[NOTES/QM-13007] Particle in A Box

The energy levels of a particle in one dimensional infinite well \begin{equation} V(x) = \begin{cases} 0, & 0\le x \le L \\ \infty & \text{outside} \label{EQ01} \end{cases} \end{equation} can be found by solving the Schr\"{o}dinger equation for $0\le x\le L$. \begin{equation} -\frac{\hbar^2}{2m}\DD[u(x)]{x} = E u(x), \qquad 0\le x\le L. \end{equation} Here the particle is like a free particle and the most general solution can written as \begin{equation} u(x) = A \sin kx + B \cos kx, \qquad k^2 =\frac{2mE}{\hbar^2}.\label{EQ02} \end{equation} Out side the box, the potential is infinity and the solution vanishes: \begin{equation} u(x) = 0,\qquad \text{ if } x< 0, \text{ or } x >L.\label{EQ03} \end{equation} The boundary conditions to be imposed on the solution are \begin{equation} u(0)= u(L) = 0 ,\label{EQ04} \end{equation} and no restriction on the derivatives at the boundary points $x=0 , x=L$. This gives \begin{eqnarray} u(0)=0 &\Rightarrow& B=0, \label{EQ05}\\ u(L)=0 &\Rightarrow& \sin kL =0.\label{EQ06} \end{eqnarray} because \(A\ne0\). The solutions of this equation are $k_n=n \pi/L , n=1,2,\ldots$. The energy levels are given by \begin{equation} E_n = \frac{\hbar^2 k_n^2}{2m}, = \frac{\hbar^2 n^2\pi^2}{2mL^2}\label{EQ07} \end{equation} and the corresponding normalized wave functions are \begin{equation}\label{EQ08} u_n(x) =\begin{cases} \sqrt{\frac{2}{L} }\sin\big(\frac{n\pi x}{L}\big) & 0\le x \le L\\ 0 & x < 0 \text{ or } x >L. \end{cases} \end{equation} and $n$ takes all positive integral values.{\it It should be noted that for $k=0$ the solution vanishes identically and therefore $n=0$ is unacceptable.} The wave functions in \eqref{EQ08} obey the normalization \begin{equation} \int_0^L u_n(x) u_m(x) \, dx = \delta_{mn}. \end{equation}