P460 - Spin 1
Spin and Magnetic Moments
• Orbital and intrinsic (spin) angular momentum
produce magnetic moments
• coupling between moments shift atomic energies
· Look first at orbital (think of current in a loop)
· the “g-factor” is 1 for orbital moments. The Bohr
magneton is introduced as natural unit and the “-”
sign is due to the electron’s charge
LgL
mvrLbutr
areacurrentAI
b
llmq
rqv
2
22
eb m
e
2
lblzlbll mgllg
llL
)1(
)1(2
P460 - Spin 2
Spin • Particles have an intrinsic angular momentum - called spin
though nothing is “spinning”
• probably a more fundamental quantity than mass
• integer spin --> Bosons half-integer--> Fermions
Spin particle postulated particle
0 pion Higgs, selectron
1/2 electron photino (neutralino)
1 photon
3/2 delta
2 graviton
• relativistic QM uses Klein-Gordon and Dirac equations for spin
0 and 1/2.
• Solve by substituting operators for E,p. The Dirac equation
ends up with magnetic moment terms and an extra degree of
freedom (the spin)
22222 :: mpEDmpEKG
P460 - Spin 3
Spin 1/2 expectation values • similar eigenvalues as orbital angular momentum (but SU(2))
• Dirac equation gives g-factor of 2
• non-diagonal components (x,y) aren’t zero. Just
indeterminate. Can sometimes use Pauli spin matrices to
make calculations easier
• with two eigenstates (eigenspinors)
200232.2
,,
||...||,)1(
21
212
432
23
212
21
22
sSg
s
z
z
g
SSSfor
ssSssS
bs
0
0
01
10
10
01
10
012224
32 2
i
iSSSS yxz
2
2
1
0
0
1
eigenvalueS
eigenvalueS
z
z
P460 - Spin 4
Zeeman Effect • Angular momentum->magnetic moment->energy shifts
• additional terms in S.E. do spin-orbit later. Right now
assume atom in external magnetic field
• look at ground state of H. L=0, S=1/2
dVEVwithVEE
shiftonperturbatiBE
nBnnnnn
B
*
2
21*
BgV
dVSdiagonalS
BSg
BSg
E
bSnn
zZ
Zbsbs
B
P460 - Spin 5
Spin 1/2 expectation values • Let’s assume state in a combination of spin-up and spin-down
states (it isn’t polarized).
• Can calculate some expectation values. Griffiths Ex. 4.2. Z-
component
• x-component
• as normalized, by inspection
1|||| 22
bawithb
aba
322
2
642
2622
2
61
)(
2
1
baS
bSaS
ilet
z
zz
361
2**
2
2
2**
)1(22)1()(
0
0
iiabba
b
abaSS x
tx
31
21
61
65
61
2
65
2
)()(
)(
x
x
Syprobabilit
Syprobabilit
P460 - Spin 6
• Griffiths Prob. 4.28. For the most general normalized spinor
find expectation values:
• just did x and z
• repeat for other
• note x and y component will have non-zero “width” for their
distributions as not diagonalized
222 ,,,,,, zyxzyx SSSSSSfindba
)(0
0
)(10
01
**2
2
2**
**2
**
abbab
abaS
bbaab
abaS
x
z
2
31222
4**
4**
42
**2
2
2**
222
)(01
10
01
10
)(0
0
SSSS
bbaab
abaS
baabb
abaS
zyx
x
ii
i
y
P460 - Spin 7
• Can look at the widths of spin terms if in a given eigenstate
• z picked as diagonal and so
• for off-diagonal
0)11()(
0
1
10
01
10
0101
4
222
442
2
22
zzz
z
SSS
S
Widths
0
1
4
222
442
2
2
2
22
)(
0
1
01
10
01
1001
00
1
0
001
xxx
x
x
SSS
S
S
P460 - Spin 8
• Assume in a given eigenstate
• the direction of the total spin can’t be in the same direction as the z-component
(also true for l>0)
• Example: external magnetic field. Added energy
puts electron in the +state. There is now a torque
which causes a precession about the “z-axis” (defined by the magnetic field)
with Larmor frequency of
Components, directions, precession
0
1
31
232
2
cos
S
S z BS
BE s
BSB bsgs
Bg bs
z
P460 - Spin 9
• Griffiths does a nice derivation of Larmor precession but at the 560 level
• to understand need to solve problem 4.30.
• Construct the matrix representing the component of spin angular
momentum along an arbitrary radial direction r. Find the eigenvalues and
eigenspinors.
• Put components into Pauli spin matrices
• and solve for its eigenvalues
Angles
kjir ˆcosˆsinsinˆcossinˆ
cossinsincossin
sinsincossincos
i
iSr
10|| ISSr
P460 - Spin 10
• Go ahead and solve for eigenspinors.
• Phi phase is arbitrary. gives
• if r in z,x,y-directions
kjir ˆcosˆsinsinˆcossinˆ
cossinsincossin
sinsincossincos
i
iSr
)tan(cos
sin
sin
)cos1(
)sin(cossincos
1
2sincos1
2
2
useeae
ab
aiba
b
aforS
ii
r
2
2
2
2
cos
sin1
sin
cos
i
ri
r efor
e
21
2
2
21
22
21
21
212
1
2
,,:
,0,:
1
0,
0
10:
i
iy
x
z
P460 - Spin 11
Combining Angular Momentum • If have two or more angular momentum, the
combination is also eigenstate(s) of angular momentum. Group theory gives the rules:
• representations of angular momentum have 2 quantum numbers:
• combining angular momentum A+B+C…gives new states G+H+I….each of which satisfies “2 quantum number and number of states” rules
• trivial example. Let J= total angular momentm
stateslllllm
l
12,1...1,
......,1,,0 23
21
221sinsin
,,0 21
21
21
gletdoubletglet
JJSLif
SSLLSLJ
z
ii
P460 - Spin 12
Combining Angular Momentum • Non-trivial example.
• Get maximum J by maximum of L+S. Then all possible combinations of J (going down by 1) to get to minimum value |L-S|
• number of states when combined equals number in each state “times” each other
• the final states will be combinations of initial states. The “coefficiants” (how they are made from the initial states) can be fairly easily determined using group theory (and step-up and step-down operaters). Called Clebsch-Gordon coefficients
2423
,,,
1,0,1,1
21
21
21
23
23
21
21
doubletquartetdoublettriplet
JJANDJJ
SLwithSLif
zz
zz
P460 - Spin 13
• Same example.
• Example of how states “add”:
• Note Clebsch-Gordon coefficients
23
23
21
21
23
21
21
21
23
21
21
21
23
21
21
21
23
21
21
23
23
21
21
1
0
1
1
0
1
JJJSL zzz
21
31
21
32
21
21
21
32
21
31
21
23
, 0 , 1
,
, 0 , 1
,
z z z z
z
z z z z
z
S L S L
J J
S L S L
J J
3
2,
3
1
2 terms
P460 - Spin 14
• Clebsch-Gordon coefficients for different J,L,S