Spin and Magnetic Mon\ments
Download
Report
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
r 2
but L m vr
b
q
2m L l gl L
qv
2r
e
b
2me
· 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
L2 l (l 1)
l g l b l (l 1)
zl g l b ml
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
delta
2
graviton
• relativistic QM uses Klein-Gordon and Dirac
equations for spin 0 and 1/2.
KG : E 2 p 2 m 2 D : E
p 2 m2
• Solve by substituting operators for E,p. The Dirac
equation ends up with magnetic moment terms and
an extra degree of freedom (the spin)
P460 - Spin
2
Spin 1/2 expectation values
• similar eigenvalues as orbital angular momentum
(but SU(2)). No 3D “function”
S , S
i
j
2
2
S
S
s
(
s
1
)
, S z | s | ... | s |
ijk
k
for S 12 S 2 12 23 2 43 2 , S z 12 , 12
g s b S
s
g s 2.00232 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
1 0
0 1
1 0
0 1
0 i
S z 2
S x 2
S y 2
0
1
1
0
i
0
Si i S 2
2
3 2
4
• with two eigenstates (eigenspinors)
1
S z eigenvalue 2
0
0
S z eigenvalue 2
1
P460 - Spin
3
Spin 1/2 expectation values
• “total” spin direction not aligned with any
component.
• can get angle of spin with a component
1
Sz
1
2
cos
3
3
S
4
P460 - Spin
4
Spin 1/2 expectation values
• Let’s assume state in an arbitrary combination of
spin-up and spin-down states.
a
a b with | a |2 | b |2 1
b
• expectation values. z-component
S z | S z | a
*
b
*
2
1 0 a
0 1 b
2 ( a 2 b2 )
• x-component
S x a*
0
2 a
*
2 (a*b b*a )
b
2 0 b
• y-component
S y a*
0
2 i a
*
i 2 (a*b b*a )
b
0 b
2i
P460 - Spin
5
Spin 1/2 expectation values
example
• assume wavefunction is
1
6
• expectation values. z-component
S z 2 a 2
S z 2 b2
2
6
1 i
2
4
6
S z 2 (a 2 b2 ) 3
• x-component
0
2 a
t
*
*
S x S x a b
2 0 b
2 (a *b b*a) 2 16 (1 i )2 2(1 i )
3
• Can also ask what is the probability to have
different components. As normalized, by inspection
probability ( S x 2 )
5
6
probability ( S x 2 ) 16 ( 56 16 ) 12 13
• or could rotate wavefunction to basis where x is
diagonal
P460 - Spin
6
•
Can also determine
2
4
a
*
S
2
y
2
4
a
*
S
2
z
2
4
a
*
S
2
x
0
b
1
* 0
b
i
* 1
b
0
1 0 1 a 2 *
*
2
4 ( a a b b) 4
0 1 0 b
i 0 i a 2 *
*
2
4 ( a a b b) 4
0 i 0 b
0 1 0 a 2 *
*
2
4 ( a a b b) 4
1 0 1 b
*
S x2 S y2 S z2
1
3
S2
• and widths
2
( S x ) 2 S x2 S x (1 ( a *b b*a ) 2 )
4
2
2
( S y ) 2 S y2 S y (1 ( a *b b*a ) 2 )
4
2
2
( S z ) 2 S z2 S z (1 ( a *a b*b) 2 )
4
2
P460 - Spin
7
Widths- example
• Can look at the widths of spin terms if in a given
1
eigenstate
0
• z picked as diagonal and so
1 0 1 0 1
2
4
S z 4 1 0
0 1 0 1 0
2
(S z ) S
2
2
2
z
Sz
2
2
4
(1 1) 0
• for off-diagonal
0 2 1
0
S x 1 0
2 0 0
0 1 0 1 1
2
2
S x 4 1 0
1 0 1 0 0
(S x ) 2 S x2 S x
2
P460 - Spin
2
4
2
4
8
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)
Sz
z
2
S2
3
2
cos
1
3
B
S
• Example: external magnetic field. Added energy
E s B
puts electron in the +state. There is now a torque
g s b
s B S B
which causes a precession about the “z-axis”
(defined by the magnetic field) with Larmor
g s b B
frequency of
P460 - Spin
9
Precession - details
• Hamiltonian for an electron in a magnetic field
eg
H
B
S
4m
2
• assume solution of form
(t )
(t )
H i
d
dt
• If B direction defines z-axis have Scr.eq.
