Periodic vs Hard-Wall Boundary Conditions and Mode Counting

To do statistical mechanics on a quantum gas you must first count states: convert a sum over the discrete single-particle levels of a box into a smooth integral over energy. But the discrete levels depend on a modelling choice you seem free to make — what happens at the walls. Two idealizations dominate. Periodic boundary conditions (Born–von Kármán) pretend the box wraps onto a torus, giving travelling plane waves $e^{i\mathbf k\cdot\mathbf r}$ with $k_i=2\pi n_i/L$ and both signs of $n_i$ allowed. Hard-wall (Dirichlet) conditions force the wavefunction to vanish at the surface, giving standing sine waves with $k_i=\pi n_i/L$ and only positive $n_i$ . These look completely different — running waves versus standing waves, all of $\mathbf k$ -space versus one octant of it. The reassuring and load-bearing fact is that the bulk density of states is identical for both. The hard-wall cells are $2^d$ times smaller in $\mathbf k$ -space, but hard-wall modes live in only $1/2^d$ of it (the positive octant); those two factors cancel exactly. What survives is the boundary-independent statement that each quantum state occupies a $\mathbf k$ -space volume $(2\pi)^d/V$ , equivalently a phase-space volume $h^d$ . This is the measure that aq-sm-density-of-states-k-space establishes and aq-sm-density-of-states-energy pushes through the dispersion, and it is what feeds the grand-canonical ensemble, the Bose–Einstein and Fermi–Dirac distributions, and the thermal de Broglie wavelength. The equivalence is exact only in the thermodynamic limit. At finite size the two spectra genuinely differ, and the difference is a surface effect: the Dirichlet count carries a negative correction that scales as area over volume, $\sim1/L$ (Weyl's law), while periodic conditions have no surface at all. For a macroscopic sample this washes out; for a nanostructure it is the physics. This article derives both quantization rules, proves their bulk equivalence, and isolates the surface term that distinguishes them.

Key takeaways

Confining a free particle discretizes its wavevectors; the spacing is set by the boundary condition, not the bulk physics.
Periodic conditions give $e^{i\mathbf k\cdot\mathbf r}$ with $k_i=2\pi n_i/L$ ( $n_i\in\mathbb Z$ ); hard walls give $\sin$ standing waves with $k_i=\pi n_i/L$ ( $n_i\in\mathbb Z^+$ ).
The hard-wall cells are $2^d$ smaller but occupy only the positive octant ( $1/2^d$ of $\mathbf k$ -space); the two factors of $2^d$ cancel, so both give the same bulk density $V/(2\pi)^d$ per mode.
The universal measure is $\mathrm dN=g_s\,V\,\mathrm d^d k/(2\pi)^d=g_s\,\mathrm d^d r\,\mathrm d^d p/h^d$ — one state per phase-space volume $h^d$ .
Periodic conditions are the working default: they keep momentum a good quantum number, kill surface terms, and make Fourier methods trivial.
Boundary-dependent corrections scale as surface/volume $\sim1/L$ (Weyl's law) and vanish as $L\to\infty$ ; they matter only for finite or mesoscopic systems.
Key idea: the number of single-particle states per momentum-space volume is a boundary-independent bulk quantity; different boundary conditions reshuffle which discrete $\mathbf k$ are allowed, but not how densely states pack per unit volume.

The core insight

The count of states per $\mathbf k$ -space volume, $\mathrm dN=g_s V\,\mathrm d^d k/(2\pi)^d=g_s\,\mathrm d^d r\,\mathrm d^d p/h^d$ , is a bulk quantity independent of the boundary condition. Periodic and hard-wall schemes redistribute the allowed $\mathbf k$ (all of $\mathbf k$ -space with cell $(2\pi/L)^d$ versus the positive octant with cell $(\pi/L)^d$ ), but the density of states per unit volume is identical because $2^d$ (finer spacing) times $1/2^d$ (octant) equals one. The only difference is a surface term $\sim1/L$ that vanishes in the thermodynamic limit.

