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Wavefunction & Probability Density (Normalisation)

Key idea: Use |ψ|^2 as a probability density and calculate normalisation constants for square and sinusoidal wavefunctions (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Interpret wavefunctions, probability density and superposition.
  • Apply uncertainty and infinite-square-well energy quantisation.

1. Definitions (Must Know)

  • Wavefunction, ψ(x): a mathematical function describing the quantum state of a particle (in 1D position representation).
  • Probability density, |ψ(x)|²: probability per unit length of finding the particle at position x.
  • Normalisation condition: the total probability of finding the particle somewhere is 1.

For a small interval dx and for the whole position axis respectively: P(x → x + dx) = |ψ(x)|² dx ∫_(-∞)^∞|ψ(x)|² dx = 1

2. Key Ideas (What Earns Marks)

Wavefunction and probability densityA sinusoidal wavefunction changes sign across a one-dimensional box, while its squared magnitude forms two non-negative probability-density lobes.ψ(x)0L|ψ(x)|²area = probability0L
Scroll diagram horizontally to read all labels.
The wavefunction may be positive or negative, but the probability density |ψ|² is non-negative. Probability in an interval is the area under |ψ|² over that interval.
  • |ψ|² is a density, not a probability:
    • |ψ|² has units of m⁻¹ in 1D, so that |ψ|² dx is dimensionless.
  • Probabilities come from integrals over the stated interval.
  • Normalisation constants make the total integral equal to 1.

P(a ≤ x ≤ b) = ∫ₐ^b|ψ(x)|² dx ∫|ψ|² dx = 1

  • The syllabus expects normalising:
    • a square wavefunction (constant in a region)
    • a sinusoidal wavefunction (standing-wave form)

3. Detailed Explanations

A. Why we normalise

If |ψ(x)|² dx is a probability, then the probability of finding the particle anywhere must be 1: ∫_(-∞)^∞ |ψ(x)|² dx = 1

If a wavefunction is given with an unknown constant A, you find A by applying the condition above over the region where ψ ≠ 0.

B. Normalising a square wavefunction (constant)

A common exam model is:

ψ(x) = { A 0 < x < L; 0 otherwise

Normalisation:

∫_(-∞)^∞|ψ|² dx = ∫₀^L |A|² dx; = |A|²L = 1; |A| = 1/(square root of L)

C. Normalising a sinusoidal wavefunction (standing wave)

Another common model is:

ψ(x) = { A sin((nπ x)/L) 0 < x < L; 0 otherwise

Use the identity sin² (·) averages to 1/2 over full half-waves. The integral result you need is: ∫₀^L sin² ((nπ x)/L)dx = L/2

So:

1 = |A|²∫₀^L sin² ((nπ x)/L)dx; = |A|²(L/2); |A| = square root of (2/L)

4. Common Mistakes

  • Treating ψ as a probability (it can be negative; use |ψ|²).
  • Forgetting to square A when normalising.
  • Integrating over the wrong limits (use the region where ψ ≠ 0).
  • Forgetting dx (probability needs |ψ|² dx or an integral).

5. Exam Tips

  • Write the normalisation condition first: ∫ |ψ|² dx = 1.
  • State the limits clearly (e.g. “ψ = 0 outside 0 < x < L”).
  • If the wavefunction is real-valued, |ψ|² = ψ² (saves time).

6. Worked Examples

Modelled example 1

Normalising a square wavefunction

Core

Problem

For ψ(x) = A on 0 < x < 2.0 m and ψ(x) = 0 otherwise, find the positive normalisation constant A.
Study the worked solution

  1. Write the normalisation condition

    Method

    The integral of |ψ|² over all positions must equal one.

    Reason

    Total position probability is one, and the wavefunction is non-zero only between 0 and 2.0 m.

    Working

    1 = ∫₀^(2.0)A² dx

  2. Evaluate the constant integral

    Method

    1 = A²(2.0 m).

