SV-InclusionSOcouplinginNaCs

Including spin-orbit coupling of the first excited electronic states in the
photoassociation and rovibrational relaxation of NaCs
v. 2.6
Goal To find equations for the probability amplitudes of the relevant rovibrational states in my process
(see PhD proposal) when including the spin-orbit coupling interaction between the A1
Σ+ state and the
b3
Π state of NaCs.
1 Introduction
In my PhD proposal [1, §4.2], I mentioned the necessity to include the effect of spin-orbit (SO) coupling
between the A1
Σ+ and the b3
Π electronic states of NaCs into my calculation. Zaharova et al. [2] deter-
mined experimentally the strength of the SO interaction for the A1
Σ+ − b3
Π manifold of NaCs. After
learning about the fundamentals of SO coupling in diatomic molecules1, I now have the necessary under-
standing to include the SO interaction in the model Hamiltonian of my problem.
Section 2 presents the major steps involved in the derivation. Section 3 discusses how the inclusion of
the SO coupling affects the various operators appearing in the total Hamiltonian, and proposes ways to
deal with the consequences. Section 4 outlines the next steps to take toward a solution.
2 Theoretical Description
2.1 The problem
Given a gaseous mixture of Na and Cs atoms at ultracold temperature (T = 200µK), I study the transfer of
population from a scattering wave packet |χX0 above the asymptote of the X1Σ+electronic state (JX = 0)
of NaCs, to a low-lying rovibrational state |X,vX,JX of the X1Σ+ state (JX = 0 or 2), using high-lying
rovibrational state(s) |A ∼ b in the A1
Σ+—b3
Π manifold as intermediate state(s).
The key point of the present notes is to find a way to specify the actual physical content of the state(s)
|A ∼ b : which high-lying states are useful, whether they are rather of singlet or triplet character, or given
the energy these states have, whether the singlet/triplet distinction is meaningful.
One gaussian laser pulse linearly-polarized (the pump pulse) binds the scattering atoms into a di-
atomic molecule in a superposition of rovibrational states of the A1
Σ+ −b3
Π manifold. A second gaussian
laser pulse linearly-polarized (the Stokes pulse) allows stimulated emission of a photon from the molecule.
The Stokes pulse is designed so that the molecule ends up in a rovibrational state of the X1Σ+ state with
a rovibrational energy as low as possible.
1The references I used are [3, §9.3], [4, §3.1 & 3.4], and [5].
1

Stéphane
Valladier
Inclusion of Spin-Orbit coupling in my PhD project
v. 2.6
22nd December, 2013
2.2 Approximations
The intensities of the two laser pulses are sufficiently low (≤ 300kW.cm−2
) to treat the laser fields semi-
classically. Also, the wavelengths of the lasers range from ≈850nm to ≈900nm, and the spatial extension
of the molecule remains below 3nm, justifying the application of the Long Wave Approximation (LWA):
the spatial dependence of the laser electromagnetic field can be neglected. The lasers interact with the
molecules and the gas of atoms through the electric dipole interaction (EDA): other electric moments of
the molecule are also neglected.
When using wave functions in this problem, I will use the Born-Oppenheimer approximation: the
dynamics of the nuclei are decoupled from the dynamics of the electrons. This decoupling has two
consequences2[4, p. 90]:
1. For a given set α of quantum numbers describing the electrons, the wave function of the molecule
can be written as a product of two wave functions, one for the nuclei and one for the electrons
ψBO
α,v = χv(R,θ,ϕ)Φel
α (#r ;R),
where #r denotes the set of coordinates of all electrons, and the semi-colon indicates that R is a
parameter,
2. The internuclear separation R being a parameter for the electronic wave function, the radial part
Tn(R) of the nuclear kinetic energy operator Tn does not act upon the electronic wave function
Φel
α Tn(R) Φel
α = Φel
α |Φel
α Tn(R) = δαα Tn(R) .
2.3 The model
2.3.1 Hamiltonian of the system
The system consists of Na and Cs atoms governed by the Hamiltonian
H (t) = Tn + Vnn + Vne + Vee + Te
He
+HSO −
#
d ·
#
E (t), (1)
where
Tn – kinetic energy operator for the nuclei,
Vnn – nucleus-nucleus Coulomb interaction,
Vne – nucleus-electron Coulomb interaction,
Vee – electron-electron Coulomb interaction,
2See my notes on the Born-Oppenheimer approximation, Hund’s cases, and adiabaticity
SV-InclusionSOcouplinginNaCs.tex 2