Where this lives today

Boundary-condition bookkeeping is invisible in macroscopic thermodynamics but decisive in computation and nanoscience. Every lattice Monte Carlo and molecular-dynamics simulation uses periodic boundary conditions precisely because they have no surface: a cube of spins wraps onto a torus, every site is equivalent, and momentum-space (FFT) methods apply directly, with finite-size corrections that scale as known powers of $1/L$ for controlled extrapolation. In the opposite regime — quantum dots, nanowires, thin films — the level spacing $\sim\hbar^2\pi^2/2mL^2$ becomes comparable to $k_BT$ , the confinement is genuinely Dirichlet-like, and the discrete sine-mode spectrum is directly observable in transport and optical experiments. The universal density of states no longer suffices there; the actual finite spectrum, boundary and all, is the physics.

The physical picture

Think of the box as a musical instrument. Hard walls are a guitar string clamped at both ends: only standing waves with nodes at the walls fit, so the allowed wavelengths are $\lambda=2L/n$ . Periodic conditions are a circular loop with no edge: a travelling wave fits provided it closes on itself after going around, so $\lambda=L/n$ . Different instruments — but in a huge concert hall, the number of distinguishable pitches per unit frequency is the same. That "per unit frequency" density is the density of states, and it does not care which instrument you built. Now picture $\mathbf k$ -space tiled by cells. Periodic conditions fill all of $\mathbf k$ -space (positive and negative $k_i$ ) with cubic cells of side $2\pi/L$ . Hard walls fill only the positive octant with cells of side $\pi/L$ — cells that are $2^d$ times smaller. The octant is $1/2^d$ of the full space; the cells are $1/2^d$ of the volume. Multiply the two together and you get exactly one, leaving one state per $(2\pi/L)^d$ of full $\mathbf k$ -space either way. This is not a coincidence to memorize but a cancellation to see: the extra states you seem to gain from finer spacing are exactly the states you lose by throwing away seven of the eight octants.

Figure

Periodic uses all of k-space with wide cells; hard-wall uses one octant with narrow cells — same bulk density

The punchline: the wide full grid (periodic) and the narrow octant grid (hard wall) pack states at the same density per unit volume. The finer spacing on the right is exactly compensated by keeping only one quadrant.

Figure

Hard-wall count approaches the continuum from below with a surface deficit that shrinks as 1/L

The owned result made numerical: the hard-wall count sits below the continuum by a surface (Weyl) deficit that scales as $1/L$ — halving each time the box doubles — while the periodic count has no surface term. Both reach the same bulk density as $L\to\infty$ .

The core equations

Periodic (Born–von Kármán) conditions demand the wavefunction be unchanged under translation by $L$ , which quantizes $\mathbf k$ to a lattice covering all of $\mathbf k$ -space.

\psi_{\mathbf k}(\mathbf r)=\frac{1}{\sqrt V}\,e^{i\mathbf k\cdot\mathbf r},\qquad k_i=\frac{2\pi n_i}{L},\ n_i\in\mathbb Z \qquad (1)

where: $\psi_{\mathbf k}$ — normalized plane-wave eigenfunction; $V=L^d$ ; $n_i$ — any integer, positive, negative, or zero. Physical reading: momentum $\hbar\mathbf k$ stays a good quantum number, and left- and right-movers ( $\pm n_i$ ) are distinct states — which is why periodic conditions fill all of $\mathbf k$ -space. Hard-wall (Dirichlet) conditions demand the wavefunction vanish on every wall, selecting standing sine waves in the positive octant.

\psi_{\mathbf n}(\mathbf r)=\left(\frac{2}{L}\right)^{d/2}\prod_{i=1}^d\sin\!\left(\frac{\pi n_i x_i}{L}\right),\qquad k_i=\frac{\pi n_i}{L},\ n_i\in\{1,2,3,\dots\} \qquad (2)

where: $n_i$ — strictly positive integers (a node sits at each wall); $x_i$ — Cartesian coordinate, $0\le x_i\le L$ . Physical reading: $\sin$ (not $\cos$ ) forces $\psi=0$ at the walls, and $n_i=0$ gives the trivial zero, so only positive $n_i$ survive — the octant restriction. Despite the different spectra, both collapse to one universal continuum measure — the object that every ideal-gas calculation uses.

\sum_{\mathbf k}\;\longrightarrow\;\frac{g_s V}{(2\pi)^d}\int\mathrm d^d k=\frac{g_s}{h^d}\int\mathrm d^d r\,\mathrm d^d p \qquad (3)

where: $h=2\pi\hbar$ — Planck constant; $\mathbf p=\hbar\mathbf k$ ; $g_s=2s+1$ — spin degeneracy. Physical reading: one quantum state per phase-space volume $h^d$ — a boundary-independent, statistics-independent bulk fact. The cancellation that makes this work is a one-line identity in the cell sizes and the octant fraction.

