The 2D and 3D Box and Degeneracy

The 2D and 3D box is not merely a classroom toy; it is the qualitative skeleton behind quantum-dot engineering, where semiconductor nanocrystals of tunable, roughly cuboidal or spherical shape are synthesized specifically because confining an electron and hole in a nanometer-scale box shifts the absorption and emission energies in a controllable way — the additive-energy, quantum-number-squared scaling derived here is the zeroth-order explanation for why smaller dots absorb bluer light. Anisotropic "nanorod" quantum dots are engineered precisely to break the cubic degeneracy pattern and produce polarized emission, a direct real-world echo of the tetragonal-splitting calculation above. In molecular spectroscopy, the free-electron model treats linear conjugated polyenes and cyanine dyes as effectively 1D or quasi-2D boxes; the additive, product-wavefunction structure explored here is the conceptual seed for understanding why extending conjugation redshifts absorption, and why breaking planarity or symmetry (twisting a dye, substituent effects) removes accidental near-degeneracies and reorganizes absorption bands. In atomic and molecular orbital theory, the same symmetry-degeneracy logic — states related by a symmetry operation of the potential must be degenerate — is the deep reason $p_x,p_y,p_z$ orbitals are threefold degenerate in a free atom (spherical symmetry) and the reason that degeneracy splits in crystal-field or ligand-field environments of lower symmetry, exactly analogous to the cubic-to-tetragonal box splitting worked out above. In solid-state physics, the 3D box (with periodic rather than hard-wall boundaries) underlies the free-electron model of metals and the counting of density-of-states via $k$-space integer triples, a direct descendant of the $n_x^2+n_y^2+n_z^2$ counting problem solved here. This connects forward to the treatment of Particle on a Ring, where a different symmetry (rotational) produces a different, and in some ways richer, degeneracy structure, and to the broader machinery of Separation of Variables in Multidimensional Problems, of which the 2D/3D box is the cleanest possible instance.

Ask the tutor: How would the degeneracy pattern change if the box had periodic boundary conditions instead of hard infinite walls?

Picture an electron trapped in a rectangular quantum dot of dimensions $L_x \times L_y$, with potential energy zero inside the rectangle and infinite at its walls. The time-independent Schrödinger equation is

for $0 < x < L_x$, $0 < y < L_y$, with $\psi = 0$ on the boundary and everywhere outside. There is no term in the Hamiltonian that couples $x$-motion to $y$-motion — no $xy$ cross term, no coupling potential — so the kinetic energy operator is simply the sum of an $x$-piece and a $y$-piece. This structural fact, that the Hamiltonian is a sum of commuting operators acting on separate variables, is what makes the equation separable and is the single mechanical reason the whole apparatus below works. It is the same logic that will later organize the treatment of any multidimensional separable Hamiltonian. Because the box has straight, mutually perpendicular walls, the boundary conditions themselves separate: $\psi(0,y)=\psi(L_x,y)=0$ for all $y$, and $\psi(x,0)=\psi(x,L_y)=0$ for all $x$. A product ansatz $\psi(x,y) = X(x)Y(y)$ turns the partial differential equation into two ordinary 1D box equations, each identical in form to the equation you already know from the 1D box, just with its own box length and its own quantum number. The result is that every stationary state of the rectangular box is labeled by an independent pair of positive integers $(n_x, n_y)$, and its energy is literally the arithmetic sum of two 1D box energies. The chemically loaded question is what happens when $L_x = L_y$ — a square box, or in three dimensions a cube. Now the two 1D problems are identical except for which axis they run along, so swapping $n_x \leftrightarrow n_y$ leaves the energy completely unchanged while producing a genuinely different wavefunction (different nodal pattern, different orientation of the standing wave). This exchange symmetry is the geometric origin of degeneracy: the box's square (or cubic) shape possesses a symmetry — a 90° rotation maps the box onto itself — and states related by that symmetry operation are forced to share an energy. Break the symmetry (make $L_x \ne L_y$) and the states are no longer related by any symmetry of the container, so nothing forces their energies to coincide; the degeneracy splits into two distinct levels.

The picture to hold onto: the $x$-part of the wavefunction has $n_x - 1$ internal nodal lines running parallel to the $y$-axis, and the $y$-part has $n_y - 1$ nodal lines running parallel to the $x$-axis. The full 2D nodal pattern is a grid — literally the superposition of the two 1D nodal patterns — because the wavefunction is a product. This is a direct, visual consequence of separability: nodes don't get more complicated in higher dimensions, they simply accumulate additively from each independent factor.

That equals sign in the figure is the entire point: two visibly different standing-wave patterns, one with its nodal line vertical and one with its nodal line horizontal, produce numerically identical energies solely because the box is square. Stretch the box even slightly in one direction and that equals sign becomes a "$\ne$" — the degeneracy is not fundamental, it is contingent on the box's shape, and this contingency is exactly what makes it a useful teaching model for degeneracy in atoms and molecules, where $p_x, p_y, p_z$ orbitals are degenerate only because free atoms are spherically symmetric.

