26 Jul 2025

Boundary Conditions, Rigid-Wall Box, and Square Wells

boundary-conditions particle-in-a-box infinite-square-well finite-square-well

The stationary one-dimensional Schrödinger equation is

\[-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi.\]

It is a second-order differential equation, so two boundary conditions are needed. Those conditions, together with normalizability, turn an apparently continuous energy parameter into a discrete set for a bound system.

Continuity at a finite potential step

Integrate the equation across a small interval containing a finite step at $x=x_0$:

\[-\frac{\hbar^2}{2m} \left[\psi'(x_0+\epsilon)-\psi'(x_0-\epsilon)\right] +\int_{x_0-\epsilon}^{x_0+\epsilon}(V-E)\psi\,dx=0.\]

If $V$ and $\psi$ are finite, the integral tends to zero as $\epsilon\to0$. Hence

\[\boxed{\psi'(x_0^+)=\psi'(x_0^-)}.\]

The wavefunction itself must also be continuous. A finite jump in $\psi$ would produce a delta function in $\psi’$ and a derivative of a delta function in $\psi’’$, but the equation contains no term able to cancel it for a finite step. Thus

\[\boxed{\psi(x_0^+)=\psi(x_0^-)}.\]

At an infinite wall this finite-step argument does not apply. The accessible wavefunction must vanish at the wall, while derivative continuity with the forbidden exterior is not required.

Free particle in a rigid one-dimensional box

Let

\[V(x)= \begin{cases} 0,&0<x<L,\\ \infty,&x\leq0\ \text{or}\ x\geq L. \end{cases}\]

Inside the box the particle is free, so

\[\psi''+k^2\psi=0, \qquad k^2=\frac{2mE}{\hbar^2}.\]

The general interior solution is

\[\psi(x)=A\sin kx+B\cos kx.\]

The rigid walls impose $\psi(0)=0$ and $\psi(L)=0$. The first condition gives $B=0$. A non-zero state then requires

\[A\sin kL=0 \quad\Longrightarrow\quad kL=n\pi,\]

where $n=1,2,3,\ldots$; $n=0$ would make the wavefunction identically zero. Therefore

\[\boxed{k_n=\frac{n\pi}{L}}, \qquad \boxed{E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}}.\]

The energy has units $\hbar^2/(mL^2)=\mathrm J$. Normalization gives

\[1=\lvert A\rvert^2\int_0^L\sin^2\!\left(\frac{n\pi x}{L}\right)dx =\lvert A\rvert^2\frac L2,\]

so a real phase convention gives

\[\boxed{ \psi_n(x)=\sqrt{\frac2L} \sin\!\left(\frac{n\pi x}{L}\right) },\qquad 0<x<L.\]

The boundary conditions allow only integer half-wavelengths, $L=n\lambda_n/2$. This is the direct emergence of discrete energy levels from confinement.

Finite one-dimensional square well

Use a symmetric well of half-width $a$ and depth $V_0>0$:

\[V(x)= \begin{cases} -V_0,&\lvert x\rvert<a,\\ 0,&\lvert x\rvert\geq a. \end{cases}\]

A bound state must satisfy $-V_0<E<0$. Define positive real wave numbers

\[k=\frac{\sqrt{2m(E+V_0)}}{\hbar}, \qquad \kappa=\frac{\sqrt{-2mE}}{\hbar}.\]

Both have units $\mathrm{m^{-1}}$, and

\[k^2+\kappa^2=\frac{2mV_0}{\hbar^2}.\]

Inside the well, $\psi’‘+k^2\psi=0$, so the solutions oscillate. Outside, $\psi’’-\kappa^2\psi=0$. Normalizability removes the exponentials that grow as $x\to\pm\infty$.

Because $V(x)=V(-x)$, bound eigenfunctions may be chosen even or odd.

Even states

Choose

\[\psi_e(x)= \begin{cases} A\cos kx,&\lvert x\rvert<a,\\ A\cos(ka)e^{-\kappa(\lvert x\rvert-a)},&\lvert x\rvert\geq a. \end{cases}\]

Continuity of $\psi$ at $x=a$ is already built into this expression. Derivative continuity gives

\[-Ak\sin(ka)=-A\kappa\cos(ka),\]

and hence

\[\boxed{k\tan(ka)=\kappa}.\]

Odd states

Choose

\[\psi_o(x)= \begin{cases} A\sin kx,&\lvert x\rvert<a,\\ A\,\operatorname{sgn}(x)\sin(ka)e^{-\kappa(\lvert x\rvert-a)},&\lvert x\rvert\geq a. \end{cases}\]

At $x=a$, derivative continuity gives

\[Ak\cos(ka)=-A\kappa\sin(ka),\]

so

\[\boxed{-k\cot(ka)=\kappa}.\]

Introduce dimensionless quantities

\[z=ka,\qquad z_0=\frac{a\sqrt{2mV_0}}{\hbar}.\]

Then $\kappa a=\sqrt{z_0^2-z^2}$, and the allowed bound energies are the isolated roots

\[\boxed{z\tan z=\sqrt{z_0^2-z^2}} \quad\text{(even)},\] \[\boxed{-z\cot z=\sqrt{z_0^2-z^2}} \quad\text{(odd)}.\]

Once a root $z$ is found,

\[\boxed{E=-\frac{\hbar^2\kappa^2}{2m} =-V_0+\frac{\hbar^2z^2}{2ma^2}}.\]

The matching equations have solutions only at discrete intersections. Unlike the rigid box, a finite well has exponentially decaying tails outside and only finitely many bound levels. In the limit $V_0\to\infty$, $\kappa\to\infty$ and the boundary values approach zero. Measured upward from the well bottom, $E+V_0$, the levels approach the rigid-wall spectrum for the full width $2a$.

The editable TikZ source generates the figure. Schrödinger-equation and matching residuals are checked in the Maxima worksheet.