\underbrace{\frac{1}{(\pi/L)^d}}_{\text{hard-wall cell density}}\times\underbrace{\frac{1}{2^d}}_{\text{positive octant}}=\frac{1}{(2\pi/L)^d}=\underbrace{\frac{V}{(2\pi)^d}}_{\text{periodic density}} \qquad (4)

where: the first factor counts hard-wall states per unit $\mathbf k$ -volume, the second restricts to $k_i>0$ . Physical reading: the $2^d$ from finer Dirichlet spacing exactly cancels the $1/2^d$ from keeping one octant — the heart of the boundary-independence. At finite size the two schemes differ by a surface term captured in the Weyl expansion of the integrated count.

N(\varepsilon)=\frac{g_s V}{(2\pi)^d}\,\frac{\Omega_d}{d}\,k^d\;-\;\frac{1}{4}\,\frac{g_s S}{(2\pi)^{d-1}}\,\Omega_{d-1}\,k^{d-1}+\cdots \qquad (5)

where: $N(\varepsilon)$ — number of states up to energy $\varepsilon$ (i.e. $|\mathbf k|\le k$ ); $S$ — surface area; the surface term has coefficient $-\tfrac14$ for Dirichlet walls and is absent for periodic conditions. Physical reading: the leading correction is $\propto S/V\sim1/L$ — a boundary effect that vanishes in the thermodynamic limit, leaving the bulk term universal.

Derivations

Derivation AThe two quantization rules from single-valuedness and vanishing walls

The goal is to derive both spectra from their defining boundary condition — periodic $k_i=2\pi n_i/L$ over all of $\mathbf k$ -space, and hard-wall $k_i=\pi n_i/L$ in the positive octant — and to see exactly where the factor-of-two in spacing and the octant restriction come from, because those two facts are what will cancel in the next card. We take the box spectrum's structure as given by quantum mechanics (aq-qm-particle-in-3d-box owns the standing-wave eigenfunctions $\sin(n_1\pi x/L)$ ); here we extract only the counting consequences.

Step 1. Periodic phase condition. Requiring $\psi(\mathbf r+L\hat e_i)=\psi(\mathbf r)$ for the plane wave $e^{i\mathbf k\cdot\mathbf r}$ gives $e^{ik_iL}=1$ , so $k_iL=2\pi n_i$ and $k_i=2\pi n_i/L$ with $n_i\in\mathbb Z$ . Why: the phase accrued crossing the box must be a whole number of $2\pi$ ; both signs of $n_i$ solve it, so left- and right-movers are distinct.

Step 2. Hard-wall node condition. Requiring $\psi=0$ at $x_i=0$ forces $\sin(k_ix_i)$ (not $\cos$ ); requiring $\psi=0$ at $x_i=L$ then gives $\sin(k_iL)=0$ , so $k_iL=\pi n_i$ and $k_i=\pi n_i/L$ . Why: the far wall quantizes the wavelength; $n_i=0$ gives $\sin0\equiv0$ everywhere — not a state — so it is excluded, unlike the periodic $n_i=0$ which gives the valid uniform $\mathbf k=0$ mode.

Step 3. Sign of $k_i$ for hard walls. Since $\sin(-k_ix_i)=-\sin(k_ix_i)$ is the same physical state up to an overall sign, negative $n_i$ add nothing new. Only $n_i>0$ counts: the allowed $\mathbf k$ fill the positive octant of $\mathbf k$ -space. Why: the standing wave has no direction, so $\pm k_i$ are one state, not two — the origin of the $1/2^d$ octant factor.

Step 4. Cell sizes. Periodic states sit on a lattice of spacing $2\pi/L$ over all $\mathbb R^d$ , one state per cell $(2\pi/L)^d=(2\pi)^d/V$ . Hard-wall states sit on a lattice of spacing $\pi/L$ in the octant, one per cell $(\pi/L)^d$ — a cell $2^d$ times smaller, but confined to $1/2^d$ of $\mathbf k$ -space. Why: these are precisely the two competing factors of $2^d$ whose cancellation is proved next. Verified: $e^{ikL}=1$ at $k=2\pi n/L$ , $\sin(kL)=0$ at $k=\pi n/L$ , the $n_i=0$ exclusion for hard walls versus its validity for periodic conditions, and the sign identity are checked symbolically in verify_boundary-conditions-mode-counting.py.

