Schrödinger's

Question 1

TISE

Answer 1

-ħ²/2m * d²Ψ/dx² + ΨV(x) = ΨE

Question 2

general solution for an infinite potential well

Answer 2

Ψ(x) = Acos(kx) + Bsin(kx)
k: wavenumber

Question 3

General way to solve the TISE

Answer 3

for the limits of 0 –> a
apply the boundary conditions so Ψ(a) = 0
hence k = nπ/a (from sin(ka) = 0)

since V(x) = 0, write TISE as:
d²Ψ/dx² = -2m/ħ² * E * Ψ
compare to d²Ψ/dx² = -k² * Ψ
hence: k² = 2m/ħ² * E

sub in for k and solve for E
E = n²ħ²π² /2ma²

Question 4

define the tunnelling probability

Answer 4

T = |Aₜ|²/|Aᵢ|²
T: tunnelling probability
Aₜ: transmission coefficient
Aᵢ: incident coefficient

Question 5

derive the energy stored across one wavelength of a wave on a string

Answer 5

draw a right-angle triangle. δx is the bottom. δy is the side, and δs is the hypotenuse. the string’s original length is δx so its m = µδx.

K.E = ½ µδx (∂y/∂t)²

P.E = T * (δs - δx)
δs = (δx² + δy²)^½ = δx (1 + (δy/δx)²)^½
hence: δs -δx ≈ ½(δy/δx)² δx
P.E = T * ½(δy/δx)² δx = ½v²µ(δy/δx)² δx (from v² = T/µ)

dE = K.E + P.E = ½µ δx[(∂y/∂t)² v²(δy/δx)²]
sub in y= Asin(kx-ωt) then integrate. (evaluated at t=0, between the limits 0 and λ)

E= ½µA²ω²λ

Question 6

use the method of separation constants on the 2D wave equation

Answer 6

∂²f/∂y² + ∂²f/∂x² = 1/v² ∂²f/∂t²
assume a solution of f(x,y,t) = f(x)f(y)f(t) and sub that in
fᵧfₜ ∂²fₓ/∂x² + fₓfₜ ∂²fᵧ/∂y² = 1/v² fₓfᵧ ∂²fₜ/∂t²
divide through by fₓfᵧfₜ

1/fₓ ∂²fₓ/∂x² + 1/fᵧ ∂²fᵧ/∂y² = 1/v² 1/fₜ ∂²fₜ/∂t²
must also be = -k² since the left side has no t dependence

1/fₓ ∂²fₓ/∂x² = -k² - 1/fᵧ ∂²fᵧ/∂y²
which must also = -kₓ² as the left side has no y dependence
1/fᵧ ∂²fᵧ/∂y² = kₓ² -k² = -kᵧ²

hence there are 3 O.D.Es:
∂²fₓ/∂x² = -kₓ² fₓ
∂²fᵧ/∂y² = -kᵧ² fᵧ
1/v² ∂²fₜ/∂t² = -k² fₜ
and k² = kₓ² + kᵧ²

this results in:
fₓ = Aₓcos(Kₓx) +Bₓsin(kₓx)
fᵧ = Aᵧcos(kᵧy) + Bᵧsin(kᵧy)
fₜ = Aₜcos(ωt) + Bₜsin(ωt)

Question 7

TDSE

Answer 7

-ħ²/2m * ∂²Ψ/∂x² + ΨV(x,t) = iħ ∂Ψ/∂t

Question 8

tunnelling probability dependencies

Answer 8

1. barrier width
2. particle mass
3. barrier height compared to particle energy, V(sub B) - E

Question 9

TDSE general solution

Answer 9

Ψ(x,t) = A * sin(nπx/a) * e^(-i/ħ Et)
A: normalisation constant