4.1: Infinite Potential Well

Consider a particle of mass \(m\) and energy \(E\) moving in the following simple potential:

\[V(x) = \left\{\begin{array}{lcl} 0&\mbox{\hspace{1cm}}&\mbox{for $0\leq x\leq a$}\\[4pt] \infty&&\mbox{otherwise} \end{array}\right.. \nonumber \]

It follows from Equation \ref{e5.2} that if \(d^2\psi/d x^2\) (and, hence, \(\psi\)) is to remain finite then \(\psi\) must go to zero in regions where the potential is infinite. Hence, \(\psi=0\) in the regions \(x\leq 0\) and \(x\geq a\). Evidently, the problem is equivalent to that of a particle trapped in a one-dimensional box of length \(a\). The boundary conditions on \(\psi\) in the region \(0<x<a\) are

\[ \psi(0) = \psi(a) = 0. \label{e5.4} \]

Furthermore, it follows from Equation \ref{e5.2} that \(\psi\) satisfies

\[ \frac{d^2 \psi}{d x^2} = - k^2\,\psi \label{e5.5} \]

in this region, where

\[ k^2 = \frac{2\,m\,E}{\hbar^2}. \label{e5.6} \]

Here, we are assuming that \(E>0\). It is easily demonstrated that there are no solutions with \(E<0\) which are capable of satisfying the boundary conditions \ref{e5.4}.

The solution to Equation \ref{e5.5}, subject to the boundary conditions \ref{e5.4}, is

\[\psi_n(x) = A_n\,\sin(k_n\,x), \nonumber \]

where the \(A_n\) are arbitrary (real) constants, and

\[ k_n = \frac{n\,\pi}{a}, \label{e5.8} \]

for \(n=1,2,3,\cdots\). Now, it can be seen from Equations \ref{e5.6} and \ref{e5.8} that the energy \(E\) is only allowed to take certain discrete values: that is,

\[ E_n = \frac{n^2\,\pi^2\,\hbar^2}{2\,m\,a^2}. \label{eenergy} \]

In other words, the eigenvalues of the energy operator are discrete. This is a general feature of bounded solutions: that is, solutions for which \(|\psi|\rightarrow 0\) as \(|x|\rightarrow\infty\). According to the discussion in Section [sstat], we expect the stationary eigenfunctions \(\psi_n(x)\) to satisfy the orthonormality constraint

\[\int_0^a \psi_n(x)\,\psi_m(x)\,dx = \delta_{nm}. \nonumber \]

It is easily demonstrated that this is the case, provided \(A_n = \sqrt{2/a}\). Hence,

\[ \psi_n(x) = \sqrt{\frac{2}{a}}\,\sin\left(n\,\pi\,\frac{x}{a}\right) \label{e5.11} \]

for \(n=1,2,3,\cdots\).

Finally, again from Section [sstat], the general time-dependent solution can be written as a linear superposition of stationary solutions:

\[\psi(x,t) = \sum_{n=0,\infty} c_n\,\psi_n(x)\,\rm e^{-{\rm i}\,E_n\,t/\hbar}, \nonumber \]

where

\[ c_n = \int_0^a\psi_n(x)\,\psi(x,0)\,dx. \label{e5.13} \]

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