Chapter 17 - Quantum Rigid Rotor - Grandinetti · Chapter 17 Quantum Rigid Rotor P. J. Grandinetti Chem. 4300 Would be useful to review Chapter 4 on Rotational Motion. P. J. Grandinetti

Chapter 17Quantum Rigid Rotor

P. J. Grandinetti

Chem. 4300

Would be useful to review Chapter 4 on Rotational Motion.

P. J. Grandinetti Chapter 17: Quantum Rigid Rotor

Quantum Rigid RotorFrom classical mechanics of rotational motion we found that kinetic energy can be written interms of principal moments of inertia tensor as

K =J2

a2Ia

+J2

b2Ib

+J2

c2Ic

Translating to QM we look for solutions of Schrödinger equation

𝜓(𝜃, 𝜙, 𝜒) =[

J2a

2Ia+

J2b

2Ib+

J2c

2Ic

]𝜓(𝜃, 𝜙, 𝜒) = E𝜓(𝜃, 𝜙, 𝜒)

𝜓(𝜃, 𝜙, 𝜒), is function of molecule orientation.

Euler angles, (𝜃, 𝜙, 𝜒), define orientation of molecule’s moment of inertia tensor PAS relativeto some non-rotating (but translating) space-fixed frame.


Quantum Rigid RotorCommon to express Hamiltonian as

= 2𝜋ℏ

[AJ2

a + BJ2b + CJ2

c]

whereA ≡ ℏ

4𝜋IaB ≡ ℏ

4𝜋IbC ≡ ℏ

4𝜋Ic

are called rotational constants where A ≥ B ≥ C. Rotational constants havedimensionality of frequency.Also common to express Hamiltonian as

=2𝜋c0ℏ

[AJ2

a + BJ2b + CJ2

c]

whereA ≡ A

c0= ℏ

4𝜋c0IaB ≡ B

c0= ℏ

4𝜋c0IbC ≡ C

c0= ℏ

4𝜋c0Ic

are called wavenumber rotational constants, and have dimensionality of wave numbers.P. J. Grandinetti Chapter 17: Quantum Rigid Rotor

Classification of moleculesMolecules grouped into 5 classes based on principal components.

Name Diagonal values ExamplesSpherical Ia = Ib = Ic = I CH4

Prolate Symmetric I|| = Ia < Ib = Ic = I⟂ CH3F

Oblate Symmetric I⟂ = Ia = Ib < Ic = I|| CHF3

Asymmetric Ia < Ib < Ic CH2Cl2, CH2CHCl

Linear Ia = 0, Ib = Ic = I OCS, CO2

All molecules with one 3-fold or higher rotational symmetry axis are symmetric topsbecause principal moments about two axes normal to n-fold rotational symmetry axis(n ≥ 3) are equal.Molecules with two or more 3-fold or higher rotational symmetry axes are spherical tops.


Quantum Rigid Rotor - Angular Momentum OperatorsJa, Jb, and Jc are angular momentum components in body-fixed frame (moment of inertia PAS).

Jx, Jy, and Jz are angular momentum components in space-fixed frame.

Body-fixed frame are related to space-fixed frame components

J2 = J2x + J2

y + J2z = J2

a + J2b + J2

c

Commutators in body-fixed frame: [Ja, Jb] = −iℏJc [Jb, Jc] = −iℏJa [Jc, Ja] = −iℏJb

Commutators in space-fixed frame: [Jx, Jy] = iℏJz [Jy, Jz] = iℏJx [Jz, Jx] = iℏJy

Each space fixed component commutes with any body-fixed component

[J2, J𝛼] = 0 [J2, Ji] = 0 and [Ji, J𝛼] = 0 where{

i = x, y, z𝛼 = a, b, c

Uncertainty principle says you can simultaneously know angular momentum vector length,1 body-fixed frame component, and 1 space-fixed frame component.


Spherically Symmetric Molecules


Spherically Symmetric MoleculeFor spherical symmetric molecule we have Ia = Ib = Ic = I

= K =J2

a + J2b + J2

c

2I= J2

2ISchrödinger equation becomes

𝜓(𝜃, 𝜙, 𝜒) = J2

2I𝜓(𝜃, 𝜙, 𝜒) = E𝜓(𝜃, 𝜙, 𝜒)

Eigenvalue of J2 is J(J + 1)ℏ2 and J = 0, 1, 2,….Energy is

EJ = J(J + 1)ℏ2

2I= hc0BJ(J + 1)

B is wavenumber rotational constantB ≡ ℏ

4𝜋c0INo zero point energy. Why?Free rotation, like free translation has no zero point energy.


