Particle-in-a-Box Spectrum

Purpose

For a particle of mass $m$ confined to $0\lt x\lt L$ by impenetrable walls,

$$V(x)= \begin{cases} 0, & 0\lt x\lt L,\\ \infty, & x\leq0\text{ or }x\geq L, \end{cases}$$

the normalized energy eigenfunctions and energies are

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\!\left(\frac{n\pi x}{L}\right),$$ $$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad n=1,2,3,\ldots.$$

The quantization comes from the Dirichlet boundary conditions

$$\psi(0)=\psi(L)=0,$$

not from a nonzero force inside the box. The symbol $V=\infty$ is shorthand for this ideal boundary-value problem.

The full derivation, domain discussion, revival structure, and numerical benchmark are at Infinite Square Well.

At a glance

Quantity	Formula
Allowed wave number	$k_n=n\pi/L$
De Broglie wavelength	$\lambda_n=2L/n$
Normalized eigenfunction	$\psi_n(x)=\sqrt{2/L}\sin(n\pi x/L)$
Energy	$E_n=n^2\pi^2\hbar^2/(2mL^2)$
Adjacent spacing	$E_{n+1}-E_n=(2n+1)E_1$
Interior nodes	$n-1$
Mean position	$\langle x\rangle_n=L/2$
Mean-square momentum	$\langle p^2\rangle_n=(n\pi\hbar/L)^2$
Time factor	$e^{-iE_nt/\hbar}$
Revival time	$T_{\rm rev}=4mL^2/(\pi\hbar)$

The label starts at $n=1$. The apparent $n=0$ solution is the zero function and cannot represent a normalized state.

Hamiltonian and domain

The Hilbert space is $L^2([0,L])$ with

$$\langle\phi\vert\psi\rangle = \int_0^Ldx\, \phi(x)^*\psi(x).$$

The Hamiltonian is

$$H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$$

on the Dirichlet domain. A standard self-adjoint realization has

$$\mathcal D(H) = H^2(0,L)\cap H_0^1(0,L).$$

The boundary conditions are part of the operator. Replacing them by periodic conditions produces a particle on a ring, with different eigenfunctions, degeneracies, and momentum structure.

For functions in the Hamiltonian domain,

$$\langle\psi\rvert H\lvert\psi\rangle = \frac{\hbar^2}{2m} \int_0^Ldx\, \lvert\psi'(x)\rvert^2 \geq0.$$

Thus the hard-wall Hamiltonian has no negative- or zero-energy eigenstate.

Spectrum and level spacing

Define the natural scales

$$k_1=\frac{\pi}{L}, \qquad p_1=\frac{\pi\hbar}{L}, \qquad E_1=\frac{\pi^2\hbar^2}{2mL^2}.$$

Then

$$k_n=nk_1, \qquad E_n=n^2E_1.$$

The spectrum is not equally spaced:

$$E_{n+1}-E_n =(2n+1)E_1.$$

The absolute spacing grows with $n$, while the relative spacing shrinks:

$$\frac{E_{n+1}-E_n}{E_n} = \frac{2n+1}{n^2} \sim \frac{2}{n}.$$

All levels are nondegenerate in one dimension. The $n$th state has nodes at

$$x_j=\frac{jL}{n}, \qquad j=1,\ldots,n-1.$$

Normalization and completeness

The basis obeys

$$\int_0^Ldx\, \psi_m(x)^*\psi_n(x) = \delta_{mn},$$

and the distributional completeness relation

$$\sum_{n=1}^{\infty} \psi_n(x)\psi_n(x')^* = \delta(x-x')$$

for interior coordinates. Any state in $L^2([0,L])$ can be expanded as

$$\psi(x,0) = \sum_{n=1}^{\infty}c_n\psi_n(x),$$

with

$$c_n = \int_0^Ldx\, \psi_n(x)^*\psi(x,0).$$

For a normalized state,

$$\sum_{n=1}^{\infty}\lvert c_n\rvert^2=1.$$

Square integrability is enough for norm convergence of the expansion. It is not enough to justify applying $H$ term by term or to guarantee finite mean energy; those require additional regularity or weighted coefficient sums.

Time evolution

An energy eigenstate evolves by a phase:

$$\psi_n(x,t) = \psi_n(x)e^{-iE_nt/\hbar}.$$

A general initial state evolves as

$$\psi(x,t) = \sum_{n=1}^{\infty} c_n\psi_n(x)e^{-iE_nt/\hbar}.$$

The energy probabilities $\lvert c_n\rvert^2$ remain constant, but relative phases can make the position density time dependent. For two components, interference oscillates at the Bohr frequency

$$\omega_{mn} = \frac{E_m-E_n}{\hbar} = \frac{(m^2-n^2)\pi^2\hbar}{2m_{\rm p}L^2},$$

where $m_{\rm p}$ denotes the particle mass in this line to avoid confusing it with the state index $m$.

