Transcript Spin and Magnetic Mon\ments
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)
I
A
current
area
2
qv
r q
2
m
r
2
L
l but
L g l
b
mvr
L
b
e
2
m e
· 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
l L
2
g l
b l
(
l
1 )
l
(
l
1 )
zl
g l
b m l
P460 - Spin 1
• • •
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 D 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)
KG
:
E
2
p
2
m
2
D
:
E
p
2
m
2 P460 - Spin 2
Spin 1/2 expectation values
• similar eigenvalues as orbital angular momentum (but SU(2)). No 3D “function”
S i
,
S j
ijk
S k S
2
s
(
s
1 ) 2 ,
S z
|
s
| ...
|
s
|
for S
S
2 1 2 1 2 3 2 2 3 4 2 ,
S z
1 2 , 1 2
s
g s
b
S
g s
2 .
00232
• Dirac equation gives g-factor of 2
2 P460 - Spin 3
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
S i
2
i
S
2 3 2 4
S z
2 1 0 0 1
S x
2 0 1 1 0 1 0 0 1
S y
2 0
i
i
0 • with two eigenstates (eigenspinors) 1 0 0 1
S z eigenv alue
2
S z eigenv alue
2 P460 - Spin 4
Spin 1/2 expectation values
• “total” spin direction not aligned with any component. • can get angle of spin with a component cos
S
S z
1 2 3 4 1 3 P460 - Spin 5
Spin 1/2 expectation values
• Let’s assume state in an arbitrary combination of spin-up and spin down states.
a
• expectation values. z-component
b
a b with
|
a
| 2 |
b
| 2 1
S z
|
S z
|
a
*
b
* 2 1 0 0 1
a b
2 (
a
2
b
2 ) • x-component
S x
a
*
b
* 0 2 2 0
a b
2 (
a
*
b
b
*
a
) • y-component
S y
a
*
b
* 0 2
i
0 2
i
a b
i
2 (
a
*
b
b
*
a
) P460 - Spin 6
Spin 1/2 expectation values example
1 • assume wavefunction is 1 6
i
2 • expectation values. z-component
S z S z
2 2 (
a
2
a
2 2 6
b
2 )
S z
3 2
b
2 4 6 • x-component
S x
2
t S x
(
a
*
b
b
*
a
)
a
* 2 1 6
b
* 0 2
a
2 ( 1
i
) 2 0
b
2 ( 1
i
) 3 • Can also ask what is the probability to have different components. As normalized, by inspection
probabilit y
(
S x probabilit y
(
S x
2 ) 2 ) 5 6 1 6 ( 5 6 1 6 ) 1 2 1 3 • or could rotate wavefunction to basis where x is diagonal P460 - Spin 7
• Can also determine
S x
2
S y
2
S z
2
S x
2 • and widths 4 2
a
* 4 2
a
* 2 4
a
*
S
2
y
b
*
b
*
S z
2 0 1 0
i b
* 1 0 1 0 0 1 0
i
0
i
1 0
a b
2 4 0
i
a b
(
a
*
a
b
*
b
) 4 2 (
a
*
a
b
*
b
) 2 4 4 2 1 3 0 1 1 0
S
2 0 1
a b
2 4 (
a
*
a
b
*
b
) 2 4 ( ( D
S y
) 2 ( D
S
D
S x z
) ) 2 2
S S z
2
x S y
2 2
S S x S y z
2 2 ( 1 (
a
*
b
b
*
a
) 2 ) 4 2 2 4 2 4 2 ( 1 (
a
*
b
( 1 (
a
*
a
b
*
a
) 2 )
b
*
b
) 2 ) P460 - Spin 8
Widths- example
• Can look at the widths of spin terms if in a given eigenstate • z picked as diagonal and so 1 0
S z
2 2 4 1 ( D
S z
) 2
S z
2 0 1 0
S z
0 1 1 0 2 2 4 0 1 1 0 ( 1 1 ) 0 2 4 • for off-diagonal
S x
1 0 0 2 2 0 1 0 0
S
2
x
2 4 1 ( D
S x
) 2
S
2
x
0 0 1
S x
1 0 0 1 2 2 4 1 0 1 0 2 4 P460 - Spin 9
Components, directions, precession
• Assume in a given eigenstate 1 0 • the direction of the total spin can’t be in the same direction as the z component (also true for l>0)
S z
2
S
S
2 3 2 cos 1 3 • Example: external magnetic field. Added energy D
E
s
B
z B
puts electron in the +state. There is now a torque
s
B
g s
b
S
B
which causes a precession about the “z-axis” (defined by the magnetic field) with Larmor frequency of
g s
b B
P460 - Spin 10
Precession - details
• Hamiltonian for an electron in a magnetic field
H
eg
4
m
B
• assume solution of form (
t
) (
t
)
S H
2
i
d
dt
• If B direction defines z-axis have Scr.eq.
