Normalizing the Eigenfunctions

The energy eigenfunctions for a particle confined to the interval $0 \le x \le L$ take the un-normalized form

$\psi_n(x) = A\sin\!\left(\frac{n\pi x}{L}\right), \qquad n = 1, 2, 3, \ldots$

The constant $A$ is not yet determined by the time-independent Schrödinger equation alone — the equation is linear, so any scalar multiple of a solution is also a solution. To make the wavefunction represent a genuine probability, we impose the normalization condition

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

Because $\psi_n$ is zero outside $[0, L]$ , the integral collapses to

$A^2\int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right)dx = 1.$

Evaluating the integral

Apply the half-angle identity $\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$ :

= \frac{1}{2}\int_0^L dx - \frac{1}{2}\int_0^L \cos\!\left(\frac{2n\pi x}{L}\right)dx.$$ The first piece gives $L/2$. The second piece is $$\frac{1}{2}\cdot\frac{L}{2n\pi}\left[\sin\!\left(\frac{2n\pi x}{L}\right)\right]_0^L = \frac{L}{4n\pi}\bigl(\sin(2n\pi) - \sin(0)\bigr) = 0,$$ because $\sin(2n\pi) = 0$ for every positive integer $n$. Therefore $$A^2 \cdot \frac{L}{2} = 1 \quad\Longrightarrow\quad A = \sqrt{\frac{2}{L}}.$$ The fully normalized eigenfunctions are $$\boxed{\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right)}, \qquad 0 \le x \le L.$$ <Callout type="tip"> The factor $\sqrt{2/L}$ has units of $\text{m}^{-1/2}$ in SI, so that $|\psi_n|^2$ has units of $\text{m}^{-1}$ — a probability **density**, not a dimensionless probability. The probability of finding the particle in a small interval $dx$ is $|\psi_n(x)|^2\,dx$, which is dimensionless. </Callout> ## Physical interpretation Normalization is not just bookkeeping. Once $A$ is fixed, $|\psi_n(x)|^2$ is an honest probability density: the particle is most likely to be found where the sine squared is large (the antinodes) and has zero probability of being at the walls $x = 0$ and $x = L$, where $\psi_n = 0$. You can verify orthogonality as a bonus: a similar integral shows $\int_0^L \psi_m^*\psi_n\,dx = 0$ whenever $m \neq n$, so the normalized eigenfunctions form an **orthonormal** set — the quantum analogue of perpendicular unit vectors. ## Try it This is a numerical exercise — your code should `return` a number, not a circuit. Compute the normalization constant $A = \sqrt{2/L}$ for a box of length $L = 1\,\text{m}$. <RunnableExample />