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Next: The square potential barrier Up: One-dimensional potentials Previous: Introduction   Contents


The infinite potential well

Consider a particle of mass $m$ and energy $E$ moving in the following simple potential:
\begin{displaymath}
V(x) = \left\{\begin{array}{lcl}
0&\mbox{\hspace{1cm}}&\mbox...
...leq a$}\\ [0.5ex]
\infty&&\mbox{otherwise}
\end{array}\right..
\end{displaymath} (284)

It is clear, from Eq. (283), 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 $x\geq a$. Clearly, 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
\begin{displaymath}
\psi(0) = \psi(a) = 0.
\end{displaymath} (285)

Furthermore, it follows from Eq. (283) that $\psi$ satisfies
\begin{displaymath}
\frac{d^2 \psi}{d x^2} = - k^2\,\psi
\end{displaymath} (286)

in this region, where
\begin{displaymath}
k^2 = \frac{2\,m\,E}{\hbar^2}.
\end{displaymath} (287)

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 (285).

The solution to Eq. (286), subject to the boundary conditions (285), is

\begin{displaymath}
\psi_n(x) = A_n\,\sin(k_n\,x),
\end{displaymath} (288)

where the $A_n$ are arbitrary (real) constants, and
\begin{displaymath}
k_n = \frac{n\,\pi}{a},
\end{displaymath} (289)

for $n=1,2,3,\cdots$. Now, it is clear, from Eqs. (287) and (289), that the energy $E$ is only allowed to take certain discrete values: i.e.,
\begin{displaymath}
E_n = \frac{n^2\,\pi^2\,\hbar^2}{2\,m\,a^2}.
\end{displaymath} (290)

In other words, the eigenvalues of the energy operator are discrete. This is a general feature of bounded solutions: i.e., solutions in which $\vert\psi\vert\rightarrow 0$ as $\vert x\vert\rightarrow\infty$. According to the discussion in Sect. 4.12, we expect the stationary eigenfunctions $\psi_n(x)$ to satisfy the orthonormality constraint
\begin{displaymath}
\int_0^a \psi_n(x)\,\psi_m(x)\,dx = \delta_{nm}.
\end{displaymath} (291)

It is easily demonstrated that this is the case, provided $A_n = \sqrt{2/a}$. Hence,
\begin{displaymath}
\psi_n(x) = \sqrt{\frac{2}{a}}\,\sin\left(n\,\pi\,\frac{x}{a}\right)
\end{displaymath} (292)

for $n=1,2,3,\cdots$.

Finally, again from Sect. 4.12, the general time-dependent solution can be written as a linear superposition of stationary solutions:

\begin{displaymath}
\psi(x,t) = \sum_{n=0,\infty} c_n\,\psi_n(x)\,{\rm e}^{-{\rm i}\,E_n\,t/\hbar},
\end{displaymath} (293)

where
\begin{displaymath}
c_n = \int_0^a\psi_n(x)\,\psi(x,0)\,dx.
\end{displaymath} (294)


next up previous contents
Next: The square potential barrier Up: One-dimensional potentials Previous: Introduction   Contents
Richard Fitzpatrick 2006-12-12