Rigid rotator

Question 1

Basic theory

Answer 1

2D, particle on a ring

Question 2

What can it help us understand?

Answer 2

Cyclic molecules like aromatics

Question 3

Angular momentum equation

Answer 3

J = Iw
Where J is angular momentum, I is the moment of Inertia and w is the angular velocity

Question 4

Moment of Inertia equation

Answer 4

I = mr^2
m is mass and r is distance from axis of rotation

Question 5

What is the axis of rotation for rigid rotator?

Answer 5

Jz as particle is confined in the x and y plane so we only have rotation about the z axis

Question 6

What values can the quantum number m_l take

Answer 6

m_l = 0,±1,±2,….,±l

Question 7

Jz simplification

Answer 7

Jz=±Iw=±mr^2w=±mvr=±pr

Question 8

Why can Jz = ± value

Answer 8

Jz will vary based on direction of rotation, if clockwise Jz>0, if anticlockwise Jz<0

Question 9

Boundary condition for rigid rotator

Answer 9

Cyclic motion, so for constructive interference of our wavefunction we must have our Ψ repeat after 2pi radians of rotation.
If there is deconstructive interference particle will destroy itself

Question 10

Boundary condition equation

Answer 10

Ψ(ϕ=0)=Ψ(ϕ=2pi)

Question 11

Derivation to show quantised wavelength

Answer 11

λ*m_l =2pir
λ=2pir/m_l
m_l is the quantum number so wavelength is quantised

Question 12

Derivation to show quantised energy

Answer 12

Jz=±pr
p=h/λ
Jz=±hr/λ
λ=2pir/m_l
Jz=±m_lℏ
E=Jz^2/2I
E=m_l^2ℏ^2/2I
So energy is quantised

Question 13

Double degeneracy

Answer 13

Since quantum number which can be + and - is squared then our energy can have 2 of the same values when m_l =|value| >0

Question 14

Zero point energy

Answer 14

Can be 0 as m_l is can be 0, which means we have no idea for the probability density which fits with schrodingers

Question 15

Rigid rotator Schrodinger equation

Answer 15

(-(ℏ^2)/2m)(∇^2Ψ) =EΨ

Question 16

Laplacian for rigid rotator

Answer 16

∇^2 for 2d= d^2/dx^2+ d^2/dy^2
Convert to spherical coordinates:
d^2/dr^2 +(1/r)d/dr+(1/r2)d^2/dϕ^2
For us r is constant so
∇^2 = (1/r^2)d^2/dϕ^2
Where ϕ is our angle of rotation
(remember these are partial derivatives)

Question 17

Simplified schrodinger equation for rigid rotator

Answer 17

(-(ℏ^2)/2I)(d^2Ψ/dϕ^2) =EΨ
Where I is the moment of inertia

Question 18

General solution for wavefunction

Answer 18

Ψ =Ae^im_lϕ
Where i is the (-1)^1/2
m_l is our quantum number and ϕ is our angle of rotation
A is just a constant

Question 19

Calculating constant A process

Answer 19

Use normalisation integral:
∫ΨΨdx = 1
Remember we will be in spherical so change x to ϕ
And remember that Ψ will be - as it is a complex number so cancels out.
Integral is between 0 and 2pi, will eventually give
A =(1/2pi)^1/2

Question 20

Calculating Energy

Answer 20

Sub in general equation for Ψ into simplified Schrodinger’s, solve for E.
Will get E = (ℏ^2m_l^2)/2I =Jz^2/2I