$$\require{cancel}$$

# 4.1: Infinite Potential Well

Consider a particle of mass $$m$$ and energy $$E$$ moving in the following simple potential: $V(x) = \left\{\begin{array}{lcl} 0&\hspace{1cm}&\mbox{for }0\leq x\leq a\\[0.5ex] \infty&&\mbox{otherwise} \end{array}\right..$ It follows from Equation ([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 $\label{e5.4} \psi(0) = \psi(a) = 0.$ Furthermore, it follows from Equation ([e5.2]) that $$\psi$$ satisfies $\label{e5.5} \frac{d^{\,2} \psi}{d x^{\,2}} = - k^{\,2}\,\psi$ in this region, where $\label{e5.6} k^{\,2} = \frac{2\,m\,E}{\hbar^{\,2}}.$ 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 ([e5.4]).

The solution to Equation ([e5.5]), subject to the boundary conditions ([e5.4]), is $\psi_n(x) = A_n\,\sin(k_n\,x),$ where the $$A_n$$ are arbitrary (real) constants, and

$\label{e5.8} k_n = \frac{n\,\pi}{a},$ for $$n=1,2,3,\cdots$$. Now, it can be seen from Equations ([e5.6]) and ([e5.8]) that the energy $$E$$ is only allowed to take certain discrete values: that is, $\label{eenergy} E_n = \frac{n^{\,2}\,\pi^{\,2}\,\hbar^{\,2}}{2\,m\,a^{\,2}}.$

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

$\int_0^a \psi_n(x)\,\psi_m(x)\,dx = \delta_{nm}.$ It is easily demonstrated that this is the case, provided $$A_n = \sqrt{2/a}$$. Hence,

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

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

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

# Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)
