Chapter 3: Solving the Schrödinger Equation — The Infinite Square Well and Other Exactly Solvable Problems

"So far, the electron is merely a hypothetical particle. To demonstrate that it is a fact of nature, we must find solutions that correspond to real physical systems." — adapted from Erwin Schrödinger, 1926

In Chapter 2, we wrote down the rules of the game: the time-dependent Schrödinger equation (TDSE) governs how quantum states evolve, and the time-independent Schrödinger equation (TISE) gives us the stationary states — the energy eigenstates that form the building blocks for any quantum state. We established that $|\Psi(x,t)|^2$ gives probability density and we learned to normalize.

But we have not yet solved anything.

This chapter changes that. We begin with the simplest possible quantum system — a particle trapped in a box — and work our way through a gallery of exactly solvable one-dimensional potentials. Along the way we will discover quantized energy levels emerging naturally from the mathematics, encounter the genuinely strange phenomenon of quantum tunneling, and learn what to do when exact solutions are impossible.

Learning paths:

• 🏃 Fast Track: If you already know the infinite square well from a modern physics course, skim Section 3.2 for notation and move to Section 3.4 onward. But do not skip the normalization details — they matter more than you think.
• 🔬 Deep Dive: Section 3.7 introduces numerical methods that will power the rest of the course. If you enjoy computation, spend extra time there and explore the code in code/project-checkpoint.py.

3.1 The Art of Solving Differential Equations in Physics

Before we dive into specific problems, let us establish the conceptual framework. Solving the Schrödinger equation is not simply a mathematical exercise — it is the fundamental method by which we extract physics from the theory. Every prediction of quantum mechanics — every spectral line, every tunneling rate, every scattering cross-section — begins with solving the Schrödinger equation for the appropriate potential.

The number of exactly solvable problems in quantum mechanics is remarkably small: the free particle, the infinite and finite square wells, the harmonic oscillator (Chapter 4), the hydrogen atom (Chapter 5), and a handful of others. Yet these solutions form the foundation for all of quantum mechanics, because every realistic problem can be understood as a perturbation of, or approximation to, one of these exactly solvable cases. This is why we study them so carefully.

The time-independent Schrödinger equation,

$$\hat{H}\psi(x) = E\psi(x) \quad \Longrightarrow \quad -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi(x) = E\psi(x),$$

is an eigenvalue equation: we seek the functions $\psi(x)$ and the values $E$ for which this holds. The potential $V(x)$ is the input — it specifies the physical situation. Everything else follows.

The general strategy is always the same:

1. Specify the potential $V(x)$ in each region of space.
2. Write the TISE in each region. Rearrange to a recognizable form.
3. Solve the ODE (ordinary differential equation) in each region.
4. Apply boundary conditions — continuity of $\psi$ and $d\psi/dx$ at interfaces; physical behavior at $x \to \pm\infty$.
5. Normalize the resulting wavefunction.
6. Interpret — what do the allowed energies and wavefunctions tell us about the physics?

Steps 4 and 5 are where quantization comes from. Not from any assumption we impose, but from the requirement that the solutions be physically sensible: single-valued, continuous, normalizable, and smooth (except where the potential has infinite discontinuities).

🔗 Connection: Recall from Chapter 2 that the general solution to the TDSE can be built from stationary states: $\Psi(x,t) = \sum_n c_n \psi_n(x) e^{-iE_n t/\hbar}$. Finding those $\psi_n$ and $E_n$ is exactly our task here.

Let us begin with the simplest conceivable problem.

3.2 The Infinite Square Well: Hello World of Quantum Mechanics

3.2.1 Setting Up the Problem

Consider a particle of mass $m$ confined to a region $0 \le x \le a$ by infinitely high walls:

$$V(x) = \begin{cases} 0, & 0 < x < a \\ \infty, & \text{otherwise} \end{cases}$$

This is the infinite square well (or "particle in a box"). The potential is zero inside the box and infinite outside. Since the particle would need infinite energy to exist outside the box, the wavefunction must vanish there:

$$\psi(x) = 0 \quad \text{for } x \le 0 \text{ and } x \ge a.$$

These are our boundary conditions.

Inside the box, $V(x) = 0$, so the TISE becomes:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \quad \Longrightarrow \quad \frac{d^2\psi}{dx^2} = -\frac{2mE}{\hbar^2}\psi.$$

Define the wave number $k$ by:

$$k^2 \equiv \frac{2mE}{\hbar^2} \quad \Longrightarrow \quad k = \frac{\sqrt{2mE}}{\hbar}.$$

Note that $E$ must be positive. If $E < 0$, the solutions would be exponentials rather than oscillatory, and we could not satisfy the boundary conditions at both walls. (This is your first taste of how boundary conditions constrain the energy.) If $E = 0$, we get $\psi = 0$ everywhere — no particle. So $E > 0$.

The ODE is:

$$\frac{d^2\psi}{dx^2} = -k^2\psi.$$

The general solution is:

$$\psi(x) = A\sin(kx) + B\cos(kx).$$

3.2.2 Applying Boundary Conditions

At $x = 0$:

$$\psi(0) = A\sin(0) + B\cos(0) = B = 0.$$

So $B = 0$ and $\psi(x) = A\sin(kx)$.

At $x = a$:

$$\psi(a) = A\sin(ka) = 0.$$

Since $A \neq 0$ (otherwise $\psi = 0$ everywhere — no particle), we need:

$$\sin(ka) = 0 \quad \Longrightarrow \quad ka = n\pi, \quad n = 1, 2, 3, \ldots$$

We exclude $n = 0$ because that gives $\psi = 0$, and we exclude negative $n$ because $\sin(-n\pi x/a) = -\sin(n\pi x/a)$ is just the same physical state with a different overall sign (recall that $\psi$ and $-\psi$ represent the same physical state since $|\psi|^2$ doesn't change).

3.2.3 Quantized Energies

From $k_n = n\pi/a$ and $k^2 = 2mE/\hbar^2$:

$$\boxed{E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}, \quad n = 1, 2, 3, \ldots}$$

This is extraordinary. The energy is quantized. The particle cannot have just any energy — only these specific values are allowed. And notice:

• The ground state energy is $E_1 = \pi^2\hbar^2/(2ma^2) > 0$. The particle is never at rest. This is the zero-point energy — a consequence of the uncertainty principle. Confining a particle to a region of width $a$ forces $\Delta x \lesssim a$, which by $\Delta x\,\Delta p \ge \hbar/2$ requires a minimum momentum, hence minimum energy.

• The energies grow as $n^2$, not $n$. The gap between adjacent levels, $E_{n+1} - E_n = (2n+1)E_1$, increases with $n$. Higher states are more widely spaced.

• For a macroscopic particle ($m \sim 1\,\text{kg}$, $a \sim 1\,\text{m}$), $E_1 \sim 10^{-68}\,\text{J}$ — so small that quantization is utterly unobservable. The quantum number $n$ for a room-temperature particle would be enormous. This is why you don't notice quantization in everyday life.

💡 Key Insight: Quantization did not come from any special assumption. It emerged automatically from requiring the wavefunction to satisfy physical boundary conditions. This is a recurring pattern in quantum mechanics: the mathematics itself enforces discrete energy levels whenever a particle is confined.

🔄 Spaced Review (Ch 1): The de Broglie relation $\lambda = h/p$ connects momentum to wavelength. In the infinite well, only certain wavelengths "fit" in the box — just like standing waves on a guitar string. The allowed wavelengths are $\lambda_n = 2a/n$, which gives $p_n = h/\lambda_n = n\pi\hbar/a$, matching $E_n = p_n^2/2m$. Wave-particle duality is not a metaphor here — it is the quantization mechanism.

Let us pause and appreciate the structure of this result. The energy spectrum has several remarkable features:

Scaling analysis. The ground state energy $E_1 = \pi^2\hbar^2/(2ma^2)$ involves the fundamental constants $\hbar$ and $m$, and the geometric parameter $a$. You can recover this scaling from dimensional analysis: the only energy you can form from $\hbar$, $m$, and $a$ is $\hbar^2/(ma^2)$, so $E_1$ must be proportional to this. The factor $\pi^2/2$ comes from the specific boundary conditions. Different geometries (cylinder, sphere, etc.) give different numerical prefactors, but the scaling $E \propto \hbar^2/(ma^2)$ is universal for confined quantum particles.

