6. QM in 3D

Question 1

Why is the 1D harmonic potential limited (be specific)?

Answer 1

Because simple diatomic molecules are 3D

- They cacn rotate and stretch

Question 2

What coordinate system do we use for diatomic molecules in 3D?

Answer 2

Spherical coordinates

Question 3

State the spherical volume element

Answer 3

dV = r^2 sin(θ)dθdϕdr

Question 4

State the normalisation condition

Answer 4

The integral everywhere of the modulus squared of the wave function is equal to 1

Question 5

See page 3 of the document for the normalisation calculation for the wave function

Answer 5

Do it lol

Question 6

What can we say about angular momentum?

Answer 6

It is quantised - number of values and states is restricted

Question 7

What is the equation for angular momentum?

Answer 7

L = r x p

Question 8

State the z component for the angular momentum

Answer 8

L_z = xp_y - yp_x

Question 9

Translate from Cartesian coordinates to spherical

Answer 9

r = (x,y,z) = (r,θ,ϕ)

Question 10

Describe the components of the spherical coordinates and their ranges

Answer 10

r - distance from the origin (0 to infinity)
θ - Polar angle (0 to pi)
ϕ - Azinuthal angle (0 to 2 pi)

Question 11

What type of potential are we only interested in?

Answer 11

Central potential

Question 12

Mathematically, how are central pontentials described?

Answer 12

V(r) not V(r) or V(r,θ,ϕ)

Question 13

What is m_l ?

Answer 13

The magnetic quantum number

Question 14

What is l?

Answer 14

The orbital quantum number

Question 15

State how we separate the wavefunction u(r)

Answer 15

u(r) = u(r,θ,Φ) = R(r) (Θ(θ) Φ(ϕ)

Question 16

What can we say about the total energy of the wave funciton for a central potential?

Answer 16

It only depends on the radial equation because it is the only one involting V(r) following the separation of variables

Question 17

What can we say about the solution of the potential?

Answer 17

V(r) is proportional to 1/r

Question 18

Describe the proposed solution for solving Φ(ϕ)

Answer 18

Try Φ = A exp(im_lϕ)

Φ’’ + (m_l)^2 Φ = 0

Question 19

What conditions can we put on Φ which is part of the wavefunction?

Answer 19

• Φ is continuous
• u(r) is allowed to be complex
• Φ is single valued -> doesn’t change around 2pi (it is well behaved)