S18 Ch. 5 Rigid Rotator - University of Minnesota

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Lecture 17: Chap. 5, Sect. 8: Rigid Rotator - model for molecular rotation Chap. 4, Sect. 2: Angular momentum properties MathChap. C, pp. 111-2 Angular momentum vectors MathChap. D, pp. 147-8 Spherical coordinates

Lecture 18: Chap. 5, Section 9: Rigid Rotator, continued energy levels microwave spectroscopy

Chem 4502 Quantum Mechanics & Spectroscopy (Jason Goodpaster)

Chapter 5 (continued): Rigid Rotator

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�Trying to capture the physicists� precise mathematical description of the quantum world with our crude words and mental images is like playing Chopin with a boxing glove on one hand and a catcher�s mitt on the other.�

George Johnson, New York Times, 1996

Quantum quote of the day, submitted by 3502 student David:

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Assuming that H79Br has a vibrational frequency of

∼2560 cm-1, predict its gas phase IR absorption spectrum.

Figure 13.2 p. 500

There is a gap centered at ∼2560 cm-1.

On either side, absorption lines are spaced by ∼17 cm-1.

These are due to rotational transitions (plus ∆v = 1). 4

Classical Rotational Motion

Consider a mass, m, rotating about a point at radius r:

r Moment of inertia (I)

I = m r2

I corresponds to m in linear motion.

Consider 2 masses separated by r, rotating about their center of mass:

Fig. 5.9 p. 174

I = µ r2

µ = reduced mass = m1 m2 m1 + m2 (for 2 masses)

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Center of mass (CM) : defined by

m1 m2

XCM

X1 X2

m1 > m2

XCM (m1+m2) = m1X1+m2X2

If m1 = m2, then XCM = ½ (X1+X2) right in the middle

and µ = ½ (m1) = ½ (m2)

mass-weighted average position of all of the masses

XCM position of the center of mass

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Angular momentum (magnitude)

L = I ω

L = angular momentum

I = moment of inertia (µr2)

ω (�omega�) = rotational speed or angular velocity radians/sec

In the absence of an external torque (force causing rotation), angular momentum is conserved.

Example: figure skater brings arms in

(reducing I) → L remains constant so ω increases (spins faster).

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Angular momentum (direction)

MathChapter C pp. 111-112

right hand rule

p = mv = linear momentum

L = r x p → → →

→ →

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Relate our two expressions for angular momentum, L:

L = r x p → → →

Since r and p are perpendicular for circular motion, magnitude of L is:

L = r p = r mv

→ →

ω (rad/s) = v (m/s) 2 π (rad/cycle) 1 / (2 π r) (cycle/m) = v/r

Substituting,

L = m r2 (v / r) = m r v Same as above

L = m r v

L = I ω = m r2 ω

We also had (for one rotating mass):

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Rotational kinetic energy

Erot = ½ I ω2 Analogous to: E = ½ mv2

= ½ (I ω)2 / I = L2 / 2I = p2 / 2m

Is skater�s Erot conserved when brings arms in?

No: Erot = L2 / 2I

L is the same, but I is reduced so Erot is increased

To bring arms in, skater has to do work against the �centrifugal force� pulling arms outward.

This compensates for the increased rotational kinetic energy.

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p. 120

r

Lz

rotation in x-y plane


QM Rotation:

Complementary observables (uncertainty relation) x, px Φ, Lz

where Φ is (azimuthal) angle in xy plane

(v)

mv2

(ω=v/r) (I = mr2)

; L= r p = r mv

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Quantum Mechanical Rotational Motion

Extend to 3-D. Applications:

Particle constrained to move on surface of a sphere

Gas phase molecule rotating freely about its center of mass

2-D (planar) motion (in an inherently 3-D system) forbidden

Uncertainty Principle: if molecule were to rotate only in the xy plane, both z and pz would be zero

Model: Rigid (fixed r) rotator

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Schrödinger equation for the rigid rotator:

Ĥ Ψ = E Ψ All the energy is kinetic

∇2 LaPlacian Operator in Cartesian coordinates

�Convenient� to convert to spherical coordinates

(since r is constant)

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Spherical Coordinates (MathChapter D p. 147)

Φ (�phi�) �azimuthal� angle (in xy plane)

θ �theta� angle from z axis

r distance from origin

0 ≤ Φ ≤ 2 π 0 ≤ θ ≤ π

0 ≤ r < ∞ generally but here r is fixed

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Using the chain rule, convert ∇2 from Cartesian to spherical coordinates:

