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The Schrodinger Equation And Wave Function

Key idea: Understand what the wavefunction represents, how probability density works, and how to normalise simple wavefunctions; Schrödinger equation is included as optional context (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use photon energy and momentum and analyse the photoelectric effect.
  • Apply de Broglie wavelength and wave-particle evidence.
  • Interpret wavefunctions, probability density and superposition.
  • Apply uncertainty and infinite-square-well energy quantisation.
  • Analyse atomic energy levels and emission or absorption spectra.
Extension scope

The meanings of ψ and |ψ|², normalisation and superposition are core. The Schrödinger equation itself is included here as mathematical context beyond the named 9478 outcomes.

1. Definitions (Must Know)

A. Wavefunction, ψ

The wavefunction, ψ, is a mathematical function used to represent the state of a particle (e.g. an electron).

In general, ψ can be positive/negative (and often complex). It is not itself a probability.

B. Probability density, |ψ|²

The probability density is: |ψ(x,t)|²

In 1D, the probability of finding the particle between x and x + dx is: |ψ(x,t)|² dx

C. Normalisation

A wavefunction is normalised if the total probability of finding the particle somewhere is 1.

In 1D: ∫_(-∞)^∞|ψ(x,t)|² dx = 1

Where this fits in the syllabus

The 9478 learning outcomes require using |ψ|² as probability density and calculating normalisation factors for simple wavefunctions. The Schrödinger equation itself is not explicitly required, so it is treated as optional context in this lesson.

2. Key Ideas (What Earns Marks)

  • In quantum physics, you do not usually model motion with a definite trajectory x(t); instead you predict probabilities using |ψ|².
  • Probabilities come from areas under the |ψ|² curve: P(x₁ ≤ x ≤ x₂) = ∫_x₁^x₂|ψ(x,t)|² dx
  • Always normalise (or use a given normalised form) before calculating probabilities.
  • For “particle in a box” problems: ψ = 0 outside the well, and boundary conditions fix the allowed standing waves.

3. Detailed Explanations

A. Classical vs quantum descriptions

In classical mechanics, a particle’s state can be described by its trajectory x(t). From this, you can determine:

  • v = dx/dt,
  • momentum p = mv,
  • kinetic energy K = 1/2 mv².

For microscopic particles, experiments like single-particle interference suggest we should not treat x(t) as always well-defined. Instead, we use a wavefunction and extract measurable predictions from |ψ|².

B. Probability from |ψ|² (1D)

If a normalised wavefunction is given, the probability of finding the particle in a region is the integral of |ψ|² over that region.

C. How normalisation works (workflow)

  1. Write |ψ|².
  2. Integrate over the allowed region (where ψ ≠ 0).
  3. Set the integral equal to 1 and solve for the constant.

4. Common Mistakes

  • Treating ψ as a probability (it is |ψ|² that links to probability).
  • Forgetting the dx (probability is an integral/area, not the value at a point).
  • Normalising over the wrong region (e.g. not restricting to where the particle is allowed).

5. Exam Tips

  • If a wavefunction includes an unknown constant (e.g. A), normalise first before doing anything else.
  • Use units as a quick sanity check: in 1D, |ψ|² has units of m⁻¹ so that |ψ|²dx is dimensionless.
  • For a particle in a box from 0 to L, your limits are 0 to L.

6. Worked Examples

Modelled example 1

Normalising a constant wavefunction in a box

Core

Problem

A particle is confined to 0 ≤ x ≤ L with ψ = A inside and zero outside. Find the real positive normalisation constant A.
Study the worked solution

  1. Choose the region

    Method

    Integrate only from 0 to L.

    Reason

    The wavefunction is zero everywhere else.

    Working

    ∫₀^L|ψ|²dx = 1

  2. Square the amplitude

    Method

    |ψ|² = |A|² is constant.

    Reason

    Probability density, not ψ itself, is normalised.

    Working

    |A|²L = 1

  3. Solve

    Method

    |A| = 1/square root of L; choose the stated real positive representative.

    Reason

    An overall phase or sign does not change |ψ|².

    Working

    A = 1/(square root of L)

Guided practice 2

Probability of being in the left half of the box

About 4 min

Problem

Using ψ = 1/square root of L in 0 ≤ x ≤ L, find P(0 ≤ x ≤ L/2).

Try this before viewing the solution

Hints

Hint 1: square before integrating
The probability density is |ψ|² = 1/L.
View solution step by step

  1. Write probability density

    Method

    Square the magnitude of the normalised wavefunction.

    Reason

    ψ is an amplitude, not a probability.

    Working

    |ψ|² = 1/L

  2. Integrate the requested region

    Method

    Use limits 0 to L/2.

    Reason

    Probability is area under the density curve.

    Working

    P = ∫₀^(L/2)(1/L)dx = 1/2

Challenge 3

Normalising the n = 1 standing wave (given shape)

Minimal support

Independent transfer

For 0 ≤ x ≤ L, let ψ = A sin(π x/L) and zero elsewhere. Find the real positive A and explain why it differs from the constant-wavefunction value.

Try this before viewing the solution

Hints

Hint 1: use the mean of sine squared
Orient with ∫₀^L sin² (π x/L)dx = L/2.
View solution step by step

  1. Set the integral

    Method

    Normalise over the box.

    Reason

    The wavefunction vanishes outside.

    Working

    A²∫₀^L sin² ((π x)/L)dx = 1

  2. Evaluate

    Method

    The sine-squared integral is L/2.

    Reason

    The standing wave is nonuniform and averages to one-half in squared magnitude.

    Working

    A²L/2 = 1

  3. Solve and compare

    Method

    A = square root of (2/L), larger than 1/square root of L.

    Reason

    The density is zero at the boundaries, so a larger peak amplitude is needed for total area one.

    Working

    A = square root of (2/L)

7. Mind Stretchers

Mind stretcher 1: Why does localisation imply momentum spread?Extension

Show Answer

To make |ψ|² concentrated in a small region of x, you need a wavefunction made from many wavelengths (many momenta). That connects to the uncertainty idea that a smaller Δ x requires a larger spread Δ p.

8. Optional (Enrichment)

A. Where the Schrödinger equation fits

The Schrödinger equation is the quantum analogue of an “equation of motion”: it determines how ψ changes in space and time for a given potential. Solving it with boundary conditions is what produces discrete energy levels in bound systems.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027