Complete Quantum Mechanics

Part I: Theory

Core ideas

Quantum mechanics is built upon a set of postulates that replace the deterministic trajectories of classical mechanics with a probabilistic framework based on Hilbert space.

Postulates of Quantum Mechanics.

• The State Vector: The state of a physical system is completely described by a vector $|\psi(t)\rangle$ in a complex Hilbert space.
• Observables and Operators: Every physical observable (e.g., position, momentum, energy) corresponds to a Hermitian operator acting on this Hilbert space.
• Measurement (Born’s Rule): If an observable $A$ is measured, the outcome must be one of the eigenvalues $a_n$ of the corresponding operator $\hat A$. The probability of obtaining $a_n$ is $P(a_n) = |\langle a_n|\psi\rangle|^2$.
• State Collapse: Immediately after a measurement of $A$ that yields $a_n$, the state of the system “collapses” to the corresponding eigenvector $|a_n\rangle$.
• Time Evolution: Between measurements, the state evolves according to the Time-Dependent Schrodinger Equation:

$$i\hbar\frac{d}{dt}|\psi(t)\rangle = \hat H|\psi(t)\rangle,$$

where $\hat H$ is the Hamiltonian (total energy) operator.

Superposition Principle. Because the Schrodinger equation is linear, any linear combination of valid states is also a valid state: $|\psi\rangle = c_1|\psi_1\rangle + c_2|\psi_2\rangle$. This leads to interference effects, where amplitudes (not probabilities) add.

Unitary Evolution. For a time-independent Hamiltonian, the evolution operator is $\hat U(t) = e^{-i\hat Ht/\hbar}$. Since $\hat H$ is Hermitian, $\hat U$ is unitary ($\hat U^\dagger \hat U = \hat I$), which ensures that the total probability $\langle\psi|\psi\rangle = 1$ is conserved over time.

Mathematical spine

$$\begin{aligned} i\hbar\frac{d}{dt}|\psi(t)\rangle &= \hat H|\psi(t)\rangle & \text{(Schrodinger Equation)} \\ \hat A|a_n\rangle &= a_n|a_n\rangle, \quad P(a_n) = |\langle a_n|\psi\rangle|^2 & \text{(Eigenvalues and Probabilities)} \\ \langle \hat A \rangle &= \langle\psi|\hat A|\psi\rangle & \text{(Expectation Value)} \end{aligned}$$

Example: The Qubit. The simplest quantum system is a two-level system (qubit). A general state is $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$. Measuring in the $\{|0\rangle, |1\rangle\}$ basis yields 0 with probability $|\alpha|^2$.

Worked example

A qubit is prepared in the state $|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}e^{i\pi/4}|1\rangle$, where $|0\rangle$ and $|1\rangle$ are eigenstates of $\hat\sigma_z$ with eigenvalues $+1$ and $-1$. We measure $\sigma_z$.

The probabilities are:

• $P(+1) = |\langle 0|\psi\rangle|^2 = (1/\sqrt{2})^2 = 1/2$
• $P(-1) = |\langle 1|\psi\rangle|^2 = (1/\sqrt{2})^2 = 1/2$

The expectation value is

$$\langle \hat\sigma_z \rangle = \frac{1}{2}(+1) + \frac{1}{2}(-1) = 0.$$

If instead we measure $\sigma_x$, we rewrite $|\psi\rangle$ in the $\sigma_x$ basis where $|+\rangle_x = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and $|-\rangle_x = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$. The expansion gives $P(+1) = \cos^2(\pi/8) \approx 0.854$ and $P(-1) = \sin^2(\pi/8) \approx 0.146$.

Problems with Solutions

Problem 1. Show that the time-evolution operator $\hat U(t) = e^{-i\hat H t/\hbar}$ is unitary for a time-independent Hermitian Hamiltonian.

Solution. For $\hat H$ Hermitian, $\hat H^\dagger = \hat H$. Then

$$\hat U^\dagger(t) = e^{+i\hat H^\dagger t/\hbar} = e^{+i\hat H t/\hbar} = \hat U(-t).$$

Therefore

$$\hat U^\dagger(t)\hat U(t) = e^{+i\hat H t/\hbar}e^{-i\hat H t/\hbar} = e^{0} = \hat I,$$

so $\hat U(t)$ is unitary. Unitarity guarantees conservation of total probability.

Problem 2. A system is in state $|\psi\rangle = \frac{1}{\sqrt{5}}(2|1\rangle + |2\rangle + i|3\rangle)$, where $\{|1\rangle,|2\rangle,|3\rangle\}$ is an orthonormal basis. Normalize the state and find the probability of measuring the system in $|1\rangle$.

Solution. First check normalization:

$$\langle\psi|\psi\rangle = \frac{1}{5}(4 + 1 + 1) = \frac{6}{5} \neq 1.$$

The normalized state is $|\psi_{\text{norm}}\rangle = \sqrt{\frac{5}{6}}|\psi\rangle = \frac{1}{\sqrt{6}}(2|1\rangle + |2\rangle + i|3\rangle)$. The probability of $|1\rangle$ is

$$P(1) = |\langle 1|\psi_{\text{norm}}\rangle|^2 = \left|\frac{2}{\sqrt{6}}\right|^2 = \frac{4}{6} = \frac{2}{3}.$$

Problem 3. If a measurement of observable $\hat A$ on state $|\psi\rangle$ yields eigenvalue $a_n$, what is the state immediately after measurement?

Solution. By the postulate of state collapse, immediately after obtaining $a_n$ the state collapses to the corresponding normalized eigenvector $|a_n\rangle$:

$$|\psi_{\text{after}}\rangle = \frac{|a_n\rangle\langle a_n|\psi\rangle}{\sqrt{|\langle a_n|\psi\rangle|^2}} = |a_n\rangle.$$

The phase is arbitrary and usually omitted.

Section summary. Quantum mechanics describes systems using state vectors in Hilbert space, with Hermitian operators representing observables and unitary operators representing time evolution.

The Wave Function

Core ideas

The state of a quantum particle is completely specified by a complex-valued wave function $\psi(\bm r, t)$. It is a probability amplitude; the physical interpretation is given by Born’s rule.

Probability Interpretation and Normalization. The probability of finding the particle in a volume $d^3r$ is $P(\bm r, t)d^3r = |\psi(\bm r, t)|^2 d^3r$. Since the particle must be somewhere in space, the wave function must be normalized:

$$\int_{-\infty}^{\infty} |\psi(\bm r, t)|^2 d^3r = 1.$$

For a wave function to be physically meaningful, it must be square-integrable (belong to the $L^2$ Hilbert space).

Expectation Values and Operators. Observables are represented by Hermitian operators $\hat A$. The average result of many measurements of $A$ on identically prepared states is the expectation value:

$$\langle A \rangle = \int \psi^* \hat A \psi \, d^3r.$$

Key operators in the position representation:

• Position: $\hat{\bm r} = \bm r$
• Momentum: $\hat{\bm p} = -i\hbar \nabla$

Probability Current. Local conservation of probability is expressed by the continuity equation:

$$\frac{\partial \rho}{\partial t} + \nabla \cdot \bm j = 0, \quad \rho = |\psi|^2.$$

The probability current $\bm j$ represents the flow of probability density:

$$\bm j = \frac{\hbar}{2mi} (\psi^* \nabla \psi - \psi \nabla \psi^*) = \frac{1}{m} \text{Re}(\psi^* \hat{\bm p} \psi).$$

Momentum Space and Fourier Transforms. The state can also be represented in momentum space by the function $\phi(\bm p, t)$. The two representations are related by a Fourier transform:

$$\psi(\bm x) = \frac{1}{(2\pi\hbar)^{3/2}} \int \phi(\bm p) e^{i\bm p \cdot \bm x / \hbar} d^3p, \quad \phi(\bm p) = \frac{1}{(2\pi\hbar)^{3/2}} \int \psi(\bm x) e^{-i\bm p \cdot \bm x / \hbar} d^3x.$$

This duality is the origin of the Heisenberg Uncertainty Principle: $\Delta x \Delta p \geq \hbar/2$. A spatially localized wave packet (small $\Delta x$) must be composed of many momentum states (large $\Delta p$).

Mathematical spine

$$\begin{aligned} \hat{\bm p} &= -i\hbar \nabla & \text{(Momentum Operator)} \\ \bm j &= \frac{\hbar}{m} \text{Im}(\psi^* \nabla \psi) & \text{(Probability Current)} \\ \Delta A \Delta B &\geq \frac{1}{2} | \langle [\hat A, \hat B] \rangle | & \text{(General Uncertainty Relation)} \end{aligned}$$

Example: Gaussian Wave Packet. A particle at rest localized at the origin can be modeled by:

$$\psi(x, 0) = \left( \frac{1}{\pi \sigma^2} \right)^{1/4} e^{-x^2 / (2\sigma^2)}.$$

This state minimizes the uncertainty product: $\Delta x = \sigma/\sqrt{2}$ and $\Delta p = \hbar/(\sigma\sqrt{2})$, so $\Delta x \Delta p = \hbar/2$. As the wave packet evolves in time, it spreads (dispersion).

