Download A Quantum Mechanical Model for Vibration and Rotation of Molecules

3/19/2017
Translational
Quantum Mechanical Models
of
Vibration and Rotation of Molecules
Chapter 18
Vibrational
Molecular
Energy
Molecular
Motions
Rotational
Electronic
…
Vibrations of Molecules:
Model approximates molecules to atoms joined by
springs. A vibration (one type of – a normal mode of
vibration) of a CH2 moiety would look like;
Frequency of vibration – classical approach
Water (3) http://www.youtube.com/watch?v=1uE2lvVkKW0
CO2 (4)
http://www.youtube.com/watch?v=W5gimZlFY6I
O2 (1)
http://www.youtube.com/watch?v=5QC4OVadKHs
http://en.wikipedia.org/wiki/Molecular_vibration
The motions are considered as harmonic oscillators.
For a molecule of N atoms there are 3N-6 normal
modes (nonlinear) or 3N-5 (linear).
1
3/19/2017
During a molecular vibration the motion of the atoms
are with respect to the center of mass, and the center of
mass is stationary as far as the vibration is concerned.
Vibrational motion - harmonic oscillator, KE and PE
– classical approach
Center of mass, does not move.
Diatomics:
This concept is true for all normal modes of vibrations of
molecules.
Force constant of spring = k
Working with center of mass coordinates simplifies
the solution.
Center of mass coordinates
Whether pulled apart or pushed together from the
equilibrium position, the spring resists the motion
by an opposing force.
Second Law
reduced mass, µ;
Solutions (general) of the DE will be of the form;
∼
Use Euler’s Formula
µ
Spring extension of a mass µ from it’s equilibrium position.
The physical picture changes from masses (m1 and m2)
connected by a spring (force constant k) to a reduced mass,
µ, connected by a spring (same k) to an immovable wall.
where b1 = c1 + c2 and
b2 = i (c1 − c2 )
2
3/19/2017
Amplitudes are real numbers, b1 and b2 are real or = 0.
and to find T
period:
Applying the BC, at t = 0; x(0) = 0, v(0) = v0.
=1
⇒
Frequency; ν =
1
T
a
v=0
Therefore:
x=0
=
-a
v = v0
v=0
angular velocity =ω =
k
µ
where ν = frequency
= 2πν
α = phase angle
General equation;
Energy terms (KE, PE) are;
1
1
KE = µ v 2
PE = kx 2
2
2
where v and x are velocity and position
(displacement from equilibrium).
α =π / 2
α ≠0
No restriction on
E values, classically.
Study Example Problem 18.1
3
3/19/2017
4
3/19/2017
Harmonic oscillator potential function
http://www.youtube.com/watch?v=5QC4OVadKHs
Harmonic oscillator potential function
Morse curve
Similar for low energy situations ~ ground state
https://www.youtube.com/watch?v=3RqEIr8NtMI
Our interest is to model the oscillatory motion of the
diatomic molecule (relative motions of atoms).
Also – of interest kinetic and potential energy of the
diatomic molecule during the oscillatory motion.
Not necessarily the energies of individual atoms of the
molecule.
The ultimate goal is to find the (eigen) energies and the
eigen functions (wavefunction) of the vibrational states
of the diatomic by solving the Schrödinger equation;
At room temperature
the potential function
very closely follow
the quadratic
function.
Vibrational motion is
equivalent to a particle
of mass µ, vibrating
about its equilibrium
distance, r0.
How can a particle like
that be described by a
set of wave functions.
x
,x
r0
Recipe to Construct the Schrodinger equation
1. Expression for total energy (classical)
2. Construct the total energy Operator, the Hamiltonian,
by expressing KE in terms of momentum operator
and mass, PE in terms of position operator.
3. Setup the Schrodinger equation, eigenequation
ψ = Eψ .
