2.7 The one-dimensional infinite square well

V(x)=\begin{cases}0,&0<x<L\\ \infty,&\text{otherwise}\end{cases}

A particle trapped between two infinitely high walls. It is the simplest bound-state problem there is, and it already contains energy quantisation, zero-point energy, node structure, orthogonality and completeness — the entire grammar of quantum mechanics makes its first appearance here.

Solving it

The full solution: from equation to energy levelsbasic~8 min

Step 1: fix the domain.

Outside the well $V=\infty$ . The stationary equation demands $\psi''=\frac{2m(V-E)}{\hbar^2}\psi$ , and as $V\to\infty$ the only finite solution is $\psi\equiv0$ . The particle is absolutely never found outside the well.

Continuity of the wavefunction then gives the boundary conditions

\psi(0)=\psi(L)=0\tag{2.7.1}

(Note: $\psi'$ is not required to be continuous here, because $V$ is infinite — see the table at the end of section 2.2.)

Step 2: solve inside the well.

Inside, $V=0$ :

-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}=E\psi \quad\Longleftrightarrow\quad \frac{\dd^2\psi}{\dd x^2}=-k^2\psi,\qquad k\equiv\frac{\sqrt{2mE}}{\hbar}\tag{2.7.2}

(A bound state requires $E>0$ : if $E<0$ then $\psi''=\kappa^2\psi$ with solutions $\ee^{\pm\kappa x}$ , which cannot vanish at both ends.) The general solution is

\psi(x)=A\sin kx+B\cos kx\tag{2.7.3}

Step 3: impose the boundary conditions.

$\psi(0)=0$ immediately gives $B=0$ . That leaves

\psi(L)=A\sin kL=0\tag{2.7.4}

Taking $A=0$ gives the trivial everywhere-zero solution (unnormalisable, and no particle). So we must have

kL=n\pi,\qquad n=1,2,3,\dots\tag{2.7.5}

Why does $n$ start at 1? $n=0$ gives $\psi\equiv0$ ; negative $n$ merely flips the overall sign of $\psi$ , which is the same physical state (a global phase). Only the positive integers count.

Step 4: the energy levels.

From $k_n=n\pi/L$ and $E=\hbar^2k^2/2m$ :

E_n=\frac{n^2\pi^2\hbar^2}{2mL^2},\qquad n=1,2,3,\dots\tag{2.7.6}

Step 5: normalisation.

1=|A|^2\int_0^L\sin^2\frac{n\pi x}{L}\dd x=|A|^2\cdot\frac{L}{2} \;\Longrightarrow\;A=\sqrt{\frac2L}\tag{2.7.7}

(taking $A$ positive real; the phase is arbitrary). Finally

\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}\tag{2.7.8}

Reading the result

The picture

Standing waves in a box. The boundary conditions $\psi(0)=\psi(L)=0$ are exactly those of a string clamped at both ends, so the eigenstates are the string’s harmonics: the well must hold an integer number of half-wavelengths, $L=n\lambda/2$ .

Combine that with de Broglie’s $\lambda=h/p$ and you get $p_n=nh/2L$ at once; then $E=p^2/2m$ reproduces the energy formula above. Quantisation = the standing-wave condition.

Nodes. $\psi_n$ has $n-1$ zeros inside the well. Higher energy means sharper bending (curvature $\propto E$ ) and more nodes. This “number of nodes = n − 1” rule holds generally for one-dimensional bound states.

The mathematics

\psi_n(x)=\sqrt{\frac2L}\sin\frac{n\pi x}{L},\qquad E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\tag{2.7.9}

E_n=n^2E_1,\qquad E_{n+1}-E_n=(2n+1)E_1\tag{2.7.10}

Orthonormality:

\int_0^L\psi_m\psi_n\,\dd x=\delta_{mn}\tag{2.7.11}

(Verify directly with a product-to-sum identity: for $m\ne n$ both terms integrate over a whole number of periods and vanish.)

The infinite square well

Drag the width L and the quantum number n and watch the energy, the waveform and the probability distribution move together. Units: ħ = m = 1.

Energy Eₙ (n=1): 4.935
relative to E₁(L=1): 1.00 ×
⟨x⟩ / L: 0.500
Δx / L: 0.180

Well width L1.00

Shrink the box and every level is pushed up together: E ∝ 1/L²

Quantum number n1

nodes = n − 1 = 0 (endpoints excluded)

State

Show ψShow |ψ|²

Try this

Drag L from 2.5 down to 0.6 and keep your eye on the dashed line in the level diagram (pinned at the reference energy E₁(L=1)): it sinks all the way to the bottom, meaning every level has risen far above it. The "relative to E₁(L=1)" readout climbs from 0.16 to 2.78 — "the tighter you confine a particle, the more kinetic energy it has", a direct consequence of ΔxΔp ≥ ħ/2.
Change n in the stationary mode and note that |ψ|² never changes with time (the curve stands still); switch to a superposition, press play, and |ψ|² immediately starts sloshing. That is exactly what "stationary" refers to.
Leave only c₁ in the superposition (drag the rest to 0) and press play — the probability density stops moving again. However long a single eigenstate evolves, it only picks up an overall phase e^(−iEₙt/ħ), which no observable can see.
Push n above 10 and look at |ψ|²: the fringes get so fine that the distribution is nearly uniform — the classical picture of a particle equally likely to be anywhere in the box. The correspondence principle, in view.

