4.1: Infinite Potential Well
- Page ID
- 15743
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\[ c_n = \int_0^a\psi_n(x)\,\psi(x,0)\,dx. \label{e5.13} \]