B
B
1 0 0 1
i
d dt
eg
B
4
m
• And can get eigenvalues and eigenfunctions 1 0 0 1 ( 0 )
egB
4
m
a b
1 0 (
t
)
egB
4
m ae
i
t be i
t
0 1 P460 - Spin 11
Precession - details
• Assume at t=0 in the + eigenstate of S x 2 0 1 1 0 (
t
)
a b
1 2
a
2
b e
i
t e i
t
a b
1 2 1 1 • Solve for the x and y expectation values. See how they precess around the z-axis
S x S y
2
i
2 (
a
*
b
(
b
*
a
b
*
a
)
a
*
b
) 2 (
e
2
i
t
2 (
e
2
i
t
2 2
e i
2
i
t e
2
i
t
) ) 2 2 cos 2
t
sin 2
t
P460 - Spin 12
Arbitrary Angles
• 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.
r
ˆ sin cos
i
ˆ sin sin • Put components into Pauli spin matrices ˆ
j
cos
k
ˆ
S r
sin cos cos
i
sin sin sin cos
i
sin cos sin • and solve for its eigenvalues
S r
|
S
I
| 0 1 P460 - Spin 13
r
ˆ sin cos
i
ˆ sin sin
j
ˆ cos
k
ˆ
S r
• Go ahead and solve for eigenspinors . cos cos
i
sin sin sin cos
i
sin cos sin
S r a b
for
1
a b
cos
b
sin
a
( 1 sin cos
e
i
) (cos
a
sin
i
cos sin 2 2
e i
)
a
(
use
1 cos sin • Gives (phi phase is arbitrary) tan 2 )
r
cos
e i
sin 2 2
for
1
r
e
i
sin cos 2 2 • if r in z,x,y directions
z
: 0 1 0 , 0 1
x
:
y
: 2 , 2 , 0 2 1 2 1 2 , 1 2
i
2 , 1 2 1 2
i
2 1 2 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:
S i
,
S j
i
ijk S k
• representations of angular momentum have 2 quantum numbers:
l
m
0 , 1 2 , 1 ,
l
,
l
3 2 ......
1 ...
l
1 ,
l
2
l
1
states
• 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
J if
L
L
0 ,
S
S
L
1 2
J
L i
1 2
S
,
J z
sin
glet
doublet
doublet
S i
1 2 1 2 2 P460 - Spin 15
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
J z if
S
1 1 2 1 2 1 2 1 2 1 2 ,
S
2 1 0 1 2 1 2 0 1 2 1 2 1 1 2
with S z
1 1 2
S z
2 1 2
4 terms. need to split up. The two 0 mix
J
J
S
1 1 ,
S
2
J z
1 , 0 , 1
AND
J
0 ,
J z
0
doublet
2 2
doublet
3 1
triplet
sin
glet
P460 - Spin 16
Combining Angular Momentum
• add spin and orbital angular momentum
if J z
S
1 1 2 3 2 , 1 2 , , 1 2 1 2
L
, 0 1 2 1 ,
with
1 2
S z
1 1 2 1 2 , 3 2 1 2
L z
1 , 0 , 1
J
3 2 ,
J z
3 2 , 1 2
AND
J
1 2 ,
J z
1 2
triplet
3 2
doublet
4 2
quartet
doublet
P460 - Spin 17
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.
l l
m m
1
m m m
1
m
2
total total m
2
l
m
1
m m
1
m
1
m
2
m
2
m
2
l
m
m
l m
l
m
m
1
m m m
m
2 P460 - Spin 18
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
l l
m m
1
m m m
1
total total
m
1
m
1
m
2
m
2
l m
1
m m
2
m
2
m
2
l
m
1
m
2
l m
l
m
1
m
1
m m m
2
m
2 P460 - Spin 19
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 S 2
S z
, , 1 ,
S z
0
S z
1 • step down from ++ state
S z
0
S S z
C
S
1
z S
1
S
2
z S
2 (
l
m
)(
l
m
1 )
S
1
S
2
S
C
( 1 2 , 1 2 )
C
( 1 2 , 1 2 ) 2 (
S
l
1 ,
m
1 )
C
( 1 , 1 )
l
2 1 ,
m
0
l l
1 ,
m
0 0 ,
m
0 2
l
1 ,
m
1 2 ( 1 2 ( ) )
orthogonal
0 • Clebsch-Gordon coefficients 1 2 P460 - Spin 20
• • • •
Combining Ang. Momentum
check that eigenstates have right eigenvalue for S 2 first write down
S
2
S
1 2
S
1 2 (
S
S
S
2 2 2 2
S
2 2 2 ) 2
S
1
S
1
z z
S
2
S
2
S
1 2
z z
S
2 2 2
S
1
x S
1
S
2
S
2 2
S
1
x
S
2 2
S
1
y S
1
S
2
S
2
y
and then look at terms
S
2 1
X
1 2 ((
S
2 1
S
2 2
X
3 4 2
X
2
S
1
z S
2
z X
with and S
2
S
1
S
2 ) 2 ( 2 )(
S
1 ) 2
X
2 (
S
2 1
S
2 )
S
2
S
1 2 )
S
1 3 4 2
X
0
X
0 (
S
2
S
1
S
1
S
2 )
X
2
X
S
S x
iS y
1 2 ( ) putting it all together see eigenstates
S
2
X
2 ( 3 4 3 4 1 2 1 )
X
2 2 0
X
P460 - Spin 21
• L=1 + S=1/2
2 terms
L z
1 0 1 1 0 1
• Example of how states “add”:
j j m m
3 2 1 2 1 2 1 2
L z
1 3 1 2 3 1
S z
1 2 1 2 2 3 1 3 0 0
S z
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
J z
3 2 1 2 1 2 1 2 1 2 3 2
J
1 3 2 3 2 3 2 3 2 3 2 3 2
J
2 • Note Clebsch-Gordon coefficients (used in PHYS 374 class for Mossbauer spectroscopy) 1 3 , 2 3 P460 - Spin 22 1 2 1 2 1 2 1 2
• Clebsch-Gordon coefficients for different J,L,S P460 - Spin 23