The classical limit. When is quantization unobservable? When the spacing between energy levels is small compared to the total energy. For a thermal particle at temperature $T$, the typical energy is $k_BT$. The level spacing near quantum number $n$ is $\Delta E_n = (2n+1)E_1$. Setting $E_n \approx k_BT$ gives $n \approx a\sqrt{2mk_BT}/(\pi\hbar)$. For a nitrogen molecule at room temperature in a 1 cm box: $n \approx 3 \times 10^9$. The fractional level spacing $\Delta E/E \approx 2/n \approx 10^{-9}$ — utterly unresolvable. This is why classical physics works for macroscopic objects.

Degeneracy. In one dimension, the infinite well has no degeneracy — each energy level corresponds to exactly one eigenstate. This is a general theorem: in one dimension, bound states of the Schrödinger equation are always nondegenerate (the proof relies on the Wronskian of two solutions at the same energy). In higher dimensions, degeneracy becomes the norm, not the exception — as we will see for the hydrogen atom in Chapter 5.

3.2.4 Normalization and Wavefunctions

The (unnormalized) solutions are $\psi_n(x) = A\sin(n\pi x/a)$. To normalize:

$$\int_0^a |A|^2\sin^2\left(\frac{n\pi x}{a}\right)dx = 1.$$

Using the identity $\sin^2\theta = (1 - \cos 2\theta)/2$:

$$|A|^2\int_0^a \frac{1}{2}\left[1 - \cos\left(\frac{2n\pi x}{a}\right)\right]dx = |A|^2\left[\frac{x}{2} - \frac{a}{4n\pi}\sin\left(\frac{2n\pi x}{a}\right)\right]_0^a = |A|^2\frac{a}{2}.$$

So $|A|^2 = 2/a$ and (choosing $A$ real and positive):

$$\boxed{\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right), \quad n = 1, 2, 3, \ldots}$$

🔄 Spaced Review (Ch 2): The normalization condition $\int_{-\infty}^{\infty}|\psi|^2\,dx = 1$ implements the Born rule: $|\psi(x)|^2\,dx$ is the probability of finding the particle between $x$ and $x + dx$. We proved in Chapter 2 that if $\Psi$ is normalized at $t = 0$, it stays normalized for all time.

3.2.5 Orthogonality and Completeness

The infinite well eigenstates have two crucial properties.

Orthogonality: For $m \neq n$:

$$\int_0^a \psi_m^*(x)\psi_n(x)\,dx = \frac{2}{a}\int_0^a \sin\left(\frac{m\pi x}{a}\right)\sin\left(\frac{n\pi x}{a}\right)dx = \delta_{mn},$$

where $\delta_{mn}$ is the Kronecker delta. This follows from the identity $\sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]$ and the fact that $\int_0^a \cos(k\pi x/a)\,dx = 0$ for any nonzero integer $k$.

This is not a coincidence — it is a theorem. Eigenstates of a Hermitian operator (here, $\hat{H}$) corresponding to different eigenvalues are always orthogonal. We will prove this generally in Chapter 6.

Completeness: Any "reasonable" function $f(x)$ defined on $[0,a]$ (with $f(0) = f(a) = 0$) can be expanded as:

$$f(x) = \sum_{n=1}^{\infty} c_n\psi_n(x), \quad c_n = \int_0^a \psi_n^*(x)f(x)\,dx.$$

This is just a Fourier sine series. The eigenstates form a complete basis for the space of functions satisfying these boundary conditions. This means any initial state of a particle in the infinite well can be decomposed into stationary states, and its time evolution is:

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

3.2.6 Physical Interpretation: What Do the Wavefunctions Look Like?

Let us build some intuition by examining the first several wavefunctions.

The ground state $\psi_1(x) = \sqrt{2/a}\sin(\pi x/a)$ has no nodes inside the well (nodes are points where $\psi = 0$, excluding the boundaries). It is a single half-arch, peaked at the center of the well. The probability density $|\psi_1|^2$ is largest at the center, meaning the particle is most likely to be found near the middle of the box.

The first excited state $\psi_2(x) = \sqrt{2/a}\sin(2\pi x/a)$ has one node at $x = a/2$. The probability density has two peaks, at $x = a/4$ and $x = 3a/4$. The particle is most likely to be found in one of these two regions and has zero probability of being found exactly at the center.

In general, $\psi_n$ has $n - 1$ nodes. This is a universal pattern: higher-energy states oscillate more rapidly and have more nodes. The ground state is always nodeless — a theorem that holds for any well-behaved potential, not just the infinite well.

Why does this pattern hold? Each node represents a half-wavelength. More nodes means shorter effective wavelength, which by $p = h/\lambda$ means higher momentum, and by $E = p^2/2m$ means higher energy. There is a direct, physical connection between the spatial structure of the wavefunction and its energy.

✅ Checkpoint: Before proceeding to the worked examples, make sure you can: (1) derive the quantized energies from $\sin(ka) = 0$, (2) normalize the wavefunctions, and (3) explain physically why the ground state energy is nonzero. If any of these is unclear, reread Sections 3.2.2–3.2.3 before continuing.

3.2.7 Worked Examples

Worked Example 3.1: An electron in a 1 nm box.

An electron ($m = 9.109 \times 10^{-31}\,\text{kg}$) is confined to a box of width $a = 1\,\text{nm} = 10^{-9}\,\text{m}$. Find the ground state energy and the wavelength of a photon emitted in the $n = 2 \to n = 1$ transition.

Solution:

Ground state energy:

$$E_1 = \frac{\pi^2\hbar^2}{2ma^2} = \frac{\pi^2(1.055 \times 10^{-34})^2}{2(9.109 \times 10^{-31})(10^{-9})^2}$$

$$E_1 = \frac{(1.097 \times 10^{-67})}{(1.822 \times 10^{-48})} = 6.024 \times 10^{-20}\,\text{J} = 0.376\,\text{eV}.$$

The $n=2$ energy is $E_2 = 4E_1 = 1.504\,\text{eV}$.

Photon energy for the $2 \to 1$ transition: $\Delta E = E_2 - E_1 = 3E_1 = 1.128\,\text{eV}$.

Photon wavelength:

$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.128 \times 1.602 \times 10^{-19}} = \frac{1.988 \times 10^{-25}}{1.807 \times 10^{-19}} = 1.10\,\mu\text{m}.$$

This is near-infrared light — consistent with what we observe from quantum dots of this size.

📊 By the Numbers: Quantum dots (semiconductor nanocrystals) are real-world particles in boxes. A 2 nm CdSe quantum dot emits blue light (~480 nm). A 6 nm dot emits red (~620 nm). The infinite well model, despite its simplicity, captures this size-dependent color remarkably well.

Worked Example 3.2: Time evolution of a superposition state.

A particle in an infinite well at $t = 0$ has the wavefunction $\Psi(x,0) = A[\psi_1(x) + \psi_2(x)]$.

(a) Find $A$. (b) Find $\Psi(x,t)$. (c) Calculate $\langle x \rangle$ as a function of time.

Solution:

(a) By orthonormality, $\int|\Psi|^2\,dx = |A|^2(\langle\psi_1|\psi_1\rangle + \langle\psi_1|\psi_2\rangle + \langle\psi_2|\psi_1\rangle + \langle\psi_2|\psi_2\rangle) = |A|^2(1 + 0 + 0 + 1) = 2|A|^2$. So $A = 1/\sqrt{2}$.

(b) Each stationary state picks up its own time-dependent phase:

$$\Psi(x,t) = \frac{1}{\sqrt{2}}\left[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}\right].$$

(c) This requires the integral $\langle x \rangle = \int_0^a x|\Psi(x,t)|^2\,dx$. Expanding:

$$|\Psi|^2 = \frac{1}{2}\left[|\psi_1|^2 + |\psi_2|^2 + \psi_1\psi_2\left(e^{i(E_2 - E_1)t/\hbar} + e^{-i(E_2 - E_1)t/\hbar}\right)\right]$$

$$= \frac{1}{2}\left[|\psi_1|^2 + |\psi_2|^2 + 2\psi_1\psi_2\cos(\omega_{21}t)\right],$$

where $\omega_{21} = (E_2 - E_1)/\hbar = 3\pi^2\hbar/(2ma^2)$.