The spectral propagator is

$$\begin{aligned} K(x_f,t;x_i,0) &= \sum_{n=1}^{\infty} \psi_n(x_f)\psi_n(x_i)^* e^{-iE_nt/\hbar}\\ &= \frac{2}{L} \sum_{n=1}^{\infty} \sin\!\left(\frac{n\pi x_f}{L}\right) \sin\!\left(\frac{n\pi x_i}{L}\right) e^{-in^2E_1t/\hbar}. \end{aligned}$$

It satisfies the Dirichlet conditions in both coordinate arguments and tends to the interval delta distribution as $t\to0$.

Position moments

For every energy eigenstate,

$$\langle x\rangle_n=\frac{L}{2},$$ $$\langle x^2\rangle_n = L^2 \left( \frac13- \frac{1}{2n^2\pi^2} \right),$$

and

$$(\Delta x)_n^2 = L^2 \left( \frac1{12}- \frac{1}{2n^2\pi^2} \right).$$

The real standing waves give the formal first-derivative matrix element

$$\langle p\rangle_n=0,$$

while the unambiguous kinetic-energy relation gives

$$\langle p^2\rangle_n = 2mE_n = \frac{n^2\pi^2\hbar^2}{L^2}.$$

Hence

$$\Delta p_n=\frac{n\pi\hbar}{L},$$

and

$$(\Delta x)_n(\Delta p)_n = \frac{\hbar}{2} \sqrt{\frac{n^2\pi^2}{3}-2}.$$

A vanishing mean momentum does not imply vanishing kinetic energy.

Position matrix elements

For calculations involving a uniform electric field or dipole coupling,

$$x_{mn} = \langle m\rvert x\lvert n\rangle$$

is

$$x_{mn} = \begin{cases} L/2, & m=n,\\ 0, & m\neq n\text{ and }m+n\text{ even},\\ -\dfrac{8Lmn} {\pi^2(m^2-n^2)^2}, & m+n\text{ odd}. \end{cases}$$

The off-diagonal rule is a parity selection rule expressed in the shifted interval convention. For example,

$$\langle1\rvert x\lvert2\rangle = -\frac{16L}{9\pi^2}.$$

The sign depends on harmless phase choices for individual eigenfunctions, while $\lvert x_{mn}\rvert$ and physical transition probabilities do not.

Centered-well parity convention

For the equivalent interval

$$-\frac{L}{2}\lt x\lt\frac{L}{2},$$

the normalized states may be chosen as

$$\psi_n^{(c)}(x) = \sqrt{\frac{2}{L}} \begin{cases} \cos(n\pi x/L), & n\text{ odd},\\ \sin(n\pi x/L), & n\text{ even}. \end{cases}$$

Odd $n$ then labels even parity and even $n$ labels odd parity. The energies are unchanged. In this convention,

$$\langle m\rvert x\lvert n\rangle=0$$

between states of the same parity, because $x$ is parity odd. Translating the origin changes diagonal position matrix elements but not energy gaps or transition probabilities.

Exact revival scale

Since $E_n=n^2E_1$, every spectral phase returns to unity at

$$T_{\rm rev} = \frac{2\pi\hbar}{E_1} = \frac{4mL^2}{\pi\hbar}.$$

Indeed,

$$e^{-iE_nT_{\rm rev}/\hbar} = e^{-i2\pi n^2} =1.$$

Thus every initial state revives in Hilbert-space norm. At half the revival time, the state is reflected about the center up to a global phase:

$$\psi(x,T_{\rm rev}/2) = -\psi(L-x,0).$$

For a wave packet concentrated near a large $n_0$, the classical round-trip time scale is

$$T_{\rm cl} \simeq \frac{2mL^2}{n_0\pi\hbar}.$$

Classical-looking bounces and exact quantum revivals are distinct phenomena.

Three-dimensional rectangular box

For a separable box with side lengths $L_x,L_y,L_z$, the eigenfunctions are products of one-dimensional sine modes and

$$E_{n_xn_yn_z} = \frac{\pi^2\hbar^2}{2m} \left( \frac{n_x^2}{L_x^2} +\frac{n_y^2}{L_y^2} +\frac{n_z^2}{L_z^2} \right),$$

with all three quantum numbers positive integers. Degeneracy can occur when different triples give the same sum, especially for equal side lengths. The full treatment is at Three-Dimensional Box.

Infinite versus finite walls

The hard-wall result is exact only for the ideal Dirichlet problem. In a finite well:

• bound-state wavefunctions penetrate into the classically forbidden region;
• the boundary value of the wavefunction is not zero;
• the energies are lower than the corresponding hard-wall values for the same nominal width and interior potential;
• only finitely many bound states exist;
• even and odd states satisfy transcendental matching equations.

Use the forthcoming finite-square-well formula card or the canonical Finite Square Well instead of imposing hard-wall sines at finite depth.

Units and scaling

Quantity	SI units	Scaling
$L$	m	box size
$k_n$	m$^{-1}$	$n/L$
$p_n$ as a kinetic scale	kg m s$^{-1}$	$n/L$
$E_n$	J	$n^2/(mL^2)$
$\psi_n$	m$^{-1/2}$	$L^{-1/2}$
$K(x_f,t;x_i,0)$	m$^{-1}$	inverse coordinate measure

Doubling $L$ divides every energy by four. Doubling the particle mass divides every energy by two. Adding a constant potential $V_0$ inside and outside the abstract interval shifts every energy by $V_0$ without changing the eigenfunctions.