    Reason

    A is constant across the interval.

    Working

    A² = 1/(2.0 m)

  3. Choose the positive root

    Method

    A = 0.707 m^(-1/2).

    Reason

    The requested positive constant has units whose square gives probability density in m⁻¹.

    Working

    A = 1/(square root of (2.0 m)) = 0.707 m^(-1/2)

Guided practice 2

Normalising a sinusoidal wavefunction

About 6 min

Problem

For ψ(x) = A sin(π x/L) on 0 < x < L and ψ(x) = 0 otherwise, find the positive constant A in terms of L.

Try this before viewing the solution

Hints

Hint 1: set up the density integral
Write 1 = A²∫₀^L sin² (π x/L) dx.
Hint 2: use the full-half-wave result
Over this interval, the sine-squared integral is L/2.
View solution step by step

  1. Apply normalisation

    Method

    Integrate |ψ|² across the non-zero region.

    Reason

    The amplitude must be squared before it becomes probability density.

    Working

    1 = A²∫₀^L sin² ((π x)/L)dx

  2. Evaluate the shape integral

    Method

    The integral is L/2.

    Reason

    sin² (π x/L) spans one complete half-wave between its boundary nodes.

    Working

    1 = A²(L/2)

  3. Solve for amplitude

    Method

    A = square root of (2/L).

    Reason

    Take the positive root requested by the problem.

    Working

    A² = 2/L ⇒ A = square root of (2/L)

Common misconception 3

Wavefunction sign and probability density

Find and correct the mistake

Learner claim

A learner says changing a normalised real wavefunction from ψ(x) to -ψ(x) creates negative probabilities and changes the position distribution. Diagnose the claim.

Try this before viewing the solution

Effect on position probability density

View solution step by step

  1. Identify the measured quantity

    Method

    Position probability density is |ψ(x)|², not ψ(x).

    Reason

    A wavefunction can be negative or complex, whereas a probability density cannot be negative.

    Working

    ρ(x) = |ψ(x)|²

  2. Apply the sign change

    Method

    The probability density is unchanged.

    Reason

    Multiplying the whole wavefunction by -1 leaves its modulus squared unchanged.

    Working

    |-ψ(x)|² = |ψ(x)|²

  3. Check normalisation

    Method

    Total probability remains one.

    Reason

    The integrand in the normalisation condition is identical.

    Working

    ∫|-ψ|²dx = ∫|ψ|²dx = 1

Challenge 4

Probability in an interval

Minimal support

Independent transfer

A particle in a one-dimensional box has normalised wavefunction ψ(x) = square root of (2/L) sin(π x/L) for 0 < x < L. Find the probability that the particle is in the left half, 0 < x < L/2.

Try this before viewing the solution

Hints

Hint 1: integrate density, not amplitude
Form |ψ|² and integrate only over 0 to L/2.
View solution step by step

  1. Form the interval probability

    Method

    P is the integral of |ψ|² over the left half.

    Reason

    Probability density must be integrated across the requested interval.

    Working

    P = 2/L∫₀^(L/2) sin² ((π x)/L)dx

  2. Evaluate the integral

    Method

    The unscaled sine-squared integral over the left half is L/4.

    Reason

    Apply the sine-squared antiderivative at the two stated limits.

    Working

    [x/2-L/4π sin((2π x)/L)]₀^(L/2) = L/4

  3. Obtain probability

    Method

    P = 1/2.

    Reason

    The ground-state probability density is symmetric about x = L/2.

    Working

    P = (2/L)(L/4) = 1/2

7. Mind Stretchers

Mind stretcher 1: Example: Normalising a cosine standing waveExtension

Within -L/2 < x < L/2, ψ(x) = A cos((π x)/L).

Outside that interval, ψ(x) = 0. Find A.

Show Answer

Use the result ∫_(-L/2)^(L/2) cos² ((π x)/L)dx = L/2: 1 = A²(L/2); A = square root of (2/L)

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