The "diagonal" integrals give $\int x|\psi_n|^2\,dx = a/2$ (by symmetry of $\sin^2$ about the midpoint). The "cross" term:

$$\int_0^a x\psi_1\psi_2\,dx = \frac{2}{a}\int_0^a x\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)dx = -\frac{16a}{9\pi^2}.$$

(This integral can be evaluated by writing $\sin\alpha\sin\beta$ as a sum of cosines and integrating by parts.)

Therefore:

$$\langle x \rangle = \frac{a}{2} - \frac{16a}{9\pi^2}\cos(\omega_{21}t).$$

The expectation value oscillates about the midpoint $a/2$ with amplitude $16a/(9\pi^2) \approx 0.180a$ and angular frequency $\omega_{21}$. The particle "sloshes" back and forth — this is a quantum version of a classical particle bouncing between walls.

⚠️ Common Misconception: Students often expect a stationary state to have the particle sitting still at the center of the well. But $\langle x \rangle = a/2$ for every stationary state does not mean the particle is "at" $a/2$. The probability distribution is spread across the well. The expectation value is the average of many measurements, not the location of the particle.

Worked Example 3.3: Uncertainty product for the ground state.

Let us calculate $\sigma_x\sigma_p$ for $\psi_1$ and check the uncertainty principle.

We need $\langle x \rangle$, $\langle x^2 \rangle$, $\langle p \rangle$, and $\langle p^2 \rangle$.

For any stationary state, by the symmetry of $\sin^2(n\pi x/a)$ about $x = a/2$: $\langle x \rangle = a/2$.

$\langle x^2 \rangle = \frac{2}{a}\int_0^a x^2\sin^2\left(\frac{\pi x}{a}\right)dx$. Using the identity $\sin^2\theta = (1 - \cos 2\theta)/2$ and integrating by parts twice:

$$\langle x^2 \rangle = a^2\left(\frac{1}{3} - \frac{1}{2\pi^2}\right).$$

So $\sigma_x^2 = \langle x^2 \rangle - \langle x \rangle^2 = a^2\left(\frac{1}{3} - \frac{1}{2\pi^2} - \frac{1}{4}\right) = a^2\left(\frac{1}{12} - \frac{1}{2\pi^2}\right)$.

For momentum: $\langle p \rangle = 0$ (as argued above). And $\langle p^2 \rangle = 2mE_1 = \pi^2\hbar^2/a^2$ (since $\hat{p}^2/(2m) = \hat{H}$ inside the well, so $\langle p^2 \rangle = 2m\langle H \rangle = 2mE_1$ for a stationary state).

So $\sigma_p^2 = \langle p^2 \rangle = \pi^2\hbar^2/a^2$.

Therefore:

$$\sigma_x\sigma_p = a\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}} \cdot \frac{\pi\hbar}{a} = \pi\hbar\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}} = \hbar\sqrt{\frac{\pi^2}{12} - \frac{1}{2}} \approx \hbar\sqrt{0.822 - 0.500} \approx 0.568\hbar.$$

Since $0.568\hbar > \hbar/2 = 0.500\hbar$, the uncertainty principle is satisfied. The infinite well ground state is close to, but not quite at, the minimum uncertainty. The minimum-uncertainty state is the Gaussian wave packet (Section 3.3.3), not the sinusoidal eigenstate.

🔄 Check Your Understanding 3.1: For the infinite well, what is $\langle p \rangle$ for the stationary state $\psi_n$? (Answer: zero. Why? Because the probability distribution is symmetric, and the particle is equally likely to be moving left as right. Alternatively: $\psi_n$ is real, and you can show that $\langle\hat{p}\rangle = 0$ for any real wavefunction.)

3.3 The Free Particle and Wave Packets

3.3.1 Plane Wave Solutions

A free particle has $V(x) = 0$ everywhere. The TISE becomes:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \quad \Longrightarrow \quad \frac{d^2\psi}{dx^2} = -k^2\psi, \quad k = \frac{\sqrt{2mE}}{\hbar}.$$

The general solution is $\psi(x) = Ae^{ikx} + Be^{-ikx}$. Unlike the infinite well, there are no walls, no boundary conditions to impose, and no quantization. The energy $E = \hbar^2 k^2/(2m)$ can take any positive value — the spectrum is continuous.

The full time-dependent solution for a right-moving wave is:

$$\Psi_k(x,t) = Ae^{i(kx - \omega t)}, \quad \omega = \frac{\hbar k^2}{2m} = \frac{E}{\hbar}.$$

This is a plane wave — a wave of definite wavelength $\lambda = 2\pi/k$, definite momentum $p = \hbar k$, and definite energy $E = \hbar^2k^2/(2m)$. But there is a problem.

3.3.2 The Normalizability Problem

Can we normalize a plane wave?

$$\int_{-\infty}^{\infty}|Ae^{ikx}|^2\,dx = |A|^2\int_{-\infty}^{\infty}dx = \infty.$$

No! A plane wave is not normalizable. This means it cannot represent a physical particle — you cannot have a particle with a perfectly definite momentum spread uniformly over all of space.

This is not a failure of quantum mechanics. It is the uncertainty principle at work: a state of perfectly definite momentum ($\Delta p = 0$) must have completely indefinite position ($\Delta x = \infty$). A plane wave is a mathematical idealization, useful for calculations but not a physical state.

🔄 Spaced Review (Ch 2): The Born rule requires $\int|\psi|^2\,dx = 1$ for a valid probability interpretation. Non-normalizable "states" like plane waves are not in the Hilbert space of physical states. They are useful as building blocks, but not as standalone descriptions of particles.

3.3.3 Wave Packets: The Physical Solution

The solution is to build a wave packet — a superposition of plane waves that localizes the particle:

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx - \omega(k)t)}\,dk,$$

where $\phi(k)$ is the momentum-space wavefunction (the Fourier transform of $\Psi(x,0)$):

$$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Psi(x,0)e^{-ikx}\,dx.$$

A wave packet is normalizable. It has a finite spread in both position and momentum, satisfying the uncertainty principle.

Worked Example 3.3: The Gaussian wave packet.

At $t = 0$, consider a Gaussian wave packet centered at $x = 0$ with mean momentum $p_0 = \hbar k_0$:

$$\Psi(x,0) = \left(\frac{1}{2\pi\sigma_x^2}\right)^{1/4}\exp\left(-\frac{x^2}{4\sigma_x^2}\right)e^{ik_0 x}.$$

This is a "bump" of width $\sigma_x$ moving to the right with average momentum $\hbar k_0$. Its Fourier transform is also Gaussian:

$$\phi(k) = \left(\frac{2\sigma_x^2}{\pi}\right)^{1/4}\exp\left[-\sigma_x^2(k - k_0)^2\right].$$

The width in momentum space is $\sigma_k = 1/(2\sigma_x)$, so $\sigma_x\sigma_k = 1/2$, giving:

$$\Delta x\,\Delta p = \sigma_x \cdot \hbar\sigma_k = \frac{\hbar}{2},$$

which saturates the uncertainty principle. The Gaussian wave packet is the minimum-uncertainty state.

3.3.4 Group Velocity vs. Phase Velocity

For the free-particle dispersion relation $\omega = \hbar k^2/(2m)$:

• Phase velocity (speed of individual wave crests): $v_{\text{phase}} = \omega/k = \hbar k/(2m) = p/(2m)$.
• Group velocity (speed of the wave packet envelope): $v_{\text{group}} = d\omega/dk = \hbar k/m = p/m$.

The group velocity $v_{\text{group}} = p/m$ is the classical velocity. The phase velocity is half of this — the individual crests move at half the speed of the envelope. This seems paradoxical until you realize that the physically meaningful quantity is the group velocity, which determines how fast the probability distribution moves.

3.3.5 Wave Packet Spreading

A Gaussian wave packet for a free particle evolves as:

$$|\Psi(x,t)|^2 = \frac{1}{\sqrt{2\pi}\sigma(t)}\exp\left[-\frac{(x - v_g t)^2}{2\sigma(t)^2}\right],$$

where $\sigma(t) = \sigma_x\sqrt{1 + (t/\tau)^2}$ and $\tau = 2m\sigma_x^2/\hbar$.

The packet moves at the group velocity $v_g = \hbar k_0/m$ but spreads over time. After time $\tau$, the width has increased by a factor of $\sqrt{2}$. This spreading is a purely quantum effect with no classical analogue — a classical particle either moves as a point or doesn't spread.