Worked example

A free electron is described at $t=0$ by the Gaussian wave packet

$$\psi(x,0) = \left(\frac{1}{2\pi\sigma^2}\right)^{1/4}e^{-x^2/(4\sigma^2)}e^{ik_0 x},$$

with $\sigma = 1.0\,\text{nm}$ and $k_0 = 2.0\times10^{9}\,\text{m}^{-1}$ (corresponding to momentum $p_0 = \hbar k_0 \approx 2.11\times10^{-25}\,\text{kg\cdotm/s}$). The normalization constant is chosen so that $\int_{-\infty}^{\infty}|\psi|^2dx = 1$.

The probability density is

$$|\psi(x,0)|^2 = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/(2\sigma^2)},$$

which is a Gaussian centered at $x=0$ with width $\Delta x = \sigma = 1.0\,\text{nm}$.

The momentum-space wave function is also Gaussian:

$$\phi(p) = \left(\frac{2\sigma^2}{\pi\hbar^2}\right)^{1/4}e^{-\sigma^2(p-p_0)^2/\hbar^2}.$$

The momentum width is $\Delta p = \hbar/(2\sigma) \approx 5.28\times10^{-26}\,\text{kg\cdotm/s}$.

Check the uncertainty principle:

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

so this state is a minimum-uncertainty wave packet.

Problems with Solutions

Problem 1. A particle has wave function $\psi(x) = A e^{-|x|/a}$ with $a = 0.5\,\text{nm}$. Find the normalization constant $A$ and the probability of finding the particle in the region $0 < x < a$.

Solution. Normalization:

$$1 = A^2\int_{-\infty}^{\infty}e^{-2|x|/a}dx = 2A^2\int_0^{\infty}e^{-2x/a}dx = 2A^2\frac{a}{2} = A^2 a.$$

Thus $A = 1/\sqrt{a} = \sqrt{2}\,\text{nm}^{-1/2} \approx 1.41\,\text{nm}^{-1/2}$.

Probability:

$$P(0<x<a) = A^2\int_0^a e^{-2x/a}dx = \frac{1}{a}\left[-\frac{a}{2}e^{-2x/a}\right]_0^a = \frac{1}{2}(1-e^{-2}) \approx 0.432.$$

Problem 2. For a particle with $\psi(x) = \sqrt{2/L}\sin(\pi x/L)$ in an infinite well $0<x<L$, calculate the probability current $j(x)$.

Solution. Using $\bm j = \frac{\hbar}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)$:

$$\psi^* = \psi = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right), \quad \frac{d\psi}{dx} = \sqrt{\frac{2}{L}}\frac{\pi}{L}\cos\left(\frac{\pi x}{L}\right).$$

Then

$$j(x) = \frac{\hbar}{2mi}\left[\psi\frac{d\psi}{dx} - \psi\frac{d\psi}{dx}\right] = 0,$$

since $\psi$ is real. A stationary state in a bound potential has zero probability current.

Problem 3. Show that a Gaussian wave packet $\psi(x) = (2\pi\sigma^2)^{-1/4}e^{-x^2/(4\sigma^2)}$ satisfies $\Delta x\,\Delta p = \hbar/2$.

Solution. The position probability density is $|\psi|^2 = (2\pi\sigma^2)^{-1/2}e^{-x^2/(2\sigma^2)}$, a Gaussian with variance $\sigma^2$, so $\Delta x = \sigma$.

The momentum-space wave function is

$$\phi(p) = \left(\frac{2\sigma^2}{\pi\hbar^2}\right)^{1/4}e^{-\sigma^2 p^2/\hbar^2},$$

with probability density $|\phi(p)|^2 = \sqrt{2\sigma^2/(\pi\hbar^2)}\,e^{-2\sigma^2 p^2/\hbar^2}$. This is a Gaussian with variance $\hbar^2/(4\sigma^2)$, so $\Delta p = \hbar/(2\sigma)$.

Therefore $\Delta x\,\Delta p = \sigma \cdot \hbar/(2\sigma) = \hbar/2$.

Section summary. The wave function is a normalized probability amplitude whose evolution is constrained by probability conservation and whose spread is limited by the uncertainty principle.

Time-Independent Schrodinger Equation

Core ideas

When the potential $V(\bm r)$ is independent of time, the TDSE can be solved using separation of variables: $\Psi(\bm r, t) = \psi(\bm r) e^{-iEt/\hbar}$. This leads to the Time-Independent Schrodinger Equation (TISE):

$$\hat H \psi = E \psi, \quad \left[ -\frac{\hbar^2}{2m} \nabla^2 + V(\bm r) \right] \psi = E \psi.$$

This is an eigenvalue problem for the energy $E$. The solutions $\psi_n$ are stationary states because the probability density $|\Psi|^2 = |\psi|^2$ is constant in time.

Standard Solvable Models. Undergraduate quantum mechanics focuses on a few exactly solvable 1D potentials:

• Infinite Square Well: $V=0$ for $0 < x < a$, and $\infty$ otherwise.

$$E_n = \frac{n^2 \pi^2 \hbar^2}{2 m a^2}, \quad \psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right), \quad n=1, 2, \dots$$

Quantization arises from the boundary condition $\psi(0) = \psi(a) = 0$.

• Harmonic Oscillator: $V(x) = \frac{1}{2}m\omega^2 x^2$.

$$E_n = \left( n + \frac{1}{2} \right) \hbar \omega, \quad n=0, 1, 2, \dots$$

The **zero-point energy** $E_0 = \frac{1}{2}\hbar\omega$ is the minimum possible energy, a direct consequence of the uncertainty principle.

Nodes and Energy. The $n$-th excited state (for a 1D bound system) has $n$ nodes (points where $\psi=0$). Higher energy states have more nodes because more “wiggles” correspond to higher curvature ($\nabla^2 \psi$) and thus higher kinetic energy.

Tunneling and Bound States.

• Bound States: Discrete energy levels ($E < V_\infty$).
• Scattering States: Continuous energy spectrum ($E > V_\infty$).
• Tunneling: A particle has a non-zero probability of being found in classically forbidden regions ($E < V(x)$), where the wave function decays exponentially rather than oscillating.

Mathematical spine

$$\begin{aligned} \hat H \psi &= E \psi & \text{(TISE)} \\ E_n &= \frac{n^2 \pi^2 \hbar^2}{2 m a^2} & \text{(Infinite Well Energies)} \\ E_n &= \left( n + \frac{1}{2} \right) \hbar \omega & \text{(SHO Energies)} \end{aligned}$$

Example: Finite Square Well. Unlike the infinite well, the finite well has a finite number of bound states. The wave function “leaks” into the walls, leading to lower energy levels than an infinite well of the same width.

Worked example

An electron is confined in a 1D infinite square well of width $a = 2.0\,\text{nm}$. The allowed energies are

$$E_n = \frac{n^2\pi^2\hbar^2}{2m_e a^2} = n^2\frac{\pi^2(1.055\times10^{-34}\,\text{J\cdots})^2}{2(9.11\times10^{-31}\,\text{kg})(2.0\times10^{-9}\,\text{m})^2} \approx n^2(0.094\,\text{eV}).$$

For the ground state ($n=1$): $E_1 \approx 0.094\,\text{eV}$. For the first excited state ($n=2$): $E_2 = 4E_1 \approx 0.376\,\text{eV}$. The energy spacing is $\Delta E_{21} = E_2 - E_1 \approx 0.282\,\text{eV}$.

The corresponding de Broglie wavelength for $n=1$ is $\lambda_1 = 2a = 4.0\,\text{nm}$, which equals twice the well width. The wave function has nodes at the boundaries and no nodes inside.

Problems with Solutions

Problem 1. For a quantum harmonic oscillator with $m = 1.0\times10^{-26}\,\text{kg}$ and $\omega = 3.0\times10^{14}\,\text{rad/s}$, calculate the ground-state energy and the energy spacing between adjacent levels in electron-volts.

Solution. The ground-state energy is

$$E_0 = \frac{1}{2}\hbar\omega = \frac{1}{2}(1.055\times10^{-34})(3.0\times10^{14})\,\text{J} = \frac{3.165\times10^{-20}}{2}\,\text{J} = 1.583\times10^{-20}\,\text{J}.$$

Converting to eV: $E_0 = 1.583\times10^{-20}/(1.602\times10^{-19}) \approx 0.0988\,\text{eV}$.

The spacing between adjacent levels is $\Delta E = \hbar\omega = 2E_0 \approx 0.198\,\text{eV}$.

Problem 2. A particle in an infinite square well of width $a$ is in the superposition state $\psi(x) = \frac{1}{\sqrt{2}}[\psi_1(x) + \psi_2(x)]$, where $\psi_n(x) = \sqrt{2/a}\sin(n\pi x/a)$. Find the expectation value of the energy.

Solution. The expectation value is

$$\langle E \rangle = \frac{1}{2}E_1 + \frac{1}{2}E_2 = \frac{1}{2}\frac{\pi^2\hbar^2}{2ma^2} + \frac{1}{2}\frac{4\pi^2\hbar^2}{2ma^2} = \frac{5\pi^2\hbar^2}{4ma^2}.$$

With $a = 1.0\,\text{nm}$ and $m = m_e$:

$$\langle E \rangle = \frac{5}{2}E_1 \approx \frac{5}{2}(0.376\,\text{eV}) = 0.940\,\text{eV}.$$

Problem 3. Estimate the number of bound states in a finite square well of depth $V_0 = 5\,\text{eV}$ and width $a = 1.0\,\text{nm}$ for an electron.

