Harmonic Oscillator

Prerequisites: Newton's Second Law, Differential Equations. Full path (3 pages)

• Intuitive
• Concrete
• Abstract
• Why is it interesting?
• FAQ

Intuitive

A harmonic oscillator simply consists of an object attached to a spring and is one of the simplest physical systems we can study.

To describe the system we usually track the $\color{magenta}{\text{position}}$ of the object at the end of the spring. The continuous up and down movement is called an oscillation, which is where the name "oscillator" comes from.

The movement depends crucially on the $\color{red}{\text{properties of the spring}}$. When the spring is harder, the object will oscillate faster. In addition, the motion of the object depends on its $\color{blue}{\text{mass}}$. A heavier object will oscillate more slowly.

Concrete

Classical oscillator. From the Lagrangian $L = \frac{m}{2}\dot x^2 - \frac{k}{2}x^2$ the Euler-Lagrange equation gives $$m\ddot x = -kx , \qquad \text{or} \qquad \ddot x + \omega^2 x = 0, \quad \omega = \sqrt{k/m} .$$ The general solution is $x(t) = A\cos(\omega t + \varphi)$, where the amplitude $A$ and phase $\varphi$ are fixed by the initial position and velocity. The period $T = 2\pi\sqrt{m/k}$ does not depend on the amplitude: a stiffer spring makes the oscillation faster, a heavier mass makes it slower. The energy $$E = \frac{m}{2}\dot x^2 + \frac{k}{2}x^2 = \frac{1}{2}kA^2$$ is conserved and moves back and forth between kinetic and potential form.

Why it is everywhere. Any system near a stable equilibrium $x_0$ of a smooth potential $V(x)$ is approximately a harmonic oscillator. Expanding, $$V(x) = V(x_0) + \frac{1}{2}V''(x_0)(x - x_0)^2 + \mathcal O\big((x-x_0)^3\big) ,$$ since $V'(x_0) = 0$. For small displacements the effective spring constant is $k = V''(x_0) > 0$. The same argument applied to a system with many degrees of freedom gives a set of decoupled oscillators, the normal modes, after diagonalizing the matrix of second derivatives. Molecular vibrations, phonons in crystals and the modes of a free field are all treated this way.

Quantum oscillator. With $k = m\omega^2$ the Hamiltonian is $$H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2, \qquad [x, p] = i\hbar .$$ Define the ladder operators $$a = \sqrt{\frac{m\omega}{2\hbar}}\left(x + \frac{i p}{m\omega}\right), \qquad a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(x - \frac{i p}{m\omega}\right) .$$ They satisfy $[a, a^\dagger] = 1$, and the Hamiltonian becomes $H = \hbar\omega\left(a^\dagger a + \frac12\right)$. From $[H, a] = -\hbar\omega\, a$ and $[H, a^\dagger] = \hbar\omega\, a^\dagger$, the operator $a^\dagger$ raises the energy by $\hbar\omega$ and $a$ lowers it. Since $H$ is bounded below, there is a state with $a|0\rangle = 0$, and the spectrum is $$E_n = \hbar\omega\left(n + \frac12\right), \qquad |n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle, \qquad n = 0, 1, 2, \dots$$ In position space $a|0\rangle = 0$ is a first-order differential equation with solution $\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x^2/2\hbar}$. The ground state energy $\frac12\hbar\omega$ is not zero: the particle cannot sit at rest at the bottom of the well, as the uncertainty principle requires. The excited states are $\psi_n \propto H_n\!\left(\sqrt{m\omega/\hbar}\,x\right)\psi_0$, with $H_n$ the Hermite polynomials.

