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Chapter 5 • Theory & Derivations

The One-Dimensional Quantum Harmonic Oscillator

The quintessential quantum model: the harmonic oscillator potential, analytic series solution using Hermite polynomials, algebraic Dirac ladder operator formalism, zero-point energy, expectation values, and comparison with classical turning points.

§5.1 Physical Importance and Analytic Solution

Universal Significance of the Harmonic Oscillator

The harmonic oscillator is a cornerstone of theoretical physics. Any arbitrary potential $V(x)$ with a local stable minimum at $x_0$ can be Taylor-expanded about that minimum:

$$V(x) = V(x_0) + V'(x_0)(x - x_0) + \frac{1}{2}V''(x_0)(x - x_0)^2 + \dots$$

Setting $V(x_0) = 0$ as reference, and noting $V'(x_0) = 0$ at equilibrium:

$$V(x) \approx \frac{1}{2} k (x - x_0)^2 = \frac{1}{2} m \omega^2 x^2$$

where $\omega = \sqrt{k/m} = \sqrt{V''(x_0)/m}$. Consequently, any system undergoing small oscillations about stable equilibrium—such as molecular vibrations, phonons in crystal lattices, and electromagnetic field modes—behaves as a harmonic oscillator.

Analytic Solution of the Schrödinger Equation

The Time-Independent Schrödinger Equation is:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \psi = E\psi$$

Introducing the dimensionless coordinate $\xi$ and energy parameter $\epsilon$:

$$\xi = \alpha x = \sqrt{\frac{m\omega}{\hbar}} x, \qquad \epsilon = \frac{2E}{\hbar\omega}$$

The equation simplifies to:

$$\frac{d^2\psi}{d\xi^2} + (\epsilon - \xi^2)\psi = 0$$
Asymptotic Behavior and Hermite Polynomials

As $\xi \to \pm \infty$, $\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi$, which has normalizable asymptotic solutions $\psi(\xi) \propto e^{-\xi^2/2}$. We therefore factor out the Gaussian:

$$\psi(\xi) = H(\xi) e^{-\xi^2/2}$$

Substituting this into the differential equation yields the Hermite Differential Equation:

$$\frac{d^2 H}{d\xi^2} - 2\xi \frac{dH}{d\xi} + (\epsilon - 1)H = 0$$

Expressing $H(\xi)$ as a power series $H(\xi) = \sum_{j=0}^\infty a_j \xi^j$, the recurrence relation is:

$$a_{j+2} = \frac{2j + 1 - \epsilon}{(j+1)(j+2)} a_j$$

For the wavefunction to remain normalizable as $\xi \to \infty$, the series must terminate at some finite index $j = n$. Setting the numerator to zero:

$$2n + 1 - \epsilon = 0 \implies \epsilon = 2n + 1$$

Since $\epsilon = \frac{2E}{\hbar\omega}$, we obtain the Quantized Energy Eigenvalues:

$$E_n = \left( n + \frac{1}{2} \right) \hbar \omega, \quad n \in \{0, 1, 2, 3, \dots\}$$

The corresponding polynomial solutions $H_n(\xi)$ are the Hermite Polynomials:

  • $H_0(\xi) = 1$
  • $H_1(\xi) = 2\xi$
  • $H_2(\xi) = 4\xi^2 - 2$
  • $H_3(\xi) = 8\xi^3 - 12\xi$

The normalized stationary state wavefunctions are:

$$\psi_n(x) = \left( \frac{m\omega}{\pi \hbar} \right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n\left( \sqrt{\frac{m\omega}{\hbar}} x \right) e^{-\frac{m\omega x^2}{2\hbar}}$$

§5.2 The Algebraic Operator Method: Dirac Ladder Operators

Paul Dirac introduced a powerful algebraic method using non-Hermitian ladder operators:

$$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right) \quad (\text{Annihilation / Lowering Operator})$$
$$\hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} - \frac{i}{m\omega}\hat{p} \right) \quad (\text{Creation / Raising Operator})$$
Fundamental Commutation Relations

