Normalisation of the Wavefunction

The wavefunction ψ itself is not measured directly. Predictions come from the probability density |ψ|².

1. Probability density

In one dimension, the probability of finding the particle between x and x + dx is:

dP = |ψ(x)|² dx

In three dimensions:

dP = |ψ(r)|² dV

Use the normalization diagram to anchor your setup: the total shaded area under |ψ(x)|² over the allowed domain must be exactly 1.

2. Normalisation condition

The particle must be somewhere, so total probability must be 1:

∫_(-∞)^∞ |ψ(x)|² dx = 1

If the particle is only defined on a finite interval (e.g. 0 < x < L), then normalise over that interval instead:

∫₀^L |ψ(x)|² dx = 1

3. Normalisation workflow

If your solution has an unknown constant:

ψ(x) = A f(x)

then

1 = ∫ |ψ|² dx = |A|²∫ |f|² dx ⇒ |A| = 1/(square root of (∫ |f|² dx)).

Choosing A real and positive is a convenient convention. Multiplying the whole state by a constant phase e^iφ does not change |ψ|² or any measurement probability. In one dimension, a normalised wavefunction has units m^(-1/2) so that |ψ|²dx is dimensionless.

Decaying-exponential example

Suppose ψ(x) = Ae^(-ax) for x ≥ 0 and ψ(x) = 0 for x < 0, where a > 0.

1 = ∫₀^∞ |A|²e^(-2ax) dx = |A|²/2a ⇒ |A| = square root of 2a .

4. Common traps

1. Using ψ² instead of |ψ|².
2. Normalising over the wrong interval (wrong limits).
3. Forgetting that | Aψ |² = |A|²|ψ|² (you square the constant too).
4. Treating normalisation as optional when the question asks for a physical wavefunction.

5. Quick check

A. If ψ is normalised, what is ∫ |ψ|² dx?

B. If you multiply a normalised ψ by 2, what happens to the total probability?