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Chapter 4 • Theory & Derivations

One-Dimensional Potentials & Quantum Mechanical Tunneling

Analytical solutions of piecewise constant potentials: boundary conditions, potential steps (reflection and transmission), the rectangular potential barrier, quantum tunneling ($E < V_0$), the infinite square well (particle in a box), and the finite square well.

§4.1 Boundary Conditions and Piecewise Constant Potentials

Standard Boundary Conditions

For a one-dimensional Time-Independent Schrödinger Equation:

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

Integrating across an infinitesimal interval $[x_0 - \epsilon, x_0 + \epsilon]$ centered at a potential boundary $x_0$:

$$-\frac{\hbar^2}{2m} \left[ \left.\frac{d\psi}{dx}\right|_{x_0+\epsilon} - \left.\frac{d\psi}{dx}\right|_{x_0-\epsilon} \right] + \lim_{\epsilon \to 0} \int_{x_0-\epsilon}^{x_0+\epsilon} V(x)\psi(x) dx = E \lim_{\epsilon \to 0} \int_{x_0-\epsilon}^{x_0+\epsilon} \psi(x) dx$$

From this integration, two general boundary conditions emerge:

1. Continuity of the Wavefunction: $\psi(x)$ must be continuous everywhere:

$$\psi(x_0^-) = \psi(x_0^+)$$

2. Continuity of the Derivative: Provided $V(x)$ does not contain infinite Dirac delta spikes, $\frac{d\psi}{dx}$ must be continuous:

$$\left.\frac{d\psi}{dx}\right|_{x_0^-} = \left.\frac{d\psi}{dx}\right|_{x_0^+}$$

§4.2 The Potential Step

Consider a potential step defined by:

$$V(x) = \begin{cases} 0 & \text{for } x < 0 \quad (\text{Region I}) \\ V_0 & \text{for } x \ge 0 \quad (\text{Region II}) \end{cases}$$

A stream of particles of mass $m$ and energy $E$ is incident from the left ($x \to -\infty$).

Case 1: $E > V_0$

In Region I ($x < 0$): $\psi_I(x) = A e^{i k_1 x} + B e^{-i k_1 x}$, where $k_1 = \frac{\sqrt{2mE}}{\hbar}$. In Region II ($x > 0$): $\psi_{II}(x) = C e^{i k_2 x}$, where $k_2 = \frac{\sqrt{2m(E - V_0)}}{\hbar}$.

Applying boundary conditions at $x = 0$:

  1. $\psi_I(0) = \psi_{II}(0) \implies A + B = C$
  2. $\psi'_I(0) = \psi'_{II}(0) \implies i k_1 (A - B) = i k_2 C \implies A - B = \frac{k_2}{k_1} C$

Solving for reflection amplitude $B/A$ and transmission amplitude $C/A$:

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

The Reflection Coefficient $R$ and Transmission Coefficient $T$ are ratios of probability currents:

$$R = \frac{|J_{\text{refl}}|}{|J_{\text{inc}}|} = \left| \frac{B}{A} \right|^2 = \left( \frac{k_1 - k_2}{k_1 + k_2} \right)^2$$
$$T = \frac{|J_{\text{trans}}|}{|J_{\text{inc}}|} = \frac{k_2}{k_1} \left| \frac{C}{A} \right|^2 = \frac{4 k_1 k_2}{(k_1 + k_2)^2}$$

Summing these yields probability conservation: $R + T = 1$. Even when $E > V_0$, quantum mechanics predicts non-zero reflection ($R > 0$), a purely wave phenomenon with no classical counterpart.

Case 2: $E < V_0$

In Region II, the wave vector becomes imaginary: $k_2 = i \kappa$, where $\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$. The physically acceptable solution in Region II is an exponentially decaying wave:

$$\psi_{II}(x) = C e^{-\kappa x}$$

Applying boundary conditions yields $R = |\frac{k_1 - i\kappa}{k_1 + i\kappa}|^2 = 1$ and $T = 0$. All particles are eventually reflected ($R = 1$), but the wavefunction penetrates a finite distance $\delta = 1/\kappa$ into the classically forbidden region. This penetration leads directly to quantum tunneling in finite barriers.

§4.3 The Rectangular Potential Barrier and Quantum Tunneling

Consider a barrier of height $V_0$ and width $a$:

$$V(x) = \begin{cases} 0 & x < 0 \quad (\text{Region I}) \\ V_0 & 0 \le x \le a \quad (\text{Region II}) \\ 0 & x > a \quad (\text{Region III}) \end{cases}$$

For $E < V_0$:

  • Region I ($x < 0$): $\psi_I(x) = A e^{i k x} + B e^{-i k x}$, with $k = \frac{\sqrt{2mE}}{\hbar}$
  • Region II ($0 \le x \le a$): $\psi_{II}(x) = F e^{\kappa x} + G e^{-\kappa x}$, with $\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$
  • Region III ($x > a$): $\psi_{III}(x) = C e^{i k x}$

Matching $\psi$ and $\frac{d\psi}{dx}$ at $x = 0$ and $x = a$ yields the exact Transmission Coefficient Formula:

$$T = \left[ 1 + \frac{V_0^2}{4E(V_0 - E)} \sinh^2(\kappa a) \right]^{-1}$$

When the barrier is wide and high ($\kappa a \gg 1$), $\sinh(\kappa a) \approx \frac{1}{2} e^{\kappa a}$, simplifying $T$ to:

$$T \approx 16 \frac{E}{V_0} \left( 1 - \frac{E}{V_0} \right) e^{-2\kappa a}$$
Physical Applications of Quantum Tunneling

1. Alpha Decay in Nuclear Physics: Gamow (1928) explained the Geiger-Nuttall law by modeling the alpha particle as tunneling through the nuclear Coulomb barrier.

