Checkpoint: Solve the Box

This checkpoint pulls together everything from Module 11. You have all the machinery: the energy eigenvalues $E_n$ , the normalized eigenfunctions $\psi_n$ , zero-point energy, the node rule, and expectation values. The goal is to apply them fluently to a concrete physical system.

The infinite square well — a brief recap

A particle of mass $m$ confined to a one-dimensional box of width $L$ (with $V = 0$ inside and $V = \infty$ at the walls) has exactly the set of allowed energies

$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad n = 1, 2, 3, \ldots$

This formula follows from demanding that the wave function vanish at both walls. The boundary condition $\psi(0) = \psi(L) = 0$ forces the wave number to satisfy $k_n L = n\pi$ , and the energy is related to the wave number by $E = \hbar^2 k^2 / (2m)$ . Substituting $k_n = n\pi/L$ gives $E_n$ above.

The corresponding normalized eigenfunctions are

$\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right), \qquad 0 \le x \le L.$

A real electron in a nanometer-scale box

Consider a single electron ( $m_e = 9.109 \times 10^{-31}\,\text{kg}$ ) confined in a one-dimensional infinite well of width $L = 1\,\text{nm} = 1 \times 10^{-9}\,\text{m}$ . This is roughly the width of a small organic molecule or a quantum dot, so the numbers are physically meaningful.

The ground-state energy is

$E_1 = \frac{\pi^2\hbar^2}{2m_e L^2}.$

Inserting $\hbar = 1.0546 \times 10^{-34}\,\text{J\,s}$ , $m_e = 9.109 \times 10^{-31}\,\text{kg}$ , and $L = 1 \times 10^{-9}\,\text{m}$ :

= \frac{9.870 \times 1.112 \times 10^{-68}}{1.822 \times 10^{-48}} \approx 6.024 \times 10^{-20}\,\text{J}.$$ Converting to electron-volts (dividing by $1.602 \times 10^{-19}\,\text{J/eV}$): $$E_1 \approx 0.376\,\text{eV}.$$ This is non-negligible — thermal energy at room temperature is only about $0.025\,\text{eV}$, so a 1 nm confinement produces quantization that is easily observable. <Callout type="tip"> Because $E_n = n^2 E_1$, the energy of any level follows immediately once you know $E_1$. The $n = 3$ level has energy $E_3 = 9 \times E_1$. </Callout> ## Energy levels and their spacing The first few energy levels for this electron are | $n$ | $E_n = n^2 E_1$ | Energy (eV) | |-----|-----------------|-------------| | 1 | $E_1$ | 0.376 | | 2 | $4 E_1$ | 1.504 | | 3 | $9 E_1$ | 3.384 | | 4 | $16 E_1$ | 6.016 | The gap between consecutive levels grows: from $n = 1$ to $n = 2$ the spacing is $3E_1 \approx 1.13\,\text{eV}$, but from $n = 2$ to $n = 3$ it is $5E_1 \approx 1.88\,\text{eV}$. Generalizing, $E_{n+1} - E_n = (2n+1)E_1$, which increases without bound. This contrasts sharply with the hydrogen atom, where the level spacing decreases toward the continuum. ## Node structure and wave function shapes Each eigenfunction $\psi_n$ has exactly $n - 1$ interior nodes (zeros strictly inside the box). The ground state ($n = 1$) has no interior nodes and looks like a single arch of a sine wave. The $n = 3$ state has two interior nodes, at $x = L/3$ and $x = 2L/3$, and three arches. Equivalently, exactly $n$ half-wavelengths fit between the walls: $\lambda_n = 2L/n$, giving a de Broglie momentum $p_n = h/\lambda_n = n\pi\hbar/L$ and kinetic energy $p_n^2/(2m) = E_n$. ## Expectation values at a glance For any energy eigenstate $\psi_n$ the probability density $|\psi_n(x)|^2 = (2/L)\sin^2(n\pi x/L)$ is symmetric about the midpoint $x = L/2$, so the position expectation value is always $$\langle x \rangle_n = \frac{L}{2} \quad \text{for all } n.$$ The energy expectation value in an eigenstate is just the eigenvalue itself: $\langle E \rangle_n = E_n$. For a superposition $\Psi = c_1\psi_1 + c_2\psi_2$ with $|c_1|^2 + |c_2|^2 = 1$, the expectation value is $\langle E \rangle = |c_1|^2 E_1 + |c_2|^2 E_2$ — a weighted average, never a value that lies between or outside the allowed energies. ## Try it Use the energy formula $E_n = n^2\pi^2\hbar^2/(2m_e L^2)$ to compute the energy of the $n = 3$ level, in eV, for an electron in a box of width $L = 1\,\text{nm}$. <RunnableExample />