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PU-202 / L19 · lecture

The Harmonic Oscillator: Algebraic Solution

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the postulates of quantum mechanics, specifically the correspondence between physical observables and Hermitian operators, and the concept of eigenvalue equations. Familiarity with the time-independent Schrödinger equation and the basic properties of the harmonic potential \( V(x) = \frac{1}{2}m\omega^2 x^2 \) is assumed.

Why this matters

The harmonic oscillator is the Rosetta Stone of quantum mechanics, providing the first exact solution that reveals the quantization of energy and the structure of quantum states.

While the infinite square well introduced the concept of discrete energy levels, it did so through boundary conditions that felt somewhat arbitrary. The harmonic oscillator, however, offers a deeper insight into the nature of quantum systems. It is the most important potential that can be solved exactly by pure algebra, and the method reveals the underlying algebraic structure of the theory. (Other potentials, such as the Coulomb potential of hydrogen, are also exactly solvable.) By introducing ladder operators, we move beyond mere calculation to a structural understanding of how quantum states are related. This algebraic method is not just a trick for this specific problem; it is the foundation for quantum field theory and the treatment of interacting particles, making it an indispensable tool in the physicist's repertoire.

The Algebraic Intuition

Classically, a harmonic oscillator is a mass on a spring. Its energy is the sum of kinetic and potential terms, and it can have any positive value depending on how hard you pull the mass. In quantum mechanics, however, the energy is restricted to a discrete set of values. Instead of solving a differential equation directly, which involves messy integrals and special functions, we can use an algebraic approach. The key idea is to define two new operators, often called "ladder" or "raising and lowering" operators. One operator, let's call it \( a^\dagger \), takes a quantum state and moves it to a state with slightly higher energy. The other, \( a \), moves it to a state with slightly lower energy. By combining these operators, we can express the Hamiltonian (the total energy operator) in a form that makes the quantization obvious. The ground state is simply the state that cannot be lowered any further. Every other state is built by applying the raising operator repeatedly. This perspective transforms the problem from a calculus exercise into a logical construction of states, highlighting the discrete, step-like nature of quantum energy levels.

Defining the Ladder Operators

To construct these operators, we first need to express the position \( x \) and momentum \( p \) operators in terms of them. We define the lowering operator \( a \) and the raising operator \( a^\dagger \) as linear combinations of \( x \) and \( p \). The coefficients are chosen specifically to simplify the commutator relations. The definitions are:

\[ a = \sqrt{\frac{m\omega}{2\hbar}} \left( x + \frac{i}{m\omega} p \right) \]
\[ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \left( x - \frac{i}{m\omega} p \right) \]

Here, \( m \) is the mass, \( \omega \) is the angular frequency, and \( \hbar \) is the reduced Planck constant. Note that \( a^\dagger \) is the Hermitian conjugate of \( a \). By inverting these equations, we can express \( x \) and \( p \) in terms of \( a \) and \( a^\dagger \). This allows us to rewrite the Hamiltonian \( H \), which is originally \( \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 \), purely in terms of the ladder operators. The crucial step relies on the commutation relation \( [a, a^\dagger] = 1 \). When we substitute the expressions for \( x \) and \( p \) into the Hamiltonian, the cross terms cancel out due to the symmetry, leaving us with a remarkably simple form.

Quantization of Energy

Using the commutation relation, the Hamiltonian can be written as:

\[ H = \hbar\omega \left( a^\dagger a + \frac{1}{2} \right) \]

This expression is profound because it separates the energy into a part that depends on the number of "quanta" and a constant offset. We define the number operator \( N = a^\dagger a \). The eigenvalue equation for the Hamiltonian is \( H |n\rangle = E_n |n\rangle \). Substituting the algebraic form of \( H \), we get \( \hbar\omega (N + \frac{1}{2}) |n\rangle = E_n |n\rangle \). Since \( N \) is a Hermitian operator, its eigenvalues must be real. Furthermore, because \( N = a^\dagger a \), the eigenvalue \( n \) must be non-negative. If we assume a state \( |n\rangle \) is an eigenstate of \( N \) with eigenvalue \( n \), then applying the lowering operator \( a \) gives a new state \( a|n\rangle \) which is an eigenstate of \( N \) with eigenvalue \( n-1 \). This process can continue until we reach a state that is annihilated by \( a \). This lowest state, the ground state \( |0\rangle \), satisfies \( a|0\rangle = 0 \). The energy of this state is \( E_0 = \frac{1}{2}\hbar\omega \). Each application of \( a^\dagger \) increases the energy by exactly \( \hbar\omega \). Thus, the allowed energy levels are:

