Lecture 33:Quantum Mechanical Spin

Phy851 Fall 2009

Intrinsic Spin• Empirically, we have found that most

particles have an additional internaldegree of freedom, called ‘spin’

• The Stern-Gerlach experiment (1922):

• Each type of particle has a discretenumber of internal states:– 2 states --> spin _– 3 states --> spin 1– Etc….

Interpretation• It is best to think of spin as just an

additional quantum number needed tospecify the state of a particle.– Within the Dirac formalism, this is

relatively simple and requires no newphysical concepts

• The physical meaning of spin is not well-understood

• Fro Dirac eq. we find that for QM to beLorentz invariant requires particles tohave both anti-particles and spin.

• The ‘spin’ of a particle is a form ofangular momentum

Spin Operators• Spin is described by a vector operator:

• The components satisfy angularmomentum commutation relations:

• This means simultaneous eigenstates ofS2 and Sz exist:

zzyyxx eSeSeSSrrrr

++=

yxz

xzy

zyx

SiSS

SiSS

SiSS

h

h

h

=

=

=

],[

],[

],[

2222zyx SSSS ++=

ss msssmsS ,)1(, 22 += h

ssz msmmsS ,, h=

Allowed quantum numbers

• For any set of 3 operators satisfying theangular momentum algebra, the allowedvalues of the quantum numbers are:

• For orbital angular momentum, theallowed values were further restricted toonly integer values by the requirementthat the wavefunction be single-valued

• For spin, the quantum number, s, canonly take on one value– The value depends on the type of particle– S=0: Higgs– s=1/2: Electrons, positrons, protons,

neutrons, muons,neutrinos, quarks,…– s=1: Photons, W, Z, Gluon– s=2: graviton

{ }K,,1,,0 23

21∈j

{ }jjjm j ,,1, K+−−∈

{ }sssms ,,1, K+−−∈

Complete single particle basis

• A set of 5 commuting operators whichdescribe the independent observables ofa single particle are:

– Or equivalently:

• Some possible basis choices:

• When dealing with a single-particle, it ispermissible to drop the s quantumnumber

zSSR ,, 2r

zz SSLLR ,,, 2,

2

{ }smsr ,,r

{ }smsp ,,r

{ }smsmr ,,,, ll

{ }smsmn ,,,, ll

Intrinsic Magnetic DipoleMoment

• Due to spin, an electron has an intrinsicmagnetic dipole moment:

– ge is the electron g-factor– For an electron, we have:

– The is the most precisely measuredphysical quantity

• For most purposes, we can take ge≈ 2,so that

• For any charged particle we have:

Sm

eg

e

ee

rr

2−=µ

0150000000000.06220023193043.2 ±=eg

Sm

e

ee

rr−=µ

SM

qg rr

2=µ Each particle

has a differentg-factor

Hamiltonian for an electron ina magnetic field

• Because the electron is a point-particle,the dipole-approximation is always validfor the spin degree of freedom

• Any `kinetic’ energy associated with S2

is absorbed into the rest mass

• To obtain the full Hamiltonian of anelectron, we must add a single term:

)(RBSm

eHH

e

rrr⋅+→

[ ] )()()(2

1 2RBS

m

eRÖeRAeP

mH

ee

rrrrrrr⋅+−+=

Uniform Weak Magnetic Fieldas a perturbation

• For a weak uniform field, we find

• With the addition of a sphericallysymmetric potential, this gives:

– If the zero-field eigenstates are known

– The weak-uniform-field eigenstates are:

( )zzee

SLm

Be

m

PH 2

220

2

++=

( )zzee

SLm

BeRV

m

PH 2

2)(

20

2

+++=

ll ll mnEmnH n ,,,, ,00 =

{ }smmn ,,, ll

( )sBnmmn mmBEEs

20,0,, ++= llµ

smmns mmnEmmnHs

,,,,,, ,, ll lll

=

eB m

e

2

h=µ ‘Bohr Magneton’

Wavefunctions

• In Dirac notation, all spin does is addtwo extra quantum numbers

• The separate concept of a ‘spinor’ isunnecessary

• Coordinate basis:

– Eigenstate of

• Projector:

• Wavefunction:

{ }smsr ,,r

zSSR ,, 2r

ss

s

sm V

msrmsrrdIs

,,,,3 rr∑ ∫

−=

=

( ) ψψ sm msrrs

,,:rr

=

Spinor Notation:

• We think of them as components of alength 2s+1 vector, where eachcomponent is a wavefunction

• Example: s=1/2

• Spinor wavefunction definition:

• If external and internal motions are notentangled, we can factorize the spinorwavefunction:

( ) ψψ sm msrrs

,,:rr

=

( ) ψψ 21,,: srr

rr=↑

( ) ψψ 21,,: −=↓ srr

rr

[ ]( )( )( )

=

↓

↑

r

rr r

rr

ψ

ψψ :

( )( )

=

− r

rr

r

21

21

ψ

ψ Note thatdescending order

is unusual

[ ]( ) ( )rc

cr

rrψψ

=

↓

↑:

