A Review of Some INVARIANCE PRINCIPLES Hermitian Operators: R † = R can be diagonalized have real...
-
date post
19-Dec-2015 -
Category
Documents
-
view
213 -
download
0
Transcript of A Review of Some INVARIANCE PRINCIPLES Hermitian Operators: R † = R can be diagonalized have real...
A Review of Some INVARIANCE PRINCIPLES
Hermitian Operators: R† = R •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)
AA = RAR† and =R so that all computed matrix elements:
< |A| > = < |A | > = < |R†(RAR†)R| > = < |A| >
Unitary Operators: R† = R
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 = RR = 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:SO(n)SU(n)
simplerotations
• 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 cos sin 0Rzsin cos 0 0 0 1
cos sin 0sin cos 0 0 0 1
1 0 00 +1 0 0 0 +1
x
y
z
or
or
cos sin 0sin cos 0 0 0 1
then compare that to
1 0 00 1 0 0 0 +1
x
y
z
x
x
y
y
z
z
x
y
z
y
z
x
y
z
x
y
z
x
x'
y'
x'
y'
z'
PARITY TRANSFORMATIONS ALL are equivalent to a reflection (axis inversion) plus a rotation
The PARITY OPERATOR on 3-dim space vectors
every point is carried through the originto the diametrically opposite location
1 0 00 1 0 0 0 1
Wave functions MAY or MAY NOT have a well-defined parity(even or odd functions…or NEITHER)
xcos xx cos)cos(P P = +1
xsin xx sin)sin(P P = 1
but the more general
xx sincos xx sincosP
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!
azrea
z /23
100
1
aZrea
Zr
a
Z 2/23
200 21
2
1
iaZr erea
Z
sin
8
1 2/25
121
cos2
1 2/25
210aZrre
a
Z
iaZr erea
Z sin
8
1 2/25
211
aZrea
rZ
a
Zr
a
Z 3/2
2223
300 21827381
1
(r)=(r)ml () the angular part of the solutions
are the SPHERICAL HARMONICS
ml () = Pm
l (cos)eim
Pml (cos) = (1)msinm [( )m Pl (cos)]
(2l + 1)( lm)! 4( l + m)!
d d (cos)
d d (cos)Pl (cos) = [( )l (-sin2)l ] 1
2l l!
The Spherical Harmonics Y ,ℓ m(,)
ℓ = 0
ℓ = 1
ℓ = 2
ℓ = 34
100 Y
ieY sin
8
311
cos4
310
Y
ieY 2
2
sin2
15
4
122
ieY cossin
8
1521
2
12cos2
3
4
1520
Y
ieY 3
3
sin4
35
4
133
ieY 2
cos2
sin2
105
4
132
ieY 12cos5sin4
21
4
131
cos2
33cos2
5
4
730Y
then note r r means
x
y
z
and: Pml (cos) Pm
l (cos()) = Pml (-cos)
(eim=(1)m
so: eimeimeim
but d/d(cos) d/d(cos)
(-sin2)l = (1cos2l
ml () = Pm
l (cos)eim
Pml (cos) = (1)m(1-cos2m [( )m Pl (cos)]
(2l + 1)( lm)! 4( l + m)!
d d (cos)
d d (cos)Pl (cos) = [( )l (-sin2)l ] 1
2l l!
So under the parity transformation:
P:ml () =m
l (-)=(-1)l(-1)m(-1)m m
l ()
= (-1)l(-1)2m ml () )=(-1)l m
l ()
An atomic state’s parity is determined by its angular momentum
l=0 (s-state) constant parity = +1l=1 (p-state) cos parity = 1l=2 (d-state) (3cos2-1) parity = +1
Spherical harmonics have (-1)l parity.
v
P
When acting on a vector, the parity operator gives:
v
21vv P 2121
)()( vvvv
cos)cos(2121
vvvv scalars have
positive parity!