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

Structure of the Eigenstates

The algebraic method not only gives us the energies but also the states themselves. The ground state \( |0\rangle \) is defined by the condition \( a|0\rangle = 0 \). In the position representation, this differential equation yields a Gaussian wavefunction, which is the unique normalizable solution. All excited states are generated by applying the raising operator to the ground state:

\[ |n\rangle = \frac{1}{\sqrt{n!}} (a^\dagger)^n |0\rangle \]

The factor \( \frac{1}{\sqrt{n!}} \) ensures that the states remain normalized. This recursive construction is elegant and powerful. It shows that the entire spectrum of the harmonic oscillator is built from a single fundamental state. The orthogonality of these states follows directly from the properties of the ladder operators. This algebraic structure is robust; it does not depend on the specific coordinate representation, making it applicable in a wide range of physical contexts, from atomic vibrations to quantum optics.

The derivation

◆ DERIVATION EMBED · D-010 · not re-derived here

The Harmonic Oscillator via Ladder Operators

Diagonalises the harmonic oscillator algebraically, deriving the equally spaced spectrum and the zero-point energy from raising and lowering operators.

En = ( n + ½ ) ℏω,    n = 0, 1, 2, …
Open full derivation →
Tier 3 · watch out

A common misconception is that the zero-point energy \( \frac{1}{2}\hbar\omega \) is merely a mathematical artifact that can be subtracted away. While it is true that only energy differences are physically measurable in many contexts, the zero-point energy is a direct consequence of the uncertainty principle. If the particle were to have zero energy, it would have zero momentum and zero position uncertainty, violating \( \Delta x \Delta p \geq \frac{\hbar}{2} \). The algebraic derivation confirms this: the ground state is not a state of rest, but a state of minimum uncertainty, where the kinetic and potential energies are balanced at their lowest possible non-zero values. Ignoring this term can lead to incorrect predictions in systems where absolute energy levels matter, such as in the Casimir effect or in the stability of matter. C

Worked example

Problem. Consider a quantum harmonic oscillator with a mass \( m = 1.0 \times 10^{-26} \) kg and an angular frequency \( \omega = 1.0 \times 10^{13} \) rad/s. Calculate the energy of the ground state and the first excited state in electronvolts (eV). Use \( \hbar = 1.055 \times 10^{-34} \) J·s and \( 1 \text{ eV} = 1.602 \times 10^{-19} \) J.

  1. Calculate the energy quantum \( \hbar\omega \) in Joules: \( \hbar\omega = (1.055 \times 10^{-34}\ \mathrm{J\,s})(1.0 \times 10^{13} \text{ s}^{-1}) = 1.055 \times 10^{-21} \) J.
  2. Convert this energy to electronvolts: \( E_{\text{quantum}} = \frac{1.055 \times 10^{-21} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 0.006586 \) eV.
  3. Calculate the ground state energy \( E_0 = \frac{1}{2} \hbar\omega \): \( E_0 = 0.5 \times 0.006586 \text{ eV} = 0.003293 \) eV.
  4. Calculate the first excited state energy \( E_1 = \frac{3}{2} \hbar\omega \): \( E_1 = 1.5 \times 0.006586 \text{ eV} = 0.009879 \) eV.
  5. Round to appropriate significant figures based on the given data (2 significant figures): \( E_0 \approx 0.0033 \) eV and \( E_1 \approx 0.0099 \) eV.
\[ E_0 \approx 0.0033 \ \mathrm{eV}, \quad E_1 \approx 0.0099 \ \mathrm{eV} \]

Units check. The product \( \hbar\omega \) has units of \( \mathrm{J \cdot s \cdot s^{-1}} = \mathrm{J} \). Dividing by the conversion factor for eV yields eV. The final answers are in eV, which is consistent with the request.

CHECK YOURSELF

Three quick tests

  1. What is the commutator \( [a, a^\dagger] \)?
    Answer1
  2. If \( a|n\rangle \) is an eigenstate of the number operator \( N \), what is its eigenvalue?
    Answer\( n-1 \)
  3. Why must the eigenvalues of the number operator \( N \) be non-negative integers?
    AnswerBecause \( N = a^\dagger a \) is a positive semi-definite operator, and the ladder process must terminate at a ground state where \( a|0\rangle = 0 \).
PROBLEMS

Problem set for PU-202 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-010 · Threads: matter, waves, energy