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Fall 2014 Chem 356: Introductory Quantum

Chapter 5 – Vibrational ...... 65 Potential Surfaces, Rotations and ...... 65 Oscillator ...... 67 General Solution for H.O.: Operator Technique ...... 68 Vibrational Selection Rules ...... 72 Poly-atomic Molecules ...... 73 Beyond the approximation ...... 74

Chapter 5 – Vibrational Motion

Potential Energy Surfaces, Rotations and Vibrations

Suppose we assume the nuclei of a molecule are fixed, then we can solve the Schrodinger equation for the electrons ˆ Hrrrψψ(,12 ,....)NN= Errr (, 12 ,...)

This is a complicated problem, that will be discussed later (Chapter 7 and beyond)

We would get the (ground state) energy at a particular nuclear configuration { R }

Hence, assume we can solve this

We can fit a curve through the points → VR({ })

This VR({ }) is called the surface (PES) and is a crucial concept in chemistry, eg.

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Fall 2014 Chem 356: Introductory

Minima on the PES we associate with different isomers, for example with the reaction AB+→ CD+. Saddle points on the PES we associate with transition states

Obtaining the minima (+ curvature) we can get thermochemical information. From TS we get into rates of reactions (Chem254, Chem 350)

Ideally, once we have obtained PES, we should solve for the Quantum involving the nuclei. This is very complicated, and today can only be done for small molecules (up to 5 atoms).

What can be done is solving for molecular rotations + vibrations in the harmonic oscillator + Rigid Rotor approximation

The crux is to make a quadratic approximation to the PES around each minimum ( = molecular species)

! 1 V (R) ≈ V (R ) + k (R − R ) (R − R ) e 2 ∑ αβ e α e β αβ

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Fall 2014 Chem 356: Introductory Quantum Mechanics

A molecule with N atoms has 3 N degrees of freedom - 3 overall translations - 3 overall rotations or 2 rotations for linear molecules

(3N − 6)Vibrations , [35N − for linear molecule] k : , 1....3N αβ α β =

! ! ! All of the information V (R ) , R and k (R ) is obtained from an electronic structure program (like e e αβ e Gaussian, Gamess, Turbomole…)

Once we have these, one can use ‘exact’ calculations to solve for Harmonic oscillator + Rigid Rotor energies.

The harmonic (quadratic) potential is an approximation, but often works well, certainly for stiff molecules/ degrees of freedom.

What we will do next is: discuss harmonic oscillator for diatomic molecules

- Ground state of harmonic oscillator - All excited states using operator technique - Generalize to polyatomic molecules - More accurate solution for diatomic (Beyond H.O.) - Selection rules, vibrational spectroscopy

Harmonic Oscillator

In the following chapter we will go on to discuss rotations

For a general polyatomic molecule we can define H.O. Hamiltonian as !2 1 H = P2 + k (R − R ) (R − R ) ∑ 2M α 2 ∑ ab e a e b α α a,b=1...3N For a diatomic this can be reduced to 1 !2 d 2 1 Hˆ = − + kx2 2 2 2 µ dx M M Here µ = 1 2 is the [kg] reduced M M 1 + 2

xRR=()−e is deviation [m] from equilibrium k is the constant [Nm-1]

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Fall 2014 Chem 356: Introductory Quantum Mechanics

In McQuarrie this is derived using classical . I will post on the Website a general derivation to get the H.O. form for a general polyatomic, but will not discuss in class. Here I will simply use the form for H.

2 The ground state solution has the form e−αx /2 . Let us determine the constant α , and the energy. Later on I will discuss the full solution.

d 22 exe−−ααxx/2=−α /2 dx 2 d 22 2 eexe−−ααxx/2=−αα /2+ 2 2 − α x /2 dx2 2 2 2 2 ⎡ ! d 1 2 ⎤ −α x2 /2 ⎡ ! ! 2 2 1 2 ⎤ −α x2 /2 −α x2 /2 ⎢− 2 + kx ⎥e = ⎢ α − α x + kx ⎥e = E0e 2µ dx 2 2µ 2µ 2 ⎣ ⎦ ⎣ ⎦ !2 E = α 0 2 µ 1 1 !2α 2 1 k − = 0 α = µk 2 2 µ ! !2 µk 1 k 1 E = = ! = !ω 0 2µ 2 µ 2 ! k µ ω = ; α = ω µ !

General Solution for H.O.: Operator Technique (see appendix 5 in McQuarrie)

2 2 ˆ −! d 1 2 H = 2 + kx 2m dx 2 2 Solution e−αx /2 α = µω / = µk / ! ! Define qx= α

2 2 e−αx /2 → e−q /2 1 1 − ⎛ 1 ⎞ 2 ⎛ 1 ⎞ 2 q = µk x x = µk q ⎝⎜ ⎠⎟ ⎝⎜ ⎠⎟ ! !

