# Laser spectroscopy of diatomic metal-containing molecules

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```					                                   Imperial lectures, April 2007
Timothy C. Steimle,
Lecture #3
Summary from Lecture #2
a) “Effective” Hamiltonian for 2P state
Fine structure
NJ-S

Mag.Hyf. terms
“+” raising op.

2S +   state drop the spin-orbit (A), -dbl (p&q). and a & d terms

b) Expressions for Hund’s case (a) basis set matrix elements

Eq. 1

Eq. 2
Summary from Lecture #2
Near term goal: Parameters  spectrum & 

Example: Model energy levels and spectra for OH X2P

Parameters (MHz) for X2P:               Energy levels X2P:
Rotation B=555660.97
Rotation D=57.175                              Proton mag. Hyf. splitting
Spin=orbit: A=-4168639.13
F=2
Spin=rotation: g=-3574.88                                   Parity:+
-doubling: p=7053.09846                          F=1     Does not look
like simple model
-doubling: q=-4168639.13        Cm-1                       J=3/2
prediction. i.e.
Mag. hyf: a=86.1116                         -doubling    E=BJ(J+1)

Mag. hyf: bF=-73.2537
F=2        Parity:-
Mag. hyf: c=130.641
F=1
Mag. hyf: d=56.6838

Next
Summary from Lecture #2
What we have calculated:
Parity                             F DE(MHz)
+                                2       70.15
1
-doubling                                Notice large diff.
J=3/2                                                          between -doublets. Due
7799.101 MHz
to the “d” hyperfine
Rotational                                                     parameter
spacing; 3.3B       _                             2      11.99
1
J=1/2
_
1     90.53
0
4733.123 MHz
Spin-orb. A
+                             1     14.69
0

J=3/2                +                             2       55.16
1
1667.145 MHz
_                             2
1      55.06
This transition is used to detect OH in the interstellar
media. We will look at it in more detail below.
Summary from Lecture #2
How good is the calculation ? Compare the transition freq. for lowest -doublet
A B C   D
2
1

J=3/2

2
1

F
Outline for Lecture 3
A) Interpretation of spin-orbit & hyperfine parameters of OH.
i) Spin-orbit of OH from O information
ii) hyperfine parameters of OH
B) Energies and wave functions for 2S+ states

C) Prediction Relative Transition Intensities
i) OH -dbl. transitions
ii) 174YbF N=0 N=1
A) Interpretation of the parameter in X2P OH i) spin-orbit,A
“Interpretation” means obtaining of a molecular orbital level understanding of
the magnitude and sign of the parameters used to model the energy levels. I.E
is our m.o. correlation diagram and our knowledge of atomic properties
consistent with the molecular parameters B,p, q and A?
(2p0-1s)s

This requires that the “ macroscopic”
2pp+1 2pp-1
2p+1    2p0 2p-                         operators used in Heff be written in
1
H(1s)   “microscopic” form. For spin-orbit
(2p0+1s)s
interaction the two operators are: .
The Macroscopic “fitting” spin-orbit Heff=ALzSz
O 2s         2ss
OH

The way to obtain this molecular orbital level understanding is to:
a) select any non-zero matrix element of Heff .
b) write down the el as anti-symmeterized product of one-electron
function which is an eigen-function of Sz & Lz. (i.e a linear
combination of Slater determinants)
c) Evaluate the selected matrix elements using the one-elec. functions.
A) Interpretation of the parameter in X2P OH i) spin-orbit,A
a) A matrix element of the Macroscopic effective operator used to model the
data for J=3/2 =3/2 X2P of OH.

Focus on this matrix element
for a comparison between
Micro and Macro description.

