5.73 Lecture #35
35 - 1
SPIN - ORBIT: Many-Electron ζ(N, L, S) ↔ Single-Orbital
ζnl Coupling Constants
LAST TIME:
∑ e2 rij
i≥ j
death of orbital picture
expansion of 1/rij: multipoles, integrals over
AOs in nucleus-centered coordinates
SELECTION RULES: orbital and many-e–
basis sets
Gaunt Coefficients: a k , b k , c k [products of 3-j]
k
k
Slater-Condon ( F , G ) → ( Fk , Gk )
Sum Rule Method - avoid necessity to derive:
*eigenvectors
* off diagonal elements in Slater basis
Hund’s 1st and 2nd Rules
TODAY:
A.
General Importance of spin - orbit term
HSO = ∑ a ( ri )l i ⋅ s i
1 − e − operator
i
B.
C.
Trick: replace 1 - e – operator by more convenient ζ(N , L , S)L ⋅ S for ∆S = 0,
∆L = 0 special case
Pattern: Landé Interval Rule (Patterns are for breaking! Information
about "dark" states)
D.
HSO matrix elements in Slater Determinantal Basis Set
* another operator replacement
* Fundamental control parameter: n , l - scaling of ζ nl
* ζ nl ↔ ζ(N , L , S)
* off - diagonal spin - orbit matrix elements: ∆L ≠ 0, ∆S ≠ 0, ∆J = 0.
next time → Hund’s 3rd Rule and Lande gJ–factors
updated September 19,
5.73 Lecture #35
35 - 2
A. Importance of spin-orbit
1. HSO produces diagnostically significant “fine structure”
CONFIGURATIONAL ASSIGNMENTS
L,S term assignments
PATTERNS:* # components
* sign of pattern (largest splitting on top or bottom)
* statistical weight (2J + 1) of lowest vs. highest
energy component
* overall magnitude of splitting
2. for heavy atoms, HSO responsible for such large splittings and offdiagonal interactions that L-S terms “vanish”, ∆S selection rules
are violated. “Inter-System Crossing (ISC)”.
Need to “deperturb” to recover Fk, Gk parameters which should
vary smoothly from atom to atom (isoelectronic series) (shielding
rules)
3. spin-forbidden transitions provide energy linkages between
manifolds of states with different values of S. “InterSystem
Crossing (ISC)”
4. non-textbook Zeeman tuning coefficients (clues about unobserved
“dark” states)
Atoms, Molecules, Quantum Dots, solids: in an electronic transition, light
acts on single e– and operates exclusively on spatial part of ψ → spin-flips
are forbidden except when HSO mixes states of different S — forbidden
transitions “borrow” intensity from allowed transitions. In time-domain: a
short pulse prepares, at t = 0, a non-eigenstate that is a pure ∆S = 0
excitation (and ∆l = ±1) basis state. The “pluck”!
Language: the name of each eigenstate is based on dominant (i.e., “nominal”)
character. It is the name of the dominant basis states. Use of same name for
both eigenstate and basis state is a source of confusion.
updated September 19,
5.73 Lecture #35
35 - 3
B. Operator Replacement for HSO
H SO = ∑ a(ri )l i ⋅ si
a one - e − operator
i
Wigner-Eckart Theorem for vector operator — operator replacement for
special case of ∆J = 0 matrix elements.
JM J Â JM J′ = J A J JM J Ĵ JM J′
proportionality
constant
∆J = 0
matrix element of
corresponding component of Ĵ
CTDL p. 1054 use projection theorem: V = J ∑V J.
J2
Especially useful when V is an angular momentum that is included in J.
useful trick
for HSO
a(ri )l i
vector with respect to L
si
vector with respect to S
Special case of ∆L = 0, ∆S = 0 matrix elements in NLM LSM S basis set
configuration
label
H SO = ∑ a(ri )l i ⋅ si → ζ ( N , L, S )L ⋅ S
i
*
*
operator
replacement!
ζ ( N , L, S ) ≡ ∑ L a(ri )l i L S si S
i
a different spin-orbit coupling constant for EACH L-S term of the N
configuration
* convenient because it is easy to evaluate matrix elements of L⋅S without
having to resort to Slater determinatal basis set
( )
 ASIDE: a ri l i and s i are both vectors with respect to J, thus HSO is scalar 