1 0
B B
0
1
d egB 1 0
i
dt 4m 0 1
• And can get eigenvalues and eigenfunctions
egB 1
egB 0
4m
4m
0
1
aeit
a
(0) (t ) it
b
be
P460 - Spin
10
Precession - details
• Assume at t=0 in the + eigenstate of Sx
0 1 a a a 1 1
1
0
b
b
b
2
2 1
2
i t
1 e
i t
(t )
2e
• Solve for the x and y expectation values. See how
they precess around the z-axis
*
e2it e 2it
*
S x (a b b a ) (
) cos2 t
2
2
2
2
i *
e2it e 2it
*
S y ( b a a b) (
) sin 2 t
2
2
2i
2
P460 - Spin
11
Arbitrary Angles
•
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.
rˆ sin cos iˆ sin sin ˆj cos kˆ
• Put components into Pauli spin matrices
cos
sin cos i sin sin
Sr
cos
sin cos i sin sin
• and solve for its eigenvalues
Sr | S I | 0 1
P460 - Spin
12
rˆ sin cos iˆ sin sin ˆj cos kˆ
cos
sin cos i sin sin
Sr
cos
sin cos i sin sin
• Go ahead and solve for eigenspinors.
S r
a
for 1
b
a cos b sin (cos i sin ) a
a (1 cos )
sin 2 i
cos
b
a
e
(use 1sin
t an 2 )
i
sin e
cos 2
• Phi phase is arbitrary. gives
i
cos
e
sin
2
r i 2 for 1 r
e sin 2
cos 2
• if r in z,x,y-directions 1
0
z : 0 ,
0
1
x : 2 , 0
1
2
1
2
12
, 1
2
P460 - Spin
1
2
i
2
i2
, 1
2 13
y : 2 , 2
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 , S i
i
j
S
ijk k
• representations of angular momentum have 2
quantum numbers:
l 0, 12 ,1, 32 ......
m l ,l 1...l 1, l
2l 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 L S L Li S Si
1
if L 0, S 2 J 12 , J z 12
sin glet doublet doublet 1 2 2
P460 - Spin
14
Combining Angular Momentum
• Non-trivial examples. add 2 spins
1
if S1 2 , S2 12 with S z1 12 S z 2 12
J S1 S2
J 1, J z 1,0,1 AND J 0, J z 0
doublet doublet triplet sin glet
2 2 31
• add spin and orbital angular momentum
if L 1, S 12 with Lz 1,0,1 S z 12
3
3
1
J 2 , J z 2 , 2 AND J 12 , J z 12
triplet doublet quartet doublet
3 2 4 2
P460 - Spin
15
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
m total m1 m2 m1 m2
l m total m1 m2 m1 m2
m m1 m2 m1 m2
m1 m2 l
m l m
m1 m2 l m l m
P460 - Spin
16
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
, , ,
S z S1z S2 z
S z 1
Sz 0
S S1 S2
Sz 0
C (l m)(l m 1)
S z 1
• step down from ++ state
S1 C ( 12 , 12 )
S2 C ( 12 , 12 )
S 2 (
2
)
S l 1, m 1 C (1,1) l 1, m 0 2 l 1, m 0
1
( )
2
1
l 0, m 0
( ) orthogonal
2
l 1, m 0
• Clebsch-Gordon coefficients
P460 - Spin
1
2
17
Combining Ang. Momentum
• check that eigenstates have right eigenvalue for S2
• first write down
S Sx iSy
2
2 2 2
S ( S1 S2 ) S1 S2 2 S1 S2
2 2
S1 S2 2 S1z S2 z 2 S1x S2 x 2 S1 y S2 y
2 2
S1 S2 2 S1z S2 z S1 S2 S1 S2
• and then look at terms
2
2
1 2
3
S1 X
((S1 ) ( S1 ) ) 2 X
4
2
1
3
X
( )
S22 X 2 X
2
4
2 S1z S2 z X 2( )( ) X
2
2
with S2 S1 S2 S1 0
and S1 S2 2 S2 S1 2 0
( S2 S1 S1 S2 ) X 2 X
• putting it all together see eigenstates
2
2 3
2 2
3
1
S X ( 4 4 2 1) X X
0
P460 - Spin
18
• L=1 + S=1/2
Jz
J1
1 12 32
3
2
12 12
3
2
1
2
1 12 12
3
2
1
2
1 12 12
3
2
1
2
12 12
3
2
1
2
1 12 32
3
2
Lz
Sz
0
2 terms
0
J2
• Example of how states “add”:
Lz
Sz
j m
3
2
1
2
1
3
1 12
2
3
0
1
2
j m
1
2
1
2
2
3
1 12
1
3
0
1
2
• Note Clebsch-Gordon coefficients
1 2
,
3 3
P460 - Spin
19
• Clebsch-Gordon coefficients for different J,L,S
P460 - Spin
20