Stéphane
Valladier
v. 2.6
22nd December, 2013
Te – kinetic energy operator for the electrons,
HSO ≡ i âi
#
i · #si – spin-orbit interaction (see [4, Eq. (3.4.3) p. 182]. Also the sum runs only over
open shells electrons, [4, last sentence p. 183])
#
d – electric dipole operator,
#
E (t) ≡
#
EP (t) +
#
ES(t) – electric field operator (P: pump pulse, S: Stokes pulse).
The Hamiltonian H (t) governs 2 nuclei and a total of 11 + 55 = 66 electrons and is written in the center
of mass frame. The kinetic energy operator of the nuclei is defined as:
Tn(
#
R) ≡ −
2
2µ
1
R2
∂
∂R
R2 ∂
∂R
+
R2
2µR2
(2)
with the rotational energy operator defined as:
R2
≡ − 2 1
sinθ
∂
∂θ
sinθ
∂
∂θ
+
1
sin2
θ
∂2
∂ϕ2
. (3)
Figure 1 defines the angles θ and ϕ in the center of mass frame.
ˆX
ˆY
ˆZ ˆz
Cs
Na
θ
ϕ
Figure 1: Definition of angles θ and ϕ in the center of mass frame. The ˆz axis is the
molecular axis. The cesium atom being heavier than the sodium atom, the center of mass
of the diatomic molecule is closer to Cs than to Na.
2.3.2 Descriptor of the system
There are two ways to describe the system: either using a wave function or a density operator. The
treatment via the density operator is the best way to treat the initial condition (i.e. a gaseous mixture in
thermal equilibrium at ultracold temperature T = 200µK), but requires to solve the quantum Liouville-
von Neumann equation. If the density operator is expressed in a basis of the relevant Hilbert space

Stéphane
Valladier
v. 2.6
22nd December, 2013
of dimension N, then solving the quantum Liouville-von Neumann equation means solving N2 coupled
partial differential equations. When using a wave function formalism, solving the problem means solving
only N coupled partial differential equations.
The density operator also allows for the appropriate treatment of spontaneous emission, which I am
not considering in my problem.
Furthermore, there exists a way to express the initial condition for the system in the density operator
formalism using an expansion over wave packets [6, p. 013412-3]. To facilitate my understanding of the
underlying physics in my problem, I think it is best to examine the dynamics of the process for a given
wave packet, and then see how the initial spatial width, central position, and energy of said wave packet
affect the dynamics. Doing so allows to describe the system simply by a wave function, and thus to solve
only N coupled partial differential equations.
Therefore I describe the system with a wave function:
|Ψ (t) =
α
∞
J=0
J
MJ =−J
1
R
Γ α
JMJ Ω(R,t)|JMJΩ |Φel
α , (4)
where
• R ≡ internuclear separation,
• α ≡ set of quantum numbers labeling a particular electronic state,
• |Φel
α ≡ electronic wave function for electronic state α, satisfying the Born-Oppenheimer approxi-
mation (§2.2),
• The rotational energy operator R2 acts on the kets |JMJΩ as
R2
|J,MJ,Ω = (J 2
− J 2
z + L 2
− L 2
z + S 2
− S 2
z )|JMJΩ . (5)
The rotational perturbations discussed in Sec. 3.1.2.3 p. 96 and Sec. 3.2.1.1 p. 107-108 of [4] are
neglected in this calculation because the lasers involved in the problem are far off-resonance from
any transition that the rotational perturbations would allow.
• Γ α
JMJ Ω(R,t) is a superposition of rovibrational and continuum states of the electronic state α with
rotational quantum numbers J, MJ, Ω:
Γ α
JMJ Ω(R,t) =
v
a
αJ
v (t)|αvJ +
+∞
E∞
α
a
αJ
E (t)|χ
αJ
E dE (6)
where the |αvJ s are the rovibrational states in electronic state α with vibrational quantum number
v and rotational quantum number J, E∞
α the asymptotic value of the potential energy for electronic
state α (here, E∞
X = 0), and |χ
αJ
E the energy-normalized stationary scattering state with energy E
above the asymptote of the electronic state α with rotational quantum number J.