The 2D rectangular box Hamiltonian separates as

where: $\hat{H}_x$ and $\hat{H}_y$ are each identical in form to the 1D particle-in-a-box Hamiltonian, but act only on $x$ or only on $y$ respectively, and commute with each other since they involve independent variables.

Reading: because $[\hat H_x, \hat H_y] = 0$ and each acts on a different variable, they can be diagonalized simultaneously; their common eigenfunctions are products, and eigenvalues add. The normalized stationary states of the 2D rectangular box are

where: $n_x$ and $n_y$ are independent positive integer quantum numbers, one per confined direction, and each square-root prefactor is the normalization constant for a 1D box of the corresponding length.

Reading: the wavefunction is literally the product of two ordinary 1D box wavefunctions; nothing new is invented, the same function you already know is simply reused twice with different labels. The corresponding energy is

where: $h$ is Planck's constant, $m$ the particle mass, $L_x$ and $L_y$ the box edge lengths, and the two terms are exactly the 1D box energies for quantum numbers $n_x$ and $n_y$ in boxes of length $L_x$ and $L_y$.

Reading: the total energy is a sum, not a product — this additivity is the direct energetic consequence of the separable, additive Hamiltonian above. Extending to three dimensions with edges $L_x, L_y, L_z$:

where: $n_x,n_y,n_z = 1,2,3,\dots$ independently, and the three-term sum is three independent 1D box energies added together.

Reading: three independent standing waves, three independent quantum numbers, one additive total energy — the pattern from 2D simply repeats. For the special case of a cube, $L_x=L_y=L_z=L$, this collapses to

where: the energy now depends only on the sum of squares of the three quantum numbers, not on which axis carries which value.

Reading: this single fact — that energy depends on $n_x^2+n_y^2+n_z^2$ and not on the individual assignment — is the entire mathematical origin of degeneracy in the cubic box; any permutation of a fixed triple of integers gives the identical energy. The degeneracy $g$ of a level is the number of distinct quantum-number combinations giving that energy:

where: the count runs over all positive-integer triples whose squares sum to the fixed value corresponding to energy $E$.

Reading: degeneracy is fundamentally a number-theory counting question once the box is cubic — how many ways can a given integer be written as an ordered sum of three positive squares.

Example 1 — Foundational: energies and degeneracy in a square 2D box. An electron ($m = 9.109\times10^{-31}\ \text{kg}$) is confined to a square 2D box with $L_x = L_y = L = 2.00\ \text{Å} = 2.00\times10^{-10}\ \text{m}$. Find the two lowest distinct energies and their degeneracies. The 2D energy formula for a square box is

First compute the prefactor:

Numerator: $(6.626\times10^{-34})^2 = 4.390\times10^{-67}\ \text{J}^2\text{s}^2$. Denominator: $8 \times 9.109\times10^{-31} \times (2.00\times10^{-10})^2 = 8\times9.109\times10^{-31}\times4.00\times10^{-20} = 2.915\times10^{-49}\ \text{kg m}^2$.

The ground state is $(n_x,n_y)=(1,1)$, non-degenerate (the only way to write $N=2$ as a sum of two positive squares, since order doesn't matter when both are equal):

The first excited level has $N = n_x^2+n_y^2 = 5$, reached by $(n_x,n_y)=(1,2)$ and $(2,1)$ — two distinct wavefunctions (nodal line vertical vs. horizontal), same energy:

Limiting-case check: if instead $L_x \ne L_y$ (say $L_y = 1.01 L_x$, a 1% stretch), then $E_{1,2} \ne E_{2,1}$ because the $n_y^2$ term is divided by a different $L_y^2$ than the $n_x^2$ term is by $L_x^2$; the twofold degeneracy above splits into two distinct energies separated by roughly 2% of the level spacing, confirming that the degeneracy found is contingent on the exact equality $L_x=L_y$ and vanishes continuously as that equality is broken.

Example 2 — Applied: a rectangular quantum dot model for a conjugated dye. Model an elongated organic chromophore as a 2D electron confined to a rectangle with $L_x = 8.00$ Å (along the conjugated backbone) and $L_y = 3.00$ Å (across it), representing the effective delocalization dimensions of a linear polyene-like dye. Compare the lowest two energy levels and determine whether any accidental degeneracy occurs for small quantum numbers. Using the same prefactor structure but separately for each axis:

Denominator: $8\times9.109\times10^{-31}\times64.0\times10^{-20} = 4.664\times10^{-48}\ \text{kg m}^2$.

For the $y$-direction:

The general energy is

Ground state $(1,1)$:

First excited state — since the box is strongly non-square, increasing $n_x$ costs far less energy than increasing $n_y$ (because $L_x \gg L_y$ makes the $x$-direction "soft"), so the next level is $(2,1)$:

Compare to $(1,2)$, which would require exciting the "hard" $y$-direction:

These three levels ($4.766$, $6.529$, $17.30$ eV) are all numerically distinct — no accidental degeneracy appears for small quantum numbers in this rectangle, precisely because $L_x/L_y = 8.00/3.00 = 2.667$ is not a ratio that makes any pair of $(n_x^2, n_y^2)$ combinations coincide. This models a real physical trend: elongating a chromophore lowers the HOMO–LUMO-type gap along the long axis dramatically (compare $4.766\to6.529$ eV, a modest shift) while transitions requiring excitation across the short axis remain far higher in energy (17.30 eV) — consistent with why extending conjugation length is a standard strategy for red-shifting dye absorption, since the low-lying transitions are dominated by the soft, long direction.