2. Scanning Tunneling Microscopy (STM): Binnig and Rohrer (Nobel Prize 1986) developed the STM, which relies on tunneling current between an atomic tip and sample surface: $I \propto e^{-2\kappa d}$, yielding sub-angstrom spatial resolution.

Experiment with barrier height and width in Simulation 4.1 below to observe real-time evanescent decay and transmission.

§4.4 The Infinite Square Well (Particle in a Box)

Consider a particle trapped in an infinite potential well:

$$V(x) = \begin{cases} 0 & 0 \le x \le L \\ \infty & \text{otherwise} \end{cases}$$

Because $V = \infty$ outside the well, $\psi(x) = 0$ for $x \le 0$ and $x \ge L$. Inside the well ($0 < x < L$):

$$\frac{d^2\psi}{dx^2} + k^2\psi = 0, \quad k = \frac{\sqrt{2mE}}{\hbar} \implies \psi(x) = A \sin(kx) + B \cos(kx)$$

Applying boundary conditions:

  1. $\psi(0) = 0 \implies B = 0$
  2. $\psi(L) = 0 \implies A \sin(kL) = 0 \implies kL = n\pi, \quad n \in \{1, 2, 3, \dots\}$

This gives the quantized wave numbers $k_n$ and Quantized Energy Levels:

$$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}$$

Normalizing $\int_0^L |\psi_n(x)|^2 dx = 1$ yields:

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right)$$

Key properties:

  • Zero-Point Energy: The ground state ($n=1$) energy $E_1 = \frac{\pi^2 \hbar^2}{2mL^2} > 0$ is strictly positive, satisfying the uncertainty principle $\Delta p \approx \hbar / L$.
  • Nodes: The state $\psi_n(x)$ has $n-1$ interior nodes where the probability density vanishes identically.

See Simulation 4.2 below to interact with quantum numbers $n=1$ to $5$ and view standing wavefunctions alongside their energy ladder.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Medium Example 4.1: Symmetric Finite Square Well: Bound State Transcendental Equations

For a symmetric finite potential well $V(x) = -V_0$ for $|x| \le a$ and $V(x) = 0$ for $|x| > a$, derive the transcendental equations for the bound-state energy eigenvalues for even-parity states. Prove that at least one even bound state exists regardless of how shallow the well is.

Step 1: Set up wavefunctions by parity
$$\text{For } |x| \le a: \psi(x) = A \cos(l x), \quad l = \frac{\sqrt{2m(E + V_0)}}{\hbar} $$ $$\text{For } x > a: \psi(x) = C e^{-\kappa x}, \quad \kappa = \frac{\sqrt{-2mE}}{\hbar}$$

For an even parity potential $V(-x) = V(x)$, the eigenstates can be categorized into even and odd functions. Even bound states use cosine inside the well.

Step 2: Match boundary conditions at x = a
$$\psi(a^-) = \psi(a^+) \implies A \cos(l a) = C e^{-\kappa a} $$ $$\psi'(a^-) = \psi'(a^+) \implies -l A \sin(l a) = -\kappa C e^{-\kappa a} $$ $$\frac{\psi'(a)}{\psi(a)} \implies l \tan(l a) = \kappa$$

Defining dimensionless variables $\xi = l a$ and $\eta = \kappa a$, this becomes $\xi \tan\xi = \eta$ subject to the circle constraint $\xi^2 + \eta^2 = \frac{2m V_0 a^2}{\hbar^2} = R^2$. Because $\tan\xi \to 0$ as $\xi \to 0$, the curve $\eta = \xi \tan\xi$ always intersects the circle in the first quadrant for any $R > 0$, guaranteeing at least one even bound state.

An electron is accelerated from rest through an electrostatic potential difference of $V = 150 \text{ V}$.\n(a) Determine its de Broglie wavelength using non-relativistic mechanics.\n(b) At what accelerating potential does the relativistic correction to the de Broglie wavelength exceed $1\%$?
Step 1: Non-relativistic Calculation
$$K = e V = 150 \text{ eV} = 150 \times 1.602 \times 10^{-19} \text{ J} = 2.403 \times 10^{-17} \text{ J} $$ $$p = \sqrt{2 m_e K} = \sqrt{2(9.109 \times 10^{-31} \text{ kg})(2.403 \times 10^{-17} \text{ J})} = 6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s} $$ $$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 1.001 \times 10^{-10} \text{ m} = 1.001 \text{ Å}$$
For quick calculations, note that $\lambda = \sqrt{\frac{150}{V}} \text{ Å}$. For $V = 150 \text{ V}$, $\lambda = 1.00 \text{ Å}$, which corresponds to typical atomic crystal lattice spacings.
Step 2: Relativistic Condition
$$E^2 = p^2 c^2 + m_0^2 c^4 \implies p = \frac{1}{c}\sqrt{K(K + 2m_0 c^2)} $$ $$\lambda_{\text{rel}} = \frac{h c}{\sqrt{K(K + 2m_0 c^2)}} = \frac{\lambda_{\text{class}}}{\sqrt{1 + \frac{K}{2m_0 c^2}}} \approx \lambda_{\text{class}}\left(1 - \frac{K}{4 m_0 c^2}\right) $$ $$\frac{\Delta \lambda}{\lambda} \approx \frac{K}{4 m_0 c^2} \ge 0.01 \implies K \ge 0.04 m_0 c^2 = 0.04 (511 \text{ keV}) \approx 20.44 \text{ keV}$$
When accelerating potentials exceed roughly $20 \text{ kV}$ (typical in transmission electron microscopes), relativistic momentum corrections become necessary.