= -69.52 cm-1

The same matrix element using the Microscopic operator form: (Note only diagonal
terms of operator required):

= ??? cm-1

b) re-write the basis function in terms of Slater determinant of
molecular orbitals: p3 Config. with electrons arranged to have =1, S=1/2
Short-hand notation used by
“L-B&F” and “B&C”

This wave function is an
eigenfunction of Sz and Lz
with eigenvalues ½ and 1.
A) Interpretation of the parameter in X2P OH i) spin-orbit,A

Remember that              Note that lzi & szi are one-electron
operators. Use rules for evaluation of matrix
elements of one elec. op. for Slater
determinaental representations of el. L-B&F

Where because the 3 electrons are
all in the same p type orbital we
have ai(r)  ap. .

Note that p orbitals are non-
bonding and look like O-atom
p-orbitals.
Therefore
Atomic spin-orbit par for O

ap                              =153cm-1
= -153/2 =-75.8 cm-1
Our m.o. picture does a great job of predicting spin-orbit (-75.7
vs. -69.52 cm-1)!
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters

Hyperfine parameters (MHz) for X2P: a=86.1116 bF=-73.2537
c=130.641     d=56.6838

Slater Determinant

Need a value for <r-3> to proceed. Should
be  bond length because unpaired elec. is
on O whereas the spin is on H.  let r 
1.0Å <r-3> 1.010-30m-3
a 97 MHz     Simple model works well for a !
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters

The microscopic form of parameter:

Spin-density at nucleus. Only 
for electrons in orbitals that
have some “s-orbital” character.

The p-orbitals have no “s-orbital” character.  to
the first approx. it is predicted that bF =0

Exp. Value is bF=-73.2537 MHz

Need to modify the simple m.o. picture for X2P of OH.
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters
Current Model for bonding in OH

(2p0-1s)s       3s*
Proton Spin
2pp+1   2pp-1   1p
2p+1    2p0 2p-1

(2p0+1s)s      3s      H(1s)

2s                                   This interaction is
O 2s                                                             different from interaction
2ss
of up-down pair. i.e. Spin-
polarization. This is treated
OH
by mixing in a bit of a
Config: .. (2s)2 ((3)s)2(1p)3                                            higher energy configuration.
X2Pi

Config2: .. (2s)2 ((3s)1 (1p)3 ((3s*)1        2P        2P       & 4P r
i,        r

It only takes a very small admixture of Config2 because aF(H-atom) =1470 MHz)

Need the wavefunctions for the two 2P states from 3 open shell configuration:
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters
Nesbit’s table:

Eq. 1

Eq. 2

Eq. 3

Eq. 4

Eq. 4 will produce a small negative bF value.   bF  -100 MHz
B) Energy levels for    174YbF   X2S+
174Yb   (I=0)   19F   (I=1/2)
YbF X2S+        =0, S=1/2, S= 1/2

Basis set
Y(case(abJ))=|>|SS>|(JI)FMF>                             FJ        I
   S   S      J    I                                                   R
1=|0;1/2,1/2;F-1/2,1/2,F,0.5>                               S
2=|0;1/2,1/2;F+1/2,1/2,F,0.5>                    S
3=|0;1/2,-1/2;F-1/2,1/2,F,-0.5>
4=|0;1/2,-1/2;F+1/2,1/2,F,-0.5>           4x4 Matrices

Although , we will use a Hund’s “case a” basis set, S states have an energy level
pattern that is close to that of a Hund’s “case b” because the electron spin is not
tied to the internuclear axis by spin-orbit interaction. In a Hund’s case (b) Hrot
and Hspin-rot have only diagonal matrix elements:

HSpin.R. small splitting between J= N+1/2
and J=N-1/2 levels which increases approx.
linearly with N.