Domain caution for momentum

The differential expression $-i\hbar d/dx$ can be used formally to evaluate matrix elements on smooth hard-wall states, but the first-derivative operator with Dirichlet conditions at both endpoints is not a self-adjoint momentum observable on the interval. Self-adjoint interval momenta use phase-related endpoint conditions instead.

The Hamiltonian and $p^2=2mH$ are well defined for the hard-wall problem. A release-and-measure momentum experiment is instead described by extending the released wavefunction to the full line and Fourier transforming it; it does not generally yield only two delta peaks at $\pm n\pi\hbar/L$. See Hermitian vs Self-Adjoint Operators.

Common mistakes

• Starting the spectrum at $n=0$.
• Treating the infinite potential as an ordinary number rather than a boundary condition.
• Forgetting the factor $\sqrt{2/L}$.
• Imposing $\psi'=0$ at a hard wall; Dirichlet conditions require $\psi=0$.
• Using periodic eigenfunctions for a hard-wall interval.
• Calling the $n^2$ spectrum equally spaced.
• Confusing $\langle p\rangle=0$ with zero kinetic energy.
• Assuming $-i\hbar d/dx$ with two Dirichlet endpoints is self-adjoint.
• Using hard-wall energies for a finite-depth well.
• Expecting high-$n$ densities to converge pointwise to a uniform classical density; the agreement is coarse grained.
• Assuming every normalized sine series has finite mean energy.
• Forgetting that the coordinate origin changes parity and position matrix element conventions.

Related formulas

• Normalization
• Schrödinger Equation
• Time-Evolution Operator
• Propagator Composition Law
• Boundary Conditions
• Stationary States
• Fourier Series
• Sturm–Liouville Theory

Exercises

1. Normalize the eigenfunctions

Show that $A\sin(n\pi x/L)$ is normalized when $\lvert A\rvert=\sqrt{2/L}$.

Solution

Use

$$\int_0^Ldx\, \sin^2\!\left(\frac{n\pi x}{L}\right) = \frac{L}{2}.$$

Then

$$1 = \lvert A\rvert^2\frac{L}{2},$$

so

$$\lvert A\rvert = \sqrt{\frac{2}{L}}.$$

The remaining phase of $A$ is physically irrelevant.

2. Derive the dipole selection rule

Evaluate $\langle m\rvert x\lvert n\rangle$ for $m\neq n$ and show that it vanishes when $m+n$ is even.

Solution

Use the product identity for the two sine functions:

$$\begin{aligned} x_{mn} &= \frac{1}{L} \int_0^Ldx\,x \left[ \cos\!\left(\frac{(m-n)\pi x}{L}\right) -\cos\!\left(\frac{(m+n)\pi x}{L}\right) \right]. \end{aligned}$$

For nonzero integer $r$,

$$\int_0^Ldx\,x\cos\!\left(\frac{r\pi x}{L}\right) = \frac{L^2}{r^2\pi^2} \left[(-1)^r-1\right].$$

The integers $m-n$ and $m+n$ have the same parity. If that parity is even, both terms vanish. If it is odd, simplification gives

$$x_{mn} = -\frac{8Lmn}{\pi^2(m^2-n^2)^2}.$$

Thus $x_{mn}$ is nonzero only between opposite-parity centered-well states.

3. Ground-state uncertainty

Compute $(\Delta x)_1(\Delta p)_1$ and compare it with $\hbar/2$.

Solution

For $n=1$,

$$(\Delta x)_1^2 = L^2 \left( \frac1{12}-\frac{1}{2\pi^2} \right),$$

and

$$(\Delta p)_1=\frac{\pi\hbar}{L}.$$

Therefore

$$(\Delta x)_1(\Delta p)_1 = \frac{\hbar}{2} \sqrt{\frac{\pi^2}{3}-2} \approx 0.568\hbar.$$

This exceeds $\hbar/2$. The box ground state is not a minimum-uncertainty Gaussian.

4. Half revival

Show that the state at $T_{\rm rev}/2$ is the reflected initial state up to a global phase.

Solution

At half the revival time,

$$e^{-iE_nT_{\rm rev}/(2\hbar)} = e^{-i\pi n^2} = (-1)^n.$$

The eigenfunctions satisfy

$$\psi_n(L-x) = (-1)^{n+1}\psi_n(x).$$

Hence

$$(-1)^n\psi_n(x) = -\psi_n(L-x).$$

Applying this identity term by term to the spectral expansion gives

$$\psi(x,T_{\rm rev}/2) = -\psi(L-x,0).$$

The minus sign is a global phase and does not affect probabilities.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018, Ch. 2.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, Ch. 5.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1, Wiley, 1977.
• E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.
• R. W. Robinett, “Quantum wave packet revivals,” Physics Reports 392, 1–119 (2004).