Solution. A rough estimate is obtained by comparing the well depth to the energy spacing of an infinite well of the same width: $E_1^{\infty} = \pi^2\hbar^2/(2m_e a^2) \approx 0.376\,\text{eV}$. The number of bound states is approximately $N \approx \lfloor\sqrt{V_0/E_1^{\infty}}\rfloor = \lfloor\sqrt{5/0.376}\rfloor = \lfloor\sqrt{13.3}\rfloor = \lfloor 3.65 \rfloor = 3$. A more accurate numerical solution for this well gives either 3 or 4 bound states depending on the exact parameters, with the crude estimate capturing the correct order of magnitude.

Section summary. The TISE converts the search for energy states into an eigenvalue problem, yielding quantized energy levels for bound systems and predicting phenomena like zero-point energy and tunneling.

Formalism

Core ideas

The mathematical language of quantum mechanics is linear algebra in Hilbert space. States are represented by vectors, and observables by Hermitian operators.

Dirac Notation and Hilbert Space.

• Ket $|\psi\rangle$: A state vector in Hilbert space.
• Bra $\langle\psi|$: The dual vector (complex conjugate transpose).
• Inner Product $\langle\phi|\psi\rangle$: A complex number representing the amplitude of $\psi$ in state $\phi$.
• Completeness: For an orthonormal basis $\{|n\rangle\}$, the sum of projection operators equals the identity: $\sum_n |n\rangle\langle n| = \hat I$.

Hermitian Operators and Measurement. An operator $\hat A$ is Hermitian if $\hat A = \hat A^\dagger$. Its eigenvalues are real, and its eigenvectors form a complete basis. Measurement “projects” the state:

$$|\psi\rangle = \sum_n c_n |a_n\rangle, \quad c_n = \langle a_n|\psi\rangle.$$

The probability of measuring $a_n$ is $|c_n|^2$.

Commutators and CSCO. Two observables can be measured simultaneously if and only if their operators commute: $[\hat A, \hat B] = 0$. A Complete Set of Commuting Observables (CSCO) is a set of operators whose joint eigenvectors uniquely specify a state. Non-commuting operators satisfy the generalized uncertainty relation.

Schrodinger vs. Heisenberg Pictures.

• Schrodinger Picture: States evolve in time ($|\psi(t)\rangle$), operators are stationary.
• Heisenberg Picture: Operators evolve in time ($\hat A(t) = \hat U^\dagger \hat A \hat U$), states are stationary. The equations of motion are $\frac{d\hat A}{dt} = \frac{i}{\hbar}[\hat H, \hat A] + \frac{\partial \hat A}{\partial t}$.

Mathematical spine

$$\begin{aligned} \langle \hat A \rangle &= \langle\psi|\hat A|\psi\rangle & \text{(Expectation Value)} \\ \sum_n |n\rangle\langle n| &= \hat I & \text{(Completeness/Resolution of Identity)} \\ \frac{d\langle A \rangle}{dt} &= \frac{i}{\hbar} \langle [\hat H, \hat A] \rangle + \langle \frac{\partial \hat A}{\partial t} \rangle & \text{(Ehrenfest's Theorem)} \end{aligned}$$

Example: Spin-1/2. The state of a spin-1/2 particle (e.g., an electron) is a spinor in a 2D Hilbert space. The observables are the Pauli matrices $\sigma_x, \sigma_y, \sigma_z$. They do not commute, e.g., $[\sigma_x, \sigma_y] = 2i\sigma_z$, so spin components along different axes cannot be known simultaneously.

Worked example

An electron (spin-$1/2$) is prepared in the state $|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle_z + \sqrt{\frac{2}{3}}|-\rangle_z$, where $\hat S_z|\pm\rangle_z = \pm\frac{\hbar}{2}|\pm\rangle_z$ and $\hbar = 1.055\times10^{-34}\,\text{J\cdots}$. We measure $S_z$.

The probabilities are obtained from the expansion coefficients:

• $P(+\hbar/2) = |\langle +|\psi\rangle|^2 = (1/\sqrt{3})^2 = 1/3$
• $P(-\hbar/2) = |\langle -|\psi\rangle|^2 = (\sqrt{2/3})^2 = 2/3$

Check: $1/3 + 2/3 = 1$. The expectation value is

$$\langle \hat S_z \rangle = \frac{1}{3}\left(+\frac{\hbar}{2}\right) + \frac{2}{3}\left(-\frac{\hbar}{2}\right) = -\frac{\hbar}{6} \approx -1.76\times10^{-35}\,\text{J\cdots}.$$

If the measurement yields $+\hbar/2$, the state collapses to $|+\rangle_z$.

Problems with Solutions

Problem 1. Verify the commutator $[\hat x, \hat p] = i\hbar$ in the position representation.

Solution. With $\hat x = x$ and $\hat p = -i\hbar\frac{\partial}{\partial x}$, act on a test function $f(x)$:

$$[\hat x, \hat p]f = x\left(-i\hbar\frac{\partial f}{\partial x}\right) - (-i\hbar)\frac{\partial}{\partial x}(xf) = -i\hbar x\frac{\partial f}{\partial x} + i\hbar f + i\hbar x\frac{\partial f}{\partial x} = i\hbar f.$$

Since this holds for all $f(x)$, $[\hat x, \hat p] = i\hbar$.

Problem 2. A system has orthonormal basis $\{|1\rangle, |2\rangle\}$. Observable $\hat A$ has eigenvalues $a_1 = 5$ with eigenvector $|1\rangle$ and $a_2 = -1$ with eigenvector $|2\rangle$. The system is in state $|\psi\rangle = \frac{1}{2}|1\rangle + \frac{\sqrt{3}}{2}|2\rangle$. Calculate $\langle \hat A \rangle$, $\langle \hat A^2 \rangle$, and the uncertainty $\Delta A$. Solution.

$$\langle \hat A \rangle = \frac{1}{4}(5) + \frac{3}{4}(-1) = \frac{5-3}{4} = 0.5.$$ $$\langle \hat A^2 \rangle = \frac{1}{4}(25) + \frac{3}{4}(1) = 7.$$ $$\Delta A = \sqrt{\langle \hat A^2 \rangle - \langle \hat A \rangle^2} = \sqrt{7 - 0.25} = \sqrt{6.75} \approx 2.60.$$

Problem 3. Show that the Pauli matrices satisfy $\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = I$ and $[\sigma_x, \sigma_y] = 2i\sigma_z$.

Solution. Using $\sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, $\sigma_y = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$, $\sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$:

$$\sigma_x^2 = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = I,$$

and similarly for $\sigma_y^2$ and $\sigma_z^2$. For the commutator:

$$[\sigma_x, \sigma_y] = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} - \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}i & 0 \\ 0 & -i\end{pmatrix} - \begin{pmatrix}-i & 0 \\ 0 & i\end{pmatrix} = \begin{pmatrix}2i & 0 \\ 0 & -2i\end{pmatrix} = 2i\sigma_z.$$

Section summary. The formalism provides a representation-independent framework using Dirac notation, Hermitian operators, and completeness relations to describe states and measurements.

Quantum Mechanics in Three Dimensions

Core ideas

Extending quantum mechanics to 3D involves the Laplacian $\nabla^2$. For central potentials $V(r)$, the problem separates into radial and angular parts.

Spherical Coordinates and Separation of Variables. Using $\psi(r, \theta, \phi) = R(r) Y_{\ell m}(\theta, \phi)$, where $Y_{\ell m}$ are the spherical harmonics:

• Angular Part: Solutions to the angular momentum eigenvalue problem:

$$\hat L^2 Y_{\ell m} = \hbar^2 \ell(\ell+1) Y_{\ell m}, \quad \hat L_z Y_{\ell m} = \hbar m Y_{\ell m}.$$

• Radial Part: The function $u(r) = rR(r)$ satisfies a 1D-like equation with an effective potential:

$$-\frac{\hbar^2}{2m} \frac{d^2u}{dr^2} + \left[ V(r) + \frac{\hbar^2 \ell(\ell+1)}{2mr^2} \right] u = E u.$$

The term proportional to $\ell(\ell+1)/r^2$ is the **centrifugal barrier**.

Hydrogen Atom and Quantum Numbers. For the Coulomb potential $V(r) = -e^2/(4\pi\epsilon_0 r)$, the energies are quantized:

$$E_n = -\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2} \frac{1}{n^2} \approx -\frac{13.6 \text{ eV}}{n^2}.$$

The state is labeled by three quantum numbers:

• $n$ (Principal): $n = 1, 2, \dots$ (Determines energy).
• $\ell$ (Azimuthal): $\ell = 0, 1, \dots, n-1$ (Magnitude of $\bm L$).
• $m$ (Magnetic): $m = -\ell, \dots, +\ell$ ($z$-component of $\bm L$).

Spin and Addition of Angular Momentum. Spin $\bm S$ is intrinsic angular momentum. For an electron ($s=1/2$), $S_z = \pm \hbar/2$. When combining two angular momenta $\bm J_1$ and $\bm J_2$ (e.g., $L$ and $S$), the total $J$ can range from $|j_1-j_2|$ to $j_1+j_2$. The states in the combined basis are related to the individual bases by Clebsch—Gordan coefficients.