Therefore we expect that the predicted energy level pattern will look like:
B) Energy levels for   174YbF   X2S+ (cont)
3/2                   2
1                         1
1/2                      1       We want to model these energy levels exactly
0
using first a Hund’s case a basis set. We neen
matrices for F=0,1,2. There are only 2 levels
with F=0 but 4 levels for F=1 and 2 (same for
higher J values.
1
0 1/2
0
N J                        F

We will only calc. these energy levels.  need matrix rep. for F=0,1 &2
F=0.0                                         F=1
1=|0;1/2,1/2; 1/2,1/2,1,0.5>
2=|0;1/2,1/2; 1/2,1/2,0, 1/2>
2=|0;1/2,1/2;3/2,1/2,1,0.5
4=|0;1/2,-1/2; 1/2,1/2,0,-1/2>               3=|0;1/2,-1/2; 1/2,1/2,1,-0.5>
>
4=|0;1/2,-1/2; 3/2,1/2,1,-0.5>

F=2                                  The spectroscopic parameters (this group)
1=|0;1/2,1/2; 3/2,1/2,2,0.5>                     B =7233.8271     MHz
2=|0;1/2,1/2;5/2,1/2,2,0.5>                      D=0.00715        MHz
3=|0;1/2,-1/2; 3/2,1/2,2,-0.5>                   g=13.41679       MHz
4=|0;1/2,-1/2; 5/2,1/2,2,-0.5>                   bF= 170.26374    MHz
c=85.4028        MHz
B) Energy levels for        174YbF   X2S+ (cont)
Need Eq. 1 &2 first slide +

The results( Fortran program provided)
F=0
ROT AND CENT COR. HAM                   THE SPIN ROTATION HAMILTONIAN
0.00000 0.00000   0.00000 0.00000       0.00000 0.00000       0.00000 0.00000
0.00000 0.24129   0.00000 -0.24129      0.00000 0.22377D-03 0.00000 -0.22377D-03
0.00000 0.00000   0.00000 0.00000       0.00000 0.00000       0.00000 0.00000
0.00000-0.24129   0.00000 0.24129       0.00000 -0.22377D-03 0.00000 0.22377D-03
B) Energy levels for         174YbF     X2S+ (cont)
THE HYPERFINE HAMILTONIAN                         THE COMPLETE HAM
0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00   0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00
0.00000D+00-0.18946D-02 0.00000D+00-0.23649D-02   0.00000D+00 0.23962D+00 0.00000D+00-0.24388D+00
0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00   0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00
0.00000D+00-0.23649D-02 0.00000D+00-0.18946D-02   0.00000D+00-0.24388D+00 0.00000D+00 0.23962D+00

THE EIGENVALUES AND           EIGEN VECTORS
0.483506 cm-1
“-” parity
F=0                          0.0000 0.7071 0.0000         -0.7071
0.000000
1.0000 0.0000 0.0000          0.0000
0.000000
0.0000 0.0000 1.0000          0.0000
-0.004260 cm-1                                    “+” parity
0.0000 0.7071 0.0000          0.7071