Stéphane
Valladier
v. 2.6
22nd December, 2013
The goal of these notes is to find a justification for which states should be involved in the discrete sum in
Eq. (6).
Plugging the wave function (4) into the Time-Dependent Schrödinger Equation
i
∂
∂t
|Ψ (t) = H (t)|Ψ (t) , (7)
and using the fact that the lasers are linearly polarized along the ˆZ-axis (see Fig. 1) yields equations for
the Γ ’s:
∀α, J, MJ, Ω,
i
∂
∂t
Γ α
JMJ Ω(R,t) = −
2
2µ
∂2
∂R2
−
1
2R2
Φel
α | JMJΩ R2
JMJΩ |Φel
α + V BO
α Γ α
JMJ Ω(R,t)
+
α
Φel
α | JMJΩ H SO
J MJ Ω |Φel
α Γ α
J MJ Ω (R,t)
+ (−1)MJ +1
2J + 1E (t)
α
dαα (R)

 2J + 3


J 1 J + 1
−MJ 0 MJ




J 1 J + 1
0 0 0

Γ α
J+1MJ Ω(R,t)
+ 2J − 1


J 1 J − 1
−MJ 0 MJ




J 1 J − 1
0 0 0


only 0 if J 0 and MJ ±J
Γ α
J−1MJ Ω(R,t)
(8)
where the 2 × 3 matrices are Wigner 3-j symbols.
2.3.3 Initial condition
At t = 0, the system is simply a pair of atoms scattering above the asymptote of the X1Σ+electronic state,
for which Ω = 0. Thus ∀α X1Σ+, ∀{J,MJ,Ω}, Γ α
JMJ Ω(R,t = 0) = 0. The gaseous mixture of atoms is in
thermal equilibrium at T = 200µK. I calculated3 that at this temperature only the JX = 0 rotational state
is occupied, and therefore MJX
= 0. Consequently
Γ X1Σ+
000 (R,t = 0) = 1, and ∀J 0, Γ X1Σ+
JMJ 0 (R,t = 0) = 0. (9)
I will translate the initial condition Eq. (9) into conditions for the expansion coefficients—probability
amplitudes— a’s that appear in Eq. (6) later in these notes.
3See Mathematica notebook Rotational level probability of occupation.nb.

Stéphane
Valladier
v. 2.6
22nd December, 2013
2.3.4 Method of solution
To solve the problem, notice that the Hamiltonian H (t) can be split into a time-independent term H0 and
a time-dependent term Vext(t). The idea here is to first find the eigenbasis of H0, which solves for the
spatial dependence of the descriptor |Ψ (t) , and then expanding the Γ ’s over the eigenbasis of H0 will
yield equations for the time-dependent expansion coefficients a’s of Eq. (6).
2.4 Rules for the calculation
2.4.1 Choice of basis
To start the derivation, I need to choose a basis for the electronic states |Φel
α . The research published
in Zaharova et al. [2], where Hund’s case (a) potentials and spin-orbit coupling functions are reported,
suggests to begin deriving equations in the Hund’s case (a) basis.
Since we initially decided to model the population transfer using the A1
Σ+ state as an intermediate
state, and given the results of [2], the electronic states involved in the problem are the X1Σ+ state, the
A1
Σ+ state, and the b3
Π state.
2.4.2 Selection rules and allowed transitions
Electric dipole Electric dipole transitions between a singlet and a triplet electronic state are forbid-
den, so dXb = dAb = 0. Moreover, although rotational transitions within the same electronic states are
rigorously allowed, the lasers used in this project are far off resonance from any rotational transition
within the same electronic state. Therefore, for the purpose of the derivation I can safely assume that
dXX = dAA = dbb = 0.
Because the lasers used in this problem are linearly polarized and given Eq. (9), the pump pulse can
only populate rovibrational levels of the A1
Σ+ − b3
Π manifold with J = 1. The Stokes pulse can then
populate only the rotational states of the X1Σ+ state with JX = 0 or 2. Moreover, because the lasers are
linearly polarized, the selection rule for MJ is ∆MJ = 0. Since the system starts with MJ = 0, all states
involved in the problem will have MJ = 0. Therefore, I will no longer specify MJ anymore, and remember
that it remains equal to 0 throughout the whole problem.
Spin-Orbit The spin-orbit operator HSO couples only electronic states that dissociate to the same asymp-
tote, hence X1Σ+ H SO A1
Σ+ = X1Σ+ H SO b3
Π = 0. Moreover 1Σ+ states are not affected by diag-
onal spin-orbit coupling: X1Σ+ H SO X1Σ+ = A1
Σ+ H SO A1
Σ+ = 0. The only remaining non-zero
terms of the spin-orbit operator are:
A1
Σ+ H SO
b3
Π = −
√
2ξSO
Ab (R) = −
√
2ξ(R) and b3
Π H SO
b3
Π = −ηSO
b0b1
(R) = −η(R),
where the notation from [2] has been modified for clarity.