Spherically Symmetric MoleculeSolutions to Schrödinger equation have form

𝜓J,KJ ,MJ(𝜃, 𝜙, 𝜒) = eiMJ 𝜙 f (J)KJ ,MJ

(𝜃) eiKJ 𝜒

J = 0, 1, 2,…, and

KJ = −J,−J + 1,… , 0,… , J − 1, J body-fixed frame component

andMJ = −J,−J + 1,… , 0,… , J − 1, J space-fixed frame component

KJ is quantum number for rotation about c axis in PAS frame with KJℏ as projection of J ontoc axis of PAS frame.MJ is quantum number for rotation about z axis in space-fixed frame with MJℏ as projectionof J onto z axis of space-fixed frame.Energy is independent of KJ and MJ.


Energy levels and degeneracies of spherical rotor moleculedegeneracyenergy

Rotational state energy divided by hc0 gives quantity,F(J), in wave numbers,

EJhc0

= F(J) = BJ(J + 1)

Each energy level, EJ , has gJ = (2J + 1)(2J + 1) fromrange of both MJ and KJ values.

ΔE in wave numbers between adjacent levels is

EJ+1 − EJhc0

= FJ+1 − FJ

= B(J + 1)(J + 2) − BJ(J + 1)= 2B(J + 1)


Symmetric Molecules


Prolate Symmetric Moleculethink “cigar shaped”In prolate symmetric molecule: Ia < Ib = Ic. Define I|| = Ia & I⟂ = Ib = IcHamiltonian is

=J2

a2I|| +

J2b + J2

c

2I⟂= J2

2I⟂+(

12I|| −

12I⟂

)J2

a

Only operators appearing in Hamiltonian on right are total angular momentum operator,J2, and a component, Ja, and these 2 operators commute, [J2, Ja] = 0.Solutions to Schrödinger Eq are


(𝜃) eiKJ 𝜒

Here KJ is quantum number for rotation about a axis in PAS frame with KJℏ asprojection of J onto a axis of PAS frame.

If |KJ| = J prolate molecule is found rotating about axis nearly parallel to a axis.If |KJ| is small or zero prolate molecule is found rotating about b or c axes, primarily.


Prolate Symmetric Moleculesthink “cigar shaped”Rotational energy of a prolate symmetric molecule is

EJ,KJ= J(J + 1)ℏ2

2I⟂+(

12I|| −

12I⟂

)K2

Jℏ2

or with some rearranging becomes

EJ,KJ

hc0= FJ,KJ

= BJ(J + 1) + (A − B)K2J

A and B are rotational constants defined as

B ≡ ℏ4𝜋c0I⟂

and A ≡ ℏ4𝜋c0I||

Since rotational energy is independent of orientation of J in isotropic space—space-fixedframe—rotational energy of prolate symmetric molecule is independent of MJ.


Prolate Symmetric Molecule Rotational Energy Levels

3210

4

5

6J

|KJ|=0 |KJ|=1|KJ|=2

|KJ|=3

Prolate

Ener

gy

EJ,KJ

hc0= FJ,KJ

= BJ(J + 1) + (A − B)K2J

For prolate symmetric molecules(A − B) > 0 so rotational energy levels ofprolate symmetric molecule increase withincreasing KJ

KJ = 0 energy levels are (2J + 1)-folddegenerate|KJ| ≠ 0 energy levels are 2(2J + 1)-folddegenerate


Oblate Symmetric Moleculesthink “frisbee shaped”In oblate symmetric molecule: Ic > Ia = Ib. Define I|| = Ic & I⟂ = Ia = IbHamiltonian is

= J2

2I⟂+(

12I|| −

12I⟂

)J2

c

Only operators appearing in Hamiltonian on right are total angular momentum operator,J2, and c component, Jc, and these 2 operators commute, [J2, Jc] = 0.Solutions to Schrödinger Eq are


(𝜃) eiKJ 𝜒

Here KJ is quantum number for rotation about c axis in PAS frame with KJℏ asprojection of J onto c axis of PAS frame.

If |KJ| = J oblate molecule is found rotating about axis nearly parallel to c axis.If |KJ| is small or zero prolate molecule is found rotating about a or b axes, primarily.


Oblate Symmetric Moleculesthink “frisbee shaped”

Rotational energy of a oblate symmetric molecule is

FJ,K = BJ(J + 1) + (C − B)K2J

Rotational constants defined as

B ≡ ℏ4𝜋c0I⟂

and C ≡ ℏ4𝜋c0I||

Again, since rotational energy is independent of orientation of J in isotropic space—space-fixedframe—rotational energy of oblate symmetric molecule is independent of MJ.


Oblate Symmetric Molecule Rotational Energy Levels

|KJ|=0

3210

4

5

6J

|KJ|=1|KJ|=2|KJ|=3

Oblate

Ener

gyFJ,K = BJ(J + 1) + (C − B)K2

J

KJ = 0 energy levels are (2J + 1)-folddegenerate|KJ| ≠ 0 energy levels are 2(2J + 1)-folddegenerateFor oblate symmetric molecules(C − B) < 0 so rotational energy levels ofprolate symmetric molecule decrease withincreasing KJ.Opposite behavior of prolate symmetricmolecules.