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Fall 2014 Chem 356: Introductory Quantum Mechanics

1 ∂ ∂q ∂ ⎛ 1 ⎞ 2 ∂ = = µk ∂x ∂x ∂q ⎝⎜ ⎠⎟ ∂q ! 2 2 ˆ −! µk ∂ 1 ! 2 H = 2 + k q 2µ ! ∂q 2 µk

k ⎛ 1 ∂2 1 ⎞ = ! − + q2 µ ⎝⎜ 2 ∂q2 2 ⎠⎟

⎛ 1 d 2 1 ⎞ = !ω − + q2 ⎝⎜ 2 dq2 2 ⎠⎟

1 ‘ q ’ are called dimensionless coordinates (Check that q = µk x is indeed dimensionless!) !

Commutation Relation: ⎡⎤d ⎛⎞dd ⎢⎥q,1=− ⎜⎟qqfqfq−()=− () ⎣⎦dq ⎝⎠dq dq

Define new operators:

ˆ+ 1 ⎛⎞d ˆ 1 ⎛⎞d bq=⎜⎟− bq=+⎜⎟ 2 ⎝⎠dq 2 ⎝⎠dq !ω ⎛ d ⎞ ⎛ d ⎞ Then !ωb+b = q − q + 2 ⎝⎜ dq⎠⎟ ⎝⎜ dq⎠⎟

!ω ⎛ d 2 ⎡ d ⎤⎞ q2 q, = − 2 + ⎢ ⎥ 2 ⎝⎜ dq dq ⎠⎟ ⎣ ⎦ 1 = Hˆ − !ω 2 ˆ ⎛⎞ˆ+ 1 → Hbb=+hω ⎜⎟ ⎝⎠2

Commutation Relations, between b operators: ⎡⎤⎡bbˆˆ,,0== b ˆ++bˆ ⎤ ⎣⎦⎣ ⎦ 1 ⎡⎤⎛⎞⎛⎞dd ⎡⎤bbˆˆ,,+ q q ⎣⎦=+⎢⎥⎜⎟⎜⎟ − 2 ⎣⎦⎝⎠⎝⎠dq dq 1 ⎛⎞⎡⎤⎡⎤⎡⎤dddd =+⎜⎟[,]qq⎢⎥⎢⎥⎢⎥ , q−− q , , 2 ⎝⎠⎣⎦⎣⎦⎣⎦dq dq dq dq

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Fall 2014 Chem 356: Introductory Quantum Mechanics

1 =+( 1(1)1−−) = 2 Hence ⎡⎤bbˆˆ,1+ = ⎡⎤bbˆˆ+ ,1=− ⎣⎦ ⎣⎦

Using this form of the Hamiltonian and the commutation relations we can derive the eigenvalues and eigenfunctions of H.O.!!

a) If ψ n ()q is eigenfunction with eigenvalue En then i) bqˆ+ψ () is eigenfunction with E = E + !ω n n+1 n ii) bqˆψ () is eigenfunction with E = E − !ω n n−1 n Proof: ⎛ 1⎞ i) Hˆ bˆ+ψ (q) = −!ω bˆ+bˆ + bˆ+ψ (q) ( n ) ⎝⎜ 2⎠⎟ n

⎡ ⎛ 1⎞ ⎤ ˆ+ ⎡ ˆ ˆ+ ⎤ ˆ+ ˆ+ ˆ = !ω ⎢b b,b + b b b + ⎥ψ n (q) ⎣ ⎦ ⎝⎜ 2⎠⎟ ⎣ ⎦ = bˆ+ !ω + Hˆ ψ (q) ( ) n = (E + !ω )bˆ+ψ (q) n n ⎛ 1⎞ ii) Hˆ bˆψ (q) = !ω bˆ+bˆ + bˆψ (q) ( n ) ⎝⎜ 2⎠⎟ n

⎛ 1⎞ = !ω ⎡bˆ+ ,bˆ⎤ + bˆbˆ+ + bˆψ (q) ⎝⎜ ⎣ ⎦ 2⎠⎟ n

⎛ ⎛ 1⎞ ⎞ = !ω −1⋅bˆ + bˆ bˆ+bˆ + ψ (q) ⎝⎜ ⎝⎜ 2⎠⎟ ⎠⎟ n

= bˆ −!ω + Hˆ ψ (q) = E − !ω bˆψ (q) ( ) n ( n ) n

bˆ+ and bˆ are called the raising and lowering operators, or ladder operators

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Fall 2014 Chem 356: Introductory Quantum Mechanics

What about the ground state? bˆψ (q) = E − !ω bˆψ (q) n ( 0 ) n Still true, but E − !ω < E !! 0 0

ˆ Only way out: bqψ 0 ()= 0 1 ⎛⎞d ⎜⎟qq+=ψ 0 () 0 2 ⎝⎠dq 1 − q2 2 with solution ψ 0 ()qe= 11 ⎛⎞d −−qq22 qe+=22() qqe−= 0 ⎜⎟dq ⎝⎠

Putting it all together 1 1 1 4 − q2 ⎛⎞ 2 normalized ψ 0 ()qe= ⎜⎟ ⎝⎠π n 1 ˆ+ ψψn ()qbq= 0 () normalized n! ( )