Stéphane
Valladier
v. 2.6
22nd December, 2013
Rotational operator R2 matrix elements Following the recommendations of [4, pp. 96, 107-108], I
merge the electronic orbital angular momentum matrix element X1Σ+| 000 L 2 000 |X1Σ+ with the
potential V BO
X (R). Note that using the asymptotic electronic wave functions for the X1Σ+ state given in
Eq. (12) of Katô [5] yields
X1Σ+| 000 L 2 000 |X1Σ+ = 0. However, this is only an estimate of the matrix element at large in-
ternuclear separation: in diatomic molecules the electronic orbital angular momentum never commutes
with the Hamiltonian, and thus the quantum number L is not a constant of the motion. Zaharova et al.
[2, p. 012508-6] applied the van Vleck pure precession hypothesis4 to estimate the matrix elements5
A1
Σ+| 100 L 2 100 |A1
Σ+ = b3
Π| 100 L 2 100 |b3
Π = 2.
Since I neglect the rotational perturbations when defining R2 in Eq. (5), all operators in the Hamil-
tonian H (t) defined in Eq. (1) have the same selection rule for Ω, ∆Ω = 0. The system starts as a pair of
scattering atoms above the asymptote of the X1Σ+ state. Therefore Ω, like MJ, starts as 0, and keeps the
same value throughout the whole process. Thus in what follows, I will no longer specify the quantum
numbers MJ and Ω, and remember that they are always equal to zero.
3 An apparently smaller system of equations
Using all the information from Sec. 2.4 and plugging it into Eq. (8) yields the system of equations:
i
∂
∂t


Γ X
0
Γ X
2
Γ A
1
Γ b
1


=


−
2
2µ
∂2
∂R2 + V BO
X (R) 0 −dXA(R)
√
3
3 E (t) 0
0 −
2
2µ
∂2
∂R2 − 6
R2 + V BO
X (R) −dXA(R)2
√
15
15 E (t) 0
−dAX(R)
√
3
3 E (t) −dAX(R)2
√
15
15 E (t) −
2
2µ
∂2
∂R2 − 4
R2 + V BO
A (R) −
√
2ξ(R)
0 0 −
√
2ξ(R) −
2
2µ
∂2
∂R2 − 4
R2 + V BO
b (R) − η(R)




Γ X
0
Γ X
2
Γ A
1
Γ b
1


(10)
The above system of equations looks like a 4×4 system of coupled partial differential equations, which
is already not a trivial thing to solve. If I was to use the method outlined in Sec. 2.3.4 without more
input, I would have to remember that the X1Σ+ state with JX = 0 supports 87 rovibrational states, the
JX = 2 supports 86 rovibrational states, the A1
Σ+ state (JA = 1) supports 147 rovibrational states, and
the b3
Π state (JA = 1) supports 106 rovibrational states. Therefore, I would be faced with a system of
coupled, no-longer-partial, differential equations (the only remaining variable being time, t) of dimension
426×426.
Analysing transition dipole moment matrix elements (TDMME) allows to reduce drastically the num-
ber of rovibrational states to involve in the problem. Because of the spin-orbit coupling between the
4See my report 20121101-20121108-SVweeklyReport.pdf.
5 The analytic potentials reported in [2] do not contain the L 2 term, which is explicitly separated (see [2, Eqs. (4) (5)
p. 012508-6]). The OU12 potentials that were constructed for the A1Σ+ and b3Π states from the experimental results of [2], are
also devoid of the L 2 term, which is accounted for explicitly in the equations below.

Stéphane
Valladier
v. 2.6
22nd December, 2013
A1
Σ+ state and the b3
Π state, it is no longer valid to decide on which rovibrational states to include
based on examination of the TDMME between the X1Σ+ state and the A1
Σ+ state. What follows exposes
the necessary preliminary steps that lead to the relevant quantities to analyse in order to pick the proper
rovibrational state(s) in the A1
Σ+ − b3
Π manifold.
3.1 The hybrid basis
I can split the 4×4 matrix in Eq. (10) in 4 terms: the nuclear kinetic energy T, the rotational energy R, the
electric dipole-electric field interaction D, and the electronic and spin-orbit term Hel. In the Hund’s case
(a) basis A defined by the 4 kets {|X1Σ+,J = 0 ,|X1Σ+,2 , |A1
Σ+,1 , |b3
Π0,1 }, these matrices are:
TA = −
2
2µ