📊 By the Numbers: For an electron with $\sigma_x = 1\,\text{nm}$: $\tau \approx 3.5 \times 10^{-15}\,\text{s}$ — the packet doubles in width in a few femtoseconds. For a baseball ($m = 0.145\,\text{kg}$, $\sigma_x = 10^{-10}\,\text{m}$): $\tau \approx 5 \times 10^{13}\,\text{s}$ — about 1.6 million years. Quantum spreading is irrelevant for macroscopic objects.

3.3.6 Why Spreading Happens: Dispersion

Wave packet spreading is a consequence of dispersion — the fact that different frequency components of the packet travel at different speeds. For the free particle, $\omega = \hbar k^2/(2m)$, so $d\omega/dk = \hbar k/m$ depends on $k$. A broad-$k$ packet (narrow in space) has components that move at very different speeds, causing rapid spreading. A narrow-$k$ packet (broad in space) has components that move at similar speeds, causing slow spreading.

This is the same physics that causes a short pulse of light to spread as it travels through a dispersive medium (like glass). In optics, this is called group velocity dispersion (GVD). The mathematical structure is identical — only the dispersion relation changes.

For comparison, light in vacuum has $\omega = ck$ (no dispersion), so light pulses in vacuum do not spread. The de Broglie matter wave always has a quadratic dispersion relation, so matter wave packets always spread in free space. There is no non-dispersive regime for free-particle matter waves.

Worked Example 3.4: Spreading of an electron wave packet.

An electron is prepared in a Gaussian wave packet with $\sigma_x = 5\,\text{nm}$ and mean momentum corresponding to $E = 1\,\text{eV}$.

(a) Calculate the spreading time $\tau$.

$$\tau = \frac{2m\sigma_x^2}{\hbar} = \frac{2(9.109 \times 10^{-31})(5 \times 10^{-9})^2}{1.055 \times 10^{-34}} = \frac{4.555 \times 10^{-47}}{1.055 \times 10^{-34}} = 4.32 \times 10^{-13}\,\text{s} = 0.43\,\text{ps}.$$

(b) After traveling 1 cm (a macroscopic distance), how much has it spread?

The travel time is $t = L/v_g$, where $v_g = \sqrt{2E/m} = \sqrt{2(1.602 \times 10^{-19})/(9.109 \times 10^{-31})} = 5.93 \times 10^5\,\text{m/s}$.

$t = 0.01/(5.93 \times 10^5) = 1.69 \times 10^{-8}\,\text{s}$.

$t/\tau = 1.69 \times 10^{-8}/(4.32 \times 10^{-13}) = 3.91 \times 10^4$.

$\sigma(t) = \sigma_x\sqrt{1 + (t/\tau)^2} \approx \sigma_x(t/\tau) = (5\,\text{nm})(3.91 \times 10^4) = 0.20\,\text{mm}$.

The packet, originally 5 nm wide, has spread to 0.2 mm after traveling just 1 cm. Electron wave packets spread dramatically over macroscopic distances. This is why electron optics (electron microscopes, etc.) must carefully control beam coherence.

🔄 Check Your Understanding 3.2: A free particle's energy spectrum is continuous, while the infinite well's is discrete. What physical difference accounts for this? (Answer: the infinite well confines the particle, restricting which wavelengths "fit." The free particle has no such constraint. Confinement → quantization.)

3.4 The Finite Square Well

3.4.1 Setting Up the Problem

The infinite well is elegant but unphysical — no potential is truly infinite. A more realistic model is the finite square well:

$$V(x) = \begin{cases} -V_0, & -a \le x \le a \\ 0, & |x| > a \end{cases}$$

where $V_0 > 0$ is the well depth and $2a$ is the well width. (Note the different convention: the well is centered at $x = 0$ and extends from $-a$ to $a$. This exploits the symmetry of the potential.)

We seek bound states with $-V_0 < E < 0$. (States with $E > 0$ are scattering states — they correspond to particles that approach the well from far away and are not confined.)

3.4.2 Solving in Each Region

Inside the well ($|x| \le a$): $V = -V_0$, so the TISE is:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} - V_0\psi = E\psi \quad \Longrightarrow \quad \frac{d^2\psi}{dx^2} = -l^2\psi,$$

where $l = \sqrt{2m(E + V_0)}/\hbar$. Since $E > -V_0$, we have $l^2 > 0$ and the solutions are sinusoidal:

$$\psi_{\text{inside}}(x) = C\sin(lx) + D\cos(lx).$$

Outside the well ($|x| > a$): $V = 0$, so:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \quad \Longrightarrow \quad \frac{d^2\psi}{dx^2} = \kappa^2\psi,$$

where $\kappa = \sqrt{-2mE}/\hbar > 0$ (remember $E < 0$ for bound states). The solutions are real exponentials:

$$\psi_{\text{outside}}(x) = Fe^{\kappa x} + Ge^{-\kappa x}.$$

For $x > a$, we need $\psi \to 0$ as $x \to +\infty$, so $F = 0$ in the right region. For $x < -a$, we need $\psi \to 0$ as $x \to -\infty$, so $G = 0$ in the left region.

3.4.3 Even and Odd Solutions

The potential $V(x)$ is symmetric: $V(-x) = V(x)$. A theorem we will prove in Chapter 10 states that for a symmetric potential, the energy eigenstates can be chosen to have definite parity — either even ($\psi(-x) = \psi(x)$) or odd ($\psi(-x) = -\psi(x)$).

Even solutions ($\psi(-x) = \psi(x)$): Inside the well, only the cosine survives: $\psi = D\cos(lx)$. Outside:

$$\psi(x) = \begin{cases} Fe^{\kappa x}, & x < -a \\ D\cos(lx), & -a \le x \le a \\ Fe^{-\kappa x}, & x > a \end{cases}$$

Matching $\psi$ and $d\psi/dx$ at $x = a$:

$$D\cos(la) = Fe^{-\kappa a}, \quad -Dl\sin(la) = -F\kappa e^{-\kappa a}.$$

Dividing:

$$l\tan(la) = \kappa. \quad \text{(even solutions)}$$

Odd solutions ($\psi(-x) = -\psi(x)$): Inside, only the sine survives: $\psi = C\sin(lx)$. Matching at $x = a$:

$$-l\cot(la) = \kappa. \quad \text{(odd solutions)}$$

3.4.4 Transcendental Equations and Graphical Solution

These are transcendental equations — they have no closed-form solutions. We must solve them graphically or numerically. Define dimensionless variables:

$$z = la, \quad z_0 = \frac{a}{\hbar}\sqrt{2mV_0}.$$

Then $\kappa a = \sqrt{z_0^2 - z^2}$ (since $l^2 + \kappa^2 = 2mV_0/\hbar^2$), and the equations become:

$$\tan z = \frac{\sqrt{z_0^2 - z^2}}{z} \quad \text{(even)}, \qquad -\cot z = \frac{\sqrt{z_0^2 - z^2}}{z} \quad \text{(odd)}.$$

To solve graphically, plot $\tan z$ (or $-\cot z$) and $\sqrt{z_0^2 - z^2}/z$ on the same axes. The intersections give the allowed energies.

Key features:

• The number of bound states is finite. There is always at least one bound state (the ground state, which is even). Additional states appear as $z_0$ increases — as the well gets deeper or wider.

• The number of bound states is approximately $1 + \lfloor 2z_0/\pi \rfloor$, where $\lfloor \cdot \rfloor$ is the floor function.

• Each new state alternates between even and odd parity.

• In the limit $V_0 \to \infty$ (with the well centered at $x = 0$), we recover the infinite well energies.

💡 Key Insight: The wavefunctions of the finite well "leak" into the classically forbidden region ($|x| > a$). The particle has a nonzero probability of being found outside the well, in a region where classically it could never exist because its kinetic energy would be negative. This is quantum penetration, and it is the precursor to tunneling.

Worked Example 3.4: How many bound states?

A finite well has $V_0 = 50\,\text{eV}$ and $a = 0.3\,\text{nm}$, with an electron. How many bound states exist?

