin x  
However for any spherically symmetric potential, the Hamiltonian:
→
→
H(-r)
= H(r)
 H(r)
[ P, H ] = 0
So the bound states of such a system have DEFINITE PARITY!
That means, for example, all the wave functions of the hydrogen atom!
52
1 Z
- Zr / 2 a
 21-1 =
sin q e -i
  re
8  a 
32
 100 =
32
 200 =
1  z  - zr / a
  e
 a
1  Z   Zr  - Zr / 2 a
  1 -  e
  2a   2a 
52
1 Z 
- Zr / 2 a
 210 =
cos q
  re
2
a
 
52
1 Z
- Zr / 2 a
 211 =
sin q ei
  re
8  a
1
 300 =
81 3
Z
 
a
32

Zr
Z 2 r 2  - Zr / 3a
 27 - 18 + 2 2 e
a
a 

Y(r)=c(r)ml (q, )
ml (q,
)=
the angular part of the solutions
are the SPHERICAL HARMONICS
(2l + 1)( l- m)! m
im
P
(cosq)e
l
4( l + m)!
Pml (cosq) = (-1)msinmq [(
d
d (cosq)
1
d
Pl (cosq) = 2l l! [( d (cosq)
)m Pl (cosq)]
)l (-sin2q)l ]
The Spherical Harmonics Yℓ,m(q,)
ℓ=0
ℓ=1
1
Y00 =
4
3
Y11 = -
8
3
Y10 =
4
sin q
15
4
2
Y21 = -
15
8
i
e
Y32 =
2
sin q
e
2i
sin q cos q
3

Y20 =
cos2 q

4  2
15
Y33 = -
cos q
1
Y22 =
ℓ=2
ℓ=3
i
e
1
- 
2
1
35
4
4
1
105
4
2
Y31 = -
3
sin q
2
sin q cos q
1
21
4
4
e
2i


i
sin q 5 cos2 q - 1 e
5

Y30 =
cos3 q

4  2
7
3i
e
3

- cos q 
2

z
then note r  -r means
q
-q
q-q
 +
y

so: eim  eimeim
x
(eim=(-1)m
and: Pml (cosq) Pml (cos(-q)) = Pml (-cosq)
(-sin2q)l = (-1-cos2ql
but d/d(cosq)

- d/d(cosq)
ml (q,
)=
(2l + 1)( l- m)! m
im
P
(cosq)e
l
4( l + m)!
d
Pml (cosq) = (-1)m(1-cos2qm [( d (cosq)
1
d
Pl (cosq) = 2l l! [( d (cosq)
)m Pl (cosq)]
)l (-sin2q)l ]
So under the parity transformation:
P:ml (q, ) =ml (-q,+ )=(-1)l(-1)m(-1)m ml (q, )
= (-1)l(-1)2m ml (q, ) )=(-1)l ml (q, )
An atomic state’s parity is determined by its angular momentum
l=0 (s-state)
l=1 (p-state)
l=2 (d-state)
 constant
 cosq
 (3cos2q-1)
parity = +1
parity = -1
parity = +1
Spherical harmonics have (-1)l parity.
When acting on a vector, the parity operator gives:


Pv = -v
  = ( -v )  ( -v ) = + v  v
P v1  v2
1
2
1
2
scalars have
positive parity!
= v1v2 cos( -q ) = v1v2 cosq

  

 

P (c = a  b ) = (-a )  (-b ) = +a  b = + c ???
This confusion arises from the indefinite nature of cross products…
their direction is DEFINED by a convention…the right-hand rule.
and reflections (parity!) change right hands  left hands!
We call such derived/defined vectors PSEUDO-VECTORS
(or “axial” vectors)
→
as opposed to POLAR VECTORS, like v.
QUANTITY PARITY Comments
r
p
-r
-p
L

