For a particle in an infinite square well with its left-hand edge at the origin (see Figure 1.), the potential $V(x)$ is defined as

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

Once again, the wave function $\mathrm{\Psi }(x)$ outside the well is set to zero with

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

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

$$\begin{array}{}& \mathrm{\Psi }(0)=\mathrm{\Psi }(L)=0\end{array}$$

No restriction for continuity is made on $d\mathrm{\Psi }(x)/dx$ at the boundary.¹

In the majority of introductory textbooks², the general solution to the Time-Independent Schrodinger equation for the potential given by Eq.$\text{(1)}$ is given as

$$\mathrm{\Psi }(x)=A\cos kx+B\sin kxA,B\in \mathbb{R}$$

with

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

Applying the boundary condition $\mathrm{\Psi }(0)=0$ eliminates the constant $A$ which leaves

$$\begin{array}{}& \mathrm{\Psi }(x)=B\sin kx\end{array}$$

In order for $\mathrm{\Psi }(L)=0$,

$$kL=n\pi$$

thereby giving

$$k_n=\frac{n\pi }{L},\text{ with }n=1,2,3,\dots$$

From Eq.$\text{(2)}$, the energy E is quantized as

$$E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}$$

The constant $B$ is found by normalizing $\mathrm{\Psi }(x)$ as

$$\begin{array}{}& {\int }_0^L|\mathrm{\Psi }(x){|}^{2}dx=1\end{array}$$

which gives

$$\begin{array}{}& B=\sqrt{\frac{2}{L}}\end{array}$$

Finally, the wave solution for the particle in the infinite square well is given by

$${\mathrm{\Psi }}_n(x)=\sqrt{\frac{2}{L}}\sin (\frac{n\pi }{L}x)n=\pm 1,\pm 2,\pm 3,\dots$$