Particle on a chain / particle in a box

Setting up the well. Consider a particle of mass $m$ confined to $0 \le x \le L$ by infinitely high walls, so the potential is zero inside and infinite outside:

Because $V=\infty$ outside the well, the particle can never be found there, so $\phi(x)=0$ for $x\le 0$ and $x\ge L$. Continuity of $\phi$ then forces the boundary conditions $\phi(0)=0$ and $\phi(L)=0$ — these two conditions do all the work of quantization.

Solving inside the well. With $V=0$, the time-independent Schrödinger equation reduces to a simple oscillator-type equation:

where: $\hbar$ is the reduced Planck constant, $E$ is the particle's energy, and $k$ is the wavenumber the equation defines. The general solution is $\phi(x) = C\sin(kx) + D\cos(kx)$. Applying $\phi(0)=0$ forces $D=0$; applying $\phi(L)=0$ then requires $\sin(kL)=0$, which holds only for $kL = n\pi$ with $n=1,2,3,\dots$ ($n=0$ is excluded because it gives $\phi\equiv0$ everywhere — no particle at all).

Quantized energies. Since $E=\hbar^2k^2/2m$, quantizing $k$ directly quantizes the energy:

where: $n$ is the quantum number, $m$ the particle's mass, and $L$ the box width. Energy can only take these discrete values — a purely quantum result with no classical counterpart, since a classical ball bouncing between two walls can have any energy at all. The lowest allowed energy, $E_1 = \pi^2\hbar^2/2mL^2$, is strictly greater than zero. This zero-point energy is a direct consequence of confinement plus the uncertainty principle: a particle strictly localized to a region of size $L$ cannot have exactly zero momentum, so it cannot have exactly zero energy either.

The normalized wavefunctions. Normalizing $\phi_n(x)=C\sin(n\pi x/L)$ by requiring $\int_0^L|\phi_n|^2\,dx=1$ gives $C=\sqrt{2/L}$ (proved step by step below), so

Each $\phi_n$ has $n-1$ nodes strictly inside the well and $n$ lobes — exactly the standing-wave pattern of a string fixed at both ends, now reinterpreted as a probability-amplitude wave rather than a physical displacement.

Step 1 — set up the problem. An electron ($m=9.11\times10^{-31}\,\text{kg}$) is confined to a 1D box of width $L=0.20\ \text{nm}$, roughly atomic scale. Using $E_n = n^2\pi^2\hbar^2/2mL^2$ with $\hbar=1.055\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ and $L=2.0\times10^{-10}\ \text{m}$, first collect the numbers: $\hbar^2 = 1.113\times10^{-68}\ \text{J}^2\text{s}^2$, $L^2 = 4.0\times10^{-20}\ \text{m}^2$, and $2mL^2 = 2\times(9.11\times10^{-31})\times(4.0\times10^{-20}) = 7.29\times10^{-50}\ \text{kg}\,\cdot\,\text{m}^2$.

Step 2 — compute $E_1$. $E_1 = \dfrac{\pi^2 \times 1.113\times10^{-68}}{7.29\times10^{-50}} = \dfrac{1.098\times10^{-67}}{7.29\times10^{-50}} \approx 1.51\times10^{-18}\ \text{J}$. Converting to electron-volts, $E_1 = \dfrac{1.51\times10^{-18}}{1.6\times10^{-19}} \approx 9.4\ \text{eV}$.

Step 3 — find $E_2$ and the transition energy. Since $E_n \propto n^2$, doubling the quantum number quadruples the energy: $E_2 = 4E_1 \approx 37.7\ \text{eV}$. The energy released or absorbed on a transition between them is $\Delta E = E_2 - E_1 = 3E_1 \approx 28.3\ \text{eV}$.

Final answer: $E_1 \approx 9.4\ \text{eV}$, $E_2 \approx 37.7\ \text{eV}$, and a photon of energy $\Delta E \approx 28.3\ \text{eV}$ connects the $n=2$ and $n=1$ levels. This tens-of-eV spacing is enormous for a sub-nanometer box, which is exactly why quantization is unmistakable for electrons confined at atomic scales but utterly unobservable for any macroscopic box, where $L$ is large and $E_n$ spacings shrink toward zero.

Competitive stretch

Because $E_n \propto 1/L^2$, doubling the box width to $L=0.40\ \text{nm}$ (everything else unchanged) divides every $E_n$ by four, so $E_1$ drops from about $9.4\ \text{eV}$ to about $2.35\ \text{eV}$ — a quick reminder that confinement, not just the particle, controls the energy scale.

• Including $n=0$ as an allowed state — this gives $\phi\equiv0$, meaning no particle is present at all, so $n=0$ must always be excluded.
• Forgetting that $E_n \propto n^2$, not $n$ — doubling the quantum number quadruples the energy, it does not double it.
• Demanding that $d\phi/dx$ be continuous at the infinite walls — that condition does not apply there; only ordinary continuity of $\phi$ (forcing it to zero) is required, because $V\to\infty$ at the walls.
• Forgetting the explicit $L$-dependence $E_n \propto 1/L^2$ — a wider box gives closely spaced, "more classical-looking" levels, while extreme confinement produces widely separated, unmistakably quantum energy gaps.

Ask the tutor: curious how the boundary conditions and energy formula change for a finite-depth well instead of infinite walls — ask the AI tutor to walk through the comparison with you.