p460 - spin1 spin and magnetic moments (skip sect. 10-3) orbital and intrinsic (spin) angular...
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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
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)).
No 3D “function”
• 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
zkijkji
g
SSSfor
ssSssSSSS
bs
0
0
01
10
10
01
10
01
2
222
432 2
i
iSSS
SS
yxz
ii
2
2
1
0
0
1
eigenvalueS
eigenvalueS
z
z
P460 - Spin 4
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 5
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 6
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 7
• 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 8
• 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- example
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 9
• 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 10
• 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 11
• 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 12
• can look at any direction (p 160 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 13
• 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 14
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 15
Combining Angular Momentum • Non-trivial examples. add 2 spins
• add spin and orbital angular momentum
1322
sin
0,01,0,1,1
,
21
21
221
121
221
1
glettripletdoubletdoublet
JJANDJJ
SSJ
SSwithSSif
zz
zz
2423
,,,
1,0,1,1
21
21
21
23
23
21
21
doubletquartetdoublettriplet
JJANDJJ
SLwithSLif
zz
zz
P460 - Spin 16
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 17
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 18
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 19
• L=1 + S=1/2
• 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
3
2,
3
1
2 terms
21
31
21
32
21
21
21
32
21
31
21
23
01
01
mj
mj
SL zz