Mathematical spine

$$\begin{aligned} V_{\rm eff}(r) &= V(r) + \frac{\hbar^2 \ell(\ell+1)}{2mr^2} & \text{(Effective Potential)} \\ \hat S_i &= \frac{\hbar}{2} \sigma_i & \text{(Spin and Pauli Matrices)} \\ \bm J &= \bm L + \bm S, \quad [J^2, L^2] = 0 & \text{(Addition of Angular Momentum)} \end{aligned}$$

Example: The Stern—Gerlach Experiment. This experiment demonstrated the quantization of spin by passing a beam of silver atoms through a non-uniform magnetic field, splitting the beam into two distinct spots corresponding to $s_z = \pm \hbar/2$.

Worked example

A hydrogen atom is in the $n=2$, $\ell=1$, $m=0$ state. The energy is $E_2 = -13.6\,\text{eV}/2^2 = -3.40\,\text{eV}$. The Bohr radius is $a_0 = 0.529\times10^{-10}\,\text{m}$. The radial probability density for the $2p$ state is

$$P_{21}(r) = r^2|R_{21}(r)|^2 = r^2\left(\frac{1}{\sqrt{3}(2a_0)^{3/2}}\frac{r}{a_0}e^{-r/(2a_0)}\right)^2 = \frac{r^4}{24a_0^5}e^{-r/a_0}.$$

To find the most probable radius, set $dP_{21}/dr = 0$:

$$\frac{d}{dr}\left(r^4 e^{-r/a_0}\right) = 4r^3 e^{-r/a_0} - \frac{r^4}{a_0}e^{-r/a_0} = 0 \Rightarrow r = 4a_0 \approx 2.12\times10^{-10}\,\text{m} = 0.212\,\text{nm}.$$

This is twice the most probable radius of the $1s$ state ($r = a_0$), reflecting the larger spatial extent of excited states.

Problems with Solutions

Problem 1. An electron in a hydrogen atom is in the state with quantum numbers $n=3$, $\ell=2$, $m=-1$. What are the possible values of $L^2$, $L_z$, and the energy? Solution.

• $L^2 = \hbar^2\ell(\ell+1) = \hbar^2(2)(3) = 6\hbar^2 \approx 6.67\times10^{-68}\,\text{J}^2\text{\cdots}^2$
• $L_z = m\hbar = -\hbar \approx -1.055\times10^{-34}\,\text{J\cdots}$
• $E_3 = -13.6\,\text{eV}/3^2 = -1.51\,\text{eV}$

Problem 2. What are the possible total angular momentum quantum numbers $j$ when combining orbital angular momentum $\ell=1$ with spin $s=1/2$?

Solution. Using the addition rule $|j_1-j_2| \leq j \leq j_1+j_2$:

$$j = |1 - 1/2|, |1 - 1/2| + 1, \dots, 1 + 1/2 = 1/2, 3/2.$$

So there are two possible $j$ values: $j=1/2$ (doublet) and $j=3/2$ (quartet).

Problem 3. For a particle in a central potential, show that the effective potential for the radial motion contains a centrifugal barrier term.

Solution. Starting from the 3D TISE with $\psi(r,\theta,\phi) = R(r)Y_{\ell m}(\theta,\phi)$ and substituting $u(r) = rR(r)$:

$$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V(r) + \frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u = Eu.$$

The term $V_{\text{cf}}(r) = \frac{\hbar^2\ell(\ell+1)}{2mr^2}$ is the centrifugal barrier. For $\ell > 0$, it diverges as $r\to 0$, preventing the particle from reaching the origin.

Section summary. Three-dimensional systems with central symmetry are solved by separating radial and angular motion, leading to the quantization of orbital and intrinsic (spin) angular momentum.

Identical Particles

Core ideas

Identical quantum particles cannot be labeled in a physically meaningful way. Exchanging two identical particles cannot produce a new observable state. In three dimensions, many-particle wave functions are either symmetric or antisymmetric:

$$\psi(x_1,x_2)=+\psi(x_2,x_1)\quad \text{for bosons},$$ $$\psi(x_1,x_2)=-\psi(x_2,x_1)\quad \text{for fermions}.$$

Bosons have integer spin and can occupy the same one-particle state. Fermions have half-integer spin and obey the Pauli exclusion principle. The Pauli principle explains shell structure in atoms and the structure of the periodic table.

For two fermions the antisymmetric state can be written as a Slater determinant. For particles with spin, the total wave function includes spatial and spin parts. The product must have the correct exchange symmetry.

Mathematical spine

$$\psi(x_1,x_2)=\pm\psi(x_2,x_1),\qquad \Psi_{\rm fermion}(1,2)=\frac{1}{\sqrt2} \begin{vmatrix} \phi_a(1)&\phi_b(1)\\ \phi_a(2)&\phi_b(2) \end{vmatrix}.$$

Working details

Exchange symmetry applies to the complete state, including spin and spatial parts. Two electrons in the same spatial orbital can coexist only if their spin state is antisymmetric, the singlet. If the spin state is symmetric, the spatial state must be antisymmetric, reducing the probability of finding the particles close together.

For many fermions, the Slater determinant automatically changes sign when two particle labels are exchanged and vanishes if two one-particle states are identical. This is the mathematical form of Pauli exclusion. For bosons, symmetrized products allow macroscopic occupation of one state, which underlies Bose—Einstein condensation and coherent fields.

Identical-particle physics is not a small correction. It controls atomic shells, chemical bonding, the Fermi sea in metals, blackbody radiation, phonons, superfluidity, and the structure of matter. The first check in many-body problems is whether the particles are distinguishable, bosons, or fermions.

Worked example: lithium ground state $1s^2\,2s^1$. Pauli exclusion fixes the electron configuration of lithium. Two electrons fill the $1s$ shell with opposite spins, and the third must occupy the next available spin-orbital, $2s$. Writing the spin-orbitals $\chi_1=\phi_{1s}\alpha$, $\chi_2=\phi_{1s}\beta$, $\chi_3=\phi_{2s}\alpha$, the totally antisymmetric three-electron ground state is the Slater determinant

$$\Psi(1,2,3)=\frac{1}{\sqrt{3!}} \begin{vmatrix} \phi_{1s}(1)\alpha(1) & \phi_{1s}(1)\beta(1) & \phi_{2s}(1)\alpha(1)\\ \phi_{1s}(2)\alpha(2) & \phi_{1s}(2)\beta(2) & \phi_{2s}(2)\alpha(2)\\ \phi_{1s}(3)\alpha(3) & \phi_{1s}(3)\beta(3) & \phi_{2s}(3)\alpha(3) \end{vmatrix}.$$

A hypothetical $1s^3$ configuration would place two columns with identical spin-orbitals (e.g.\ a second $\phi_{1s}\alpha$); the determinant would then vanish identically. This is Pauli exclusion in action: it forces the third electron out of the $K$-shell and into $2s$, producing the alkali-metal chemistry of lithium.

Worked example: helium singlet vs.\ triplet. For two-electron helium, the antisymmetric total wave function factorises into spatial and spin parts. The ground state has both electrons in $1s$: the spatial part is symmetric, so the spin part must be the antisymmetric singlet $\frac{1}{\sqrt2}(\alpha\beta-\beta\alpha)$, giving the configuration $1s^2\,{}^1S_0$. Excited states like $1s^1 2s^1$ admit two combinations: the symmetric spatial $\frac{1}{\sqrt2}[\phi_{1s}(1)\phi_{2s}(2)+\phi_{2s}(1)\phi_{1s}(2)]$ paired with the singlet (parahelium), and the antisymmetric spatial $\frac{1}{\sqrt2}[\phi_{1s}(1)\phi_{2s}(2)-\phi_{2s}(1)\phi_{1s}(2)]$ paired with the symmetric triplet (orthohelium). The triplet lies lower in energy because the antisymmetric spatial part suppresses the probability of the two electrons being close, reducing Coulomb repulsion. This exchange splitting is a direct, measurable consequence of fermion antisymmetry.

Worked example

Two non-interacting electrons are in a 1D infinite square well of width $L = 1.0\,\text{nm}$. The single-particle energies are $E_n = n^2 E_1$ with $E_1 = \pi^2\hbar^2/(2m_e L^2) \approx 0.376\,\text{eV}$. In the ground state, both electrons occupy $n=1$ with opposite spins (singlet). The symmetric spatial wave function is

$$\psi(x_1,x_2) = \frac{2}{L}\sin\left(\frac{\pi x_1}{L}\right)\sin\left(\frac{\pi x_2}{L}\right).$$

The first excited spatial configuration has one electron in $n=1$ and one in $n=2$. Because the electrons are fermions, the total wave function must be antisymmetric. For the spin triplet (symmetric spin), the spatial part must be antisymmetric:

$$\psi_A(x_1,x_2) = \frac{1}{\sqrt{2}}\frac{2}{L}\left[\sin\left(\frac{\pi x_1}{L}\right)\sin\left(\frac{2\pi x_2}{L}\right) - \sin\left(\frac{2\pi x_1}{L}\right)\sin\left(\frac{\pi x_2}{L}\right)\right].$$

This vanishes at $x_1 = x_2$, keeping the electrons apart and reducing Coulomb repulsion. The total energy is $E_1 + E_2 = 0.376 + 1.504 = 1.880\,\text{eV}$.