↓

↑

c

c Is then apure spinor

Schrödinger's Equation

• We start from:

• Hit from left with with

• Insert the projector

• Let:

– For s=1/2:

ψψ Hdt

di =h

smr ,r

ψψ Hmrmrdt

di ss ,,

rrh =

ψψ mrmrHmrrdmrdt

di ss

s

sm V

s

s

′′′′′= ∑ ∫−=′

,,,, 3 rrrrh

[ ][ ] [ ][ ] )(2

2

RVIM

PH

r+=

[ ][ ]

=

10

01I [ ][ ]

=

↓↓↓↑

↑↓↑↑

)()(

)()()(

rVrV

rVrVrV rr

rrr

[ ]( ) [ ]( ) [ ][ ]( )[ ]( )rrVrM

rdt

di

rrrhrh ψψψ +∇−= 2

2

2

Example: Electron in a UniformField

[ ]( ) [ ]( ) [ ][ ]( )[ ]( )rrVrM

rdt

di

rrrhrh ψψψ +∇−= 2

2

2

( )( )

( )( )

( )( )

−+

∂

∂−∇−=

↓

↑

↓

↑

↓

↑

r

rB

r

rBi

mr

r

dt

di BB

e

r

r

r

rh

r

r

hψ

ψµ

ψ

ψ

φµ

ψ

ψ

10

01

2 002

2

( ) ( ) ( )

( ) ( ) ( )rBrBim

rdt

di

rBrBim

rdt

di

BBe

BBe

rrhrh

rrhrh

↓↓↓

↑↑↑

−

∂

∂−∇−=

+

∂

∂−∇−=

ψµψφ

µψ

ψµψφ

µψ

002

2

002

2

2

2

• This is just a representation of twoseparate equations:

• We would have arrived at these sameequations using Dirac notation, withoutever mentioning ‘Spinors’

Pauli Matrices

• Here we see that we have recovered oneof the Pauli Matrices:

• The other Pauli matrices are:

• Then in the basis of eigenstates ofwe have:

• If we only care about spin dynamics:

( )( )

( )( )

( )( )

−+

∂

∂−∇−=

↓

↑

↓

↑

↓

↑

r

rB

r

rBi

mr

r

dt

di BB

e

r

r

r

rh

r

r

hψ

ψµ

ψ

ψ

φµ

ψ

ψ

10

01

2 002

2

−=

10

01zσ

=

01

10xσ

−=

0

0

i

iyσ

zS

σrhr

2=S

[ ]( ) [ ]( )riBM

rdt

di zB

rhrh ψσ

φµψ

+

∂

∂−+∇−= 0

22

2

[ ] [ ]cBm

ec

dt

di

e

rr⋅= σ

2

Two particles with spin

• How do we treat a system of twoparticles with masses M1 and M2, chargesq1 and q2, and spins s1 and s2?

– Basis:

– Wavefunction:

– Hamiltonian w/out motional degrees offreedom:

– Hamiltonian w/ motional degrees offreedom:

222111 ,,;,, ss msrmsrrr

ψψ 22211121,,, ,,;,,:),(2211 ssmsms msrmsrrrss

rrrr=

)()( 222

211

1

1 RBSM

qRBS

M

qH

rrrrrr⋅−⋅−=

( )

( ) )()()(2

1

)()()(2

1

222

222222

2

111

111111

1

RBSM

qRqRAqP

M

RBSM

qRqRAqP

MH

rrrrrrr

rrrrrrr

⋅−Φ+−

⋅−Φ+−=

Example #1

• A spin _ particle is in the ↑ state withrespect to the z-axis. What is theprobability of finding it in the ↓-statewith respect to the x-axis?

• Let:

• In the basis, the operatorfor the x-component of spin is:

• By symmetry, σx must have eigenvalues+1 and -1

• The eigenvector corresponding to -1 isdefined by:

z↑=ψ

{ }zz ↓↑ ,

=

01

10xσ

xxx ↓−=↓σ

Example #1 continued:

• This implies that:

=

01

10xσ xxx ↓−=↓σ

xxzxz ↓↑−=↓↑ σ

xxx ↓−=↓ σ

xz ↓↓−=

( )zzx ↓−↑=↓2

1

1−=

2

12=↑↓= zxP

Example #2

• Two identical spin-1/2 particles are placedin a uniform magnetic field. Ignoringmotional degrees of freedom, what are theenergy-levels and degeneracies of thesystem?

• States:

– Z-axis chosen along B-field

• Hamiltonian:

• Basis states are already eigenstates:

{ }↓↓↓↑↑↓↑↑ ,,,

( )zz SSM

gqBH 21

0

2+−=

↑↑−=↑↑M

gqBH

20h

0=↓↑=↑↓ HH

↓↓=↓↓M

gqBH

20h

1;2 1

01 =−= d

M

gqBE

h

2;0 22 == dE

1;2 3

03 == d

M

gqBE

h