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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
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