∂2
∂R2 0 0 0
0 ∂2
∂R2 0 0
0 0 ∂2
∂R2 0
0 0 0 ∂2
∂R2


A
RA = −
2
2µ


0 0 0 0
0 − 6
R2 0 0
0 0 − 4
R2 0
0 0 0 − 4
R2


A
(11a)
DA = −E (t)


0 0 dXA(R)
√
3
3 0
0 0 dXA(R)2
√
15
15 0
dAX(R)
√
3
3 dAX(R)2
√
15
15 0 0
0 0 0 0


A
Hel
A =


V BO
X (R) 0 0 0
0 V BO
X (R) 0 0
0 0 V BO
A (R) −
√
2ξ(R)
0 0 −
√
2ξ(R) V BO
b (R) − η(R)


A
(11b)
Diagonalizing Hel provides a new hybrid6 basis H. Expressing the eigenvectors of Hel in the basis A gives
the passage matrix U from basis H to basis A.
The eigenvalues of Hel are
V BO
X (R) (doubly degenerate) (12a)
V1/2(R) =
1
2
VA + Vb0 − (VA − Vb0)2 + 8ξ2 (12b)
V3/2(R) =
1
2
VA + Vb0 + (VA − Vb0)2 + 8ξ2 (12c)
where all quantities are R-dependent, V BO
A = VA, and Vb0(R) = V BO
b (R) − η(R) to simplify the notation.
Looking at Fig. 2 explains the choice of labels for the eigenvalues: V1/2(R) dissociates to the Na(32S1/2)+Cs(62P1/2)
asymptote while V3/2(R) dissociates to Na(32S1/2)+Cs(62P3/2), which is consistent with the spin-orbit cou-
pling function accounting for fine structure. Note that asymptotically, the PECs V1/2(R) and V3/2(R)
should merge with the corresponding Hund’s case (c) PECs, respectively (2)0+ and (3)0+.
Expressing the eigenvectors of Hel in basis A gives the passage matrix U from H to A. The diagonal-
6This basis is hybrid because it does not correspond to any pure Hund’s case (a), neither is it diabatic or adiabatic since Tn is
almost diagonal in H for some ranges of R and definitely non-diagonal in other ranges.

Stéphane
Valladier
v. 2.6
22nd December, 2013
ization of Hel yields:
U =


|X1
Σ+
,0 |X1
Σ+
,2 |V1/2,1 |V3/2, 1
X1
Σ+
,0| 1 0 0 0
X1
Σ+
,2| 0 1 0 0
A1
Σ+
,1| 0 0 cosγ −sinγ
b3
Π0,1| 0 0 sinγ cosγ


(13)
where
cosγ =
√
2ξ(R)
(2ξ2(R) + (VA − V1/2)2)1/2
sinγ =
(VA − V1/2)
(2ξ2(R) + (VA − V1/2)2)1/2
(14)
A similar situation and set of definitions can be found in Londoño et al. [7].
NaCs
b3
0
A1
V3 2
V1 2
4 6 8 10 12 14 16
10 000
11 000
12 000
13 000
14 000
15 000
16 000
17 000
Internuclear Separation
Energycm
1
Figure 2: NaCs Hund’s case (a) potential energy curves (PECs) for the b3
Π and A1
Σ+
state, coupled by spin-orbit interactions to yield hybrid PECs V1/2 and V3/2. Note the
double-well of the V1/2 curve with a local maximum around 4.25 ˚A, and the smooth
step of the V3/2 adiabatic curve for internuclear separations around 9.27 ˚A. The PECs
are drawn using OU12 potentials.
Looking at Fig. 2 combined with Eq. (14), I notice that when V1/2 = VA, then cosγ = 1 and sinγ = 0 i.e.
|V1/2, 1 = |A1Σ+, 1 : the |V1/2, 1 state has singlet character, and conversely, |V3/2, 1 = |b3Π0, 1 , i.e. the
|V3/2, 1 state has triplet character.