∇2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2 Cartesian

Since r is constant, can omit first term:

Chap. 5 #30-32, not assigned Eqn. 5.49

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∇2 = We had:

Hamiltonian operator in Cartesian coordinates:

Factoring out 1/r2 and using I = µ r2 , obtain

Hamiltonian operator in spherical coordinates (fixed r)

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Lecture 18: Chap. 5, Sect. 9: Rigid Rotator, continued energy levels; microwave spectroscopy Lecture 19: Chapter 6: Hydrogen Atom

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Schrödinger equation for the rigid rotator:

call wave functions Y(θ, Φ) �spherical harmonics�

Ĥ E = rotational kinetic energy

So, a rotating molecule in a state characterized by one of the Y(θ,Φ) eigenfunctions will have a well-defined rot. kinetic energy.

Will any of its other properties also be well-defined?

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Ĥ Y(θ,Φ) = E Y(θ,Φ) We had:

Are the spherical harmonics also eigenfunctions of any other operator?

Recall for kinetic energy (here, call it �K�): K = ½ I ω2 = L2 / 2 I for rotational motion

Analogous to: K = ½ mv2 = p2 / 2m for linear motion

L2 = 2 I K

Write the operator corresponding to L2 :

L2 = 2 I Ĥ ˆ since have no potential energy K = Ĥ ˆ So, L2 = 2 I K ˆ ˆ

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We had:

L2 = 2 I Ĥ ˆ Also, since

ˆ L2 = -�2

The spherical harmonics, Y(θ,Φ), will also be eigenfunctions of with eigenvalues L2 = 2I E. ˆ L2

Ĥ Y(θ,Φ) = E Y(θ,Φ) Y(θ,Φ) = 2I E Y(θ,Φ) ˆ L2

unitless Natural unit of angular momentum is � (i.e., J•s).

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Ĥ Y(θ,Φ) = E Y(θ,Φ)

Rigid Rotator

Apply boundary conditions Y(θ, Φ) = Y(θ, Φ+2π)

Schrödinger equation Use separation of variables.

Solution yields 2 quantum numbers:

also determines magnitude of angular momentum

in rigid rotator, J determines total (= kinetic) energy

J = 0, 1, 2, ...

(we will call this ℓ for H atom)

m = 0, ±1, ±2, ... ±J (so, 2J+1 degenerate states for a given J)

determines direction of angular momentum vector

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Rigid Rotator

Look at the lowest energy solution:

J=0, m=0 Y(θ,Φ) = 1 / (4π)½ E=0

J=0 no rotation so rotational kinetic energy = 0

angular momentum = 0

(also linear momentum = 0)

Does this violate the uncertainty principle?

Y(θ,Φ) is a constant; spherically symmetric No.

The position (angular orientation) is completely unknown. 22

Rigid Rotator

The energy levels turn out to be:

EJ = ћ2 J (J+1) 2I

J = 0, 1, 2, ...

where I = µr2 moment of inertia

Rotational energy levels are quantized.

In contrast, classically the rotational kinetic energy can have a continuous range of values:

Erot = ½ I ω2 ω (�omega�) = rotational speed (rad/sec)

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EJ = ћ2 J (J+1) 2I

Rigid Rotator

where I = µr2

We had:

µ = m1 m2 m1 + m2

Model for rotating diatomic molecule

So, by measuring the rotational energy levels EJ ,

we can determine r, the bond length of the molecule.

For polyatomics, can also measure bond angles.

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Rigid Rotator

EJ = ћ2 J (J+1) 2I

We had:

In units of wave numbers (cm-1),

EJ = EJ / hc = ћ2 J (J+1) 2I hc

∼ ћ = h/2π

= h J(J+1) 8 π2 I c

B ∼ Rotational

constant (cm-1) EJ (cm-1) = B J (J+1) ∼ ∼

In units of E/h, B = h / ( 8 π2 I )

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Rigid Rotator

EJ (cm-1) = B J (J+1) ∼ ∼

Selection rule for absorption or emission of a photon:

∆J = ± 1

Fig. 5.10 p. 178

J

0

E ∼ (cm-1)

2B ∼

6B ∼

20B ∼

12B ∼

g

1 3

5

7

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Rotational spectrum shows absorption lines separated by 2B ∼

Degeneracy g = 2J+1

since m = 0, ±1, ..., ±J 26

Rigid Rotator

Selection rule

to absorb or emit a photon and change rotational level,

the molecule must have a permanent dipole moment.