+L
+
E
B
-E
+B
 •B
 •E
•p
+ •B
- •E
- • p
position
momentum
polar vectors
angular momentum
intrinsic “spin”
axial or
pseudovectors
electric field E = -V/r
B = ×A
pseudo-vector
magnetic dipole moment
electric dipole moment
longitudinal polarization
 •(p1×p2) + •(p1×p2) transverse polarization
Potential problems with this concept…or assuming its invariance…
Lot’s of physics uses Right Hand Rules and cross-products…
angular momentum
  
L = r  p has POSITIVE parity
the Lorentz Force Law

 v 
F = q( E + c  B )

 F

d 
E=
F =m v
dt vectors! q
 
v B
vector
Just like you can’t add vectors to scalars, cannot
(should not try) to add vectors and pseudovectors
at least for any theory that should respect parity invariance
But notice a vector crossed with a pseudo-vector:


 
 
P (v  c ) = (-v )  (+c ) = - v  c
  
where c = a  b
is a vector!
→
Might B really be a pseudo-vector quantity after all?
  
B =  A
the “vector” potential
is a true vector!
and as we have argued in quantum mechanics/high energy theory
is the more fundamental field!
In our Lagrangians we identify the currents as vectors.
Of particular interest:
→ →
J•A
of electromagnetic interactions
→
J is a vector
→
A is a vector field
i.e., the photon is a vector particle
with “odd” parity!
So electromagnetic interactions conserve parity.
By the same argument QCD Color (the STRONG force) interactions
…its Lagrangian terms all involve inner products of all 4-vectors…
conserve parity!
A mono-chromatic electromagnetic wave
is composed of n mono-energetic photons
each with
E = 


p = k
 2
k =

In addition to energy and momentum
light (the photon)
also carries angular momentum
(its spin!)
1909 Poynting predicts circularly polarized electromagnetic waves
carry angular momentum
proposes a test: if incident upon an absorber, the
absorber should rotate
1936 Richard Beth detects the angular momentum of light
transferred to matter.
photon momentum transferred to a “half-wave plate”,
a macroscopic object hung on a fiber
(but with a non-vanishing torsion constant).
R. A. Beth Phys.Rev. 50, 115 (1936).
1964 P.J.Allen updates the experiment
oil drop
glass fiber
glass
bead
1.65 cm wire
P. J. Allen Am.J.Phys. 34, 1185 (1964).
Microwave generator beamed up cavity
sets rotor in motion
E
=
LZ
If delivered by n photons, then E = nħ
with
means
LZ = n jZ
jZ = ħ
Each photon has spin 1.
Energy level diagram for Hydrogen
Transitions occur between adjacent angular momentum states (i.e. ℓ ±1).
2S
2P
2D
2F
N shell
n=4
M shell
n=3
L shell
n=2
K shell
n=1
ℓ =0
ℓ =1
ℓ =2
ℓ =3
The parity of a state must change in such an “electric dipole transition”
ℓ=±1, S=0, J=0, ±1
s, d, g, …
even parity
p, f, h, …
odd parity
The atomic transitions responsible for
the observed atomic spectra
connect these states of different parity.
When the atomic state changes, PARITY must be conserved.
The ELECTRIC DIPOLE transitions (characterized by ℓ ±1)
emit PHOTONS whose parity must therefore be NEGATIVE.
Like s which can be singly emitted (created) or absorbed (annihilated)
s are created/destroyed singly in STRONG INTERACTIONS
Their “INTINSIC PARITY” is also important.
To understand the role of parity in interactions, consider a system
(let’s start with a single pair) of initially
free (non-interacting) particles
which we can describe as a product state: | r - r |
large


2
1
 =  1 ( r1 ) 2 ( r2 )


P  =  1 ( -r1 ) 2 ( -r2 )




= p1 1 ( r1 ) p2 2 ( r2 ) = p1 p2 [ 1 ( r1 ) 2 ( r2 )]
parity is a multiplicative quantum number
If the pair is bound with orbital angular momentum 
P  = ( -1) p1 p2

If P commutes with both the free particle Hamiltonian
and the full Hamiltonian with interactions
the (parity) quantum numbers are conserved throughout the interactions
which is obviously true for
electromagnetic interactions
strong interactions
since the Lagrangian terms
incorporating these interactions
are all invariant under P
With sufficient energy,
collisions of hadrons can
produce additional particles
(frequently pions) among
the final states
But while
-d  nn
has been observed
-d  nn 0
has never been!