• A good description of the harmonic oscillator in quantum mechanics can be found at http://www.hep.caltech.edu/~fcp/physics/quantumMechanics/SHO/SHO.pdf
• No-Nonsense Classical Mechanics by Jakob Schwichtenberg, chapter "The Harmonic Oscillator"
• No-Nonsense Quantum Mechanics by Jakob Schwichtenberg, chapter "Harmonic Quantum Mechanics"

Abstract

Classical. The phase space is $\mathbb R^2$ with coordinates $(q, p)$ and symplectic form $\Omega = dp \wedge dq$. In the rescaled coordinates $Q = \sqrt{m\omega}\, q$, $P = p/\sqrt{m\omega}$ (still canonical), the Hamiltonian is $H = \frac{\omega}{2}(P^2 + Q^2)$ and its Hamiltonian flow is rigid rotation with angular velocity $\omega$. Thus $H/\omega$ is the moment map of a $U(1)$ action on phase space. Action-angle variables are $I = H/\omega$ and the rotation angle $\theta$, with $\Omega = dI \wedge d\theta$; the orbits are circles enclosing area $\oint p\, dq = 2\pi I$. For $n$ degrees of freedom with equal frequencies, $H = \frac{\omega}{2}|Z|^2$ with $Z = Q + iP \in \mathbb C^n$, and the group of linear symplectic maps preserving $H$ is $U(n)$. Quadratic Hamiltonians on $\mathbb R^{2n}$ form the Lie algebra $\mathfrak{sp}(2n, \mathbb R)$ under the Poisson bracket, and the positive-definite isotropic oscillator generates the center of its maximal compact subalgebra $\mathfrak u(n)$.

Quantum. Quantization represents the Heisenberg algebra $[a, a^\dagger] = 1$. By the Stone-von Neumann theorem, for finitely many degrees of freedom the irreducible representation with $a$ and $a^\dagger$ adjoint to each other (integrated to the Weyl relations) is unique up to unitary equivalence. A convenient model is the Bargmann-Fock space of holomorphic functions $f(z)$ with $$\langle f, g\rangle = \frac{1}{\pi}\int_{\mathbb C} \overline{f(z)}\, g(z)\, e^{-|z|^2}\, d^2 z ,$$ where $a^\dagger$ acts as multiplication by $z$, $a$ as $\partial_z$, and $z^n/\sqrt{n!}$ is the orthonormal eigenbasis $|n\rangle$ of $a^\dagger a$.

The quadratic operators $$K_0 = \frac14\left(a^\dagger a + a a^\dagger\right), \qquad K_+ = \frac12 (a^\dagger)^2, \qquad K_- = \frac12 a^2$$ satisfy $[K_0, K_\pm] = \pm K_\pm$ and $[K_-, K_+] = 2K_0$, a realization of $\mathfrak{sp}(2, \mathbb R) \cong \mathfrak{sl}(2, \mathbb R)$. The Hamiltonian is $H = 2\hbar\omega K_0$. This is the metaplectic (oscillator) representation: it exponentiates to the double cover $Mp(2,\mathbb R)$ of $Sp(2,\mathbb R)$, not to $Sp(2,\mathbb R)$ itself, and it splits into two irreducible lowest-weight representations on even and odd $n$, with lowest $K_0$ weights $\frac14$ and $\frac34$. The zero-point energy $\frac12\hbar\omega$ is tied to this double cover: in geometric quantization it appears through the half-form (metaplectic) correction, and the Bohr-Sommerfeld condition $\oint p\,dq = 2\pi\hbar\left(n + \frac12\right)$ including the Maslov correction reproduces the exact spectrum.

A free quantum field is a collection of such oscillators, one per mode, and its Fock space is built from the same algebra. For infinitely many degrees of freedom the Stone-von Neumann uniqueness fails, which is the origin of inequivalent representations in quantum field theory.

The phase space of a harmonic oscillator

Why is it interesting?

“The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.” Sidney Coleman

In a sense, the harmonic oscillator is to physics what the set of natural numbers is to mathematics. It is a simple, almost “trivial” system, but one which conceals much subtlety and beauty and from which a great deal of what is of interest in the subject evolves. Gregory L. Naber

The harmonic oscillator is an important model system pervading many areas in classical physics; it is likewise ubiquitous in quantum mechanics. The nonrelativistic Scrodinger equation with a harmonic oscillator potential is readily ¨ solved with standard analytic methods, whether in one or three dimensions. However, we will take a different tack here, and address the one-dimension problem more as an excuse to introduce the notion of “creation” and “annihilation” operators, or “step-up” and “step-down” operators. This is an example of a type of operation which will repeat itself in many contexts, including the theory of angular momentum.

The Simple Harmonic Oscillator by F. Porter

FAQ

Path to this page

1. Calculus
2. Differential Equations
3. Newton's Second Law

Leads to: Pendulum.