Solution:

$$z_0 = \frac{a}{\hbar}\sqrt{2mV_0} = \frac{0.3 \times 10^{-9}}{1.055 \times 10^{-34}}\sqrt{2(9.109 \times 10^{-31})(50 \times 1.602 \times 10^{-19})}$$

$$= (2.844 \times 10^{24})(3.817 \times 10^{-24}) = 2.80.$$

Number of bound states $\approx 1 + \lfloor 2(2.80)/\pi\rfloor = 1 + \lfloor 1.78\rfloor = 1 + 1 = 2$.

So there are 2 bound states: the ground state (even) and one excited state (odd).

⚠️ Common Misconception: Students sometimes think the finite well is "just like the infinite well but with slightly different energies." This is wrong in an important way: the finite well has exponential tails outside the well. These tails mean the particle can be found in classically forbidden regions, they reduce the effective confinement, and they lower the bound state energies below the corresponding infinite-well values. The physics is qualitatively different.

Worked Example 3.5: Wavefunction of the finite well ground state.

Let us sketch what the ground state wavefunction of the finite well looks like and understand how it differs from the infinite well.

For the ground state (even parity):

• Inside the well ($|x| \le a$): $\psi(x) = D\cos(lx)$. This is a cosine function centered at the origin, oscillating with wave number $l$.
• Outside the well ($|x| > a$): $\psi(x) = Fe^{-\kappa|x|}$. This is an exponentially decaying function.

The key difference from the infinite well: the wavefunction does not vanish at the well edges. Instead, it smoothly joins onto the exponential tails. The cosine inside has slightly less than one quarter-wavelength between the center and the edge (compared to exactly one quarter-wavelength for the infinite well ground state), which means the effective wavelength is longer, the momentum is lower, and the energy is lower than the corresponding infinite well state.

The penetration depth $\delta = 1/\kappa$ determines how far the tails extend. For a deep well ($V_0 \gg |E|$), $\kappa$ is large and the tails are short — approaching the infinite well limit. For a shallow well ($V_0$ barely above $|E|$), $\kappa$ is small and the tails extend far — the particle is barely bound and is "almost free."

⚖️ Interpretation: The exponential tails represent the particle existing in a region where, classically, its kinetic energy would be negative: $KE = E - V(x) < 0$ for $|x| > a$. This is not a violation of energy conservation. The kinetic energy operator $\hat{p}^2/(2m)$ always has a non-negative expectation value for any state, but it does not have a definite value in a region where the wavefunction is non-oscillatory. The concept of a "local kinetic energy" is subtle in quantum mechanics — we will return to this in Chapter 6 when we discuss operator formalism.

🔄 Check Your Understanding 3.3: Why must a 1D symmetric finite well always have at least one bound state? (Hint: for any $z_0 > 0$, no matter how small, the functions $\tan z$ and $\sqrt{z_0^2 - z^2}/z$ always intersect at least once in $0 < z < \pi/2$.)

3.5 The Step Potential: Reflection and Transmission

3.5.1 Setting Up the Problem

Consider a particle incident from the left on a potential step:

$$V(x) = \begin{cases} 0, & x < 0 \\ V_0, & x > 0 \end{cases}$$

with $V_0 > 0$. This models a sudden change in potential energy — a particle approaching a barrier.

There are two cases: $E > V_0$ (particle has enough energy to classically cross the step) and $E < V_0$ (particle does not). Both lead to surprising results.

3.5.2 Case 1: $E > V_0$ — Partial Reflection

In classical mechanics, a particle with $E > V_0$ would simply slow down and continue past the step. Every particle gets through. Quantum mechanics tells a different story.

Region I ($x < 0$, $V = 0$):

$$\frac{d^2\psi}{dx^2} = -k_1^2\psi, \quad k_1 = \frac{\sqrt{2mE}}{\hbar}.$$

General solution: $\psi_I = Ae^{ik_1 x} + Be^{-ik_1 x}$ (incident + reflected waves).

Region II ($x > 0$, $V = V_0$):

$$\frac{d^2\psi}{dx^2} = -k_2^2\psi, \quad k_2 = \frac{\sqrt{2m(E - V_0)}}{\hbar}.$$

General solution: $\psi_{II} = Ce^{ik_2 x} + De^{-ik_2 x}$. We set $D = 0$ because there is no wave coming from the right (nothing to reflect off of on the right side).

Boundary conditions at $x = 0$ (continuity of $\psi$ and $d\psi/dx$):

$$A + B = C, \quad ik_1(A - B) = ik_2 C.$$

Solving for $B/A$ and $C/A$:

$$\frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2}, \quad \frac{C}{A} = \frac{2k_1}{k_1 + k_2}.$$

The reflection coefficient $R$ and transmission coefficient $T$ are the ratios of probability current:

$$R = \left|\frac{B}{A}\right|^2 = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^2, \quad T = \frac{k_2}{k_1}\left|\frac{C}{A}\right|^2 = \frac{4k_1 k_2}{(k_1 + k_2)^2}.$$

The factor $k_2/k_1$ in the transmission coefficient accounts for the different velocities in the two regions. You can verify that $R + T = 1$, as it must be (probability is conserved).

🔴 Warning: A common error is to write $T = |C/A|^2$. This is wrong because the transmitted wave has a different speed than the incident wave. The probability current (probability per unit time crossing a point) is $j = (\hbar k/m)|A|^2$, so $T = j_{\text{trans}}/j_{\text{inc}} = (k_2/k_1)|C/A|^2$.

Worked Example 3.5: Reflection where there "shouldn't" be any.

An electron with kinetic energy $E = 2V_0$ hits a potential step of height $V_0$. What fraction is reflected?

Solution:

$k_1 = \sqrt{2mE}/\hbar$ and $k_2 = \sqrt{2m(E-V_0)}/\hbar = \sqrt{2mV_0}/\hbar = k_1/\sqrt{2}$.

$$R = \left(\frac{k_1 - k_1/\sqrt{2}}{k_1 + k_1/\sqrt{2}}\right)^2 = \left(\frac{1 - 1/\sqrt{2}}{1 + 1/\sqrt{2}}\right)^2 = \left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right)^2.$$

Rationalizing: $(\sqrt{2} - 1)/(\sqrt{2} + 1) = (\sqrt{2} - 1)^2/(2 - 1) = 3 - 2\sqrt{2} \approx 0.172$.

$$R \approx (0.172)^2 \approx 0.029 = 2.9\%.$$

About 3% of the probability is reflected, even though the particle has twice the energy needed to clear the step. Classically, $R = 0$ for $E > V_0$. This quantum partial reflection has no classical analogue.

💡 Key Insight: This partial reflection is the quantum analogue of the partial reflection of light at a glass-air interface. When light passes from glass (slower propagation) to air (faster propagation) or vice versa, some light is always reflected — even at normal incidence — unless the refractive indices are perfectly matched. The Fresnel equations of optics have exactly the same mathematical structure as our quantum reflection coefficients, with the refractive index ratio $n_1/n_2$ replaced by the wave number ratio $k_1/k_2$. This is not a coincidence: both phenomena arise from matching wave solutions at a discontinuity.

Worked Example 3.6: Limiting behavior of the step potential.

(a) Show that $R \to 0$ as $E \to \infty$ (high-energy limit).

As $E \gg V_0$: $k_2 \approx k_1\sqrt{1 - V_0/E} \approx k_1(1 - V_0/2E)$. Then $k_1 - k_2 \approx k_1 V_0/(2E) \ll k_1$ and $k_1 + k_2 \approx 2k_1$, so $R \approx V_0^2/(16E^2) \to 0$. A very energetic particle barely notices the step.

(b) Show that $R \to 1$ as $E \to V_0^+$ (threshold limit).

As $E \to V_0$: $k_2 \to 0$, so $R = (k_1 - k_2)^2/(k_1 + k_2)^2 \to k_1^2/k_1^2 = 1$. Just above threshold, nearly all the particle is reflected. This makes sense: the transmitted wave has nearly zero momentum and cannot carry much probability current.

3.5.3 Case 2: $E < V_0$ — Exponential Decay

Now $E < V_0$. Classically, the particle bounces back. Quantum mechanically?

In Region II, $E - V_0 < 0$, so the wave equation gives:

$$\frac{d^2\psi}{dx^2} = \kappa^2\psi, \quad \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}.$$

The solution is $\psi_{II} = Ce^{-\kappa x}$ (we discard $e^{+\kappa x}$ because it blows up as $x \to \infty$).