THE EIGENVALUES AND EIGEN VECTORS                  THE EIGENVALUES AND EIGEN VECTORS
1.449375                                            1.449375
-0.0003 0.7071    -0.0003 -0.7071                   -0.0003 0.7071    -0.0003 -0.7071
F=1       0.484722
0.5968 -0.3792   -0.5968 -0.3792         F=2
0.484722
0.5968 -0.3792   -0.5968 -0.3792
0.478313                                            0.478313
0.3792 0.5968    -0.3792 0.5968                     0.3792 0.5968    -0.3792 0.5968
0.001420                                            0.001420
0.7071 0.0003     0.7071 -0.0003                    0.7071 0.0003    0.7071 -0.0003
B) Energy levels for                    174YbF        X2S+ (cont)
Many matrix elements to calculate. Best to write program. Here is sample of code provided
SUBROUTINE HSROT12(FL,FSI,FOM,FJ,HW,PAR,S)
Rotation terms; Eq. 1
IMPLICIT REAL*8(A-H,O-Z)
DIMENSION FL(4),FOM(4),FSI(4),FJ(4),HW(4,4),PAR(6)
DIMENSION HCD(4,4)
DO 500 I=1,4
DO 500 J=I,4
HW(I,J)=0.0
IF(FJ(I).NE.FJ(J))GOTO 499
IF(FJ(I).LT.0.0)GOTO 499
IF(FJ(J).LT.0.0)GOTO 499
PAR4=0.0
DO 400 K=1,2
IF(K.EQ.1)Q=-1.0
IF(K.EQ.2)Q=1.0
X=FJ(J)-FOM(I)+S-FSI(I)+0.1
IP=IDINT(X)
PHASE=(-1.0D0)**IP
PAR1=THREEJ(FJ(I),1.0D0,FJ(I),-1.0*FOM(I),Q,FOM(J))
PAR2=THREEJ(S,1.0D0,S,-1.0*FSI(I),Q,FSI(J))
PAR3=(FJ(I)*(FJ(I)+1.0)*(2.0*FJ(I)+1.0)*S*(S+1.0)*(2.0*S+1.0))
PAR4=PAR4+PAR1*PAR2*DSQRT(PAR3)*PHASE
400            CONTINUE
PAR5=0.0
IF(FSI(I).NE.FSI(J))GOTO 480
IF(FOM(I).NE.FOM(J))GO TO 480
PAR5=FJ(I)*(FJ(I)+1.0)+S*(S+1.0)
480            CONTINUE
C      NOW CALC THE N MAT
HW(I,J)=PAR5-2.0*PAR4
499             CONTINUE
HW(J,I)=HW(I,J)
500            CONTINUE
C      NOW SQUARE N MAT. TO GET CENT. COR
DO 610 I=1,4
DO 610 J=1,4
HCD(I,J)=0.0
DO 610 K=1,4
HCD(I,J)=HCD(I,J)+HW(I,K)*HW(K,J)
610           CONTINUE                                             DO 630 J=1,4
C      NOW CALC. HROT AND HCD                                630     HCD(I,J)=HCD(I,J)*PAR(2)
DO 620 I=1,4                                          DO 640 I=1,4
DO 620 J=1,4                                         DO 640 J=1,4
620             HW(I,J)=HW(I,J)*PAR(1)                       640     HW(I,J)=-HCD(I,J)+HW(I,J)
DO 630 I=1,4                                          RETURN
END
C) Prediction of relative transition probability (OH & YbF pure rotation)
The strongest interaction between the light and a molecule is via the interaction
between the electric dipole moment op. and the electric field of the radiation:
Helec. dipoole= -mE(t) = -mE(t) cos

 transition moment.     When |> = |’’> then “transition moment” =
the permanent electric dipole moment of the
Location of charges
molecule. A typical value values are: OH ~
1.5 D; CaF ~ 3.5 D; CCCCCH ~ 10 D

When |> and |’’> are different
Internuclear Sep.
vibrational level of the same electronic state
then “transition moment” is proportional to
2nd term in expansion. This term is typically
110-4 or less smaller than the permanent
electric dipole moment. (i.e much smaller

When |> and |’’> are different electronic   states then the “transition
moment”

Franck-Condon Factor                             Typical values 0.1D- 10D
C) Prediction of relative transition probability (OH & YbF pure rotation)
Note: not necessarily a square matrix
“b”                                                                  Eigenvec for a

An “a” “b” matrix of
transition matrix
Eigenvec for b   elements calculated
using the same basis
functions used to
“a”                                        produce the eigenvecors
for “a” and “b”

Need expressions for matrix elements of dipole moment operator. First consider
no nuclear spin; Hund’s case (a):

If the spectra are recorded in the absence of a static or electric field then all of
the levels with different MJ values will be degenerate and the transition moment
will be the sum over all possible MJ values

We need expression for YbF, OH and CaF; one
nuclear spin
C) Prediction of relative transition probability (OH & YbF pure rotation)

Hund’s case (abJ)

