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Cold moleculesMike Tarbutt

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Outline

• 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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TN 2

2MAA

2 2

2MBB

2

Te i1

N2

2mi

2

V ZAZBe2

40R

e2

40i1

N ZAriA

ZBriB

e2

40 ij1

N1rij

The Hamiltonian for a diatomic molecule (*non-relativistic)

H TN Te V

Kinetic energy of nuclei

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

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RCM MARAMBRB

MA MB, R RA RB

Let’s immediately simplify the nuclear kinetic energy term using centre-of-mass coordinates and relative coordinates:

TN 2

2MAA

2 2

2MBB

2This transforms the nuclear kinetic energy from

TN 2

2MRCM

2 2

2R

2to

M MA MB MAMBMA MB

We’re interested in the internal energy of the molecule, not the translation of the centre of mass

Remove the centre-of-mass motion…

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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.

H e Te VElectronic Hamiltonian

HeqR; ri EqRqR; riElectronic wave equation

qR ; r i EqRand 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)

TN Te VR, ri E R, riNeed to solve:

q p pqN.B. The eigenfunctions form a complete set at every value of R:

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Potential curves…

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R, ri qqR; riqRLet’s now expand the complete wavefunction on the basis of the electronic eigenfunctions:

He TN ER, ri 0Substitute this expansion into the full Schrodinger equation,He TN EqqR; riqR 0

…and use the result HeqR; ri EqRqR; riqTN EqR EqR; riqR 0

n

Multiply by , integrate over electronic coordinates, and use the orthonormality condition:EnR En q dri nTNqq 0

Electronic wavefunctions Nuclear wavefunctions

…an infinite set of coupled differential equations which determine the nuclear wavefunctions

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TNqq 2

2R

2 qqAdiabatic 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)...

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2R

2 EnR E nR 0

Then we obtain a much simpler equation for the nuclear wavefunction:

Nuclear kinetic energy

Effective potential for nuclear motion

En R E q dri q TN q q 0

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Separate the nuclear wave equation…

2

2R

2 EnR E nR 0

R , , R 1f Rg , Central potential – Separate into radial and angular parts:

g, YJM, As always for a central potential, the angular functions are spherical harmonics:

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R2JJ 12R2

EnR E fR 0

Then we’re left with a fairly simple looking radial equation…

They are eigenfunctions of J2 and Jz

J2YJM JJ 1YJMJz YJM M YJM

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Solve the radial equation…2

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R2JJ 12R2

EnR E fR 0

Could solve this numerically, or find an approximate solution for R close to R0

EnR EnR0 En R

R0R R0

122En R2 R0

R R02 ...

1) Set R=R0 in the denominator of the second term. 2) Expand En(R) in a Taylor series about R0

EnR EnR0 12kR R02 ... k

2En R2 R0

where

2

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R2JJ 12R02

EnR0 12kR R02 E fR 0

Er E EnR0 Ev ErDefine Ev such that the total energy is

We’re left with a harmonic oscillator equation:

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2 2

R212k R R02 Ev f R 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

Ev v 12 km ' , k d 2EeR

d R2RR0

EJ B JJ 1B

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2I; I moment of inertia

The total energy is E EnR0 Ev Er

The total wavefunctions are products of electronic, vibrational and rotational eigenstates

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Electronic frequencies ~ 1015 Hz

Energy scales

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 notationRemember 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...

The electronic states of a molecule are labelled: (unique letter) 2S+1LW

X, A, B, C...Spin

multiplicity

Projection of orbital angular momentum

onto internuclear axis

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)