Problems with Solutions

Problem 1. Write the Slater determinant for two electrons in spin-orbitals $\chi_1 = \phi_{1s}\alpha$ and $\chi_2 = \phi_{1s}\beta$. Solution.

$$\Psi(1,2) = \frac{1}{\sqrt{2!}}\begin{vmatrix}\phi_{1s}(1)\alpha(1) & \phi_{1s}(1)\beta(1) \\ \phi_{1s}(2)\alpha(2) & \phi_{1s}(2)\beta(2)\end{vmatrix} = \frac{1}{\sqrt{2}}\phi_{1s}(1)\phi_{1s}(2)[\alpha(1)\beta(2) - \beta(1)\alpha(2)].$$

This is the antisymmetric singlet state for two electrons in the same spatial orbital.

Problem 2. Three identical bosons can each occupy one of two single-particle states $a$ or $b$. How many distinct symmetric many-body states exist?

Solution. For bosons, any number can occupy each state. The allowed occupation-number configurations are $(N_a, N_b) = (3,0), (2,1), (1,2), (0,3)$. Each corresponds to one distinct symmetric state, so there are 4 states. In contrast, three fermions could not be placed in only two spin-orbitals because of Pauli exclusion.

Problem 3. Two identical spin-$0$ bosons are in a 1D box. Write the symmetrized two-particle state if one is in $\phi_1(x) = \sqrt{2/L}\sin(\pi x/L)$ and the other in $\phi_2(x) = \sqrt{2/L}\sin(2\pi x/L)$.

Solution. Since they are bosons, the total wave function must be symmetric under exchange:

$$\psi_S(x_1,x_2) = \frac{1}{\sqrt{2}}[\phi_1(x_1)\phi_2(x_2) + \phi_2(x_1)\phi_1(x_2)] = \frac{1}{L}\left[\sin\left(\frac{\pi x_1}{L}\right)\sin\left(\frac{2\pi x_2}{L}\right) + \sin\left(\frac{2\pi x_1}{L}\right)\sin\left(\frac{\pi x_2}{L}\right)\right].$$

Section summary. Identical particles force quantum states to be symmetric or antisymmetric under exchange, producing bosons, fermions, and Pauli exclusion.

Part II: Applications

Core ideas

Few quantum systems are exactly solvable. For most realistic problems (multi-electron atoms, atoms in external fields, anharmonic vibrations) we expand around a solvable reference. The standard approach is Rayleigh—Schr”odinger perturbation theory: split the Hamiltonian into a solvable part plus a small correction, then expand energies and states in powers of a bookkeeping parameter $\lambda$.

Setup. Write $\hat H(\lambda)=\hat H_0+\lambda\hat H'$, with $\hat H_0|n^{(0)}\rangle=E_n^{(0)}|n^{(0)}\rangle$ the known eigenproblem. The unperturbed kets $\{|n^{(0)}\rangle\}$ are orthonormal and complete. We seek series

$$E_n=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+\cdots,\qquad |n\rangle=|n^{(0)}\rangle+\lambda|n^{(1)}\rangle+\lambda^2|n^{(2)}\rangle+\cdots,$$

and set $\lambda=1$ at the end. Intermediate normalisation $\langle n^{(0)}|n\rangle=1$ is conventional, so each $|n^{(k\ge1)}\rangle$ is orthogonal to $|n^{(0)}\rangle$.

Non-degenerate corrections

Substituting the series into $\hat H|n\rangle=E_n|n\rangle$ and matching powers of $\lambda$ gives the standard formulas. Define the matrix element $H'_{mn}\equiv\langle m^{(0)}|\hat H'|n^{(0)}\rangle$.

• First-order energy: $E_n^{(1)}=H'_{nn}=\langle n^{(0)}|\hat H'|n^{(0)}\rangle$, the expectation value of the perturbation in the unperturbed state.
• First-order state:

$$|n^{(1)}\rangle=\sum_{m\neq n}\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\,|m^{(0)}\rangle.$$

• Second-order energy:

$$E_n^{(2)}=\sum_{m\neq n}\frac{|H'_{mn}|^2}{E_n^{(0)}-E_m^{(0)}}.$$

For the ground state every denominator is negative, so $E_0^{(2)}\le0$: second-order corrections always lower the ground-state energy.

When does the series converge?

The expansion is sensible only when each successive term is much smaller than the previous one. A practical criterion is that, for every state $|m^{(0)}\rangle$ coupled to $|n^{(0)}\rangle$ by $\hat H'$,

$$\bigl|H'_{mn}\bigr|\ll\bigl|E_n^{(0)}-E_m^{(0)}\bigr|.$$

Practical implications.

• The relevant small parameter is the dimensionless ratio of a typical matrix element to the nearest energy gap, not just the strength of $\hat H'$ itself.
• Levels close in energy (small denominators) destroy convergence even for a tiny perturbation; this is precisely where degenerate perturbation theory is needed.
• The series is generically asymptotic rather than convergent: in many physical problems (e.g. anharmonic oscillators, QED) terms first decrease, then grow. Truncating at the smallest term gives the best estimate.
• Perturbation theory cannot describe states absent from $\hat H_0$. Bound states produced non-perturbatively (tunnelling resonances, bound states in a new potential well) are missed at every finite order.

Degenerate perturbation theory

If $E_n^{(0)}$ has a $g$-fold degenerate subspace $\mathcal D=\{|n^{(0)}_a\rangle\}_{a=1}^g$, the denominators inside $\mathcal D$ vanish. The fix is to choose the right basis inside $\mathcal D$ before applying the formulas above. Build the $g\times g$ matrix $W_{ab}=\langle n^{(0)}_a|\hat H'|n^{(0)}_b\rangle$ and diagonalise it; its eigenvalues are the first-order shifts and its eigenvectors are the “good” zeroth-order states that diagonalise $\hat H'$ inside $\mathcal D$. Couplings to states outside $\mathcal D$ then enter through the standard non-degenerate sums.

Worked example: linear Stark effect for $n=2$ hydrogen. A hydrogen atom in a uniform electric field $\bm{\mathcal E}=\mathcal E\hat{\bm z}$ sees the perturbation $\hat H'=e\mathcal E\,z$, where $e>0$ is the elementary charge and $z$ is the electron coordinate along the field. The $n=2$ shell is four-fold degenerate with basis

$$|2s\rangle=|200\rangle,\quad |2p_0\rangle=|210\rangle,\quad |2p_{+1}\rangle=|211\rangle,\quad |2p_{-1}\rangle=|21\,{-1}\rangle.$$

Selection rules from parity ($z$ is odd) and from $[\hat L_z,z]=0$ (so $\Delta m=0$) leave only one independent matrix element,

$$\langle 200|z|210\rangle=-3a_0,$$

with $a_0$ the Bohr radius. The perturbation matrix in the ordered basis $(|2s\rangle,|2p_0\rangle,|2p_{+1}\rangle,|2p_{-1}\rangle)$ is

$$W=e\mathcal E \begin{pmatrix} 0 & -3a_0 & 0 & 0\\ -3a_0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}.$$

Diagonalising the upper $2\times2$ block gives eigenvalues $\pm 3ea_0\mathcal E$ with eigenvectors $\frac{1}{\sqrt2}(|2s\rangle\mp|2p_0\rangle)$; the $m=\pm1$ states are unshifted to first order. The fourfold level therefore splits into three: two shifted by $\pm 3ea_0\mathcal E$ and one doubly degenerate level at the original energy. The shift is linear in $\mathcal E$, a signature of the accidental $\ell$-degeneracy of the Coulomb spectrum.

Mathematical spine

$$\begin{aligned} \hat H &= \hat H_0+\lambda\hat H', & H'_{mn}&=\langle m^{(0)}|\hat H'|n^{(0)}\rangle\\ E_n^{(1)} &= H'_{nn}, & E_n^{(2)}&=\sum_{m\neq n}\frac{|H'_{mn}|^2}{E_n^{(0)}-E_m^{(0)}}\\ |n^{(1)}\rangle &= \sum_{m\neq n}\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\,|m^{(0)}\rangle, & W_{ab}&=\langle n^{(0)}_a|\hat H'|n^{(0)}_b\rangle\ (\text{degenerate block}) \end{aligned}$$

Example: the Zeeman effect. For a weak external field $\bm B$ the perturbation is $\hat H'=-\bm\mu\cdot\bm B$, with $\bm\mu$ the magnetic moment. Choosing $\bm B=B\hat{\bm z}$ makes $\hat H'$ diagonal in the standard $|n\ell m\rangle$ basis (already a “good” basis), so first-order theory predicts level splittings linear in the magnetic quantum number $m$, in agreement with experiment for fields where spin—orbit coupling can be ignored.

Worked example

A 1D anharmonic oscillator has Hamiltonian $\hat H = \hat H_0 + \hat H'$ with $\hat H_0 = \frac{\hat p^2}{2m} + \frac{1}{2}m\omega^2\hat x^2$ and $\hat H' = \lambda \hat x^4$, where $\lambda = 0.01\,\text{eV/nm}^4$, $m = 1.0\times10^{-30}\,\text{kg}$, and $\hbar\omega = 0.1\,\text{eV}$. Estimate the first-order energy shift of the ground state.