H
5
3/19/2017
1. Expression for total vibrational energy,
Etot ,vib = KE + PE =
=
1 2 1 2
µ v + kx
2
2
3. Set the Schrodinger equation, eigenequation
ψ = Eψ
H
p2 1 2
+ kx
2µ 2
2. Construct the total energy Operator, the Hamiltonian
Solutions (i.e. normalized wavefunctions, eigenfunctions)
of which are;
2
= 1 p 2 + 1 k ɵx 2 = 1  iℏ d  + 1 kx 2
H
2µ
2
2 µ  dx  2
With the normalization constant
The first six Hermite polynomials
H n (α 1/ 2 x ) = Hermite polynomials
n = vibrational quantum number
Hermite polynomials :
H n ( z ) = ( −1)n e z
2
d n − z2
e
dz n
6
3/19/2017
The first few wavefunctions, ψ n (n = 0,..4)
The first few
wavefunctions
(n = 0,..3)
The respective eigen energies are;
Note:
the resemblance to 1D box non-infinite potential well results.
zpe
Equal energy gaps are equal in contrast to 1D box case.
The first few wavefunctions, ψ n2 (n = 0,..4)
The constraint imposed on the
particle by a spring results in zero point energy.
The energy values of vibrational states are precisely
known, but the position of the particles (as described
by the amplitude) is imprecise, only a probability
] ≠ 0.
density ψ2(x) of x can be stated, [ɵx, H
Note the resemblance to 1D box.
Note: the overflow of the wave function (forbidden) of the
wavefunction beyond the potential barrier wall.
7
3/19/2017
At high quantum numbers n (high energy limit) the system
gets closer to a classical system (red - probability of x in
q.m. oscillator, blue - probability of x classical oscillator).
n = 12
n >>0
high
high
low high
low high
Blue to black, a particle in a
barrel classical model.
Rotational energies of a classical rigid rotor (diatomic).
In vibrational motion, velocity, acceleration and
momentum are parallel to the direction of motion.
In the center of mass coordinate system, rigid rotor motion
equivalent to a mass of µ moving in a circle of radius
r0 (= bond length, constant) with an angular velocity ω and
a tangential velocity v. Rotation in 2D (on a plane)
y
r0
No opposing force to rotation –
no PE (stored of energy) term.
All energy = KE.
∼
x
r0
8
3/19/2017
p = linear momentum
Acceleration in the radial direction
Angular velocity vector
Angular velocity ω and angular acceleration α,
r0
Mass with constant ω makes α = 0.
RHR
Tangential linear velocity,
Angular velocity (~ momentum)
vector normal to the plane of motion.
(coming out of the plane)
QM Angular momentum (2D):
p
ω=
l=r×p
angular momentum Magnitude of l = l
= I, moment of inertia
The rotation on the x,y plane (2D) occurs freely.
Equivalently the particle of mass µ= moves on the circle
(constant radius r = r0) freely; Epot = V(x,y) = 0
φ = 90o
Etot , rot = KE + PE = KE =
r
⇒ p=
Energy
Rotational:
2E
I
l
r
p2
2µ
p 2
ℏ2  ∂2
∂2 
H=
=−
 2+ 2
2µ
2µ  ∂x ∂y 
SE:
Classical rotor - no
restriction on
l (or E.
9
3/19/2017
l
0 < φ < 2π
0 <θ <π
r2sinθ dθ dφ dr
r
p
µ
r
l
In polar coordinates
(variable is φ):
BC: Φ(φ ) = Φ(φ + 2π )
BC: Φ(φ ) = Φ(φ + 2π )
Rotational
eigenfunction.
e
iml φ
= e iml (φ + 2π )
e iml φ = e iml (φ + 2π ) = e imlφ e iml 2 π
General solution for DE), wavefunctions:
i.e. e iml 2π = 1
Using the Euler’s Formula
ml = integer
clockwise
counter-clockwise
rotation
rotation
ml = rotational quantum number, quantization, next page
Now; cos( 2π ml ) + i sin( 2π ml ) = 1
cos( 2π ml ) = 1
real number
Therefore ml = 0 , ±1, ±2,.. .