Using $[\hat{x}, \hat{p}] = i\hbar$:

$$[\hat{a}, \hat{a}^\dagger] = \frac{1}{2\hbar m\omega} [m\omega\hat{x} + i\hat{p}, m\omega\hat{x} - i\hat{p}] = \frac{1}{2\hbar m\omega} (-2i m\omega [\hat{x}, \hat{p}]) = 1$$

Rewriting the Hamiltonian in terms of ladder operators:

$$\hat{H} = \hbar \omega \left( \hat{a}^\dagger \hat{a} + \frac{1}{2} \right) = \hbar \omega \left( \hat{N} + \frac{1}{2} \right)$$

where $\hat{N} = \hat{a}^\dagger \hat{a}$ is the Hermitian Number Operator, with eigenvalues $n \ge 0$: $\hat{N}|n\rangle = n|n\rangle$.

Action on Eigenstates
$$\hat{a}|n\rangle = \sqrt{n} |n-1\rangle$$
$$\hat{a}^\dagger|n\rangle = \sqrt{n+1} |n+1\rangle$$

Since lowering the ground state must terminate the ladder: $\hat{a}|0\rangle = 0$. In position space:

$$\sqrt{\frac{m\omega}{2\hbar}} \left( x + \frac{\hbar}{m\omega}\frac{d}{dx} \right)\psi_0(x) = 0 \implies \frac{d\psi_0}{dx} = -\frac{m\omega}{\hbar} x \psi_0$$

Integrating yields the Gaussian ground state: $\psi_0(x) = (\frac{m\omega}{\pi\hbar})^{1/4} e^{-\frac{m\omega x^2}{2\hbar}}$.

Any excited state $|n\rangle$ can then be generated algebraically:

$$|n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}} |0\rangle$$

Explore the energy ladder, Hermite wavefunctions, and classical turning points in Simulation 5.1 below.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Hard Example 5.1: Expectation Values , and the Virial Theorem via Ladder Operators

Using Dirac ladder operators $\hat{a}$ and $\hat{a}^\dagger$, evaluate $\langle n | \hat{x}^2 | n \rangle$ and $\langle n | \hat{p}^2 | n \rangle$ for the $n$-th excited state of a quantum harmonic oscillator. Verify the Virial Theorem: $\langle T \rangle = \langle V \rangle = \frac{1}{2} E_n$.

Step 1: Express x and p in terms of ladder operators
$$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}} (\hat{a} + \hat{a}^\dagger), \qquad \hat{p} = -i\sqrt{\frac{m\omega\hbar}{2}} (\hat{a} - \hat{a}^\dagger) $$ $$\hat{x}^2 = \frac{\hbar}{2m\omega} (\hat{a}^2 + \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} + (\hat{a}^\dagger)^2) $$ $$\hat{p}^2 = -\frac{m\omega\hbar}{2} (\hat{a}^2 - \hat{a}\hat{a}^\dagger - \hat{a}^\dagger\hat{a} + (\hat{a}^\dagger)^2)$$

Terms with $\hat{a}^2$ and $(\hat{a}^\dagger)^2$ change the state by $\pm 2$, so their diagonal matrix elements vanish: $\langle n | \hat{a}^2 | n \rangle = 0$.

Step 2: Evaluate expectation values
$$\langle n | \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} | n \rangle = (n+1) + n = 2n + 1 $$ $$\langle x^2 \rangle_n = \frac{\hbar}{2m\omega} (2n + 1) = \left( n + \frac{1}{2} \right) \frac{\hbar}{m\omega} $$ $$\langle p^2 \rangle_n = \frac{m\omega\hbar}{2} (2n + 1) = \left( n + \frac{1}{2} \right) m\omega\hbar $$ $$\langle V \rangle = \frac{1}{2}m\omega^2 \langle x^2 \rangle = \frac{1}{2}\left(n + \frac{1}{2}\right)\hbar\omega = \frac{1}{2}E_n $$ $$\langle T \rangle = \frac{\langle p^2 \rangle}{2m} = \frac{1}{2}\left(n + \frac{1}{2}\right)\hbar\omega = \frac{1}{2}E_n$$

This verifies the quantum Virial Theorem: the average kinetic energy equals the average potential energy, each contributing half of the total energy $E_n$.

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.