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Fall 2014 Chem 356: Introductory Quantum Mechanics Chapter 5 – Vibrational Motion ................................................................................................................. 65 Potential Energy Surfaces, Rotations and Vibrations ............................................................................ 65 Harmonic Oscillator ............................................................................................................................... 67 General Solution for H.O.: Operator Technique .................................................................................... 68 Vibrational Selection Rules .................................................................................................................... 72 Poly-­‐atomic Molecules .......................................................................................................................... 73 Beyond the harmonic oscillator 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 Hˆψ (r1 , r2 ,....rN ) = Eψ (r1 , r2 ,...rN ) 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 → V ({R}) This V ({R}) is called the potential energy surface (PES) and is a crucial concept in chemistry, eg. Chapter 5 – Vibrational Motion 65 Fall 2014 Chem 356: Introductory Quantum Mechanics Minima on the PES we associate with different isomers, for example with the reaction
A + B → C + D . 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 energies 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 (Re ) + ∑ kαβ (R − Re )α (R − Re )β
2 αβ
Chapter 5 – Vibrational Motion 66 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 , [ 3N − 5 for linear molecule] kαβ : α , β = 1....3N !
!
!
All of the information V ( Re ) , Re and kαβ ( Re ) is obtained from an electronic structure program (like 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 H =∑
α
!2
1
Pα2 +
∑ k (R − Re )a (R − Re )b 2 Mα
2 a,b=1...3N ab
For a diatomic this can be reduced to 1 !2 d 2 1 2
+ kx 2 µ dx 2 2
Ĥ = −
Here µ =
x = ( R − Re ) k M1 M 2
M1 + M 2
is the [kg] reduced mass is deviation [m] from equilibrium is the force constant [Nm-­‐1] Chapter 5 – Vibrational Motion 67 Fall 2014 Chem 356: Introductory Quantum Mechanics In McQuarrie this is derived using classical equations of motion. 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. The ground state solution has the form e−α x
on I will discuss the full solution. 2
/2
. Let us determine the constant α , and the energy. Later 2
d −α x2 /2
e
= −α xe−α x /2 dx
2
2
d 2 −α x2 /2
e
= −α e−α x /2 + α 2 x 2 e−α x /2 2
dx
2
2
⎡ ! d
2
1 2 ⎤ −α x 2 /2 ⎡ ! 2
! 2 2 2 1 2 ⎤ −α x 2 /2
−
+
kx
e
=
α
−
α x + kx ⎥ e
= E0 e−α x /2 ⎢
⎥
⎢
2
2µ
2
⎣ 2 µ dx 2
⎦
⎣ 2µ
⎦
!2
α 2µ
E0 =
1
1 ! 2α 2
k−
= 0 2
2 µ
E0 =
!2 µk 1
k 1
= !
= !ω 2µ !
2 µ 2
ω=
k
; µ
α=
α =
1
µ k !
µ
ω !
General Solution for H.O.: Operator Technique (see appendix 5 in McQuarrie) −! 2 d 2 1 2
Ĥ =
+ kx 2m dx 2 2
Solution e−α x
Define q = x α e−α x /2 → 2
2
/2
α = µω / ! = µ k / ! 2
e− q /2 1
2
⎛1
⎞
q=⎜
µ k ⎟ x ⎝!
⎠
⎛1
⎞
x=⎜
µk ⎟
⎝!
⎠
−
1
2
q Chapter 5 – Vibrational Motion 68 Fall 2014 Chem 356: Introductory Quantum Mechanics 1
⎞2 ∂
∂ ∂q ∂ ⎛ 1
=
=⎜
µk ⎟
∂x ∂x ∂q ⎝ !
⎠ ∂q
Ĥ =
−! 2
2µ
µ k ∂2 1
! 2
+ k
q 2
! ∂q 2
µk
k ⎛ 1 ∂2 1 2 ⎞
−
+ q = !
µ ⎜⎝ 2 ∂q 2 2 ⎟⎠
= !ω ⎜ −
⎛ 1 d2 1 2⎞
+ q 2
⎝ 2 dq 2 ⎟⎠
‘ q ’ are called dimensionless coordinates (Check that q =
1
µ k x is indeed dimensionless!) !