Note: One boundary condition leads to one q.n., ml.
10
3/19/2017
Normalization constant A +φ;
2π
∫Φ
BC: Φ(φ ) = Φ(φ + 2π )
e
=e
i.e. e
− iml ( φ + 2 π )
− iml 2 π
=e
− i ml φ
e
(φ )Φ ml (φ )dφ = 1
0
2π
Normalization
condition
e − imlφ = e − iml (φ + 2 π )
− iml φ
*
ml
− iml 2 π
A+2φ
∫e
− imlφ + imlφ
e
dφ = 1
0
2π
=1
A+2φ
∫
dφ = 1
A+2φ ( 2π ) = 1
0
Now; cos( − 2π ml ) + i sin( − 2π ml ) = 1
thus cos( 2π ml ) = 1
A+φ =
real num ber
1
2π
Therefore ml = 0, ±1, ±2 ,...
Normalization constant A ±φ;
Φ ± (φ ) =
2π
Normalization
condition
*
∫ Φ ml (φ )Φ ml (φ )dφ = 1
1
2π
e± imlφ
Rotational wave function
0
2π
A±2φ
∓ im φ ± im φ
∫ e l e l dφ = 1
ml is an integer.
0
2π
A±2φ
∫
dφ = 1
0
A±φ
1
=
2π
real part
A±2φ ( 2π ) = 1
Φ + (φ ) =
1
2π
e + imlφ =
1
2π
( cos mlφ + i sin mlφ )
Stationary state - exist
Same regardless of ml value.
11
3/19/2017
Φ ± (φ ) =
1
2π
1
e± imlφ
2π
1
2π
1
Non-stationary state
- Unacceptable wavefunction, annihilates
Energy of rotational states:
Φ ± (φ ) =
1
2π
2π
2π
1
cos 3φ
2π
1
cos 5φ
2π
e ± imlφ
E=
ω=
ω=
Note; No non-zero ZPE.
ZPE appears if the potential
energy confinement exists,
not here ( V(φ)=0)
1
cos φ
Degenerate states
- two fold degenerate
(of equal energy)
l
2
2µ r02
=
l
2
2I
=
cos 2φ
cos 4φ
cos 6φ
1 2
Iω
2
2E
and angular momentum l = I ω
I
2 Eml
I
=
2
2 ℏ 2 ml2 ℏml
Eml =
=
I
I 2I
I
ml being an integer, rotational frequency ω will take
finite values. – discrete set of rotational frequencies !!
For 2D rotor, l = lz.
12
3/19/2017
The wavefunctions, …. of a rigid rotor are similar
to the wavefunctions of a free particle restricted to
a circle (particle in a ring)!
For a rigid rotor, angular momentum lies in the z direction.
Angular momentum z component operator;
y
ɵl z = −iℏ ∂
∂φ
z
x
,ɵl z ] = 0
Note: for 2D rigid rotor [ H
Angular Momentum – 2D rigid rotor.
Operator:
p
ɵl z = −iℏ ∂
∂φ
Both operators has the same eigenfunctions.
angular analogue of momentum
Φ ± (φ ) =
1
2π
e ± imlφ
wave functions
r
Hence angular momentum (z component) for this case can
be precisely measured.
But the position, here represented by φ, cannot be measured
precisely; the probability for any interval of dφ is the same,
[φɵ,ɵl z ] ≠ 0
Angular momentum (in z direction) is quantized!!
,ɵl z ] = 0
, ɵl z have same Φ, [ H
Note: for 2D rigid rotor both H
13
3/19/2017
QM Angular momentum (3D):
Particle on a circle (2D).
r
r0
l
r0
2D
≡
r
3D
Classical 3D rotor
Particle on the surface of a sphere (3D).
