Normalization of ψ

A wavefunction $\psi(x)$ is only physically meaningful if the total probability of finding the particle somewhere equals exactly one:

\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1.

This requirement is called normalization. It follows directly from the Born rule: if $|\psi(x)|^2$ gives the probability density, the entire probability density must integrate to one over all space.

The normalization constant

In practice, we often start with an unnormalized function $\tilde\psi(x)$ and write

\psi(x) = A \, \tilde\psi(x),

where $A$ is a real positive constant chosen so that $\psi$ is normalized. Substituting:

\int |\psi(x)|^2 \, dx = A^2 \int |\tilde\psi(x)|^2 \, dx = 1 \quad \Longrightarrow \quad A = \frac{1}{\displaystyle\sqrt{\int |\tilde\psi(x)|^2 \, dx}}.

All that remains is to compute the integral of $|\tilde\psi|^2$ and take the reciprocal square root.

A worked example

Consider the trial wavefunction defined on the interval $[0, L]$ :

\tilde\psi(x) = x(L - x), \qquad 0 \le x \le L,

and $\tilde\psi = 0$ outside that interval. We need

\int_0^L \big[x(L-x)\big]^2 dx = \int_0^L \big(x^2 L^2 - 2x^3 L + x^4\big) dx.

Integrating term by term:

= L^2 \cdot \frac{L^3}{3} - 2L \cdot \frac{L^4}{4} + \frac{L^5}{5} = \frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5} = L^5 \left(\frac{10 - 15 + 6}{30}\right) = \frac{L^5}{30}.

Setting $L = 1$ gives the integral $\frac{1}{30}$ , so the normalization constant is

A = \frac{1}{\sqrt{1/30}} = \sqrt{30}.

Substituting back confirms $\psi(x) = \sqrt{30}\, x(1-x)$ is normalized:

30 \int_0^1 x^2(1-x)^2\, dx = 30 \cdot \frac{1}{30} = 1. \checkmark

Try it

This is a numerical exercise — your code should return a number. Given $\psi(x) = A \cdot x(1-x)$ on $[0, 1]$ and zero elsewhere, compute the positive normalization constant $A$ .

Run your code to see the quantum state.