P460 - Spin 1
Spin and Magnetic Moments(skip sect. 10-3)
• 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 the natural unit and the “-” sign is due to the electron’s
charge
LgL
mvrLbutr
areacurrentAI
b
llmq
rqv
2
22 e
b 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 massinteger spin Bosons half-integer
Fermions
Spin particle postulated particle
0 pion Higgs, selectron
1/2 electron photino (neutralino)
1 photon
3/2
2 graviton
• relativistic QM 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)). No 3D “function”
• Dirac equation gives g-factor of 2
200232.2
,,
||...||,)1(
,
21
212
432
23
212
21
22
sSg
s
z
z
kijkji
g
SS
Sfor
ssSssS
SSS
bs
P460 - Spin 4
Spin 1/2 expectation values • non-diagonal components (x,y) aren’t zero. Just indeterminate. Can sometimes use
Pauli spin matrices to make calculations easier
• with two eigenstates (eigenspinors)
0
0
01
10
10
01
10
01
2
22
2
432 2
i
iSS
S
SS
yx
z
ii
2
2
1
0
0
1
eigenvalueS
eigenvalueS
z
z
P460 - Spin 5
Spin 1/2 expectation values • “total” spin direction not aligned with any component.
• can get angle of spin with a component
3
1
43
21
cos
S
Sz
P460 - Spin 6
Spin 1/2 expectation values • Let’s assume state in an arbitrary combination of spin-up and spin-down states.
• expectation values. z-component
• x-component
• y-component
1|||| 22
bawithb
aba
)(
10
01||
222
2**
ba
b
abaSS zz
)(0
0 **2
2
2** abbab
abaSx
)(0
0 **2
2
2** abbaib
a
i
ibaS y
P460 - Spin 7
Spin 1/2 expectation values example• assume wavefunction is
• expectation values. z-component
• x-component
• Can also ask what is the probability to have different components. As normalized, by
inspection
• or could rotate wavefunction to basis where x is diagonal
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
322
2
642
2622
2
)(
baS
bSaS
z
zz
2
16
1i
P460 - Spin 8
• Can also determine
• and widths
2
31222
4**
4**
42
4**
4**
42
4**
4**
42
222
222
222
)(10
01
10
01
)(0
0
0
0
)(01
10
01
10
SSSS
bbaab
abaS
bbaab
a
i
i
i
ibaS
bbaab
abaS
zyx
z
y
x
))(1(4
)(
))(1(4
)(
))(1(4
)(
2**2
222
2**2
222
2**2
222
bbaaSSS
abbaSSS
abbaSSS
zzz
yyy
xxx
P460 - Spin 9
• Can look at the widths of spin terms if in a given eigenstate
• z picked as diagonal and so
• for off-diagonal
0)11()(
40
1
10
01
10
0101
4
222
2
42
2
2
zzz
z
SSS
S
Widths- example
0
1
4)(
40
1
01
10
01
1001
00
1
0
001
2222
2
42
2
2
2
xxx
x
x
SSS
S
S
P460 - Spin 10
• 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
Sz BS
BE s
BSB bsgs
Bg bs
z
P460 - Spin 11
• Hamiltonian for an electron in a magnetic field
• assume solution of form
• If B direction defines z-axis have Scr.eq.
• And can get eigenvalues and eigenfunctions
Precession - details
ti
ti
be
aet
b
a
m
egB
m
egB
)()0(
1
0
40
1
4
Bm
egH
4
2S
)(
)(
t
t
dt
diH
10
01BB
10
01
4m
Beg
dt
di
P460 - Spin 12
• Assume at t=0 in the + eigenstate of Sx
• Solve for the x and y expectation values. See how they precess around the z-axis
Precession - details
ti
ti
e
et
b
a
b
a
b
a
2
1)(
1
1
2
1
201
10
2
ti
eebaab
iS
tee
abbaS
titi
y
titi
x
2sin2
)2
(2
)(2
2cos2
)2
(2
)(2
22**
22**
P460 - Spin 13
• can look at any direction (p 160 and problem 10-2 or see Griffiths 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
Arbitrary Angles
kjir ˆcosˆsinsinˆcossinˆ
cossinsincossin
sinsincossincos
i
iSr
10|| ISSr
P460 - Spin 14
•Go ahead and solve for eigenspinors.