ˆ+ 1 ⎛⎞d bq=⎜⎟− 2 ⎝⎠dq ⎛⎞1 Enn =+⎜⎟hω n = 0,1, 2,3.... ⎝⎠2 First of unnormalized functions in terms of q :

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Fall 2014 Chem 356: Introductory Quantum Mechanics

1 − q2 2 ψ 0 ()qe=

1 2 − q 2 ⎛⎞d 2 −q /2 ψ1 ()qq→−⎜⎟ e = 2 qe ⎝⎠dq

⎛⎞d −−qq22/2 2 /2 ψ 2 ()qq→−⎜⎟ qeqe =( 2− 1) ⎝⎠dq

⎛⎞d 2/23/2−−qq22 ψ 3 ()qq→−⎜⎟( 2 qqeqqe − 1) =( 4− 6) ⎝⎠dq Etc. We can obtain all eigenfunctions by differentiation

ψ n ()x → substitute xq= α + normalize 1 − q2 2 The functions take the form Hqen ()

Hqn () are Hermite polynomials, they are either odd or even functions. (Each polynomial contains only odd or only even terms)

Vibrational Selection Rules

Later we will discuss more generally the radiation process. For now the transition dipole is used to define the strength of a spectroscopic translation ψ*(xx )µ ( )ψ ( xdx ) mn→ ∫ nm ⎛⎞dµ µµ(xx )=+ ( ) x + ...... 0 ⎜⎟dx ⎝⎠x0 dµ µ is the dipole at equilibrium distance, is the change in dipole with changing x 0 dx (internuclear-distance) d µ Eg. N2 : = 0 dx d µ HF: ≠ 0 large dx d µ CO: ≠ 0 small dx ψ*(xx )µ ( )ψ ( xdx ) ∫ nm =µψ*(xxdx ) ψ ( )→ 0 (if nm≠ ) 0 ∫ nm

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Fall 2014 Chem 356: Introductory Quantum Mechanics

dµ + ψψ*(xx ) ( xdx ) dx ∫ nm dµ 2 → αψ*(qq ) ψ ( qdq ) dx ∫ nm

dµα ˆˆ+ →ψψnm*(qb )+ b ( qdq ) dx 2 ∫ ( ) ˆ ˆ+ b()~ψψmm−1 b ()~ψψmm+1

For Harmonic Oscillator we can only get allowed transitions if Δn =+1 or Δn =−1

ΔE = ±!ω

Poly-atomic Molecules

−!2∇2 1 Hˆ = + k (R − R ) (R − R ) ∑ 2µ 2 ∑ αβ e α e β α α α ,β By some manipulation (see notes on webpage) this can be written as a sum of H.O. Hamiltonians…….tedious derivations ⎡ 2 ⎤ ˆ 1 d 1 2 H = !ω i ⎢− 2 + qi ⎥ ∑ 2 dq 2 i ⎣ i ⎦ ⎡ 1 ⎤ = !ω bˆ+bˆ + ∑ i ⎢ i i 2 ⎥ i ⎣ ⎦

The coordinates q : are linear combinations of atomic vectors

3 N coordinate → 3 translation, 3 rotation, (3 N -6) For linear molecule, rotation around axis is not a displacement → 2 rotations, and (3 N -5) vibrations Eg. Water

Normal modes : display symmetry of molecule Chapter 5 – Vibrational Motion 73

Fall 2014 Chem 356: Introductory Quantum Mechanics

The eigenfunctions are simply products of 1 dimensional H.O. functions

ψ k ()q1 ψ l ()q2 ψ m ()q3 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ E = k + !ω + l + !ω + m + !ω k,l,m ⎝⎜ 2⎠⎟ 1 ⎝⎜ 2⎠⎟ 2 ⎝⎜ 2⎠⎟ 3

Very simple solution. One needs to diagonalize the “mass weighted hessian” 1 1 k 3N × 3N M αβ M α β Reasonable approximations to all vibrational levels (recall Chem 350)

Beyond the harmonic oscillator approximation

The true potential is not quadratic. For large molecules this is not so easy to correct (people would use low order , based on a quartic )

For smalls molecules, in particular diatomics, one can solve for the full vibrational problem

The exact energies are not equidistant. A better approximation is 2 ⎛ 1⎞ ⎛ 1⎞ E(n) = !ω n + − !ω x n + e ⎝⎜ 2⎠⎟ e e ⎝⎜ 2⎠⎟

The energy levels are usually called g(v), v = n,vibrational quantum number There are a finite number of bound levels

Another effect is that transition moments can be non-zero even when Δ≠n ±1. This leads to the observation of overtones.

An often used form for the potential for a diatomic is the Morse potential

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Fall 2014 Chem 356: Introductory Quantum Mechanics

2 −β x Vx()= De ( 1− e ) ∂V = 0 x = 0 equilibrium geometry ∂x x=0 2 1 ∂ V 2 k 2 Dke β = →β = 2 ∂x De

De can be used from experiment ( D0 measurable)

This is still an approximate potential

Today: Potential energy curves can be calculated using electronic structure programs.

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