PE confinement – none; V(θ,φ) = 0
Set PE operator set to zero. E = KE + PE = KE
Particle of mass µ on the surface of
a sphere of radius r.
Just the steps in solving the SE for rigid rotor
a. collect the constants:
KE (energy) operator for rotor in polar coordinates:
note
b. multiply by sin2θ, rearrange to get
rot =
H
SE:
rot
Note the form of H
Differentiation only w.r.t.; θ
Therefore
Wavefunction;
φ
;separation of variables
possible.
spherical harmonic functions
14
3/19/2017
c. substitute for Y(θ,φ) divide by Θ(θ)Φ(φ), we get
solutions ⇓
Because each side of the equation depends only one of
the variables and the equality exists for all values of the
variables, both sides must be equal to a constant.
BC leads to q.n.;
spherical harmonic functions written in more detail
would take the form:
solutions ⇓
BC leads to more q.n.;
Energy of degenerate states of q. n. l :
For a given q.n. l there are (2l +1), ml values. That is the
state described by a l quantum number is comprised
of (2l+1) sub-states of equal energy.
β=
2 µ r02 E 2 I
= 2 E = l (l + 1)
ℏ2
ℏ
El =
ℏ2
l (l + 1)
2I
quantization of
rotational energy
Degeneracy of state with q. n. l, is 2l+1.
15
3/19/2017
Eigen-energy:
El =
ℏ2
l (l + 1)
2I
Quantization of Angular Momentum
As will be seen later, the shapes, directional properties
and degeneracy of atomic orbitals are dependent on the
q.n. l and ml vales associated with these orbitals.
Verifiable in SE:
rot
H
Spherical harmonics are eigen functions of the total
rotational energy operator.
All ml states of a 3D rotor has the same energy, El !
(degenerate)
ɵl 2
l2
For 3D rotor; Etot =
⇒ H=
2I
2I
Constant for rotor.
Therefore both operators have a common set of
eigenfunctions. The operators should commute!
the quantum number ml determines the z-component of the
angular momentum vector l.
,ɵl 2 ] = 0
[H
Components of the angular momentum in x, y and z.
Now;
The respective operators are (Cartesian coordinates):
and therefore
Note that the above operators differ only by the
multiplication constant (1/2I), thus the eigen values differ
ɵ2
by (1/2I).
= l
H
2I
Also note the calculated (and measurable) angular
momentum quantity (of precise (eigen) value) is that of | l2 |
(therefore |l|).
l = ℏ l (l + 1)
16
3/19/2017
The following commutator relationships exist.
The respective operators are in spherical coordinates
would be:
≠0
= f (θ , φ )
The component operators do not commute with one another
= f (φ )
As a result the direction of vector l cannot be specified
(known) ‘precisely’ for rotations in 3D in QM systems as
opposed to classical 2D rigid rotors. Need to know all
vectors at the same time to specify direction of vector l.
However, angular momentum wavefunction is an
eigenfunction of the Hamiltonian too and therefore,
ɵ2
= l
H
2I
Conclusion –
Spherical harmonics are eigenfunctions of l2 and lz .
Magnitudes of both |l| and lz can be known simultaneously
and precisely, but not the x and the z component values
of the angular momentum vector, l .
Examining lz;
2
Note/
[ɵl , ɵl z ] = 0
show explicitly:
!!!
l z = ml ℏ
17
3/19/2017
The Picture: Vector model of angular momentum
, l2 and l commute and l , l and l do not commute.
H
z
x
y
z
Example l = 2 case.
l
=
ℏ
with one another. Example l = 2 case.
| l |= l (l + 1)ℏ
lz = ml ℏ
2ℏ
ℏ
ml = +2
18
3/19/2017
Spatial quantization
ml = +2
Vector model of angular momentum
Spherical harmonic wavefunctions
and their shapes:
l=0
l=1
l=2
Complex functions
- visualization –
not possible.
19