Go deeper · Derivation B · 1 minThe bulk count is identical; the difference is a surface term

Derivation BThe bulk count is identical; the difference is a surface term

The goal is the owned result: periodic and hard-wall boundary conditions yield the same bulk density of states, with the only difference a surface (Weyl) correction that scales as $1/L$ and vanishes in the thermodynamic limit. We count states below a fixed $k_F$ under each scheme in 3D and show they agree, then identify the leading correction as a boundary effect.

Step 1. Periodic count. The states with $|\mathbf k|\le k_F$ fill a full sphere of radius $k_F$ ; dividing by the cell $(2\pi/L)^3$ and including $g_s$ gives $N_{\rm PBC}=g_s\dfrac{\tfrac43\pi k_F^3}{(2\pi/L)^3}=\dfrac{g_sVk_F^3}{6\pi^2}$ . Why: the full sphere is the natural domain for a lattice covering all of $\mathbf k$ -space.

Step 2. Hard-wall count. Only the positive octant contributes ( $\tfrac18$ of the sphere), with the smaller cell $(\pi/L)^3$ : $N_{\rm HW}=g_s\dfrac{\tfrac18\cdot\tfrac43\pi k_F^3}{(\pi/L)^3}=\dfrac{g_sVk_F^3}{6\pi^2}$ . The $\tfrac18$ and the eight-times-denser lattice conspire to reproduce the periodic answer exactly. Why: this is the $2^3\times\tfrac18=1$ cancellation made concrete on the Fermi sphere.

Step 3. Bulk equivalence. $N_{\rm PBC}=N_{\rm HW}=g_sVk_F^3/6\pi^2$ , so the bulk density of states — and everything built from it ( $E_F$ , $C_V$ , $T_c$ ) — is boundary-independent. Why: the density of states counts states in the bulk, where the wavefunction does not sense the walls; the walls are a surface effect.

Step 4. The surface term. A finer count (Weyl's law) gives $N(\varepsilon)=\frac{g_sV}{(2\pi)^d}\frac{\Omega_d}{d}k^d-\frac14\frac{g_sS}{(2\pi)^{d-1}}\Omega_{d-1}k^{d-1}+\cdots$ , where the surface term has coefficient $-\tfrac14$ for Dirichlet walls (states are removed by the nodes) and $0$ for periodic conditions (no surface). A direct lattice-point count confirms the hard-wall total approaches the continuum from below, with the deficit $\propto S/V\sim1/L$ — halving each time the box doubles. Why: nodes at the walls suppress amplitude in a surface layer of thickness $\sim1/k$ , costing a count proportional to the surface area, negligible against the volume as $L\to\infty$ . Verified: $N_{\rm PBC}=N_{\rm HW}=g_sVk_F^3/6\pi^2$ symbolically, the $2^d\times2^{-d}=1$ cancellation for $d=1,2,3$ , and a brute-force lattice-point count showing the hard-wall deficit scaling as $1/L$ (the deficit at $L$ is $\approx4\times$ the deficit at $4L$ ) are checked in verify_boundary-conditions-mode-counting.py.

Limiting cases

The mode count behaves differently across scales; the boundary condition matters only when the box is small.

Regime	Behaviour	Meaning
Thermodynamic limit ( $L\to\infty$ )	$N_{\rm PBC}=N_{\rm HW}$ , surface term $\to0$	sums become integrals; DOS exact and boundary-independent
Mesoscopic ( $k_BT\sim\hbar^2\pi^2/2mL^2$ )	spectra differ measurably	quantum dots/nanostructures sense the actual walls
Surface (Weyl) correction	$N=N_{\rm bulk}-\tfrac14(S/\cdots)k^{d-1}+\cdots$	Dirichlet lowers the count $\propto S/L$ ; periodic has none
Classical limit ( $n\lambda^3\ll1$ )	cell $h^d$ → smooth measure $\mathrm d^3r\,\mathrm d^3p/h^3$	recovers Maxwell–Boltzmann counting with Gibbs $1/N!$

Worked examples

Example 1 — Verify the cancellation by counting states in a small box.

