Cold molecules
Mike Tarbutt
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Outline
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Lecture 1 – The electronic, vibrational and rotational structure of molecules.
Lecture 2 – Transitions in molecules.
Lecture 3 – Direct laser cooling of molecules.
Lecture 4 – Making cold molecules from cold atoms.
Lecture 5 – Guest lecture on magneto-association from Simon Cornish.
Lecture 6 – The Stark shift.
Lecture 7 – Decelerating, storing and trapping molecules with electric fields.
Review articles:
• “Molecule formation in ultracold atomic gases”, J.M. Hutson and P Soldan,
International Reviews in Physical Chemistry, 25, 497 (2006)
• “Production and application of translationally cold molecules”, H.L. Bethlem and G.
Meijer, International Reviews in Physical Chemistry, 22, 73 (2003)
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Electronic, vibrational and
rotational structure
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Born-Oppenheimer approximation
Nuclear mass ~104 times electronic mass
Nuclei move very slowly compared to electrons
Solve the electronic Schrodinger equation with “frozen nuclei”
Do this for many different values of internuclear separation, R
Obtain electronic energies as functions of R – potential energy curves
Then solve the Schrodinger equation for the nuclear motion
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The Hamiltonian for a diatomic molecule (*non-relativistic)
H
TN
Kinetic energy
of nuclei
Te
V
Kinetic energy
of electrons
Coulomb potential between
electrons and nuclei
I’ll use subscripts A and B to denote the two nuclei, and the index i to label all the electrons
2
TN
2 MA
N
Te
i 1
V
A
2
2
2 MB
2
2m
ZA ZB e2
4 0R
i
B
2
2
N
e2
4
0
i 1
ZA
riA
ZB
ri B
N
e2
4
0
i j 1
1
rij
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Remove the centre-of-mass motion…
Let’s immediately simplify the nuclear kinetic energy term using
centre-of-mass coordinates and relative coordinates:
RCM
MA RA MB RB
, R
MA MB
RA
RB
2
This transforms the nuclear kinetic energy from
TN
2 MA
A
2
to
TN
2M
M
MA
2
2
RCM
2 MB
2
2
2
MB
B
2
R
2
MA MB
MA MB
We’re interested in the internal energy of the molecule, not the translation of the centre of mass
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Need to solve:
TN
Te
V
R, ri
E
R, ri
Let’s first solve a different problem!
Clamp the nuclei in place at a fixed separation R,.
The equation to solve is the same, except that the nuclear kinetic energy vanishes.
Electronic Hamiltonian
Electronic wave equation
q
R; ri
and
Eq R
He
He
Te
q
V
R; ri
Eq R
R; ri
q
are a set of eigenfunctions and eigenvalues,
each corresponding to an electronic state
This electronic wave equation can be solved using the same techniques as in
the atomic case (e.g. Hartree-Fock)
N.B. The eigenfunctions form a complete set at every value of R:
q
p
pq
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Potential curves…
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Let’s now expand the complete wavefunction on the basis of the electronic eigenfunctions:
R, ri
R; ri
q
q
Substitute this expansion into the full Schrodinger equation,
…and use the result
He
TN
He
q
R; ri
Eq R
TN
Eq R
E
q
Multiply by
R
Nuclear wavefunctions
Electronic wavefunctions
E
q
q
q
He
R; ri
q
q
TN
q
E
R, ri
0
0
R
R; ri
R; ri
q
0
R
n , integrate over electronic coordinates, and use the orthonormality condition:
En R
E
n
q
dri
n
TN
q
q
0
…an infinite set of coupled differential equations which determine the nuclear wavefunctions
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Adiabatic approximation - the nuclear motion does not mix the electronic states –
then the set of equations uncouple:
Simplify...
Neglect the 2nd and 3rd terms (it turns out they are very small)...
Then we obtain a much simpler equation for the nuclear wavefunction:
2
2
R
2
En R
E
n
R
0
Nuclear kinetic Effective potential
energy
for nuclear motion
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Separate the nuclear wave equation…
2
R
2
2
En R
E
n
R
0
Central potential – Separate into radial and angular parts:
R, ,
R
1
f R g
,
As always for a central potential, the angular functions are spherical harmonics:
g
,
YJM
,
They are eigenfunctions of J2 and Jz
J2 YJM J J 1 YJM
Jz YJM
M YJM
Then we’re left with a fairly simple looking radial equation…
2
2
2
R2
J J 1
2 R2
En R
E f R
0
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Solve the radial equation…
2
2
J J 1
2 R2
R2
2
En R
E f R
0
Could solve this numerically, or find an approximate solution for R close to R0
1) Set R=R0 in the denominator of the second term. 2) Expand En(R) in a Taylor series about R0
En R
En R
En R0
En
R
En R0
1
k R
2
2
2
2
R2
R
R0
R0
J J 1
2 R02
Er
R0
2
...
En R0
2
1
2
En
R2
R
R0
R0
2
...
2
En
R2
k
where
1
k R
2
Define Ev such that the total energy is
R0
R0
2
E
En R0
2
Ev f R
E f R
0
Ev
Er
We’re left with a harmonic oscillator equation:
2
2
2
2
R
1
k R
2
R0
0
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Summary
Solve the electronic wave equation with fixed nuclei
Repeat for many different R to obtain potential curves
These appear as the potential in the nuclear wave equation
Can separate and (approximately) solve this wave equation
The total energy is
E
En R0
Ev
Er
The total wavefunctions are products of electronic,
vibrational and rotational eigenstates
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Energy scales
Electronic frequencies ~ 1015 Hz
Vibrational frequencies ~ 5 1013 Hz
Rotational frequencies ~ 1010 - 1012 Hz
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Things we’ve left out...
 Coupling terms neglected in the Born-Oppenheimer approximation
 Centrifugal distortion
 Spin-orbit interaction
 Spin-rotation interaction
 Spin-spin interaction
 Lambda doubling
 Magnetic hyperfine interactions
 Electric quadrupole interactions
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Some molecular notation
Remember how it goes for an atom: (configuration) 2S+1LJ
e.g. Ground state of sodium: 1s22s22p63s 2S1/2
What are the “good quantum numbers” for a diatomic molecule
(which operators commute with the Hamiltonian)?
The lack of spherical symmetry in a diatomic molecule means that L2 does not commute with H.
However, to a good approximation, Lz does commute with H when the z-axis is taken along the
internuclear axis. Therefore, the projection of the total orbital angular momentum onto the
internuclear axis is (approximately) a good quantum number – labelled by L, which can be S, P, D...
X, A, B, C...
The electronic states of a molecule are labelled: (unique letter)
Spin
multiplicity
2S+1L
Projection of orbital
angular momentum
onto internuclear axis
W
Projection of total
angular momentum
onto internuclear axis
Can also include the vibrational state v, and the total angular momentum J
e.g. one of the excited states of CaF is written: A 2P3/2 (v=2, J=7/2)
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