with respect to J, hence matrix elements in NJLSM J basis set are ∆J = 0, 



∆M J = 0, and independent of MJ .
CAUTION: L⋅S seems to imply ∆S = 0 selection rule, but we assumed
∆S = 0 in deriving the simplified form of HSO: ζL⋅S
updated September 19,
5.73 Lecture #35
35 - 4
C. Landé Interval Rule (useful for recognizing and assigning an isolated L-S term)
J 2 = L2 + S2 + 2 L ⋅ S
J=L+S
J 2 − L2 − S2
L⋅S =
2
NJLSM J H
SO
NJLSM J =
h
2
2
ζ ( N , L, S )[ J ( J + 1) − L( L + 1) − S( S + 1)]
can be positive,
zero, or negative
So, within an L-S term, HSO causes splitting into 2S+1 (or 2L+1 if S > L) components.
J
( h2 2)ζ ( N , L, S)[ J ( J + 1) − ( J − 1) J ] = hζ ( N , L, S) J
Landé interval
rule
J–1
3
P2
2
P1
1
3
P0
spacing is ∝ to larger of 2 J’s
largest spacing locates largest J
large spacing on top: “regular” L-S term
bottom: “inverted” L-S term
[Convince yourself that you should be able to tell 3D from 3P…]
*
*
*
regular 3P [2 intervals: 2:1]
Easy to show that degeneracy–
weighted average spin-orbit energy of
a multiplet = 0 (easiest to show from
SO
trace of H in LMLSMS basis, followed
by trace invariance).
The interval rule plus the number of Jcomponents of a multiplet determine
the values of L and S.
[4P 5:3, 2 intervals; 2P 1 interval,
4
D 7:5:3, 3 intervals]
L+S
∑ (2 J + 1)EJ = 0
J = L−S
updated September 19,
5.73 Lecture #35
35 - 5
D. Matrix Elements of HSO in Slater Determinantal Basis Set
GOALS:
*
*
∆S ≠ 0 matrix elements, ∆L ≠ 0 matrix elements
relationships between ζ(N,L,S) and ζnl
orbital
L-S term
*
no interconfigurational HSO matrix elements
[except n′
l
~ nl ∝ (n ′n)
−3 /2
]
(Rydberg scaling rule.)
NONLECTURE: alternative operator replacement for HSO that is appropriate for
orbital matrix elements
HSO =
( )
∑
a ri
i
li
⋅ si
replace a(ri )ˆl i by ζ nl ˆl i using completeness:
n′l ′m′sms′ HSO nlm sms =
l
l
∑ ∑ n′l ′m′sms′ a(ri ) n′′l ′′m′′sms′′ n′′l ′′m′′sms′′l i ⋅ s i nlm sms
l
l
l
l
′′
i
a(ri )
is scalar with respect to si → ms′ = ms′′ and ms′ independent
a(ri )
is scalar with respect to l i → l ′ = l ′′, ml′ = ml′′, ml′ independent
ˆl
can' t change l in lml → l ′′ = l
i
li
⋅ si does not operate on radial part of ψ → n′′ = n
thus n ′l ′m′sms′ HSO nlm sms = δ
l
l
( )
n ′l a ri nl = (n′n)
′
l l
−3 /2
Rydberg scaling for
inner part of orbital
so, for n ′ = n ,
( )
n ′l a ri nl
 n′ 
ζ° =  
 n
l
m′sm′ l ⋅ s lm sms
l
l
l
−3 /2
ζn
l
l
∝ n −3
nl a ( ri ) nl = ζ n = n −3ζ °
l
l
updated September 19,
5.73 Lecture #35
35 - 6
This reduction of HSO shows that, for atoms, HSO acts exclusively within a
configuration except for interconfigurational matrix elements where the two
configurations differ by a single spin-orbital of the same value of l: nl ↔ n′l
same l
Examples
A is a single Slater determinant
A HSO A = ∑
u k a ( rk )l k ss k u k
k
spin-orbitals
= ∑ ζn k lk
lk
k
m k s k ms k l ⋅ s l k m k s k ms k
l
l
diagonal element picks out lzsz
= h2 ∑ ζ n k k m k ms k
l
l
k
spin-orbitals
if A is also an eigenfunction of L2 , L z , S2 , and S z then
NLML SMS HSO NLML SMS = ζ(N , L , S) h2 ML MS
ζ(N , L , S) =
∑
k
ζ n k k m k ms k
l
l
.
ML MS
The matrix element is evaluated 2 ways in order to reduce a many-e– spin-orbit
coupling constant (ζ(N,L,S)) to a sum of one-e– orbital coupling constants (ζnl)!
updated September 19,
5.73 Lecture #35
35 - 7
Example 1. nf2
uncoupled
nf 2
3
ζ (nf
H
M L = 5 M S = 1 = 3α 2α
ζ [3(1 / 2) + 2(1 / 2)]
)
=ζ
,H =
nf
2 3
5 ⋅1
nf
Single Slater
determinant!
2
(fill in the steps!)
Example 2. nf2
coupled
nf 2
3
H6
MJ = 6
= 3α 2α
nf 2 3H6 M J = 6 HSO nf 2 3H6 6
Landé :
=
h
2
2
[ J ( J + 1) − L( L + 1) − S(S + 1)]ζ(nf 2 ,3H )
6⋅7
= h2 5ζ(nf 2 ,3H )
5⋅6
1⋅2
from many-e– form
= h2ζ nf [3(1 / 2) + 2(1 / 2)]
from orbital form
∴ ζ(nf 2 ,3H ) = ζ nf 2
Example 3. ζ(nf2,3F)
3
F is never a single Slater
determinant for any value of (ML,MS)
Evaluate 2 ways:
a. Obtain explicit linear combination of Slater determinants using ladders
and orthogonality or using L2 , S2 to get nf 2 3 F M L = 3 MS = 1 [laborious].
b. Slater sum rule method [simple].
M L = 3, MS = 1 box: 3α 0α , 2α1α
3α 0α + 2α1α = E( 3 H M L = 3, MS = 1) + E( 3 F M L = 3, MS = 1)
3
3α 0α = 3α 0α HSO 3α 0α = h2ζ nf  + 0 
2