•Gives (phi phase is arbitrary)
•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 15
Combining Angular Momentum • If have two or more angular momentum, the combination is also an 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 momentum
stateslllllm
l
12,1...1,
......,1,,0 23
21
221sin
,,0 21
21
21
doubletdoubletglet
JJSLif
SSLLSLJ
z
ii
kijkji SiSS ,
P460 - Spin 16
Combining Angular Momentum • Non-trivial examples. add 2 spins. The z-components add “linearly”
and the total adds “vectorally”. Really means add up z-component and then divide up states into SU(2) groups
1322
sin
0,01,0,1,1
1
0
0
1
,
21
21
21
21
21
21
21
21
21
21
221
121
221
1
glettripletdoubletdoublet
JJANDJJ
SSJ
J
SSwithSSif
zz
z
zz
4 terms. need to split up. The two 0 mix
P460 - Spin 17
Combining Angular Momentum • add spin and orbital angular momentum
2423
,,,
,,,,,
101
1,0,11,
21
21
21
23
23
23
21
21
21
21
23
21
21
21
21
21
doubletquartetdoublettriplet
JJANDJJ
J
LSwithLSif
zz
z
zz
P460 - Spin 18
Combining Angular Momentum • 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 “coefficients” (how they are made from the initial states) can be fairly easily determined using group theory (step-down operaters). Called Clebsch-Gordon coefficients
• these give the “dot product” or rotation between the total and the individual terms.
mlmlmm
mlmlmm
mmmmm
mmmmtotalml
mmmmtotalml
21
21
2121
2121
2121
P460 - Spin 19
Combining Angular Momentum
•Clebsch-Gordon coefficients
•these give the “dot product” or rotation between the total and the individual terms. “easy” but need to remember what different quantum number labels refer to
mlmlmm
mlmlmm
mmmmm
mmmmtotalml
mmmmtotalml
21
21
2121
2121
2121
P460 - Spin 20
Combining Angular Momentum • example 2 spin 1/2• have 4 states with eigenvalues 1,0,0,-1. Two 0 states mix to form eigenstates of S2
• step down from ++ state
• Clebsch-Gordon coefficients
1
00
1
,,,
z
zz
z
S
SS
S
)1)((
21
21
mlmlC
SSS
SSS zzz
orthogonalml
ml
mlmlCmlS
S
CS
CS
)(2
10,0
)(2
10,1
0,120,1)1,1(1,1
)2
(2
),(
),(
21
21
2
21
21
1
2
1
P460 - Spin 21
Combining Ang. Momentum • check that eigenstates have right eigenvalue for S2
• first write down
• and then look at terms
• putting it all together see eigenstates
21212122
21
21212122
21
2122
21
221
2
2
222
2)(
SSSSSSSS
SSSSSSSS
SSSSSSS
zz
yyxxzz
yx iSSS
)(2
1X
XXSSSS
SSSSand
SSSSwith
XXSS
XXS
XSSXS
zz
22112
212
221
1212
21
222
221
21
21
)(
0
0
)2
)(2
(22
4
3
4
3))()((
2
1
XXXS
0
2)1( 2
21
43
4322
P460 - Spin 22
• L=1 + S=1/2
• Example of how states “add”:
• Note Clebsch-Gordon coefficients (used in PHYS 374 class for Mossbauer spectroscopy).
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
3
)2(,
3
1 sqrt
2 terms
21
31
21
3
)2(21
21
21
3
)2(21
31
21
23
01
01
sqrt
sqrt
zz
mj
mj
SL
P460 - Spin 23
• Clebsch-Gordon coefficients for different J,L,S