)( bac P baba
)()( 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 r position
p p momentum
L L angular momentum
intrinsic “spin”
E E electric field E = V/r
B B B = ×A
•B + •B magnetic dipole moment
•E •E electric dipole moment
• p •p longitudinal polarization
•(p1×p2) + •(p1×p2) transverse polarization
polar vectors
axial or pseudo-vectors
pseudo-vector
Potential problems with this concept…or assuming its invariance…
Lot’s of physics uses Right Hand Rules and cross-products…
prL has POSITIVE parity
angular momentum
)( BEqFcv
the Lorentz Force Law
vdt
dmF
q
FE
Bv
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
vectors!
vector
But notice a vector crossed with a pseudo-vector:
)( cvP
bac where
cvcv )()(
is a vector!
Might B really be a pseudo-vector quantity after all?→
AB
the “vector” potentialis a true vector!
and as we have argued in quantum mechanics/high energy theoryis 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!
In addition to energy and momentumlight (the photon)
also carries angular momentum
A mono-chromatic electromagnetic wave
is composed of n mono-energetic photonseach with
E kp
2
k
(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
1.65 cm wire
glass fiberglass bead
oil drop
P. J. Allen Am.J.Phys. 34, 1185 (1964).
Microwave generator beamed up cavitysets rotor in motion
ELZ
=
If delivered by n photons, then
with
means
E = nħ
LZ = n jZ
jZ = ħ
Each photon has spin 1.
2S 2P 2D 2F
ℓ =0 ℓ =1 ℓ =2 ℓ =3
N shell
M shell
L shell
K shell
n=4
n=3
n=2
n=1
Transitions occur between adjacent angular momentum states (i.e. ℓ ±1).Energy level diagram for Hydrogen
The parity of a state must change in such an “electric dipole transition”ℓ=±1, S=0, J=0, ±1
s, d, g, …
p, f, h, …
even parity
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) particleswhich we can describe as a product state:
)()(2211
rr
||12rr large
)()(2211
rr P
)]()([)()(221121222111
rrpprprp
parity is a multiplicative quantum number
If the pair is bound with orbital angular momentum
P 21
)1( pp
If P commutes with both the free particle Hamiltonianand the full Hamiltonian with interactions
the (parity) quantum numbers are conserved throughout the interactions
which is obviously true for
electromagnetic interactionsstrong 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 nn0 has never been!
Studies show the process d 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 nnAn =0 ground state means the d system has J = Stotal = 1
=1 s=0 =0 s=1 =1 s=1 =2 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,,)=(-1)(r,,)
Spin part ? s = 1 ms = 1s = 1 ms = 0s = 1 ms = -1s = 0 ms = 0
?
So the final nn state must also have J = 1?
d nn
Space part P :(r,,)=(-1)(r,,)
Spin part ? s = 1 ms = +1s = 1 ms = 0s = 1 ms = -1s = 0 ms = 0
P +1 under exchange
P 1 under exchange
d nn
Space part P :(r,,)=(-1)(r,,)
Spin part ? s = 1 ms = +1s = 1 ms = 0s = 1 ms = -1s = 0 ms = 0
P +1 under exchange
P 1 under exchange
1/2( + )
1/2( )
=(1)s+1
=(1)s+1
So overall must have (1)+s+1 = 1 (1)+s = 1+s even
=1 s=0=0 s=1=1 s=1=2 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)
P = 1
so P: d = (1)=0 PPd = (+1)P(+1)
Intrinsic Parity Assignments
p proton +n neutron +d deuteron + charged pion Photon
some
so far…
Neutral Pions
98.798% 0 1.198% 0 e+e3.14105 % 0 e+ee+e u
u
0
0
A
B
C
D
msTOTAL
0
2
2
0
0 has 0 spinso 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
))()((21 tEitE
yx
+ right circularly polarized ms=+1
left circularly polarized ms=+1
yx EiE
2
1
A ± D = R1R2 ± L1L2each component here is PLANE polarized
yxyxyxyx EiEEiEEiEEiE 221122112
1
xyyxyyxx
xyyxyyxx
EEEEiEEEE
EEEEiEEEE
21212121
21212121
21
A + D
A D
yyxx EEEE 2121
xyyx EEEEi 2121
polarized planes PARALLEL
polarized planes PERPENDICULAR
In a direct product representation I can write:
The plane defined by the trajectories of e+e in pair productionare 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:
0 90oThe π0 has ODD PARITY (1)