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)