-d
Studies show the process
nn
•often accompanied by an X-ray spectrum
•reveals the calculable excited states of a “mesonic atom”
•pion orbitals around a deuteron nucleus.
This suggests
•the deuteron “captures” the pion
•the strong interaction proceeds only following
cascade decays to a GROUND (=0) state.
-d  nn
An =0 ground state means the  d system has J = Stotal = 1
?
So the final
nn state must also have J = 1?
=1
 =0
 =1
 =2
s=0
s=1
s=1
s=1
Note the final state is 2 Dirac FERMIONS
overall wave function must be ANTISYMMETRIC to particle exchange
Space part described by the spherical harmonics (interchanging
the neutrons would reverse their relative positions, i.e. r  -r
P :(r,,q)=(-1)(r,,q)
Spin part
?
s=1
s=1
s=1
s=0
ms = 1
ms = 0
ms = -1
ms = 0
-d  nn
Space part
Spin part
?
P :(r,,q)=(-1)(r,,q)
s=1
s=1
s=1
s=0
m s = +1
ms = 0
ms = -1
ms = 0
P +1 under
exchange
P -1 under
exchange
-d  nn
Space part
Spin part
?
P :(r,,q)=(-1)(r,,q)
s=1
s=1
s=1
s=0
So overall must have
m s = +1
ms = 0 1/2( +
ms = -1
ms = 0 1/2( -
(-1)+s+1 = -1
+s even
=1
=0
=1
=2
s=0
s=1
s=1
s=1
P +1 under
) exchange
=(-1)s+1
P -1 under
) exchange
=(-1)s+1
(-1)+s = +1
Thus the parity of the
final nn system
Must be (-1) = -1
-d  nn
Since strong interactions conserve parity we must also have
P: -d = -1
d is a bound np s-state (-1) PpP =(+1)(+1)(+1)
so P: -d
= (-1)=0 PPd = (+1)P(+1)
P = -1
some
Intrinsic Parity Assignments
so far…
p
n
d
proton
neutron
deuteron
 charged pion
 Photon
+
+
+
-
Neutral Pions
 0  
1.198%
 0  e+e-
3.1410-5 %  0  e+e-e+e-

98.798%

u
0
A
0
u
msTOTAL
0
B
2
C
-2
D
0
0 has 0 spin
so can’t produce net
angular momentum!
0
A
B
C
D
Recall, under PARITY (inversion/reflections),
R-handed spin→L-handed spin
The sign of HELICITY reverses under PARITY.
P: A→D
P: D→A
Note:
So the parity invariant states must be
either A  D with  parity
We can for convenience, decompose circularly-polarized photons into
1
2


( E x (t )  iE y (t ) )
+ right circularly polarized ms=+1
- left circularly polarized ms=+1

In a direct product representation I can write:
A ± D = R1R2 ± L1L2


1 
Ex iE y
2
each component here is PLANE polarized

 


 


 

= E1x + iE1 y E2 x + iE2 y  E1x - iE1 y E2 x - iE2 y
 
 
 
 
 E1 E2 - E1 E2 + i E1 E2 + E1 E2

x
x
y
y
x
y
y
x
= 12   
 
 
  
 E E - E E - i E E + E E 
1x 2 x
1y 2 y
1x 2 y
1y 2x 

1
2


A+D
A-D



 
 
= E1 x E2 x - E1 y E2 y
 
 
= i E1x E2 y - E1 y E2 x






polarized planes PARALLEL
polarized planes PERPENDICULAR
The plane defined by the trajectories of e+e- in pair production
are in the plane of the E-vector of the initial gamma ray.
Look at events where both s from the π0 decay both pair produce
and compare the orientation
of the two planes:
The π0 has ODD PARITY (-1)
0

90o