Step 1. Setup. Take a 3D box and count orbital states with $|\mathbf k|\le k_F$ for a physical $k_F$ such that the periodic sphere radius is $R=k_FL/2\pi=8$ . The continuum answer is $N=Vk_F^3/6\pi^2=(k_FL)^3/6\pi^2=(2\pi R)^3/6\pi^2$ .

Step 2. Continuum count. $(2\pi\cdot8)^3/6\pi^2=(50.27)^3/59.22=1.270\times10^5/59.22=\mathbf{2145}$ orbital states.

Step 3. Periodic lattice count. Summing integer points (including zero and negatives) with $n_x^2+n_y^2+n_z^2\le8^2$ gives $2109$ — a ratio $2109/2145=\mathbf{0.983}$ , already within 2% of the continuum with no surface deficit.

Step 4. Hard-wall lattice count. Summing positive integers ( $n_i\ge1$ ) with $n_x^2+n_y^2+n_z^2\le(k_FL/\pi)^2=16^2$ gives $1849$ — a ratio $1849/2145=\mathbf{0.862}$ , below the continuum by a $\sim14\%$ surface deficit. Doubling the box to $R=16$ (hard-wall radius 32) raises the hard-wall ratio to $0.930$ , and to $R=32$ gives $0.965$ : the deficit $0.138\to0.070\to0.035$ halves each doubling, exactly the $1/L$ Weyl scaling. Both schemes converge to the same bulk density.

Example 2 — The Fermi energy is boundary-independent.

Step 1. Setup. Use the boundary-independent count $N=g_sVk_F^3/6\pi^2$ to find the Fermi wavevector and energy of a free-electron gas at density $n=N/V$ with $g_s=2$ .

Step 2. Invert for $k_F$ . $n=\dfrac{2k_F^3}{6\pi^2}=\dfrac{k_F^3}{3\pi^2}$ , so $k_F=(3\pi^2 n)^{1/3}$ .

Step 3. Fermi energy. $E_F=\dfrac{\hbar^2k_F^2}{2m}=\dfrac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ — identical whether the metal's surface is modelled by hard walls or periodic conditions, because the surface term is $\sim1/L$ and vanishes for a macroscopic crystal.

Step 4. Number. For copper ( $n=8.49\times10^{28}\ \mathrm{m^{-3}}$ ) this gives $E_F=\mathbf{7.0\ eV}$ (worked in aq-sm-density-of-states-energy) — a bulk property no experiment attributes to the choice of boundary condition. This is why lattice simulations universally adopt periodic conditions: they eliminate surface artifacts while giving the exact bulk answer.

Origins · 1 minboundary conditions and how this was discovered

boundary conditions

the wall rules that quantize k · Born and von Kármán introduced periodic boundary conditions in 1912 to count lattice vibrations without the complications of real surfaces — the trick of identifying opposite faces of the crystal into a torus. The competing Dirichlet (hard-wall) conditions predate them, being the original particle-in-a-box quantization. The deep lesson, sharpened by Weyl's 1911 asymptotic law for the eigenvalue count of the Laplacian, is that the leading term depends only on the volume, while the boundary contributes a calculable surface correction — the mathematical statement of why bulk thermodynamics ignores the walls.

History

1911

— Weyl proves the asymptotic law for the eigenvalue count: leading term set by volume

1912

— Born and von Kármán introduce periodic boundary conditions for lattice dynamics

1926

— Schrödinger's particle-in-a-box gives the Dirichlet standing-wave spectrum

1928

— Sommerfeld and Bethe systematize the free-electron mode count for metals

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §1.4, §6.1; Kardar, Statistical Physics of Particles, Ch. 4, 7; Reif, Fundamentals of Statistical and Thermal Physics, §1.10, §9.9; Sethna, Statistical Mechanics, Ch. 7; Landau & Lifshitz, Statistical Physics Part 1, §37–38.
Quantum-mechanics owner: the standing-wave box eigenfunctions $\sin(n_1\pi x/L)$ and the degeneracy structure are owned by aq-qm-particle-in-3d-box; the $\mathbf k$ -space cell $(2\pi/L)^d$ and $\mathrm dN=\frac{V}{(2\pi)^3}\mathrm d^3k$ appear in aq-qm-density-of-final-states.
Leads to: the universal measure feeds the $\mathbf k$ -space density (aq-sm-density-of-states-k-space), the energy density of states (aq-sm-density-of-states-energy), the grand-canonical ensemble (aq-sm-grand-canonical-ideal-quantum-gases), and the Bose–Einstein and Fermi–Dirac distributions.
Provenance: AllQFT `sm` export (Pathria & Beale), aq-sm-boundary-conditions-mode-counting, ref STAT.QSTAT.BCMODE.1213. The quantization rules, the $2^d\times2^{-d}=1$ cancellation, the identical bulk count $N=g_sVk_F^3/6\pi^2$ , and the $1/L$ Weyl surface scaling are python-verified (checks all PASS).