Stéphane
Valladier
v. 2.6
22nd December, 2013
3.2 Rotational matrix in the hybrid basis
The particular shape of the matrices U and RA renders the transformation of RA into the hybrid basis H
rather trivial:
RH = U−1
RAU = U†
RAU = −
2
2µ


0 0 0 0
0 − 6
R2 0 0
0 0 − 4
R2 0
0 0 0 − 4
R2


H
= RA. (15)
3.3 The dipole-field interaction in the hybrid basis
The point of theses notes is to obtain the relevant TDMME to determine which rovibrational states should
be used in my process. The electric transition dipole moment function between the X1Σ+ state and the
A1
Σ+ state published by Aymar and Dulieu [8] is a real function, and so I can simplify the notation by
defining d(R) ≡ dAX(R) = dXA(R). The transformation of the dipole-field interaction matrix from basis A
to basis H gives
DH = U−1
DAU = U†
DAU =


|X1
Σ+
,0 |X1
Σ+
,2 |V1/2,1 |V3/2, 1
X1
Σ+
,0| 0 0 −
√
3
3 cosγ
√
3
3 sinγ
X1
Σ+
,2| 0 0 −
√
3
3 cosγ
√
3
3 sinγ
V1/2,1| −
√
3
3 cosγ −2
√
15
15 cosγ 0 0
V3/2,1|
√
3
3 sinγ 2
√
15
15 sinγ 0 0


H
d(R)E (t). (16)
Thus the relevant transition dipole moment functions to consider when accounting for the spin-orbit
interaction are
dX1Σ+↔V1/2
(R) = d(R)cosγ and dX1Σ+↔V3/2
(R) = d(R)sinγ. (17)
(The Clebsch-Gordan coefficients are left out of the definitions, as they are not necessary for a qualitative
discussion.) Technically, the TDMME I need to examine are the matrix elements vX d(R)cosγ v1/2
and vX d(R)sinγ v3/2 , where |v1/2 and |v3/2 are bound states of the V1/2(R) and the V3/2(R) adiabatic
PECs respectively. The next section explains how to obtain the |v1/2 ’s and |v3/2 ’s.

Stéphane
Valladier
v. 2.6
22nd December, 2013
3.4 The kinetic energy operator expressed in the adiabatic basis
Because the transformation U depends on the internuclear separation R, the kinetic energy matrix T is no
longer diagonal in the adiabatic basis H (as expected, see [4, p. 94]):
TH = U−1
TAU = U†
TAU = −
2
2µ


∂2
∂R2 0 0 0
0 ∂2
∂R2 0 0
0 0 −
dγ
dR
2
+ ∂2
∂R2 −
d2
γ
dR2 − 2
dγ
dR
∂
∂R
0 0
d2
γ
dR2 + 2
dγ
dR
∂
∂R
−
dγ
dR
2
+ ∂2
∂R2


H
. (18)
Let’s examine the functions
dγ
dR ,
dγ
dR
2
, and
d2
γ
dR2 , plotted in figures 3, 4, and 5 respectively.
All three figures show that except in two rather narrow regions, the derivatives of γ with respect to R
are essentially 0. In the regions where the derivatives are significantly different than zero, the high-lying
rovibrational wave functions these derivatives act upon are likely to have small amplitudes and oscillate
a lot [7, Fig. 1]. Therefore, I expect diagonal and off-diagonal matrix elements of the derivatives of γ
with respect to R between the rovibrational wave functions of V1/2(R) and V3/2(R) to be small. Thus I
am inclined to solve for the rovibrational bound states of V1/2(R) and V3/2(R) by considering the various
derivatives of γ with respect to R as perturbations to the problem. Once I obtain such rovibrational
bound states, I will effectively calculate the matrix elements of the perturbation and quantify whether
the perturbative treatment is justified in the first place.
4 What’s next?
There are n things that need to be done7 in light of the current notes:
1. Compare the mixing angle matrix elements
vi
dγ
dR
2
vi , i = 1/2, 3/2, (19a)
v1/2
d2
γ
dR2
v3/2 , (19b)
v1/2
dγ
dR
∂
∂R
v3/2 , and v3/2
dγ
dR
∂
∂R
v1/2 , (19c)
7As of 15 May 2013

Stéphane
Valladier
v. 2.6
22nd December, 2013
4 6 8 10 12 14
4
3
2
1
0
Internuclear Separation R
dΓ
dR
1
Figure 3: First derivative of the mixing angle γ(R) with respect to the internuclear separation.
The extrema occur at R ≈ 4.25 ˚A and R ≈ 9.27 ˚A with respective values (dγ/dR)max ≈ −4.59 ˚A
−1
and
(dγ/dR)min ≈ 0.35 ˚A
−1
to the radial kinetic energy matrix elements
v1/2
∂2
∂R2
v1/2 and v3/2
∂2
∂R2
v3/2 . (20)
If the matrix elements of Eq. (19) are small compared to those of Eq. (20), then the perturbation
treatment is justified, and I can proceed to the following step, otherwise, I need to find a way to
solve the coupled differential equation for the rovibrational wave functions of the J = 1 V1/2 and
V3/2 PECs
Appendix A Getting the derivative of the mixing angle from its tangent
It is easy to obtain the tangent of γ from Eq. (14):
tanγ =
sinγ
cosγ
=
VA − V1/2
√
2ξ
.