updated September 19,
5.73 Lecture #35
2α1α
35 - 8
= 2α1α HSO 2α1α
=
h
 1
ζ nf 1 + 
 2
2
trace of M L = 3, M S = 1 box is 3h2ζ nf
E( 3 H M L = 3 MS = 1) =
3
H M L = 3 MS = 1 HSO 3 H 3 1 = ζ(nf 2 ,3H )h2 3 ⋅ 1
(
)
but already showed ζ nf 2 ,3 H = ζ nf 2
E( 3 H M L = 3, MS = 1) = h2ζ nf (3 2)
∴ E( 3 F ML = 3 , MS = 1) = 3 h2ζ nf − ( 3 2) h2ζ nf = ( 3 2) h2ζ nf
=
3
F 3 1 HSO 3 F 3 1 = ζ( nf 2 ,3 F)( 3 ⋅ 1) h2
1
∴ ζ( nf 2 ,3 F) = ζ nf
2


1
2
2 3
 actually would find, for nf ,ζ nf , L = ζ nf for all L


2
[not true for all configurations]
(
)
We are not done. There are some off-diagonal matrix elements between the L-S-J
terms of the same configuration. nf 2 ,1 I , 3 H , 1G, 3 F , 1 D, 3 P, 1S,
6
6
5
4
HSO is diagonal in J
must diagonalize 2 3 × 3
and 2 2 × 2 matrices.
4
4
3
2
2
2
1
0
0
updated September 19,
5.73 Lecture #35
35 - 9
set up J = 6 matrix because it is simplest
1
3
1
I6 6 = 3α 3β
H 6 6 = 3α 2α
I6 6 HSO 3 H 6 6 = 3α 3β HSO 3α 2α
Mismatch is in 2nd spin-orbital.
Needs 1/2 l+s– operator to give nonzero spin-orbital matrix element.
= 3β
=
h
1
l + s − 2α ζ nf
2
ζ nf
2
1 /2
1[
3 ⋅ 4 − 2 ⋅ 3] =
2
h
ζ nf  3 
 2
1 /2
2
zero for all singlet states
I6  0
=3 
1 /2
H 6  (3 / 2)
1
HSO
J =6
3
SO 3
H6 6 H
(3 / 2)1/2  h2ζ

5/2 
=
[J (J + 1) − L(L + 1) − S(S + 1)]ζ (nf
2
5ζ (nf H ) =
5/2 ζ
h
H6 6 =
nf
h
2
2
2 3
h
2
, 3H
)
2
nf
for more complex configurations such as (nl)a(n′l′)b → ζnl and ζn′l′: two ζ parameters
needed for the two open subshell orbitals.
But can use the value of ζnl determined from some other configuration:
e.g. ζ3d from 3d64s2 can be used to predict the 3d part of HSO in 3d6 4s4p. Unexpected
inter-relationships between superficially unrelated observables. Small number of
control parameters.
Hund’s 3rd Rule: lowest energy J of lowest energy L- S term is J = L − S
if subshell is less than 1 / 2 full, is J = L + S if subshell is
more than 1 / 2 full, and J = S because L = 0 for half filled
subshell. Sign of ζ ( N , L , S ) as diagnostic!
NEXT TIME!
updated September 19,