Homonuclear diatomics (O2, N2)

or symmetric molecules (e.g., benzene, methane)

do not have a microwave (pure rotational) spectrum

(but they do have discrete rotational levels).

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Rigid Rotator

The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location.

Engel p. 143

Selection rule (classical picture):

a molecule with a permanent dipole moment can absorb rotational energy from the oscillating electric field of the (microwave) radiation.

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Rigid Rotator

Calculate the J=0 to J=1 rotational energy spacing of 12C16O, and its rotational constant, in cm-1. The equilibrium bond length is 1.13 Å.

EJ (cm-1) = B J (J+1) ∼ ∼ B = h / 8 π2 I c

I = µr2

µ = m1 m2 / (m1+m2)

Answer: µ = m1 m2 / (m1+m2) = [12•16/28] amu = 6.9 amu = 1.14 x 10-26 kg

1.66 x 10-27 kg/amu

B = h / 8 π2 I c = 6.63 x 10-34 Js / ((8 π2 1.46x10-46 kg m2)( 3.00 x 1010 cm/s)) = 1.9 cm-1 = rotational constant

∼

∼ h = 6.63 x 10-34 Js

I = µr2 = 1.14 x 10-26 kg (1.13 x 10-10 m)2 = 1.46 x 10-46 kg m2

c = 3.00 x 108 m/s

For J=0, E=0 cm-1; for J=1, E = 2B = 3.8 cm-1 = energy spacing ∼

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Rigid Rotator

Calculate the J=0 to J=1 rotational energy spacing of 12C16O, and its rotational constant, in cm-1. The equilibrium bond length is 1.13 Å.

We had:

2B = 3.8 cm-1 ∼

Rotational transitions occur in the microwave (and far IR) regions of the electromagnetic spectrum.

If measure rotational energy spacings, can deduce precise bond length (12C16O re = 1.128229 ± 0.000001 Å).

Rot. spacings are typically much smaller than vibrational: (12C16O vibrational frequency: 2169.7559 cm-1)

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Microwave spectroscopy (�radio astronomy�)

of interstellar molecules

Almost all of the molecules known to occur in the interstellar medium were detected by rotational spectroscopy.

�As of 2001, over 120 different chemical compounds have been identified in interstellar clouds, circumstellar matter, and comets.�

(Nummelin article, WebCT)

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Ken�s group�s Fourier transform microwave spectrometer, in 29 Smith Hall http://www.chem.umn.edu/groups/kleopold/

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Ken Leopold�s Research Program: Microwave Spectroscopy of Gas Phase Complexes (trimethylamine trimethylborane, a Lewis acid-base complex)

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Rigid Rotator

Energy level diagram showing the rotational energy levels associated with each vibrational state for a diatomic molecule

Figure 13.2 p. 500

H79Br

Fig. 13.1 p. 499

The IR absorption spectrum of a gas phase molecule shows transitions from a specific rot. level of the lower vib. state to a specific rot. level of the upper vib. state

(For HBr, ∆ J = ± 1) 34

Rigid Rotator

Does rigid rotor �obey� the Correspondence Principle:

expect energy levels to be nearly continuous at high quantum numbers

EJ (cm-1) = B J (J+1) ∼ ∼

Spacings between adjacent levels:

EJ+1 – EJ = B (J+1)(J+2) - B J(J+1) = B (2J +2) ∼ ∼ ∼ ∼ ∼

Energy spacing 2B (J+1) 2 Energy B J(J+1) J

= = → 0 as J → ∞

Yes: energy spacings are ≈ continuous at high J

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Rigid Rotator

Recall: The spherical harmonics, Y(θ,Φ), are also eigenfunctions of with eigenvalues L2 = 2I E. ˆ L2

Ĥ Y(θ,Φ) = E Y(θ,Φ) Y(θ,Φ) = 2I E Y(θ,Φ) ˆ L2

Angular momentum

Angular momentum is also quantized.

(Classically, L = I ω = r mv can have a continuum of values.)

E = ћ2 J (J+1) 2I

L2 = ћ2 J (J+1)

For J=1, L2 = 2 ћ2 and the length of the angular momentum vector │L│ = √2 ћ. →

L →

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S18 Ch. 5 Rigid Rotator - University of Minnesota