Commutation Relation: ⎡ d ⎤
⎢ q, dq ⎥ = −1 ⎣
⎦
⎛ d
d ⎞
⎜ q − q ⎟ f (q) = − f (q) ⎝ dq dq ⎠
Define new operators: 1 ⎛
d ⎞
bˆ+ =
⎜ q − ⎟ dq ⎠
2⎝
1 ⎛
d ⎞
bˆ =
⎜ q + ⎟ dq ⎠
2⎝
!ω ⎛
d ⎞⎛
d⎞
+
q − ⎟ ⎜ q + ⎟ Then !ω b b =
⎜
2 ⎝
dq ⎠ ⎝
dq ⎠
!ω ⎛ 2 d 2 ⎡ d ⎤⎞
=
q − 2 + ⎢ q, ⎥⎟ 2 ⎜⎝
dq ⎣ dq ⎦⎠
1
= Ĥ − !ω 2
→ 1⎞
⎛
Hˆ = hω ⎜ bˆ + b + ⎟ 2⎠
⎝
Commutation Relations, between b operators: ⎡bˆ, bˆ ⎤ = ⎡bˆ + , bˆ + ⎤ = 0 ⎣ ⎦ ⎣
⎦
⎡
⎤
⎡bˆ, bˆ+ ⎤ = 1 ⎢⎛⎜ q + d ⎞⎟ , ⎛⎜ q − d ⎞⎟ ⎥ ⎣
⎦ 2
dq ⎠ ⎝
dq ⎠ ⎦
⎣⎝
⎡ d ⎤ ⎡ d ⎤ ⎡ d d ⎤⎞
1⎛
= ⎜ [q, q] + ⎢ , q ⎥ − ⎢ q, ⎥ − ⎢ , ⎥ ⎟ 2⎝
⎣ dq ⎦ ⎣ dq ⎦ ⎣ dq dq ⎦ ⎠
Chapter 5 – Vibrational Motion 69 Fall 2014 Chem 356: Introductory Quantum Mechanics Hence 1
( +1 − (−1) ) = 1 2
⎡bˆ, bˆ + ⎤ = 1 ⎡bˆ + , bˆ ⎤ = −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)
ii)
bˆ+ψ n (q) is eigenfunction with En+1 = En + !ω bˆψ (q) is eigenfunction with E = E − !ω n
n−1
n
Proof: i)
(
)
⎛
1⎞
Ĥ b̂+ψ n (q) = −!ω ⎜ b̂+ b̂ + ⎟ b̂+ψ n (q) 2⎠
⎝
⎡
⎛
1⎞ ⎤
= !ω ⎢ b̂+ ⎡⎣ b̂, b̂+ ⎤⎦ + b̂+ ⎜ b̂+ b̂ + ⎟ ⎥ψ n (q) 2⎠ ⎦
⎝
⎣
(
)
= b̂+ !ω + Ĥ ψ n (q) = (En + !ω )b̂+ψ n (q) ii)
(
)
⎛
1⎞
Ĥ b̂ψ n (q) = !ω ⎜ b̂+ b̂ + ⎟ b̂ψ n (q) 2⎠
⎝
⎛
1⎞
= !ω ⎜ ⎡⎣ b̂+ , b̂ ⎤⎦ + b̂b̂+ + ⎟ b̂ψ n (q) 2⎠
⎝
⎛
⎛
1⎞ ⎞
= !ω ⎜ −1⋅ b̂ + b̂ ⎜ b̂+ b̂ + ⎟ ⎟ ψ n (q) 2⎠ ⎠
⎝
⎝
(
)
= b̂ −!ω + Ĥ ψ n (q) = ( En − !ω ) b̂ψ n (q) b̂ + and b̂ are called the raising and lowering operators, or ladder operators Chapter 5 – Vibrational Motion 70 Fall 2014 Chem 356: Introductory Quantum Mechanics What about the ground state? b̂ψ n (q) = ( E0 − !ω ) b̂ψ n (q) Still true, but E0 − !ω < E0 !! Only way out: bˆψ 0 (q) = 0 Differential equation with solution ψ 0 (q) = e
1
1
− q2
⎛
d ⎞ − 2 q2
⎜ q +
= ( q − q )e 2 = 0
⎟e
dq
⎝
⎠
1 ⎛
d ⎞
⎜ q + ⎟ψ 0 (q) = 0 dq ⎠
2⎝
1
− q2
2 Putting it all together 1
1
⎛ 1 ⎞ 4 − q2
ψ 0 (q) = ⎜ ⎟ e 2 normalized ⎝π ⎠
1 ˆ+ n
normalized ψ n (q) =
b ψ 0 (q) n!