Using $\langle 0|\hat x^4|0\rangle = \frac{3}{4}\left(\frac{\hbar}{m\omega}\right)^2$, we first compute the characteristic length $a = \sqrt{\hbar/(m\omega)}$. With $\hbar\omega = 0.1\,\text{eV}$ and $m = 10^{-30}\,\text{kg}$:

$$\omega = \frac{0.1\times1.602\times10^{-19}}{1.055\times10^{-34}} \approx 1.52\times10^{14}\,\text{rad/s},$$ $$a = \sqrt{\frac{1.055\times10^{-34}}{(10^{-30})(1.52\times10^{14})}} \approx 8.3\times10^{-10}\,\text{m} = 0.83\,\text{nm}.$$

Then

$$E_0^{(1)} = \lambda\langle 0|\hat x^4|0\rangle = 0.01\,\text{eV/nm}^4 \times \frac{3}{4}(0.83\,\text{nm})^4 \approx 0.01 \times \frac{3}{4}(0.474) \approx 3.6\times10^{-3}\,\text{eV}.$$

The unperturbed ground-state energy is $E_0^{(0)} = \frac{1}{2}\hbar\omega = 0.05\,\text{eV}$, so the relative shift is $\approx 7\%$.

Problems with Solutions

Problem 1. A particle in an infinite square well ($0<x<a$) is subject to perturbation $\hat H' = V_0$ for $a/3 < x < 2a/3$ and zero elsewhere. Find the first-order energy shift of the $n$th level.

Solution. The unperturbed states are $\psi_n^{(0)}(x) = \sqrt{2/a}\sin(n\pi x/a)$. The first-order shift is

$$E_n^{(1)} = \langle n^{(0)}|\hat H'|n^{(0)}\rangle = V_0\int_{a/3}^{2a/3}\frac{2}{a}\sin^2\left(\frac{n\pi x}{a}\right)dx.$$

Using $\sin^2\theta = \frac{1}{2}(1-\cos 2\theta)$:

$$E_n^{(1)} = \frac{2V_0}{a}\left[\frac{x}{2} - \frac{a}{4n\pi}\sin\left(\frac{2n\pi x}{a}\right)\right]_{a/3}^{2a/3} = V_0\left[\frac{1}{3} - \frac{1}{2n\pi}\left(\sin\frac{4n\pi}{3} - \sin\frac{2n\pi}{3}\right)\right].$$

Problem 2. For a two-level system with unperturbed energies $E_1^{(0)} = 1\,\text{eV}$ and $E_2^{(0)} = 3\,\text{eV}$, the perturbation has matrix elements $H'_{11} = 0.1\,\text{eV}$, $H'_{22} = -0.1\,\text{eV}$, and $H'_{12} = H'_{21} = 0.2\,\text{eV}$. Calculate the second-order correction to $E_1^{(0)}$. Solution.

$$E_1^{(2)} = \frac{|H'_{21}|^2}{E_1^{(0)} - E_2^{(0)}} = \frac{(0.2\,\text{eV})^2}{1 - 3} = \frac{0.04}{-2}\,\text{eV} = -0.02\,\text{eV}.$$

The second-order correction lowers the ground-state energy, as required by the variational principle.

Problem 3. A two-fold degenerate level has degenerate subspace spanned by $|1\rangle$ and $|2\rangle$. The perturbation matrix in this subspace is $W = \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}\,\text{eV}$. Find the first-order energy shifts and the “good” zeroth-order states.

Solution. Diagonalize $W$. The characteristic equation is $(2-\lambda)^2 - 1 = 0$, giving $\lambda = 3$ or $\lambda = 1$. The first-order shifts are $E^{(1)} = 3\,\text{eV}$ and $1\,\text{eV}$. The eigenvectors are $|+\rangle = \frac{1}{\sqrt{2}}(|1\rangle + |2\rangle)$ for $\lambda=3$ and $|-\rangle = \frac{1}{\sqrt{2}}(|1\rangle - |2\rangle)$ for $\lambda=1$. These are the good zeroth-order states.

Section summary. Perturbation theory expresses corrections as power series in a small parameter, with first- and second-order energies built from the matrix elements $H'_{mn}$ divided by unperturbed gaps. Convergence requires every $|H'_{mn}|$ to be much smaller than the corresponding gap $|E_n^{(0)}-E_m^{(0)}|$; when that fails because of degeneracy, diagonalising $\hat H'$ inside the degenerate subspace, as in the linear Stark effect of $n=2$ hydrogen, restores predictive power.

Time-Dependent Perturbation Theory

Core ideas

Time-dependent perturbation theory computes transition probabilities caused by a weak time-dependent interaction:

$$\hat H(t)=\hat H_0+\hat V(t).$$

Write the state as a superposition of unperturbed energy eigenstates. The perturbation changes the coefficients.

To first order, the transition amplitude from $|i\rangle$ to $|f\rangle$ is proportional to

$$\int_0^t e^{i\omega_{fi}t'}V_{fi}(t')\,dt', \qquad \omega_{fi}=\frac{E_f-E_i}{\hbar}.$$

If the perturbation oscillates at a frequency close to $\omega_{fi}$, transitions are enhanced. This gives selection rules and resonance behavior in light-matter interaction.

For long times and dense final states, the transition rate becomes Fermi’s golden rule:

$$\Gamma_{i\to f}=\frac{2\pi}{\hbar}|V_{fi}|^2\rho(E_f).$$

It is one of the most important formulas in atomic, nuclear, and condensed-matter physics.

Mathematical spine

$$P_{i\to f}(t)\approx \frac{1}{\hbar^2}\left|\int_0^t e^{i\omega_{fi}t'}V_{fi}(t')\,dt'\right|^2,\qquad \Gamma_{i\to f}=\frac{2\pi}{\hbar}|V_{fi}|^2\rho(E_f).$$

Working details

The perturbation drives transitions most efficiently when its frequency matches an energy difference. This is the origin of absorption and emission lines. If $\hat V(t)$ comes from an electromagnetic wave, the matrix element contains both the field strength and a transition dipole or multipole matrix element.

Selection rules arise when a matrix element vanishes by symmetry. For electric dipole transitions in atoms, common rules are $\Delta \ell=\pm1$ and $\Delta m=0,\pm1$, with details depending on polarization. A forbidden transition is often not absolutely impossible; it may be allowed by a weaker magnetic dipole, electric quadrupole, or symmetry-breaking interaction.

Fermi’s golden rule assumes weak coupling, long times compared with microscopic oscillations, and a continuum or dense set of final states. It is a rate formula, not a formula for coherent two-level oscillations. Strong resonant driving instead leads to Rabi oscillations.

Worked example: Rabi oscillations of a resonantly driven two-level atom. Consider a two-level atom with ground state $|g\rangle$ and excited state $|e\rangle$ of energies $E_g$ and $E_e$, so $\omega_0=(E_e-E_g)/\hbar$. Place it in a classical monochromatic field $\bm E(t)=E_0\hat{\bm\epsilon}\cos\omega t$, giving the dipole interaction $\hat V(t)=-\hat{\bm d}\cdot\bm E(t)$. Writing $|\psi(t)\rangle=c_g(t)e^{-iE_g t/\hbar}|g\rangle+c_e(t)e^{-iE_e t/\hbar}|e\rangle$, the Schr”odinger equation gives coupled equations for $(c_g,c_e)$. Defining the Rabi frequency

$$\Omega=\frac{d_{eg}E_0}{\hbar},\qquad d_{eg}=\langle e|\hat{\bm d}\cdot\hat{\bm\epsilon}|g\rangle,$$

and dropping rapidly oscillating $\sim e^{\pm i(\omega+\omega_0)t}$ terms (rotating-wave approximation) leaves only slow terms $\sim e^{\pm i(\omega-\omega_0)t}$. On exact resonance $\omega=\omega_0$ and with $c_g(0)=1$, $c_e(0)=0$, the solution is

$$c_g(t)=\cos\!\frac{\Omega t}{2},\qquad c_e(t)=-i\sin\!\frac{\Omega t}{2},$$

so the excited-state population oscillates coherently as

$$P_e(t)=|c_e(t)|^2=\sin^2\!\frac{\Omega t}{2}.$$

The population swaps fully between $|g\rangle$ and $|e\rangle$ with period $2\pi/\Omega$; a pulse of area $\Omega t=\pi$ (a “$\pi$-pulse”) inverts the atom, while $\Omega t=\pi/2$ prepares an equal superposition. Off resonance, the generalised Rabi frequency $\tilde\Omega=\sqrt{\Omega^2+(\omega-\omega_0)^2}$ replaces $\Omega$ and the maximum excitation drops to $\Omega^2/\tilde\Omega^2<1$. This non-perturbative result complements first-order perturbation theory: short times or weak driving ($\Omega t\ll1$) recover $P_e\approx(\Omega t/2)^2$, the linear-response limit.

Worked example

A hydrogen atom initially in the ground state ($|1s\rangle$, $E_1 = -13.6\,\text{eV}$) is exposed to a time-dependent electric field $\mathcal{E}(t) = \mathcal{E}_0 e^{-t^2/\tau^2}$ polarized along $z$, with $\mathcal{E}_0 = 10^5\,\text{V/m}$ and $\tau = 1.0\times10^{-14}\,\text{s}$. Estimate the transition probability to the $|2p_z\rangle$ state ($E_2 = -3.4\,\text{eV}$) to first order.