The wavefunction penetrates into the classically forbidden region with a characteristic penetration depth $\delta = 1/\kappa = \hbar/\sqrt{2m(V_0 - E)}$. The probability of finding the particle at position $x > 0$ falls off as $e^{-2\kappa x}$.

For the step potential with $E < V_0$, the reflection coefficient is $R = 1$. The particle is always reflected — eventually. But it does penetrate a distance $\sim \delta$ into the forbidden region before turning around.

🧪 Experiment: This penetration is not just mathematics. In the attenuated total internal reflection (ATR) spectroscopy technique, light undergoes total internal reflection at a glass-sample interface, but the evanescent wave penetrating into the sample (~100 nm) interacts with the material and provides spectroscopic information. The physics is identical: wave penetration into a classically forbidden region.

🔄 Check Your Understanding 3.4: For the step potential with $E < V_0$, we found $R = 1$ (total reflection). Where does the transmitted probability current go? (Answer: there is no net probability current in Region II. The wavefunction $Ce^{-\kappa x}$ is real, and the probability current for a real wavefunction is zero. The particle "samples" the forbidden region but always returns.)

3.6 Quantum Tunneling: Through the Classically Forbidden

🧩 Productive Struggle: Before reading this section, consider the following setup: a particle approaches a barrier of height $V_0$ and width $d$, with energy $E < V_0$. Based on what you learned in Section 3.5, the particle penetrates into the barrier with an exponentially decaying wavefunction. But what if the barrier is thin enough that the wavefunction hasn't decayed to zero by the time it reaches the other side? What do you think happens?

Take a moment. Draw a picture. Then read on.

3.6.1 The Rectangular Barrier

Consider a potential barrier:

$$V(x) = \begin{cases} 0, & x < 0 \\ V_0, & 0 \le x \le d \\ 0, & x > d \end{cases}$$

with $E < V_0$. We need three regions.

Region I ($x < 0$): $\psi_I = Ae^{ik x} + Be^{-ikx}$, where $k = \sqrt{2mE}/\hbar$.

Region II ($0 \le x \le d$): $\psi_{II} = Ce^{\kappa x} + De^{-\kappa x}$, where $\kappa = \sqrt{2m(V_0 - E)}/\hbar$.

Region III ($x > d$): $\psi_{III} = Fe^{ikx}$ (transmitted wave only; no wave coming from the right).

Boundary conditions at $x = 0$ and $x = d$ (four equations in five unknowns — we solve for ratios relative to $A$):

At $x = 0$: $$A + B = C + D, \quad ik(A - B) = \kappa(C - D).$$

At $x = d$: $$Ce^{\kappa d} + De^{-\kappa d} = Fe^{ikd}, \quad \kappa(Ce^{\kappa d} - De^{-\kappa d}) = ikFe^{ikd}.$$

After considerable algebra (matching four boundary conditions, eliminating $B$, $C$, $D$ in favor of $F/A$), the transmission coefficient is:

$$\boxed{T = \frac{1}{1 + \frac{V_0^2\sinh^2(\kappa d)}{4E(V_0 - E)}}}$$

where $\sinh(\kappa d) = (e^{\kappa d} - e^{-\kappa d})/2$.

3.6.2 The Tunneling Approximation

For a "thick" barrier ($\kappa d \gg 1$), $\sinh(\kappa d) \approx e^{\kappa d}/2$, and:

$$T \approx \frac{16E(V_0 - E)}{V_0^2}e^{-2\kappa d}.$$

The transmission probability decreases exponentially with barrier width $d$ and with $\kappa = \sqrt{2m(V_0 - E)}/\hbar$. The dominant factor is $e^{-2\kappa d}$ — everything in front is order unity.

This is quantum tunneling: a particle passes through a barrier that it classically could not surmount. It is not going "over" the barrier. It is not boring a hole through it. The wavefunction simply does not vanish inside the barrier — it decays exponentially, and if the barrier is thin enough, a nonzero amplitude emerges on the other side.

💡 Key Insight: Tunneling is not a metaphor. It is a direct, measurable consequence of the wave nature of matter. The particle does not "borrow" energy to get over the barrier (a common but incorrect description). At no point does the particle have energy greater than $E$. The tunneling probability depends on three things: (1) the barrier height $V_0 - E$, (2) the barrier width $d$, and (3) the particle mass $m$. Heavier particles tunnel less — this is why protons tunnel much less readily than electrons.

Worked Example 3.6: Tunneling probability for an electron.

An electron with energy $E = 5\,\text{eV}$ encounters a barrier of height $V_0 = 10\,\text{eV}$ and width $d = 0.5\,\text{nm}$. Calculate the tunneling probability.

Solution:

$$\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar} = \frac{\sqrt{2(9.109 \times 10^{-31})(5 \times 1.602 \times 10^{-19})}}{1.055 \times 10^{-34}}$$

$$= \frac{\sqrt{1.460 \times 10^{-48}}}{1.055 \times 10^{-34}} = \frac{1.208 \times 10^{-24}}{1.055 \times 10^{-34}} = 1.145 \times 10^{10}\,\text{m}^{-1}.$$

$$\kappa d = (1.145 \times 10^{10})(0.5 \times 10^{-9}) = 5.73.$$

$$\sinh^2(\kappa d) = \sinh^2(5.73) \approx (153.0)^2 = 23{,}400.$$

$$T = \frac{1}{1 + \frac{(10)^2 \times 23{,}400}{4(5)(5)}} = \frac{1}{1 + 23{,}400} \approx 4.3 \times 10^{-5}.$$

About 1 in 23,000 electrons gets through. Small, but measurably nonzero.

Worked Example 3.7: Why protons don't tunnel (easily).

Repeat the calculation for a proton ($m_p = 1.673 \times 10^{-27}\,\text{kg}$, about 1836 times heavier than an electron) with the same barrier.

Solution:

$\kappa$ scales as $\sqrt{m}$, so $\kappa_p = \kappa_e \sqrt{m_p/m_e} = (1.145 \times 10^{10})\sqrt{1836} = (1.145 \times 10^{10})(42.85) = 4.91 \times 10^{11}\,\text{m}^{-1}$.

$\kappa_p d = 245$. Then $T \approx e^{-490} \approx 10^{-213}$.

Essentially zero. The proton's larger mass makes its wavefunction decay so rapidly inside the barrier that tunneling is negligible. This exponential sensitivity to mass is why tunneling is primarily an electron phenomenon in most contexts.

3.6.3 Tunneling in Nature and Technology

Tunneling is not a curiosity — it is everywhere.

Alpha decay: In the 1920s, George Gamow (and independently Condon and Gurney) explained alpha decay as tunneling. An alpha particle inside a nucleus is trapped by the nuclear potential (attractive, strong force) behind a Coulomb barrier (repulsive, electromagnetic). The alpha particle tunnels through this barrier. The exponential sensitivity of the tunneling probability to barrier width explains why alpha decay half-lives range over 24 orders of magnitude — from microseconds to $10^{17}$ years — even though the alpha particle energies vary by only a factor of 2.

🔵 Historical Note: Gamow's tunneling explanation of alpha decay (1928) was one of the first triumphs of quantum mechanics applied to the nucleus. It demonstrated that QM was not just about atomic spectra — it governed nuclear physics too.

Scanning Tunneling Microscope (STM): A sharp conducting tip is brought within ~1 nm of a surface. Electrons tunnel between tip and surface, and the tunneling current $I \propto e^{-2\kappa d}$ is exquisitely sensitive to the tip-surface distance $d$. A change of 0.1 nm changes the current by about an order of magnitude. By scanning the tip across the surface and keeping the current constant, you map the surface with atomic resolution.

Flash memory: Data storage in flash memory relies on electrons tunneling through a thin oxide barrier (~10 nm) onto a floating gate. The presence or absence of stored charge represents a bit. Writing and erasing both involve tunneling (Fowler-Nordheim tunneling or hot-carrier injection).

Nuclear fusion in stars: The Sun's core temperature is about 15 million kelvin. The thermal energy of protons at this temperature is $k_BT \approx 1.3\,\text{keV}$. But the Coulomb barrier between two protons is about 550 keV at the nuclear radius. Classically, fusion should be impossible at the Sun's core temperature — protons simply do not have enough energy to overcome the electrostatic repulsion. Quantum tunneling resolves this: although the tunneling probability per collision is tiny ($\sim 10^{-10}$), the enormous number of collisions per second ($\sim 10^{38}$) makes fusion proceed at a rate sufficient to power the Sun. Without tunneling, the Sun would not shine.