Stéphane
Valladier
v. 2.6
22nd December, 2013
4 6 8 10 12 14
0
5
10
15
20
dΓ
dR
2
2
Figure 4: Square of the first derivative of the mixing angle γ(R) with respect to the internuclear
separation. The extrema occur at the same R values as in Fig. 3.
Defining u(R) =
VA − V1/2
√
2ξ
, then γ = arctanu. Remembering now that
d
dR
arctanu =
u
1 + u2
,
one gets
dγ
dR
=
d
dR
arctanu =
1
1 +
VA−V1/2√
2ξ
2
d
dR
VA − V1/2
√
2ξ
,
which is an expression for dγ/dR without ever calculating γ explicitly. Substituting the definition for
V1/2 from Eq. (12) leads to
dγ
dR
=
1
1 +


VA−Vb0
2
√
2ξ
+ VA−Vb0
2
√
2ξ
2
+ 1


2
d
dR


VA − Vb0
2
√
2ξ
+
VA − Vb0
2
√
2ξ
2
+ 1


,
showing that the derivative of the mixing angle can be expressed solely from the Hund’s case (a) potentials
and the relevant spin-orbit coupling terms. This latter expression was used to obtain figures 3, 4 and 5.

Stéphane
Valladier
v. 2.6
22nd December, 2013
4 6 8 10 12 14
20
10
0
10
20
d2
Γ
dR2
2
Figure 5: Second derivative of the mixing angle γ(R) with respect to the internuclear separation.
What looks like a discontinuity around R ≈ 4.25 ˚A is not, d2
γ/dR2 just varies very rapidly around
R ≈ 4.25 ˚A, but remains smooth and continuous.
Appendix B Checking hermicity of the kinetic energy operator
All operators defined in Eqs. (11a) are hermitian. This property is obvious for all operators that do not
involve a derivative with respect to R: R, D, and Hel. A hermitian operator remains hermitian under a
unitary transformation. Thus the change of basis defined by U conserves the hermicity of R, D, and Hel
whether they are expressed in basis A or H.
However it is not trivial that the kinetic energy operator T is hermitian in the first place, and remains
so after the transformation U. Let’s prove that T is indeed hermitian, no matter what basis it is expressed
in.
First consider matrix elements of the form
vα −
2
2µ
d2
dR2
vα ,
where α denotes any of the electronic states, and |vα is any rovibrational state belonging to the electronic
state |Φel
α . The rovibrational state |vα satisfies the time-independent Schrödinger equation (TISE):
−
2
2µ
d2
dR2
|vα + V total
α |vα = Evα
|vα ,

Stéphane
Valladier
v. 2.6
22nd December, 2013
where V total
α is the sum of the rotational energy and all other potential energies. Then
vα −
2
2µ
d2
dR2
vα = Evα
δvαvα
− vα V total
α vα (21a)
= Evα
δvαvα
− vα V total
α vα (21b)
= vα −
2
2µ
d2
dR2
vα , (21c)
since V total
α is purely multiplicative and given the properties of the Kronecker δ. Matrix elements of the
type described in the previous equation occur both in the A and H basis. Equations 21 show that TA and
the parts of TH that contain d2
/dR2 are indeed hermitian.
The function dγ/dR is purely multiplicative, therefore
vα
dγ
dR
2
vα = vα
dγ
dR
2
vα ,
so all diagonal blocks of TH are hermitian.
Let’s focus now on the off-diagonal blocks of TH. To finish proving that TH is hermitian, I need to
prove that
v3/2
d2
γ
dR2
+ 2
dγ
dR
∂
∂R
v1/2 = v1/2 −
d2
γ
dR2
− 2
dγ
dR
∂
∂R
v3/2 . (22)
Let’s recall the rule of integration by parts for the product of three well-behaved functions f ,g, and h:
b
a
f gh dR = [f gh]b
a −
b
a
f ghdR −
b
a
f g hdR,
and apply this expression to
f (R) = R|v3/2 = ψv3/2
(R) = ψv3/2
, g(R) =
dγ
dR
, h(r) = R|v1/2 = ψv1/2
(R) = ψv1/2
.
Starting from part of the matrix element on the left hand side of Eq. (22):
v3/2
dγ
dR
∂
∂R
v1/2 =
+∞
0
ψv3/2
dγ
dR
ψv1/2
dR (23a)
= ψv3/2
dγ
dR
ψv1/2
R=+∞
R=0
−
+∞
0
ψv3/2
dγ
dR
ψv1/2
dR −
+∞
0
ψv3/2
dγ
dR
ψv1/2
dR, (23b)
where the quantity between square brackets is zero, since the wave functions are zero at R = 0 and R = +∞.