If the spectra are recorded in the absence of a static or electric field then all of
the levels with different MF values will be degenerate and the transition moment
will be the sum over all possible MF values

For the two nuclear spin case :
C) Prediction of relative transition probability (OH & YbF pure rotation)
Need to write program: Hund’s case (abJ)
SUBROUTINE OHINT12(nfl,nfu,basisu,basisl,vut,vlt,enut,enlt,              Transfer eigen vectors (“vut” &“vlt”), energies
c
1            FLT,ELo,FUT,EUP,BF,tf,fin,ic,TK)
calc transisiton int.
(“enut” & “enlt”) and basis (“basisu” & “Basisl”
implicit real*8(a-h,o-z)
DIMENSION FLL(8),FSIL(8),FJL(8),FOML(8),TF(200)
DIMENSION FLU(8),FSIU(8),FJU(8),FOMU(8),fin(200)
DIMENSION VU(8),VL(8),vut(10,8,8),vlt(10,8,8)
DIMENSION TOM(8,8),basisl(10,6,8),basisu(10,6,8)
DIMENSION TEMP(8),Enut(10,8),Enlt(10,8),BF(200)
Dimension EUP(200),ELo(200),FUT(200),FLT(200),IDU(200)
dimension IDl(200)
OPEN(3,FILE='int12.PUT')
c                speed of light =sol
sol =29979.2458D0
ic=1
elowest=0.0d0                                        Find lowest energy for evaluation of Boltz. Factor
do 601 i=1,nfl
do 601 j=1,8
601              if(enlt(i,j).lt.elowest)elowest=enlt(i,j)
WRITE(3,250)
250              FORMAT(1x,'low F',3x,'En Low',5x,'up. F',5x,'En up',
1                    5x,'tran. freq', 5x,'Boltz. Fac.',5x,'intensity')
do 1000 ik=1,nfl
do 1 kk=1,8
s=basisl(ik,1,kk)
fll(kk)=basisl(ik,2,kk)
fsiL(kk)=basisL(ik,3,kk)
fomL(kk)=basisL(ik,4,kk)
Put basis functions into individual arrays
fjL(kk)=basisL(ik,5,kk)
fL=basisl(ik,6,kk)
1                continue
c                loop over upper number of f values                                                   Loop over upper and Lower
do 999 ikK2=1,nfu
do 2 kk=1,8
C                PAUSE
s=basisu(ikK2,1,kk)
flu(kk)=basisu(ikK2,2,kk)
fsiu(kk)=basisu(ikK2,3,kk)
fomu(kk)=basisu(ikK2,4,kk)        Put basis functions into individual arrays
fju(kk)=basisu(ikK2,5,kk)
fu=basisu(ikK2,6,kk)
2                continue
do 225 ii=1,8
do 225 jj=1,8
TOM(II,JJ)=0.0D0
C     SKIP FALSE ROOTS
IF(FJU(II).LT.0.0)GOTO 225
IF(FJL(JJ).LT.0.0)GOTO 225
IF(fsiu(ii).ne.fsil(jj))GOTO 225
fphase = fju(ii) + 0.5d0 + fL + 1.01
i3 = dint(fphase)
phasf = (-1.00d0)**i3
aa=sixJ(fju(ii),fu,0.5d0,fl,fjl(jj),1.0d0)
Generate transition moment Matrix
ultj2=dsqrt((2.0*fu+1.0)*(2.0*fL+1.0))
aaa=fju(ii)-fomu(ii)
`
i2=dint(aaa)
phas=(-1.00d0)**I2
ultj=dsqrt((2.0*fju(ii)+1.0)*(2.0*fjL(jj)+1.0))
fiq=0.0d0
bb=threej(fju(ii),1.0d0,fjl(jj),
-1.0*fomu(ii),fiq,foml(jj))
224   continue
a4 = phasf*ultj2*aa*phas*ultj*bb
tom(ii,jj)= a4
225   Continue