Symmetric Molecules

|KJ|=0

3210

4

5

6J

|KJ|=0|KJ|=1 |KJ|=1|KJ|=2

|KJ|=2

|KJ|=3

|KJ|=3

Oblate Prolate

Ener

gy


Asymmetric Molecules

caffeineP. J. Grandinetti Chapter 17: Quantum Rigid Rotor

Asymmetric Molecules

For asymmetric molecule we have Ia ≠ Ib ≠ Ic.

Most molecules become asymmetric as the number of its atoms increase.

Hamiltonian is

=J2

a2Ia

+J2

b2Ib

+J2

c2Ic

Most challenging case to solve as eigenstate for this Hamiltonian is NOT an eigenstate of J2

and single component of angular momentum vector.

It is possible to solve for the wave function analytically, but unfortunately it is too involved forus to explore here.


Linear Molecules


Linear MoleculesFor linear molecule we have Ia = 0 and Ib = Ic = I. Hamiltonian is

=J2

b + J2c

2I= J2

2I, or FJ = BJ(J + 1)

Appears to be same as spherical molecule but it is different in a key way.

The 3 principal components are non-zero and equal, i.e., Ia = Ib = Ic, for a spherical molecule.

Only 2 principal components are non-zero and equal and one component is zero, Ia = 0, for alinear molecule

There can be no rotational angular momentum along a axis, and thus only KJ = 0 is allowed forlinear molecule.

May be helpful to think of linear molecule as more like prolate molecule with Ia = 0,In that case we require KJ = 0 to prevent 1∕I|| term in prolate energy expression from going to infinity.


Linear Molecules Rotational Wave Function & Energy

degeneracyenergy

0

0

0

0

0KJ

Linear molecule rotational energy is

FJ = BJ(J + 1)

Linear molecule rotational wave functions are thespherical harmonics:

𝜓J,MJ(𝜃, 𝜙) = YJ,MJ

(𝜃, 𝜙)

Remember only KJ = 0 allowed

MJ can have 2J + 1 integral values giving eachenergy level a degeneracy of 2J + 1


Rotational Transition Selection Rules


Rotational Transition Selection RulesFor transitions between rotational energy states through absorption and emission of light we require anon-zero electric dipole transition moment,

⟨𝜇⟩J,MJ ,KJ ,J′,M′J ,K

′J= ∫V

𝜓∗J,MJ ,KJ

(𝜙, 𝜃, 𝜒) 𝜇 𝜓J′,M′J ,K

′J(𝜙, 𝜃, 𝜒)d𝜏

Orientation of permanent electric dipole, 𝜇, is constant in body-fixed frame.

By symmetry 𝜇 must be parallel to figure axis (principal axis with highest rotational symmetry) inspherical, linear, and prolate or oblate molecules.

KJ is associated with rotation about the figure axis.

Changes in KJ in spherical, linear, and prolate or oblate molecules does not change direction of 𝜇,so transition moment is not influenced by changes in KJ and we have ΔKJ = 0.


Transition Selection RulesLinear, Spherical, and Prolate and Oblate Symmetric Molecules

ΔJ = ±1, ΔKJ = 0, ΔMJ = 0 for ��z, ΔMJ = ±1 for ��x and ��y


Centrifugal Distortion


Centrifugal DistortionMolecules stretch at they rotate faster, i.e., bonds lengths increase with increasing J causingprincipal moments of inertia increase and in turn causing E to decrease from values expected forideal rigid rotor.Experimentally, energy spacing found to decrease with increasing J.Correction term can be added. In linear molecule case energy expression becomes

FJ = BJ(J + 1) − DJ2(J + 1)2

D is centrifugal distortion constantD = 4B3

��2

�� is vibrational frequency of bond in wave numbers.Vibrational frequency is directly related to bond force constant. In diatomic case

�� =𝜔0

2𝜋c0where 𝜔0 =

√𝜅f∕𝜇


Centrifugal Distortion ConstantsBonds w/stronger force constants have less centrifugal distortion

D = 4B3

��2 and �� =√𝜅f∕𝜇

2𝜋c0

Molecule B/cm−1 D/10−6 cm−1

H2 60.853 47100D2 30.444 11410HF 20.953712 2150HCl 10.5933002 531.94HBr 8.46488 345.8HI 6.4263650 206.9I2 0.03737 0.0043N2 1.998236 5.737CO 1.931280985 6.1216O2 1.446 4.839


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Chapter 17 - Quantum Rigid Rotor - Grandinetti · Chapter 17 Quantum Rigid Rotor P. J. Grandinetti Chem. 4300 Would be useful to review Chapter 4 on Rotational Motion. P. J. Grandinetti