The perturbation is $\hat V(t) = e\mathcal{E}(t) z$. The transition frequency is

$$\omega_{21} = \frac{E_2-E_1}{\hbar} = \frac{-3.4-(-13.6)\,\text{eV}}{6.582\times10^{-16}\,\text{eV\cdots}} \approx 1.55\times10^{16}\,\text{rad/s}.$$

The matrix element $\langle 2p_z|z|1s\rangle = \frac{512\sqrt{2}}{243\sqrt{3}}a_0 \approx 0.74\,a_0$ with $a_0 = 0.529\times10^{-10}\,\text{m}$, so

$$V_{21}(t) = e\mathcal{E}_0(0.74\,a_0)e^{-t^2/\tau^2} \approx (1.602\times10^{-19})(10^5)(0.74\times0.529\times10^{-10})e^{-t^2/\tau^2} \approx 6.3\times10^{-25}\,e^{-t^2/\tau^2}\,\text{J}.$$

The first-order amplitude is

$$c_2^{(1)}(\infty) = -\frac{i}{\hbar}\int_{-\infty}^{\infty}e^{i\omega_{21}t}V_{21}(t)dt = -\frac{i}{\hbar}(6.3\times10^{-25})\sqrt{\pi}\tau e^{-\omega_{21}^2\tau^2/4}.$$

With $\omega_{21}\tau \approx 155 \gg 1$, the Gaussian factor is extremely small ($\sim e^{-6000}$), so the transition probability is essentially zero. This illustrates the adiabatic limit: when the perturbation varies slowly compared with the transition period, no transitions occur.

Problems with Solutions

Problem 1. A two-level system with $\hat H_0 = E_g|g\rangle\langle g| + E_e|e\rangle\langle e|$ is driven by a constant perturbation $\hat V = V_0(|g\rangle\langle e| + |e\rangle\langle g|)$ turned on at $t=0$. If the system starts in $|g\rangle$, find the transition probability to $|e\rangle$ at time $t$ to first order.

Solution. To first order:

$$c_e^{(1)}(t) = -\frac{i}{\hbar}\int_0^t e^{i\omega_{eg}t'}V_{eg}(t')dt' = -\frac{iV_0}{\hbar}\int_0^t e^{i\omega_{eg}t'}dt' = -\frac{iV_0}{\hbar}\left[\frac{e^{i\omega_{eg}t}-1}{i\omega_{eg}}\right] = \frac{V_0}{\hbar\omega_{eg}}(1-e^{i\omega_{eg}t}).$$

The transition probability is

$$P_{g\to e}(t) = |c_e^{(1)}(t)|^2 = \frac{4V_0^2}{\hbar^2\omega_{eg}^2}\sin^2\left(\frac{\omega_{eg}t}{2}\right).$$

Problem 2. In a scattering experiment, the initial state $|i\rangle$ couples to a continuum of final states $|f\rangle$ with constant matrix element $V_{fi} = 0.01\,\text{eV}$. The density of final states at the relevant energy is $\rho(E_f) = 2.0\,\text{eV}^{-1}$. Calculate the transition rate using Fermi’s golden rule.

Solution. Fermi’s golden rule gives

$$\Gamma_{i\to f} = \frac{2\pi}{\hbar}|V_{fi}|^2\rho(E_f) = \frac{2\pi}{6.582\times10^{-16}\,\text{eV\cdots}}(0.01\,\text{eV})^2(2.0\,\text{eV}^{-1}) \approx 1.91\times10^{12}\,\text{s}^{-1}.$$

The lifetime of the initial state is $\tau = 1/\Gamma \approx 5.2\times10^{-13}\,\text{s}$.

Problem 3. A spin-$1/2$ particle in a magnetic field $\bm B = B_0\hat{\bm z}$ has Hamiltonian $\hat H_0 = -\gamma B_0 \hat S_z$ with $\gamma$ the gyromagnetic ratio. A weak transverse field $B_1$ is applied along $x$ and oscillates as $B_1(t) = B_1\cos\omega t$. Write the perturbation and identify the resonance condition.

Solution. The perturbation is

$$\hat V(t) = -\gamma B_1(t)\hat S_x = -\gamma B_1\cos(\omega t)\frac{\hbar}{2}\sigma_x = -\frac{\gamma\hbar B_1}{2}\cos(\omega t)\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}.$$

In the basis $\{|+\rangle_z, |-\rangle_z\}$, this couples the two spin states. The energy splitting is $\Delta E = \gamma\hbar B_0$, so the resonance frequency is $\omega_0 = \Delta E/\hbar = \gamma B_0$. When $\omega = \omega_0$, the driving field is on resonance and induces Rabi oscillations between the spin states with Rabi frequency $\Omega = \gamma B_1$.

Section summary. Time-dependent perturbation theory computes transition probabilities and rates caused by weak time-dependent interactions.

The Variational Principle

Core ideas

The variational principle provides a powerful way to estimate the ground-state energy $E_0$ of a system when an exact solution is impossible.

The Variational Theorem. For any normalized trial wave function $|\psi\rangle$, the expectation value of the Hamiltonian is always greater than or equal to the true ground-state energy:

$$E_0 \leq \langle \psi | \hat H | \psi \rangle.$$

For an unnormalized trial state $\psi_\alpha$ depending on a parameter $\alpha$, we minimize the Rayleigh quotient:

$$\epsilon(\alpha) = \frac{\langle \psi_\alpha | \hat H | \psi_\alpha \rangle}{\langle \psi_\alpha | \psi_\alpha \rangle}.$$

The best estimate for $E_0$ is the minimum value of $\epsilon(\alpha)$.

Success of the Method. The method is highly successful because even a rough guess for the wave function often yields a very accurate upper bound for the energy. This is because the error in energy is of second order in the error of the wave function.

Mathematical spine

$$\begin{aligned} E_0 &\leq \frac{\langle \psi | \hat H | \psi \rangle}{\langle \psi | \psi \rangle} & \text{(Variational Bound)} \\ \frac{d\epsilon}{d\alpha} &= 0 & \text{(Minimization Condition)} \end{aligned}$$

Example: Ground State of Helium. The Helium atom has two electrons and a nucleus with charge $Z=2$. The Hamiltonian includes an electron-electron repulsion term that makes it unsolvable. A simple trial wave function is a product of two hydrogenic states with an effective charge $Z_{\rm eff}$ as the variational parameter. Minimizing the energy yields $Z_{\rm eff} = 2 - 5/16 = 1.6875$, showing how one electron “screens” the nucleus for the other.

Worked example

Estimate the ground-state energy of a 1D harmonic oscillator $V(x) = \frac{1}{2}m\omega^2 x^2$ using the trial function $\psi_\alpha(x) = \left(\frac{\alpha}{\pi}\right)^{1/4}e^{-\alpha x^2/2}$, where $\alpha > 0$ is a variational parameter. The exact answer is $E_0 = \frac{1}{2}\hbar\omega$.

First compute the normalization: $\langle\psi_\alpha|\psi_\alpha\rangle = 1$ by construction. The kinetic energy expectation is

$$\langle T \rangle = -\frac{\hbar^2}{2m}\int_{-\infty}^{\infty}\psi_\alpha\frac{d^2\psi_\alpha}{dx^2}dx = \frac{\hbar^2\alpha}{4m}.$$

The potential energy expectation is

$$\langle V \rangle = \frac{1}{2}m\omega^2\int_{-\infty}^{\infty}x^2|\psi_\alpha|^2dx = \frac{m\omega^2}{4\alpha}.$$

The Rayleigh quotient is

$$\epsilon(\alpha) = \frac{\hbar^2\alpha}{4m} + \frac{m\omega^2}{4\alpha}.$$

Minimizing: $\frac{d\epsilon}{d\alpha} = \frac{\hbar^2}{4m} - \frac{m\omega^2}{4\alpha^2} = 0 \Rightarrow \alpha = \frac{m\omega}{\hbar}$.

Substituting back:

$$\epsilon_{\min} = \frac{\hbar^2}{4m}\frac{m\omega}{\hbar} + \frac{m\omega^2}{4}\frac{\hbar}{m\omega} = \frac{\hbar\omega}{4} + \frac{\hbar\omega}{4} = \frac{1}{2}\hbar\omega.$$

The variational estimate equals the exact ground-state energy because the trial function has the same Gaussian form as the true ground state.

Problems with Solutions

Problem 1. Use the variational principle with trial function $\psi(x) = N(a^2 - x^2)$ for $|x|<a$ and zero otherwise to estimate the ground-state energy of an infinite square well of width $L$ ($-L/2<x<L/2$). Use $a$ as the variational parameter with $a \leq L/2$.

Solution. Normalize:

$$1 = N^2\int_{-a}^{a}(a^2-x^2)^2dx = N^2\frac{16a^5}{15} \Rightarrow N = \sqrt{\frac{15}{16a^5}}.$$

Kinetic energy:

$$\langle T \rangle = \frac{\hbar^2}{2m}\int_{-a}^{a}\left(\frac{d\psi}{dx}\right)^2dx = \frac{\hbar^2}{2m}N^2\int_{-a}^{a}(2x)^2dx = \frac{\hbar^2}{2m}\frac{15}{16a^5}\frac{8a^3}{3} = \frac{5\hbar^2}{4ma^2}.$$

Since $\psi=0$ outside $|x|<a$, the potential is zero in the support. Minimize $\epsilon(a) = 5\hbar^2/(4ma^2)$ subject to $a \leq L/2$. The minimum occurs at $a = L/2$, giving

$$\epsilon = \frac{5\hbar^2}{mL^2} = \frac{10}{\pi^2}\frac{\pi^2\hbar^2}{2mL^2} \approx 1.013\,E_1^{\text{exact}}.$$

This is only $1.3\%$ above the exact ground-state energy.