Why do physicists almost always use periodic boundary conditions?

Three reasons. First, momentum $\hbar\mathbf k$ stays a good quantum number, so plane-wave (Fourier) methods work directly and the Hamiltonian is diagonal in $\mathbf k$ . Second, there is no surface, so there is no surface correction to worry about — the count is $g_sVk_F^3/6\pi^2$ exactly, and finite-size effects scale as clean powers of $1/L$ . Third, in a simulation a torus has no edges, so every site is equivalent and translation invariance is exact. Hard walls are used only when the physical confinement really is a wall — quantum dots, nanowires, thin films.

If the spectra are different, how can the thermodynamics be the same?

Because thermodynamics depends only on the density of states in the bulk, not on the individual level positions. The two spectra reshuffle which discrete $\mathbf k$ are allowed, but they pack states at the same density per unit volume — that is the $2^d\times2^{-d}=1$ cancellation. Any quantity built from an energy integral $\int g(\varepsilon)(\cdots)\mathrm d\varepsilon$ sees only that density. The individual levels differ, but no macroscopic measurement resolves a single level out of $\sim10^{23}$ ; it sees the smooth $g(\varepsilon)$ , which is boundary-independent.

Where does the factor of h^d per state come from?

From the $\mathbf k$ -space cell. One state occupies $(2\pi/L)^d$ in $\mathbf k$ -space; multiplying by the real-space volume $V=L^d$ gives $(2\pi)^d$ in $\mathbf k$ -space per state per unit real volume, i.e. one state per $(2\pi)^d/V$ of $\mathbf k$ -space, or one state per $\mathrm d^dr\,\mathrm d^dk/(2\pi)^d$ of phase space. Since $\mathbf p=\hbar\mathbf k$ and $h=2\pi\hbar$ , the momentum-space cell is $h^d$ : one quantum state per phase-space volume $h^d$ . This is the same $h^{3N}$ that Gibbs and Sackur–Tetrode insert to make the classical partition function dimensionless (aq-sm-gibbs-correction-boltzmann-counting).

Does the boundary condition affect Bose-Einstein condensation?

Only through which state is the "ground state," not through the transition itself. Periodic conditions have a clean $\mathbf k=0$ mode that becomes macroscopically occupied below $T_c$ ; hard walls have no $\mathbf k=0$ mode (it would be $\sin0\equiv0$ ), so the lowest state is the standing mode $n_i=1$ . The critical temperature $k_BT_c$ , set by the bulk density and the $\sqrt{\varepsilon}$ DOS, is the same either way. This is one reason periodic conditions are preferred for BEC calculations: they isolate the condensate mode cleanly.

What is Weyl's law, exactly?

Weyl's law (1911) is the asymptotic count of eigenvalues of the Laplacian in a domain: the number of modes with $|\mathbf k|\le k$ is $N(k)=\frac{V}{(2\pi)^d}\frac{\Omega_d}{d}k^d-\frac14\frac{S}{(2\pi)^{d-1}}\Omega_{d-1}k^{d-1}+\cdots$ . The leading (volume) term is universal — this is why bulk thermodynamics ignores the walls. The subleading (surface) term depends on the boundary: it is negative for Dirichlet walls (nodes remove states) and absent for periodic conditions. The famous question "can one hear the shape of a drum?" is precisely about reading geometry off these correction terms.

How small must a box be before the boundary condition matters?

When the level spacing near the relevant energy becomes comparable to $k_BT$ . For a free particle the spacing is $\sim\hbar^2\pi^2/2mL^2$ ; setting this equal to $k_BT$ gives a crossover length $L^*\sim\hbar\pi/\sqrt{2mk_BT}$ , which is a few times the thermal de Broglie wavelength (aq-sm-thermal-de-broglie-wavelength-sm4). For electrons at room temperature that is nanometers — hence quantum dots and nanostructures are boundary-sensitive, while any macroscopic sample is firmly in the universal-DOS regime.