Limiting-case check: as $L_y \to L_x$ (dot becomes square), the coefficients $9.412\times10^{-20}$ and $6.694\times10^{-19}$ converge to the same value, and $E_{2,1} \to E_{1,2}$ — recovering exactly the degenerate pair found in Example 1, confirming the formula continuously interpolates between the non-degenerate rectangular case and the degenerate square case.

Example 3 — Exam/competition level: degeneracy and its splitting in a 3D box under anisotropic distortion. A particle is confined in a 3D box. In the perfectly cubic case ($L_x=L_y=L_z=L$), find the degeneracy of the level with $N = n_x^2+n_y^2+n_z^2 = 9$, list every contributing state, and then determine, to first order, how that degeneracy splits if the box is compressed slightly along $z$ so that $L_z = L(1-\varepsilon)$ with $\varepsilon = 0.02$, while $L_x=L_y=L$ remain fixed.

Part A — cubic degeneracy at N=9. We need all ordered triples of positive integers with $n_x^2+n_y^2+n_z^2=9$. Systematically:

• $(1,2,2)$: $1+4+4=9$ ✓ — and its permutations $(2,1,2)$, $(2,2,1)$: three states.
• $(3,...)$: $9 + n_y^2+n_z^2 = 9$ requires $n_y=n_z=0$, not allowed (quantum numbers start at 1). So no contribution from any triple containing a 3 together with two more positive integers ≥1 unless the other two vanish, which is forbidden.
• Check $(1,1,\sqrt7)$: not an integer, discard.
• No other combinations of three positive squares reach 9 besides permutations of $(1,2,2)$.

So $g(N=9) = 3$, contributed entirely by permutations of $(1,2,2)$: states $(1,2,2)$, $(2,1,2)$, $(2,2,1)$, all with

Part B — splitting under compression. With $L_x=L_y=L$ and $L_z = L(1-\varepsilon)$, the energy becomes

To first order in small $\varepsilon$, $(1-\varepsilon)^{-2} \approx 1+2\varepsilon$, so

Now evaluate the correction term $2\varepsilon n_z^2$ for each of the three formerly degenerate states, using $\varepsilon = 0.02$:

• $(n_x,n_y,n_z) = (1,2,2)$: $n_z = 2$, correction $= 2(0.02)(4) = 0.160$.
• $(n_x,n_y,n_z) = (2,1,2)$: $n_z = 2$, correction $= 2(0.02)(4) = 0.160$.
• $(n_x,n_y,n_z) = (2,2,1)$: $n_z = 1$, correction $= 2(0.02)(1) = 0.040$.

So compression along $z$ splits the threefold-degenerate cubic level into two levels: a doubly-degenerate level at $E = \frac{h^2}{8mL^2}(9.160)$ (states with $n_z=2$, i.e., $(1,2,2)$ and $(2,1,2)$, which remain degenerate with each other because they still differ only by swapping $x\leftrightarrow y$, and $L_x=L_y$ is untouched), and a non-degenerate level at $E = \frac{h^2}{8mL^2}(9.040)$ (the state $(2,2,1)$, which is pushed to lower relative energy because compressing $z$ makes it energetically cheap to put the larger quantum number along the now-shorter, "stiffer" axis only if $n_z$ is small — here $n_z=1$ costs the least extra confinement energy). This is exactly the group-theoretic pattern found in real triply-degenerate levels (e.g., $T_{2g}$ or $t_{2u}$ orbital sets) under a tetragonal (uniaxial) distortion: a symmetry lowering from cubic ($O_h$-like) to tetragonal splits a threefold degeneracy into a $1+2$ pattern, never into three fully separate levels from a single first-order uniaxial distortion, because the $x$–$y$ symmetry ($L_x=L_y$) is preserved.

Limiting-case check: setting $\varepsilon = 0$ collapses both split energies back to $9.000$ in units of $h^2/8mL^2$, recovering the exact threefold-degenerate cubic result of Part A, confirming the perturbative splitting formula reduces correctly to the unperturbed case.

The particle-in-a-box model, in one, two, and three dimensions, is a direct application of the wave mechanics developed by Erwin Schrödinger in 1926, but its systematic chemical use — as a model for π-electron delocalization, quantum confinement, and orbital degeneracy — was popularized through mid-20th-century quantum chemistry pedagogy, most influentially in the treatments by Linus Pauling and E. Bright Wilson, and later canonized in the textbooks of Ira Levine and Donald McQuarrie that generations of chemistry students have used to connect this exactly solvable toy model to real molecular orbital theory.