📊 By the Numbers: The tunneling probability for solar proton-proton fusion is roughly $T \sim 10^{-10}$. But there are approximately $10^{57}$ protons in the Sun's core, each colliding about $10^{10}$ times per second. The net fusion rate — about $4 \times 10^{38}$ reactions per second — converts 600 million tons of hydrogen into helium every second, releasing $3.8 \times 10^{26}$ watts. All powered by tunneling.

🔗 Connection: We will revisit tunneling in Chapter 20 (WKB approximation), where we develop a general formula for tunneling through arbitrary barrier shapes — not just rectangular ones. The WKB tunneling formula is:

$$T \approx \exp\left[-\frac{2}{\hbar}\int_{x_1}^{x_2}\sqrt{2m(V(x) - E)}\,dx\right],$$

where $x_1$ and $x_2$ are the classical turning points. For a rectangular barrier, this reduces to $T \approx e^{-2\kappa d}$, consistent with what we found above.

🔄 Check Your Understanding 3.5: If you double the barrier width, what happens to the tunneling probability (in the thick-barrier limit)? (Answer: $T \propto e^{-2\kappa d}$, so doubling $d$ gives $T \to T^2$. If $T$ was $10^{-3}$, it becomes $10^{-6}$. Exponential sensitivity!)

3.7 Numerical Methods: When Exact Solutions Don't Exist

The problems in Sections 3.2–3.6 are among the very few quantum mechanics problems with exact analytical solutions. Most realistic potentials — the Morse potential for molecular vibrations, the Lennard-Jones potential for intermolecular forces, the double-well potential for ammonia inversion — do not yield to analytical methods.

This is where numerical methods become essential. And this is not a failure — it is how real physics gets done.

Theme 4 preview: "Approximation is not failure." The exactly solvable problems in this chapter are not the "real" problems — they are the training ground. Real quantum systems require numerical or approximate methods. Mastering these computational tools is just as important as understanding the analytical solutions.

3.7.1 The Finite Difference Method

The idea is simple: replace the continuous wavefunction $\psi(x)$ with its values on a discrete grid $\{x_0, x_1, \ldots, x_N\}$ with spacing $\Delta x = (x_N - x_0)/N$.

The second derivative is approximated by the central difference formula:

$$\frac{d^2\psi}{dx^2}\bigg|_{x_i} \approx \frac{\psi_{i+1} - 2\psi_i + \psi_{i-1}}{(\Delta x)^2}.$$

Substituting into the TISE:

$$-\frac{\hbar^2}{2m}\frac{\psi_{i+1} - 2\psi_i + \psi_{i-1}}{(\Delta x)^2} + V_i\psi_i = E\psi_i,$$

where $V_i = V(x_i)$. Rearranging:

$$-\frac{\hbar^2}{2m(\Delta x)^2}(\psi_{i+1} + \psi_{i-1}) + \left[\frac{\hbar^2}{m(\Delta x)^2} + V_i\right]\psi_i = E\psi_i.$$

This is a matrix eigenvalue equation. Let us be explicit about how this works. Suppose our grid has $N$ interior points (the boundary points at $x_0$ and $x_{N+1}$ are fixed by boundary conditions, e.g., $\psi_0 = \psi_{N+1} = 0$ for an infinite well). Then we have $N$ unknowns $(\psi_1, \psi_2, \ldots, \psi_N)$ and $N$ equations — one for each grid point. In matrix form, this is $\mathbf{H}\boldsymbol{\psi} = E\boldsymbol{\psi}$, where $\boldsymbol{\psi} = (\psi_1, \psi_2, \ldots, \psi_N)^T$ and $\mathbf{H}$ is an $N \times N$ tridiagonal matrix (nonzero only on the main diagonal and the two adjacent diagonals).

Define the $N \times N$ tridiagonal matrix:

$$H_{ij} = \begin{cases} \frac{\hbar^2}{m(\Delta x)^2} + V_i, & i = j \\ -\frac{\hbar^2}{2m(\Delta x)^2}, & |i - j| = 1 \\ 0, & \text{otherwise} \end{cases}$$

Then $\mathbf{H}\boldsymbol{\psi} = E\boldsymbol{\psi}$. Standard eigenvalue solvers (available in numpy, scipy, LAPACK) find all eigenvalues and eigenvectors efficiently. The computational cost is $O(N^2)$ for a tridiagonal matrix, making this method practical even for $N = 10{,}000$ grid points.

The matrix $\mathbf{H}$ is real and symmetric (Hermitian), which guarantees that all eigenvalues are real — as they must be, since they represent physical energies. This is the discrete analogue of the theorem (proved in Chapter 6) that Hermitian operators have real eigenvalues. The eigenvectors are orthogonal, corresponding to the orthogonality of the analytical eigenfunctions.

🔴 Warning: The finite difference method gives $N$ eigenvalues, but only the lowest ones are physically meaningful. The higher eigenvalues are contaminated by discretization error — they correspond to oscillations on the scale of the grid spacing, which are not resolved. A good rule of thumb: trust the lowest $N/5$ eigenvalues and treat the rest with suspicion.

Worked Example 3.8: Numerical infinite well.

Let us verify our analytical solution by solving the infinite well numerically.

Set $\hbar = m = a = 1$ (natural units for the problem). The grid runs from $x = 0$ to $x = 1$ with $N = 100$ interior points. Boundary conditions: $\psi(0) = \psi(a) = 0$.

The exact energies are $E_n = n^2\pi^2/2$: $E_1 = 4.935$, $E_2 = 19.74$, $E_3 = 44.41$.

Running the finite difference code (see code/example-01-solvers.py) gives: $E_1 = 4.934$, $E_2 = 19.74$, $E_3 = 44.38$ — agreement to better than 0.1% with just 100 grid points.

Worked Example 3.10: The double well potential.

As a demonstration of the power of numerical methods, consider a potential that has no analytical solution: the double well,

$$V(x) = \frac{V_0}{a^4}(x^2 - a^2)^2,$$

which has two minima at $x = \pm a$ separated by a barrier of height $V_0$ at $x = 0$. This potential models a wide range of physical situations: the ammonia molecule (which "inverts" between two equivalent configurations), proton transfer in hydrogen bonds, and the two-state systems that appear throughout quantum mechanics.

Using the finite difference method with $V_0 = 10$ and $a = 2$ (natural units), we find something remarkable: the two lowest energy eigenstates come in a nearly degenerate pair, split by a tiny amount $\Delta E$. The ground state (symmetric, even parity) has the wavefunction peaked in both wells. The first excited state (antisymmetric, odd parity) looks almost identical but with opposite signs in the two wells.

The splitting $\Delta E$ is related to the tunneling rate between the two wells: $\Delta E = 2\hbar/\tau_{\text{tunnel}}$, where $\tau_{\text{tunnel}}$ is the tunneling time. As the barrier height increases, $\Delta E$ decreases exponentially — consistent with the tunneling formula's exponential dependence on barrier parameters.

This doublet structure is the simplest example of tunneling splitting, and it will return in Chapter 17 (perturbation theory for the double well) and Chapter 11 (two-state systems in general).

3.7.2 The Shooting Method

An alternative approach for finding individual eigenstates is the shooting method:

1. Pick a trial energy $E$.
2. Integrate the TISE forward from the left boundary ($\psi(0) = 0$, $\psi'(0) = 1$ or some nonzero value).
3. Check whether the solution satisfies the right boundary condition ($\psi(a) = 0$ for the infinite well).
4. If not, adjust $E$ and try again (bisection or Newton's method).

This converts the eigenvalue problem into a root-finding problem: find $E$ such that $\psi_{\text{right boundary}} = 0$.

The shooting method is particularly useful when you want one specific eigenstate rather than all of them, or when the grid is too large for matrix methods.

Worked Example 3.9: Shooting method for the finite well.

For the finite square well with $V_0 = 50$ (natural units), we integrate from $x = -5$ (where $\psi \approx 0$) toward $x = 0$. We use the symmetry: for even states, $\psi'(0) = 0$; for odd states, $\psi(0) = 0$. We scan energy values and look for solutions that decay (rather than blow up) as $x \to +\infty$.