C   NOW LOOP OVE EIGENVECTORS
DO 500 KK=1,8                     BEGIN        MATRIX MULTIPLICATION
C PULL OUT LOWER eIGEN VECTOR
eLOO=EnLt(ik,KK)                                                                                Eigenvec for a
DO 237 ID=1,8
237 Vl(ID)=vlt(ik,ID,KK)                                                      An “a” “b” matrix
DO 499 KK2=1,8                                                            of transition
C PULL OUT UPPER IGEN VECTOR                                   Eigenvec for b matrix elements
eupP=Enut(ikK2,KK2)
DO 239 ID2=1,8
239 VU(ID2)=vut(ikK2,ID2,KK2)
do 240 kkK=1,8
temp(kkK)=0.0d0
do 240 k=1,8
temp(kkK)=temp(kkK)+tom(kkK,k)*vl(k)
Multiply TM by lower eigenvec,
240 continue
C) Prediction of relative transition probability (OH & YbF pure rotation)
tint=0.0d0
do 242 k=1,8
tint = tint + vu(k)*temp(k)
Multiply by upper    eigen vec
242 continue
tint2=tint*tint
C     store if greater than 0        Sq. trans.moment
eUP(ic)=EUPP
eLO(ic)=ELOO
C    BF=BOLTZMANN FAC
el=ELO(IC)-elowest
el2=(ELO(IC)-elowest)/sol
Calc. Boltzmann factor
kt=TK*0.695d0
BF(ic)=DEXP(-(el2/kt))
C    DF=DEGENERCY FACTOR
DF=2.0*FL+1.0                      Degeneracy factor (the sum over the 3j involving MF)
fint=(tint2)*BF(ic)
tftt=EUP(IC)-ELO(IC)
C get only positive trans. freq.           Calc. intensity.
if(tftt.le.0.0d0) go to 247
C only keep non-zero intensites            Calc. Transition freq.
if(fint.lt.0.01d0) go to 247
fin(ic)=fint
tf(IC)=tftt
fut(ic)=fu
flt(ic)=fl
idu(ic)=KK2
idl(ic)=KK
Print out results
write(3,243)fu,elo(ic)/sol,fu,eup(ic)/sol,tf(IC),bf(ic),fint
243 format(f6.2,F12.3,f6.2,2F12.3,2f14.4)
ic=ic+1
247 continue
499 CONTINUE
500 CONTINUE
999 continue
1000 continue
Lowest -doublet

-38.120

F”          F’             Freq.   Boltz. Int.
-38.140                2
1.0 -38.201 1.0 -38.145   1665.411 1.000 1.440
+            1
1.0 -38.201 2.0 -38.143   1720.481 1.000 0.288
-38.160
2.0 -38.199 1.0 -38.145   1612.291 1.000 0.288

2.0 -38.199 2.0 -38.143   1667.361 1.000 2.592
-38.180
Output from program:

2
-38.200
-            1
J= 3/2   F
-38.220

1600.0       1650.0        1700.0
Microwave Freq (MHz)
1    F”         F’            Freq.   Boltz. Int.
88.250
-            0   0.0 88.093 1.0 88.252   4765.581 0.533 0.178

1.0 88.093 0.0 88.249   4660.192 0.533 0.178

88.200                    1.0 88.093 1.0 88.252   4750.676 0.533 0.356

88.150

88.100                1
+            0
J= 1/2   F
4650.0        4675.0        4750.0
Microwave Freq (MHz)
C) Prediction of relative transition probability (OH & YbF pure rotation)
3/2                       1     G=1
2
0
1/2
N=1                 1    G=0
174YbF N=0- N=1
2.00

1
1/2                                        1.50
0
J    N=0                 F

Int
1.00

0.50

0.00
14450.0 14455.0 14460.0 14465.0 14470.0 14475.0 14480.0 14485.0 14490.0
FREQ
Thank You !

```
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