Particle in Infinite and Finite Square Wells

The time-independent Schrödinger equation becomes a solvable ordinary differential equation once a potential $V(x)$ is fixed. The most transparent case is a particle confined to a box with perfectly rigid walls, the infinite square well. It admits exact solutions with no difficult mathematics, reproduces the standing-wave quantization of a string fixed at both ends, and displays every qualitative feature of a bound quantum system.¹

The infinite square well

A particle is free inside a region of width $L$ and cannot escape it:

A macroscopic image is a bead sliding on a frictionless wire between two massive stops; a physical realization is an electron between charged grids whose repelling fields become the walls as the voltage is raised. Because the potential is infinite outside, the wave function must vanish there — the particle is certainly inside — and by continuity $\psi$ must reach zero at each wall:

Quantization from standing waves

Inside the well $V=0$, so the equation is the free-particle equation

with general solution $\psi (x)=A\sin kx+B\cos kx$. The wall at $x=0$ forces $\psi (0)=0$; since $\cos 0=1$, the cosine term must drop, $B=0$. The wall at $x=L$ then requires

which holds only when $kL$ is an integer multiple of $\pi$:

Written through $\lambda =2\pi /k$, this is $n(\lambda /2)=L$: an integer number of half-wavelengths fits the well, exactly the standing-wave condition for a string of length $L$ clamped at both ends. The value $n=0$ gives $\psi \equiv 0$, no particle, and is excluded.

The energy follows from $E={\hslash }^2{k}^2/2m$ with $k=k_n$:

The integer $n$ is a quantum number: it fixes both the energy and the wave function.

The eigenfunctions

With $B=0$ and $k=k_n$ the wave functions are sines. The constant $A$ is fixed by normalization; since $\psi =0$ outside, only the interval $[0,L]$ contributes:

The lowest state $n=1$ is the ground state; $n=2,3,\dots$ are excited states. Every ${\psi }_n$ has the same peak amplitude $\sqrt{2/L}$, and every probability density $P_n={\psi }_n^2$ peaks at $2/L$.

The zero-point energy $E_1>0$ is not an accident of the algebra. A particle pinned exactly at rest would have definite position and definite (zero) momentum, forbidden by the uncertainty principle. Confining the particle to width $\Delta x\sim L$ forces a momentum spread $\Delta p\sim \hslash /L$ and thus a minimum kinetic energy of order ${\hslash }^2/2m{L}^2$ — the same size as $E_1$.

Classical correspondence

Classically a particle in the box moves at constant speed, reflecting off each wall, and is equally likely to be found anywhere: the position distribution is flat, $P_C(x)=1/L$. The quantum ground state looks nothing like this — it is a single hump peaked at the center with zero probability at the walls. The two pictures reconcile at large $n$. In state $n$ the density ${\psi }_n^2=(2/L){\sin }^2(k_{n}x)$ has $n$ peaks, and averaged over a small window containing several oscillations $⟨{\sin }^2⟩=\frac{1}{2}$, so

This is Bohr's correspondence principle: the quantum distribution reproduces the classical one when $n$ is large. A measurement with finite resolution $\Delta x$ cannot resolve the individual peaks and returns the classical average.²

A worked spectrum. Take $L=0.1\text{nm}$, about the size of an atom. Writing $E_1=(hc{)}^2/8m{c}^2{L}^2$ with $hc=1240\text{eV}\cdot \text{nm}$ and $m{c}^2=0.511\text{MeV}$,

comparable to the $13.6\text{eV}$ binding of hydrogen. The higher levels are $E_2=4E_1=150.4\text{eV}$ and $E_3=9E_1=338.4\text{eV}$. Photons emitted in downward transitions carry the level differences:

These are soft-X-ray wavelengths, the scale expected when an electron is confined to atomic dimensions.³

The finite square well

Real wells have finite depth. Lowering the walls to a height $V_0$,

changes the boundary behavior. For a bound state, $E<V_0$. Inside the well $V=0$ and the equation is again ${\psi }^{\prime \prime }=-k^2\psi$ with $k^2=2mE/{\hslash }^2$, so $\psi$ oscillates. Outside the well $E<V(x)$, and the equation becomes

The exterior solution is a real exponential $\psi \sim e^{-\alpha x}$, decaying into the barrier. The wave function no longer vanishes at the wall; instead $\psi$ and ${\psi }^{\prime }$ must join continuously to the interior sine, and the exterior tail must go to zero at infinity.

Origin of energy quantization

The curvature rule does the selecting. Inside, ${\psi }^{\prime \prime }$ and $\psi$ have opposite signs, so $\psi$ curves toward the axis and oscillates. Outside, they share a sign, so $\psi$ curves away from the axis. For a generic energy, the exterior solution that matches the interior at the wall curves away and diverges to $\pm \infty$ — not normalizable, not physical. Only for special energies does the interior emerge at the wall with exactly the slope that lets the exterior decay to zero. Those energies are the bound-state levels.

The finite well differs from the infinite well in three ways.⁴

• Penetration. $\psi$ is nonzero in the regions $x<0$ and $x>L$, where classically $E<V$ and the kinetic energy would be negative. There is a real probability of finding the particle in the classically forbidden region, falling off as $|\psi {|}^2\sim e^{-2\alpha x}$.
• Fewer, lower levels. Because the wave function extends into the walls, its half-wavelength inside is slightly longer than in the infinite well, so each energy is somewhat lower than the corresponding infinite-well level. Only a finite number of bound states exist, set by the depth $V_0$; a very shallow well holds just one.
• Guaranteed ground state. However shallow, a one-dimensional well always binds at least one state.

The penetration does not let a measurement return a negative kinetic energy. Localizing the particle in the tail region, of extent $\Delta x\sim 1/\alpha$, injects a momentum uncertainty $\Delta p\sim \hslash \alpha$ and a minimum kinetic energy $\sim (\Delta p{)}^2/2m=V_0-E$, exactly enough to forbid the measurement of a negative value. The same tail is the seed of barrier tunneling.

The graphical eigenvalue condition

For most finite potentials the matching conditions produce a transcendental equation with no closed-form solution, but a symmetric well admits a clean graphical reading. Center the well on the origin with half-width $a$, so it runs from $-a$ to $+a$. The symmetric potential $V(-x)=V(x)$ splits the solutions by parity: the interior is either a cosine (even) or a sine (odd). Matching $\psi$ and ${\psi }^{\prime }$ at $x=a$ gives

Introduce the dimensionless variables $u=ka$ and $v=\alpha a$. Their squares sum to a constant fixed by the well:

because $k^2+{\alpha }^2=2m{V}_0/{\hslash }^2$. The allowed states are the intersections of this circle of radius $R_0$ with the curves $v=u\tan u$ (even) and $v=-u\cot u$ (odd) in the first quadrant. Each intersection is one bound state.

Reading off the geometry: a small radius $R_0<\pi /2$ meets only the first even branch, so a shallow well holds exactly one (even, nodeless) ground state. As $V_0$ grows, $R_0$ increases and the circle sweeps across successive branches, adding an odd state, then another even state, and so on. The number of bound states is the number of half-$\pi$ intervals the radius spans,

always at least one.⁵ The same qualitative analysis applies to any well-type potential: wherever $E>V(x)$ the wave function oscillates, wherever $E<V(x)$ it curves away, and only discrete energies give a solution that decays at infinity. The harmonic oscillator is the next such potential.