The shooting method finds the bound state energies by detecting sign changes in $\psi(x_{\text{far}})$ as we sweep $E$. Between any two energies where $\psi(x_{\text{far}})$ changes sign, there is an eigenvalue.

The shooting method has a beautiful physical interpretation: you are literally "shooting" a wavefunction from one boundary and seeing if it hits the target at the other boundary. If it overshoots (blows up to $+\infty$), your trial energy is too high; if it undershoots (blows up to $-\infty$), your trial energy is too low. Bisection narrows in on the correct energy — the one where the wavefunction gently decays to zero at the far boundary.

Both methods — finite difference and shooting — are implemented in the toolkit code (code/project-checkpoint.py). The finite difference method is generally preferred for finding many eigenvalues at once, while the shooting method is better for finding individual high-precision eigenvalues or for very large grids where the full matrix would be impractical.

3.7.3 Convergence and Accuracy

The finite difference method has systematic errors that decrease as the grid is refined:

• Second-order accuracy: the error in eigenvalues scales as $(\Delta x)^2$.
• Doubling the number of grid points reduces the error by a factor of 4.
• Lower-energy eigenvalues are more accurate than higher-energy ones (higher states oscillate faster and need finer grids).

A useful convergence test: solve with $N$ points, then with $2N$ points. If the eigenvalues change by less than your tolerance, you are done.

💡 Key Insight: The beauty of the numerical approach is that once you have the code, you can solve any 1D potential — just change the function $V(x)$. The analytical solutions in this chapter become benchmarks for validating your code, not the end goal. In research and engineering, the computer is your primary tool for solving the Schrödinger equation.

🔄 Check Your Understanding 3.6: The finite difference method converts the differential equation into a matrix equation. What determines the size of the matrix? (Answer: the number of grid points $N$. More grid points means a larger matrix, more accuracy, and more computation. Typical calculations use $N = 100$ to $N = 10{,}000$.)

3.8 Comparison of One-Dimensional Potentials

Before we summarize, let us step back and compare the systems we have studied. The following table organizes the key features:

Notice the pattern: confinement produces discrete spectra; lack of confinement produces continuous spectra. The finite well lies between the two extremes — it has both bound states (discrete) and scattering states (continuous).

🔗 Connection: This pattern will repeat in three dimensions. The hydrogen atom (Chapter 5) has discrete bound states ($E_n = -13.6\,\text{eV}/n^2$) and a continuum of scattering states ($E > 0$). The 1D potentials in this chapter are the warm-up for that three-dimensional problem. The 1D infinite well, in particular, is analogous to the radial equation for the hydrogen atom with $l = 0$ — the math is structurally identical, just with a $1/r$ potential replacing the flat bottom.

There is a deep lesson in this comparison table that is worth making explicit. Every one of these problems is governed by the same equation — the TISE. The only difference between them is the potential $V(x)$. Yet the physical behavior varies enormously: standing waves in the infinite well, spreading packets for the free particle, exponential tails in the finite well, partial reflection at the step, tunneling through the barrier. All of this richness emerges from one equation and different boundary conditions.

This is the power — and the beauty — of the Schrödinger equation. A single mathematical framework describes phenomena as diverse as atomic spectra, semiconductor physics, nuclear decay, and molecular bonding. The equation is simple. The solutions are rich. And the physics is always in the boundary conditions.

3.9 Summary and Project Checkpoint

What We Learned

This chapter solved the Schrödinger equation for five one-dimensional potentials. The key results:

Infinite square well: - Energies: $E_n = n^2\pi^2\hbar^2/(2ma^2)$ - Wavefunctions: $\psi_n = \sqrt{2/a}\sin(n\pi x/a)$ - Orthogonal and complete — any state can be expanded in this basis.

Free particle: - Continuous spectrum. Plane waves are not normalizable. - Physical states are wave packets. They move at the group velocity and spread over time.

Finite square well: - Finite number of bound states. Wavefunctions penetrate into classically forbidden regions. - Transcendental equations for energies — solved graphically or numerically.

Step potential: - Partial quantum reflection even when $E > V_0$ — no classical analogue. - Total reflection for $E < V_0$, but with exponential penetration.

Quantum tunneling (rectangular barrier): - $T \approx e^{-2\kappa d}$ for thick barriers. Exponentially sensitive to width and mass. - Real applications: alpha decay, STM, flash memory.

Numerical methods: - Finite difference method: convert TISE to matrix eigenvalue problem. - Shooting method: integrate ODE and root-find for eigenvalues. - These methods work for any potential.

Themes Touched

• "Mathematics and physics are inseparable": Boundary conditions — a mathematical requirement — produce quantized energies. The physics (discrete spectra, zero-point energy) is inseparable from the math (eigenvalue equations, normalization).
• "Approximation is not failure" (preview): Numerical methods are not a lesser form of physics. They are the primary tool for real-world quantum problems.
• "QM is strange but not arbitrary": Tunneling is genuinely weird — a particle passes through a region where it "shouldn't" be. But the tunneling probability is precisely calculable, and the predictions are confirmed by experiment.
• "Physical intuition must be rebuilt": Partial reflection at a step when $E > V_0$ has no classical analogue. We must accept that waves behave differently from particles.

Project Checkpoint: Quantum Toolkit v0.3

Your toolkit (code/project-checkpoint.py) now includes:

• solve_1d(V, x_grid, hbar, m) — finite difference Schrödinger solver for arbitrary 1D potentials
• infinite_well(n, a, num_points) — returns analytical wavefunctions and energies
• tunneling_probability(E, V0, d, m) — calculates tunneling through a rectangular barrier

These build on the Wavefunction class from Chapter 2 and the constants module from Chapter 1. In Chapter 4, we will add the quantum harmonic oscillator — the most important exactly solvable problem in all of physics.

The Big Picture: What This Chapter Really Taught Us

Beyond the specific solutions, this chapter has revealed something profound about the structure of quantum mechanics. The TISE is an eigenvalue equation, and nature's allowed energy values are its eigenvalues. Boundary conditions — the requirement that wavefunctions be physically sensible — act as a selection filter, admitting only certain eigenvalues and rejecting the rest. This is the deep origin of quantization.

In classical mechanics, a particle in a box can have any energy. In quantum mechanics, it cannot. The difference is not a new postulate or an arbitrary rule — it is a mathematical consequence of requiring the wavefunction to satisfy the boundary conditions imposed by the physical setup. The discrete energy levels of atoms, the tunneling probability through a barrier, the spreading of wave packets — all emerge from the same equation and the same procedure.

We have also seen the first hints of a powerful theme: exactly solvable problems are rare, but they serve as the foundation for everything else. The infinite well teaches us about quantization. The finite well teaches us about penetration and the transition to the continuum. The free particle teaches us about wave packets and dispersion. The step potential teaches us about reflection and transmission. The rectangular barrier teaches us about tunneling. And numerical methods allow us to solve everything else.

This hierarchy — exact solutions as benchmarks, approximate and numerical methods as the workhorses — is how professional quantum mechanics is done. We will return to this theme throughout the book, especially in Part IV (Approximation Methods).

Looking Ahead

The infinite well and its cousins are the foundation, but they are also the training wheels. In Chapter 4, we tackle the quantum harmonic oscillator — a potential that is quadratic rather than piecewise constant, and whose solution introduces Hermite polynomials and the extraordinary algebraic method of ladder operators. The harmonic oscillator is more than a textbook exercise: it describes molecular vibrations, electromagnetic field modes, and (as we will see in Chapter 34) the very structure of quantum fields themselves.

If the infinite well was the "Hello World" of quantum mechanics, the harmonic oscillator is the "universal building block." Everything from molecular spectroscopy to quantum field theory is built from oscillators. Chapter 4 is where the real power of quantum mechanics begins to emerge.

🔄 Check Your Understanding 3.7: Without looking back, write down (a) the energy eigenvalues for the infinite square well, (b) the tunneling formula for a rectangular barrier, and (c) the physical reason why the free particle has a continuous spectrum but the infinite well has a discrete one. If you can do all three, you are ready for Chapter 4.

Chapter 3 of "Quantum Mechanics: From Wavefunctions to Qubits — A Complete Modern Treatment" Part I: Foundations — Wave Mechanics