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Time-Independent Schrödinger Equation

Key idea: H3 Quantum Mechanics: the time-independent Schrödinger equation (TISE), how solutions behave for E>V vs E<V, and the boundary conditions used in 1D problems.

  • Advanced Physics
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Learning objectives

  • Use state functions, boundary conditions, stationary states, and measurement to analyse introductory quantum systems.

The time-independent Schrödinger equation (TISE) is the main tool for 1D “particle in a potential” problems. For many potentials, only certain energies E produce acceptable (normalisable) solutions — that’s where quantisation comes from.

Prerequisites + Path

1. Equation in one dimension

The standard 1D form is:

-(ħ²/2m)d²ψ/dx² + V(x)ψ = Eψ

where:

  • ψ(x) is the (spatial) wavefunction
  • V(x) is the potential energy (often written as U(x) in some notes)
  • E is the energy eigenvalue

2. Meaning of a stationary state

If the potential is time-independent, solutions can be separated as:

Ψ(x,t) = ψ(x)e^(-iEt/ħ)

So for a single energy eigenstate:

|Ψ(x,t)|² = |ψ(x)|²

The probability density does not change with time (hence “stationary”).

3. Solutions in a constant-potential region

In many H3 problems, you solve the equation in regions where V(x) is constant (say V(x) = V₀).

Classically allowed region (E > V₀)

d²ψ/dx² = -k²ψ where k² = (2m(E-V₀))/ħ²

General solution (any equivalent form is fine):

ψ(x) = A sin(kx) + B cos(kx)

Classically forbidden region (E < V₀)

d²ψ/dx² = κ²ψ where κ² = (2m(V₀-E))/ħ²

General solution:

ψ(x) = Ce^(κ x) + De^(-κ x)

This is the mathematical basis of tunnelling: even when E < V, the wavefunction can extend into the “forbidden” region (usually as a decaying exponential).

The two-panel comparison below is the standard region-test in H3 working: use oscillatory forms for E > V and exponential forms for E < V.

Constant-potential-region solution typesTwo-panel comparison of oscillatory wavefunction behaviour for classically allowed energy and exponential behaviour for classically forbidden energy.E > V (classically allowed)E < V (classically forbidden)VEψ = A sin(kx) + B cos(kx)VEψ = C e^(κx) + D e^(-κx)
Scroll diagram horizontally to read all labels.
In a constant-potential region, E > V gives sinusoidal solutions while E < V gives exponential solutions (with unphysical growing terms rejected by boundary conditions).

4. Physical conditions

In typical 1D H3 questions, acceptable wavefunctions must be:

  1. finite everywhere
  2. single-valued
  3. continuous
  4. have continuous dψ/dx wherever V(x) is finite
  5. normalisable (so ∫ |ψ|² dx is finite)
Where does quantisation come from?

For bound systems, you must satisfy all conditions at once: solve in each region, match at boundaries, and keep the wavefunction normalisable. Typically this is only possible for discrete values of E.

5. Problem workflow

  1. Sketch V(x) and mark regions where it is constant/simple.
  2. Write the correct general solution in each region (sin/cos vs exponentials).
  3. Apply boundary/matching conditions to relate constants and restrict E.
  4. Normalise if asked.

6. Quick check

A. If V(x) = V₀ is constant and E < V₀, which term must be rejected as x → + ∞?

Answer

Reject the growing exponential e^(+κ x) and keep the decaying term e^(-κ x).

Continue with the next resource in this course.

Course and syllabus information
Course
Advanced Physics
Edition
Advanced Physics