Quantum Harmonic Oscillator

Question 1

Derive energy of the QHO.

Answer 1

Consider mass m on a spring.
F = -kx
Use newtons 2nd law
mẍ = -kx
Which has solutions
x = Asin ωt + Bcos ωt, ω = √(k/m)
Potential energy
V(x) = 1/2 kx²
Total energy
E = p² / 2m + 1/2 kx² = 1/2 mω²(A²+B²) = const.
Replace k, k=mω²
E = p² / 2m + 1/2 mω²x²

Question 2

TISE of QHO

Answer 2

Ĥ = p̂²/2m + 1/2 mω²x̂²
TISE is Ĥψ
-ħ/2m d²ψ/dx² + 1/2 mω²x²ψ = Eψ
with boundary conditions ψ(±∞) = 0 (as integral of |ψ|² over all x must be finite)

Question 3

Solution of QHO

Answer 3

https://online.manchester.ac.uk/webapps/blackboard/execute/content/file?cmd=view&content_id=_15161021_1&course_id=_78887_1&framesetWrapped=true
see week 3 of notes

Question 4

Properties of QHO solution

Answer 4

https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#/media/File:HarmOsziFunktionen.png

Parabolic curve shows potential energy as a function of displacement, x
Horizontal lines show energy levels of the solutions. Lowest, zero-point energy E = E₀ = 1/2 ħω. A particle in a potention can nver be at rest. Successive energy spacing is constant ħω.
Other curves are individual solutions. nth excited level has n nodes.
QHO can transition between adjacent states by emission or absorbtion of a photon with energy ħω.
Remember that all solutions are orthogonal. Easy to show if n and m are odd and even. Must integrate fully if both are odd or both are even.

Question 5

Graph of interatomic potential for diatomic molecules

Answer 5

https://en.wikipedia.org/wiki/Morse_potential

Question 6

Taylor expansion around equilibrium position for diatomic molecules

Answer 6

V(r) ≃ V(r₀) + (r-r₀) ∂V/∂r + ½(r-r₀)² ∂²V/∂r²
= -V₀ + 1/2 kx²

Question 7

Total energy at minimum potential for diatomic molecules

Answer 7

E = potential + kinetic
E = -V₀ + 1/2 kx² + p₁² / 2m₁ + p₂² / 2m₂
Centre of mass frame, total momentum is zero, so p₁=p₂=p
E = -V₀ + 1/2 kx² + p² / 2μ, where μ = m₁m₂/m₁+m₂ (reduced mass)
Constants dont affect the physics, so can remove that term.
E = p² / 2μ + 1/2 kx²
QHO good approx for diatomic molecules at small displacements. Expect photons at classical frequency ω = √(k/μ) to be emitted/absorbed.

Question 8

Potential energy of 2D SHO

Answer 8

V(x, y) = 1/2 k (x²+y²)
= 1/2 mω² (x² + y²)

Question 9

Kinetic energy of 2D SHO

Answer 9

T = p²/2m = pₓ² + pᵧ² /2m

Question 10

TISE of 2D QHO

Answer 10

T̂ + V̂
-ħ²/2m (∂²ψ(x, y)/∂x² + ∂²ψ(x, y)/∂y²) + 1/2 mω²(x² + y²)ψ(x, y) = Eψ(x, y)