Problem 2. For a particle in a potential $V(x) = \lambda x^4$, estimate the ground-state energy using a Gaussian trial function $\psi_\alpha(x) = (\alpha/\pi)^{1/4}e^{-\alpha x^2/2}$.

Solution. With $\langle x^4 \rangle = 3/(4\alpha^2)$ for a Gaussian:

$$\epsilon(\alpha) = \frac{\hbar^2\alpha}{4m} + \frac{3\lambda}{4\alpha^2}.$$

Minimizing: $\frac{d\epsilon}{d\alpha} = \frac{\hbar^2}{4m} - \frac{3\lambda}{2\alpha^3} = 0 \Rightarrow \alpha = \left(\frac{6m\lambda}{\hbar^2}\right)^{1/3}$. Substituting:

$$\epsilon_{\min} = \frac{\hbar^2}{4m}\left(\frac{6m\lambda}{\hbar^2}\right)^{1/3} + \frac{3\lambda}{4}\left(\frac{\hbar^2}{6m\lambda}\right)^{2/3} = \frac{3}{8}\left(\frac{6\hbar^4\lambda}{m^2}\right)^{1/3} \approx 0.68\left(\frac{\hbar^4\lambda}{m^2}\right)^{1/3}.$$

The exact numerical coefficient is $\approx 0.668$, so the variational estimate is about $2\%$ high.

Problem 3. Explain why the variational principle always gives an upper bound to the true ground-state energy.

Solution. Expand the trial state in the exact energy eigenbasis: $|\psi\rangle = \sum_n c_n|n\rangle$ with $\sum_n|c_n|^2 = 1$. Then

$$\langle\psi|\hat H|\psi\rangle = \sum_n |c_n|^2 E_n \geq E_0\sum_n|c_n|^2 = E_0,$$

since $E_n \geq E_0$ for all $n$. Equality holds only if $|\psi\rangle = |0\rangle$, the exact ground state.

Section summary. The variational principle allows for estimating ground-state energies by optimizing trial wave functions, always providing a guaranteed upper bound.

The WKB Approximation

Core ideas

The WKB approximation is a semiclassical method for one-dimensional problems where the potential changes slowly compared with the de Broglie wavelength. Define the local classical momentum

$$p(x)=\sqrt{2m(E-V(x))}.$$

In the classically allowed region $E>V(x)$, the wave function is approximately oscillatory:

$$\psi(x)\approx \frac{C}{\sqrt{p(x)}}\exp\left(\pm\frac{i}{\hbar}\int^x p(x')\,dx'\right).$$

In the forbidden region $E<V(x)$, the momentum is imaginary and the wave function grows or decays exponentially. This gives a simple estimate of tunneling probabilities:

$$T\sim \exp\left[-\frac{2}{\hbar}\int_{x_1}^{x_2}|p(x)|\,dx\right].$$

At turning points $E=V(x)$, the naive WKB form fails and connection formulas are needed. For bound motion between two turning points, WKB gives the quantization rule

$$\int_{x_1}^{x_2}p(x)\,dx=\left(n+\frac12\right)\pi\hbar.$$

Mathematical spine

$$\psi(x)\approx \frac{C}{\sqrt{p(x)}}\exp\left(\frac{i}{\hbar}\int^x p(x')\,dx'\right),\qquad p(x)=\sqrt{2m(E-V(x))},$$ $$\int_{x_1}^{x_2}p(x)\,dx=\left(n+\frac12\right)\pi\hbar.$$

Working details

WKB works when the local wavelength changes slowly:

$$\left|\frac{d\lambda_{\rm dB}}{dx}\right|\ll 1,\qquad \lambda_{\rm dB}=\frac{h}{p(x)}.$$

It fails near turning points because $p(x)\to0$ and the amplitude factor diverges. Connection formulas repair the solution by matching to Airy-function behavior near the turning point.

The quantization rule is a semiclassical action condition. It says that the phase accumulated over a classical round trip must be compatible with a standing wave, with a correction from the turning points. It reproduces the exact large-$n$ behavior of many bound systems and gives good intuition for why classical action appears in quantum mechanics.

For tunneling, the exponential factor is usually more important than the prefactor. This makes WKB useful for alpha decay, field emission, scanning tunneling microscopy, and barrier penetration in semiconductor devices. The approximation should be checked by asking whether the barrier varies slowly on the de Broglie-wavelength scale.

Worked example

WKB quantization of a particle in a linear potential well. Consider a particle of mass $m$ in a triangular well defined by $V(x)=Fx$ for $x>0$ and $V(x)=\infty$ for $x<0$, where $F>0$ is a constant force. We want to estimate the energy levels using the WKB quantization rule.

The turning points are $x_1=0$ (hard wall) and $x_2=E/F$ (where $E=V(x)$). The WKB quantization condition with one hard wall and one soft turning point is

$$\int_0^{E_n/F} \sqrt{2m(E_n - Fx)}\,dx = \left(n - \frac{1}{4}\right)\pi\hbar, \quad n=1,2,\dots$$

Evaluating the integral:

$$\int_0^{E/F} \sqrt{2m(E - Fx)}\,dx = \frac{2\sqrt{2m}}{3F} E^{3/2}.$$

Setting this equal to $(n-1/4)\pi\hbar$ and solving for $E_n$ gives

$$E_n = \left[\frac{3\pi\hbar F}{2\sqrt{2m}}\left(n-\frac{1}{4}\right)\right]^{2/3}.$$

For an electron ($m=9.11\times10^{-31}\,\text{kg}$) in a field corresponding to $F=1.0\times10^{-9}\,\text{N}$ (a very weak force), the ground-state energy ($n=1$) is

$$E_1 = \left[\frac{3\pi(1.055\times10^{-34}\,\text{J\,s})(1.0\times10^{-9}\,\text{N})}{2\sqrt{2(9.11\times10^{-31}\,\text{kg})}}(0.75)\right]^{2/3} \approx 1.8\times10^{-22}\,\text{J} \approx 1.1\times10^{-3}\,\text{eV}.$$

This demonstrates how WKB converts a classical action integral into quantized energy levels without solving the differential equation exactly.

Problems with Solutions

Problem 1. A particle of mass $m$ tunnels through a rectangular barrier of height $V_0>E$ and width $a$. Use the WKB approximation to estimate the transmission coefficient $T$.

Solution. In the forbidden region $0<x<a$, the local momentum is imaginary: $p(x)=i\sqrt{2m(V_0-E)}$. The WKB tunneling factor is

$$T \sim \exp\!\left[-\frac{2}{\hbar}\int_0^a \sqrt{2m(V_0-E)}\,dx\right] = \exp\!\left[-\frac{2a}{\hbar}\sqrt{2m(V_0-E)}\right].$$

For an electron ($m=9.11\times10^{-31}\,\text{kg}$) with $E=1\,\text{eV}$ tunneling through a barrier $V_0=2\,\text{eV}$ of width $a=1\,\text{nm}$:

$$\sqrt{2m(V_0-E)} = \sqrt{2(9.11\times10^{-31})(1.6\times10^{-19})} \approx 5.40\times10^{-25}\,\text{kg\,m/s},$$ $$\frac{2a}{\hbar}\sqrt{2m(V_0-E)} = \frac{2(1.0\times10^{-9})(5.40\times10^{-25})}{1.055\times10^{-34}} \approx 10.2,$$

so $T\sim e^{-10.2}\approx 3.7\times10^{-5}$. The transmission is exponentially sensitive to both barrier width and height.

Problem 2. Use the WKB quantization rule to estimate the large-$n$ energy levels of the infinite square well $V(x)=0$ for $0<x<a$, $\infty$ otherwise.

Solution. The WKB rule for two hard walls is $\int_0^a p(x)\,dx = n\pi\hbar$ with $p=\sqrt{2mE}$. Thus

$$\sqrt{2mE_n}\,a = n\pi\hbar \quad\Longrightarrow\quad E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}.$$

Remarkably, this matches the exact result for all $n$ because the WKB wave function happens to be exact for a constant zero potential between hard walls.

Problem 3. A harmonic oscillator has $V(x)=\frac{1}{2}m\omega^2 x^2$. Apply the WKB quantization rule and compare with the exact answer.

Solution. The classical turning points are $x=\pm x_n$ with $x_n=\sqrt{2E_n/(m\omega^2)}$. The action integral is

$$\int_{-x_n}^{x_n}\sqrt{2m\left(E_n-\frac{1}{2}m\omega^2 x^2\right)}\,dx = \frac{\pi E_n}{\omega}.$$

Setting this equal to $(n+1/2)\pi\hbar$ gives

$$E_n = \left(n+\frac{1}{2}\right)\hbar\omega,$$

which is exactly the known spectrum of the quantum harmonic oscillator. The WKB approximation is exact here because the turning-point corrections sum to the Maslov index $1/2$.

Section summary. The WKB approximation connects quantum waves to classical action when the potential varies slowly.