( )
1 ⎛
d ⎞
bˆ+ =
⎜ q − ⎟ dq ⎠
2⎝
1⎞
⎛
n = 0,1, 2,3.... En = ⎜ n + ⎟ hω 2⎠
⎝
First couple of unnormalized functions in terms of q : Chapter 5 – Vibrational Motion 71 Fall 2014 Chem 356: Introductory Quantum Mechanics 1
− q2
2 ψ 0 (q) = e
1
2
⎛
d ⎞ − 2 q2
ψ 1 (q) → ⎜ q − ⎟ e
= 2qe − q /2 dq ⎠
⎝
Etc. We can obtain all eigenfunctions by differentiation ψ n ( x) → substitute x = q α + normalize ⎛
d ⎞
ψ 2 (q) → ⎜ q − ⎟ qe− q /2 = ( 2q 2 − 1) e− q /2 dq ⎠
⎝
⎛
d ⎞
ψ 3 (q) → ⎜ q − ⎟ ( 2q 2 − 1) qe− q /2 = ( 4q3 − 6q ) e− q /2 dq ⎠
⎝
2
2
2
The wave functions take the form H n (q)e
1
− q2
2
2
H n (q) 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 moment is used to define the strength of a spectroscopic translation ∫ψ
⎛ dµ ⎞
µ ( x) = µ0 ( x) + ⎜
⎟ x + ....... ⎝ dx ⎠ x
n
* ( x) µ ( x)ψ m ( x)dx m → n 0
µ 0 is the dipole at equilibrium distance, dµ
is the change in dipole with changing x dx
(internuclear-­‐distance) dµ
= 0 dx
dµ
≠ 0 dx
dµ
≠ 0 dx
Eg. N2 : HF: CO: ∫ψ
= µ0 ∫ ψ n * ( x)ψ m ( x)dx → 0 (if n ≠ m )
n
large small * ( x)µ ( x)ψ m ( x)dx Chapter 5 – Vibrational Motion 72 Fall 2014 Chem 356: Introductory Quantum Mechanics dµ
dx ∫
dµ
→
dx
dµ
→
dx
+
ψ n * ( x) xψ m ( x)dx For Harmonic Oscillator we can only get allowed transitions if Δn = +1 or Δn = −1 ΔE = ±!ω α
2
∫ψ
α
∫ψ
2
n
n
* (q)qψ m (q)dq (
)
* (q) bˆ+ + bˆ ψ m (q)dq bˆ(ψ m ) ~ ψ m−1 bˆ+ (ψ m ) ~ ψ m+1 Poly-­‐atomic Molecules −! 2∇ 2 1
Ĥ = ∑
+ ∑ k αβ (R − Re )α (R − Re )β 2 µα
2 α ,β
α
By some manipulation (see notes on webpage) this can be written as a sum of H.O. Hamiltonians…….tedious derivations ⎡ 1 d2 1 2 ⎤
Ĥ = ∑ !ω i ⎢ −
+ qi ⎥ 2
i
⎣ 2 dqi 2 ⎦
=
∑ !ω
i
i
⎡ +
1⎤
⎢ b̂i b̂i + 2 ⎥ ⎣
⎦
The coordinates q : are linear combinations of atomic displacement vectors 3 N coordinate → 3 translation, 3 rotation, (3 N -­‐6) vibration 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 ) Ek ,l ,m = ⎜ k + ⎟ !ω 1 + ⎜ l + ⎟ !ω 2 + ⎜ m + ⎟ !ω 3 2⎠
2⎠
⎝
⎝ 2⎠
⎝
Very simple solution. One needs to diagonalize the “mass weighted hessian” Reasonable approximations to all vibrational levels Statistical Mechanics (recall Chem 350) ⎛
1⎞
1
Mα
kαβ
⎛
1
Mβ
1⎞
⎛
1⎞
3N × 3N Matrix 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 perturbation theory, based on a quartic force field) 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 e ⎜ n + ⎟ − !ω e xe ⎜ n + ⎟ 2⎠
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 Δn ≠ ±1 . This leads to the observation of overtones. An often used form for the potential for a diatomic is the Morse potential Chapter 5 – Vibrational Motion 74 Fall 2014 Chem 356: Introductory Quantum Mechanics V ( x) = De (1 − e − β x ) ∂V
∂x
2
= 0 x = 0 equilibrium geometry x =0
1 ∂ 2V
De β 2 = k 2 ∂x 2
→β =
k
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. Chapter 5 – Vibrational Motion 75