This is an interactive simulation (infinite-well) and needs JavaScript to run. The derivations and conclusions in the text do not depend on it, so feel free to read on.

Expanding an arbitrary initial state

Because $\{\psi_n\}$ is complete, any initial state satisfying the boundary conditions can be expanded:

\Psi(x,0)=\sum_{n=1}^{\infty}c_n\psi_n(x),\qquad c_n=\sqrt{\frac2L}\int_0^L\sin\frac{n\pi x}{L}\,\Psi(x,0)\,\dd x\tag{2.7.12}

Example: a particle initially uniform over the left halfadvanced~7 min

Take

\Psi(x,0)=\begin{cases}\sqrt{2/L},&0<x<L/2\\0,&\text{otherwise}\end{cases}

(already normalised: $\frac{2}{L}\cdot\frac{L}{2}=1$ ). The coefficients are

c_n=\sqrt{\frac2L}\cdot\sqrt{\frac2L}\int_0^{L/2}\sin\frac{n\pi x}{L}\dd x =\frac2L\cdot\frac{L}{n\pi}\left[1-\cos\frac{n\pi}{2}\right] =\frac{2}{n\pi}\left[1-\cos\frac{n\pi}{2}\right]\tag{2.7.13}

Term by term:

$n$	$\cos(n\pi/2)$	$c_n$	$\lvert c_n\rvert^2$
1	0	$2/\pi$	0.405
2	$-1$	$2/\pi$	0.405
3	0	$2/3\pi$	0.045
4	1	0	0
5	0	$2/5\pi$	0.016

The first two terms already account for 81%. Check: $\sum|c_n|^2=\frac{4}{\pi^2}(1+1+\frac19+0+\frac1{25}+\cdots)=1$ (use $\sum_{k\,\text{odd}}1/k^2=\pi^2/8$ ).

What does an energy measurement give? $E_1$ with probability 40.5%, $E_2$ with 40.5%, $E_3$ with 4.5%, and so on. Note $c_4=0$ : the overlap of $\psi_4$ with this initial state happens to vanish exactly.

Evolution. The $E_1$ and $E_2$ components carry almost half the weight each, so the probability density sloshes between the two halves at

\omega_{21}=\frac{3\pi^2\hbar}{2mL^2}\tag{2.7.14}

Set $c_1=c_2$ in the simulation above and watch it happen.

The correspondence principle

The large-n limit

A classical particle bounces back and forth at constant speed, so it spends equal time everywhere — the classical probability density is uniform, $P_\text{cl}(x)=1/L$ .

The quantum answer is $|\psi_n|^2=\frac2L\sin^2\frac{n\pi x}{L}$ , oscillating between 0 and $2/L$ . The two look nothing alike. But note:

\overline{|\psi_n|^2}=\frac2L\cdot\overline{\sin^2}=\frac2L\cdot\frac12=\frac1L=P_\text{cl}\tag{2.7.15}

The average of the oscillation is exactly the classical result. And the larger $n$ is, the finer the fringes, so any measurement of finite resolution sees only the average. That is the concrete mechanism of the correspondence principle: quantum effects do not “disappear”, they get averaged away.

In the simulation, push $n$ up to 12 and imagine a detector with resolution $L/10$ — you would measure a flat line.

The relative level spacing supports the same conclusion:

\frac{E_{n+1}-E_n}{E_n}=\frac{2n+1}{n^2}\xrightarrow{\ n\to\infty\ }0\tag{2.7.16}

At large $n$ the relative gap between neighbouring levels goes to zero and the spectrum looks continuous — precisely the classical picture.

Key formulas

Eigenstates
$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L},\quad n=1,2,3,\dots$
Nodes = n − 1; n = 0 gives the trivial solution and is excluded
Energy levels
$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}=n^2E_1$
Spacing grows with n: E_{n+1}−E_n=(2n+1)E₁
Expansion coefficients
$c_n=\sqrt{\frac2L}\int_0^L\sin\frac{n\pi x}{L}\,\Psi(x,0)\,\dd x$
|cₙ|² is the probability of measuring Eₙ
Classical limit
$\overline{|\psi_n|^2}=\frac{1}{L}=P_{\text{cl}}(x)$
At large n the quantum oscillation is averaged away by finite resolution

Self-check4 questions

What comes next

Lower the walls from infinite to finite and three interesting things happen: the number of bound states becomes finite, the wavefunction leaks outside the well, and scattering states appear.

The one-dimensional infinite square well

Recommended first

After this section you should be able to

Solving it

Reading the result

The picture

The mathematics

The infinite square well

Expanding an arbitrary initial state

The correspondence principle

Key formulas

Self-check4 questions

What comes next

Solving it

Reading the result

◑The picture

∑The mathematics

Expanding an arbitrary initial state

The correspondence principle

∎Key formulas

What comes next

The picture

The mathematics

Key formulas