For a particle in an infinite square well of length $L$ centered at the origin (see Figure 2), the potential $V(x)$ is given by

$$\begin{array}{}\text{(1)}& V(x)=\{\begin{array}{l}0,\text{ if }-L/2\le x\le L/2\\ \mathrm{\infty },\text{ otherwise.}\end{array}\end{array}$$

Since the particle cannot be found outside the well, the wave function $\mathrm{\Psi }(x)$ outside the well is set to zero with

$$\mathrm{\Psi }(x)=0\text{ when }x<-L/2\text{ or }x>L/2$$

In order to have $\mathrm{\Psi }(x)$ remain continuous, the Dirichlet boundary conditions are set as

$$\begin{array}{}\text{(2)}& \mathrm{\Psi }(-L/2)=\mathrm{\Psi }(L/2)=0\end{array}$$

As with the asymmetric well (see Asymmetric Well), no requirement for continuity is made on $d\mathrm{\Psi }(x)/dx$ at the boundary.¹

In the majority of introductory textbooks,² the general solution to the Time-Independent Schrodinger equation for the potential described by Eq.$\text{(1)}$ is given as

$$\begin{array}{}\text{(3)}& \mathrm{\Psi }(x)=A\cos kx+B\sin kxA,B\in \mathbb{R}\end{array}$$

with

$$\begin{array}{}\text{(4)}& k=\sqrt{\frac{2mE}{{\hslash }^2}}\end{array}$$

Applying the boundary conditions Eq.$\text{(2)}$ to Eq.$\text{(3)}$ results in

$$\begin{array}{}\text{(5)}& A\cos k(\frac{L}{2})+B\sin k(\frac{L}{2})=0\end{array}$$

Satisfying the requirements for Eq.$\text{(5)}$ results in two unique solutions given by

$${\mathrm{\Psi }}_{n=\text{odd}}(x)=(2/L{)}^{1/2}\cos (\frac{n\pi }{L}x)n=1,3,5,\dots$$

and

$${\mathrm{\Psi }}_{n=\text{even}}(x)=(2/L{)}^{1/2}\sin (\frac{n\pi }{L}x)n=2,4,6,\dots$$

with

$$\begin{array}{}\text{(6)}& k_n=\frac{n\pi }{L}\end{array}$$

From Eq.$\text{(4)}$ and Eq.$\text{(6)}$, the energy $E$ is quantized as

$$E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}n=1,2,3,\dots$$