St´ephane
Valladier
v. 2.6
22nd December, 2013
Permuting the order of the products in the remaining integrals yields
v3/2
dγ
dR
∂
∂R
v1/2 = −
+∞
0
ψv1/2
dγ
dR
ψv3/2
dR −
+∞
0
ψv1/2
dγ
dR
ψv3/2
dR (23c)
= − v1/2
dγ
dR
∂
∂R
v3/2 − v1/2
d2
γ
dR2
v3/2 (23d)
Let’s now combine Eq. (23d) with Eq. (22)
v3/2
d2
γ
dR2
+ 2
dγ
dR
∂
∂R
v1/2 = v3/2
d2
γ
dR2
v1/2 + 2 v3/2
dγ
dR
∂
∂R
v1/2 (24a)
= v1/2
d2
γ
dR2
v3/2 − 2 v1/2
dγ
dR
∂
∂R
v3/2 − 2 v1/2
d2
γ
dR2
v3/2 (24b)
= v1/2 −
d2
γ
dR2
− 2
dγ
dR
∂
∂R
v3/2 , (24c)
which completes the proof that TH is hermitian, as it should.
First, verifying that TH is hermitian allows to check whether I did any algebraic mistake when passing
from basis A to basis H8. Second, notice that the V1/2 state holds 146 rovibrational states, and the V3/2
holds 114. If I did not recall that T is hermitian, I would have had to calculate (146 + 114)2 = 67600
matrix elements. Thanks to hermicity, I now only have to calculate
146 × (146 + 1)/2 = 10731 elements of the form v1/2
dγ
dR
2
−
2
2µ
∂2
∂R2
v1/2 ,
114 × (114 + 1)/2 = 6555 v3/2
dγ
dR
2
−
2
2µ
∂2
∂R2
v3/2 ,
114 × 146 = 16644 v3/2
d2
γ
dR2
+ 2
dγ
dR
∂
∂R
v1/2 ,
that is 33930 matrix elements, about half what I was about to calculate before I remembered (and
checked!) the hermicity of T.
8 In versions of these notes prior to 2.4, TH was not hermitian, because I dropped a minus sign along the way.

Stéphane
Valladier
v. 2.6
22nd December, 2013
References
[1] S. Valladier, Photoassociation and rovibrational cooling of NaCs using chirped laser pulses and Stimulated Raman
Adiabatic Passage (2011).
[2] J. Zaharova, M. Tamanis, R. Ferber, A. N. Drozdova, E. A. Pazyuk, and A. V. Stolyarov, Solution of the fully-
mixed-state problem: Direct deperturbation analysis of the A1Σ+–b3Π complex in a NaCs dimer, Physical Review A,
79(1), 012508 (2009).
[3] P. F. Bernath, Spectra of atoms and molecules, Oxford University Press, New York, 2nd edition (2005).
[4] H. Lefebvre-Brion and R. W. Field, The spectra and dynamics of diatomic molecules, Elsevier Academic Press,
Amsterdam; Boston (2004).
[5] H. Katô, Energy Levels and Line Intensities of Diatomic Molecules. Application to Alkali Metal Molecules, Bulletin
of the Chemical Society of Japan, 66(11), 3203 (1993).
[6] J. Vala, O. Dulieu, F. Masnou-Seeuws, P. Pillet, and R. Kosloff, Coherent control of cold-molecule formation through
photoassociation using a chirped-pulsed-laser field, Phys. Rev. A, 63, 013412 (2000).
[7] B. E. Londoño, J. E. Mahecha, E. Luc-Koenig, and A. Crubellier, Resonant coupling effects on the photoassociation
of ultracold Rb and Cs atoms, Phys. Rev. A, 80, 032511 (2009).
[8] M. Aymar and O. Dulieu, Calculations of transition and permanent dipole moments of heteronuclear alkali dimers
NaK, NaRb and NaCs, Molecular Physics, 105(11-12), 1733 (2007).

SV-InclusionSOcouplinginNaCs

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (12)

Similar to SV-InclusionSOcouplinginNaCs

Similar to SV-InclusionSOcouplinginNaCs (20)

SV-InclusionSOcouplinginNaCs