The Wavefunction and Schrodinger

Question 1

TDSE with meanings of terms

Answer 1

(-ħ²/2m ∂²/∂x² + V(x,t) )Ψ(x,t)=iħ ∂/∂t Ψ(x,t)

Space derivative term is kinetic energy
V term is potential energy

Question 2

TDSE in 3D

Answer 2

(-ħ²/2m ∇² + V(𝗿,t) )Ψ(𝗿,t)=iħ ∂/∂t Ψ(𝗿,t)

Question 3

Born interpretation of the wavefunction

Answer 3

|Ψ(x,t)|² is probability density
aka |Ψ|² dx is probability that particle is between x and x + dx at time t

Question 4

5 properties of the wavefunction

Answer 4

1. Normalization: ∫|Ψ(𝗿,t)|² dV = 1 (over all space)
2. Ψ is single-valued, as |Ψ|² is physical. Only thing that can vary is phase
3. Ψ is finite everywhere (infinite would mean certainty of particle being at that point)
4. Ψ is continuous (no sudden changes of probability) and smooth (continuous derivative, but only if V(x) doesnt have an infinite discontinuity)
5. Ψ is fragile, meaning measuring any physical property of the particle changes Ψ, changing future measurements of all properties.

Question 5

Solve TDSE through separation of variables

Answer 5

(-ħ²/2m ∂²/∂x² + V(x,t) )Ψ(x,t)=iħ ∂/∂t Ψ(x,t)
assume Ψ(x,t) = ψ(x)T(t)
therefore ∂²Ψ/∂x² = T d²ψ/dx²
and ∂Ψ/∂t = ψ dT/dt
sub into TDSE and divide by T(t)ψ(x)
1/ψ(x) (-ħ²/2m d²ψ/dx² + V(x)ψ(x) ) = iħ/T(t) dT/dt
each side is equal to E
space: (-ħ²/2m d²/dx² + V(x)) ψ(x) = Eψ(x)
time: iħ dT/dt = ET(t) solved by T(t) = Aexp( -iEt/ħ )
general solution: Ψ(x,t) = ΣAᵢ ψᵢ(x) exp(-iEᵢt/ħ)

Question 6

TISE with explanations

Answer 6

(-ħ²/2m ∂²/∂x² + V(x) )Ψ(x) = E ψ(x)

Eigenfunctions Ψ with eigenvalues E.
Derivative term is kinetic energy, V term is potential energy.
Usually has discrete solutions, known as stationary states.