Coupled Oscillators: First Encounter

Coupled oscillators show why the harmonic oscillator is more than a one-particle model. A system with several interacting quadratic degrees of freedom can often be rewritten as a set of independent normal modes. Quantizing those normal modes gives independent quantum oscillators.

This is the first bridge from the single Quantum Harmonic Oscillator to phonons, photons, lattice vibrations, and free-field modes.

Two Coupled Oscillators

Consider two equal masses $m$ with coordinates $x_1,x_2$ and momenta $p_1,p_2$. Each mass is bound by a harmonic restoring force with frequency $\omega_0$, and the two masses are coupled by a spring-like term:

$$H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \frac12m\omega_0^2(x_1^2+x_2^2) + \frac12\kappa(x_1-x_2)^2.$$

The coupling is controlled by $\kappa\ge0$. If $\kappa=0$, the two oscillators are independent. If $\kappa\gt 0$, displacement of one oscillator affects the other.

The potential is still quadratic, so the problem should remain exactly solvable after the right change of variables.

Normal Coordinates

Define symmetric and antisymmetric coordinates

$$Q_+ = \frac{x_1+x_2}{\sqrt2}, \qquad Q_- = \frac{x_1-x_2}{\sqrt2},$$

with corresponding momenta

$$P_+ = \frac{p_1+p_2}{\sqrt2}, \qquad P_- = \frac{p_1-p_2}{\sqrt2}.$$

These are not arbitrary definitions. They preserve the canonical commutators:

$$[Q_+,P_+]=i\hbar, \qquad [Q_-,P_-]=i\hbar, \qquad [Q_+,P_-]=[Q_-,P_+]=0.$$

In these variables,

$$x_1^2+x_2^2=Q_+^2+Q_-^2,$$

and

$$(x_1-x_2)^2=2Q_-^2.$$

The Hamiltonian becomes

$$H = \frac{P_+^2}{2m} + \frac12m\omega_+^2Q_+^2 + \frac{P_-^2}{2m} + \frac12m\omega_-^2Q_-^2,$$

where

$$\omega_+=\omega_0, \qquad \omega_-=\sqrt{\omega_0^2+\frac{2\kappa}{m}}.$$

The coupled system has become two independent oscillators.

Physical Meaning of the Modes

The $+$ mode is the in-phase mode:

$$x_1=x_2.$$

The coupling spring is not stretched, so its frequency remains $\omega_0$.

The $-$ mode is the out-of-phase mode:

$$x_1=-x_2.$$

The coupling spring is stretched strongly, so the restoring force is larger and the frequency is higher:

$$\omega_-\gt \omega_+.$$

Normal modes are collective coordinates. The simple independent motions are not usually the original coordinates $x_1$ and $x_2$, but the combinations $Q_+$ and $Q_-$.

Quantization of Normal Modes

Each normal mode is an ordinary harmonic oscillator. Define ladder operators

$$a_\pm = \frac{1}{\sqrt2} \left( \frac{Q_\pm}{\ell_\pm} + \frac{i\ell_\pm}{\hbar}P_\pm \right),$$

with oscillator lengths

$$\ell_\pm = \sqrt{\frac{\hbar}{m\omega_\pm}}.$$

Then

$$H = \hbar\omega_+ \left( a_+^\dagger a_+ + \frac12 \right) + \hbar\omega_- \left( a_-^\dagger a_- + \frac12 \right).$$

The energy eigenstates are labeled by two occupation numbers:

$$\lvert n_+,n_-\rangle.$$

The spectrum is

$$E_{n_+,n_-} = \hbar\omega_+ \left(n_+ + \frac12\right) + \hbar\omega_- \left(n_- + \frac12\right),$$

where

$$n_+,n_-=0,1,2,\ldots.$$

The ground-state energy is the sum of the zero-point energies of both modes:

$$E_0 = \frac12\hbar\omega_+ + \frac12\hbar\omega_-.$$

What Is Being Quantized?

The quanta are excitations of the normal modes, not little particles attached to oscillator $1$ or oscillator $2$. A state with one $+$ quantum is a collective in-phase excitation. A state with one $-$ quantum is a collective out-of-phase excitation.

This distinction matters. When many masses are coupled in a chain, the normal modes are labeled by wavenumber. Quantizing those modes gives phonons; Phonons as Many-Body Excitations carries out that lattice-wide construction. In a field theory, the field decomposes into infinitely many oscillator modes, and quantizing those modes gives field quanta. The same oscillator algebra reappears, but the labels change from $+$ and $-$ to mode labels such as $k$.

For the QFT bridge version, see Harmonic Oscillator to Fields.

General Quadratic Pattern

A general system of coupled quadratic degrees of freedom has the schematic Hamiltonian

$$H = \frac12\mathbf p^T M^{-1}\mathbf p + \frac12\mathbf x^T K\mathbf x,$$

where $M$ is a mass matrix and $K$ is a stiffness matrix. After an appropriate mass-weighted change of variables, the problem reduces to an eigenvalue problem for normal-mode shapes and frequencies.

The result is a sum of independent oscillators:

$$H = \sum_\alpha \left[ \frac{P_\alpha^2}{2} + \frac12\omega_\alpha^2Q_\alpha^2 \right],$$

up to the chosen normalization of the normal coordinates. Quantization then gives

$$H = \sum_\alpha \hbar\omega_\alpha \left( a_\alpha^\dagger a_\alpha+\frac12 \right).$$

This is the algebraic skeleton behind many-body oscillator systems and free quantum fields.

Common Mistakes

• Quantizing $x_1$ and $x_2$ as if they remained independent after coupling.
• Transforming coordinates but forgetting to transform momenta.
• Dropping the zero-point energy of each normal mode.
• Thinking a normal-mode quantum belongs to only one original oscillator.
• Assuming the coupling always lowers a frequency; in the simple spring-coupled model, the out-of-phase mode is stiffened.
• Treating normal-mode diagonalization as a quantum trick rather than a classical linear-algebra step followed by quantization.

Exercises

1. Show that $x_1^2+x_2^2=Q_+^2+Q_-^2$ and $(x_1-x_2)^2=2Q_-^2$.

Solution

Invert the definitions:

$$x_1=\frac{Q_+ + Q_-}{\sqrt2}, \qquad x_2=\frac{Q_+ - Q_-}{\sqrt2}.$$

Then

$$x_1^2+x_2^2 = \frac12 \left[ (Q_+ + Q_-)^2 + (Q_+ - Q_-)^2 \right] = Q_+^2+Q_-^2.$$

Also,

$$x_1-x_2 = \sqrt2 Q_-,$$

so

$$(x_1-x_2)^2=2Q_-^2.$$

2. Derive the two normal-mode frequencies for the Hamiltonian in this page.

Solution

After substituting the normal coordinates, the potential becomes

$$V = \frac12m\omega_0^2(Q_+^2+Q_-^2) + \kappa Q_-^2.$$

The $Q_+$ coefficient is

$$\frac12m\omega_0^2,$$

so $\omega_+=\omega_0$. The $Q_-$ coefficient is

$$\frac12m\omega_0^2+\kappa = \frac12m \left( \omega_0^2+\frac{2\kappa}{m} \right).$$

Thus

$$\omega_-=\sqrt{\omega_0^2+\frac{2\kappa}{m}}.$$

3. For weak coupling, $\kappa\ll m\omega_0^2$, expand $\omega_-$ to first order in $\kappa$.

Solution

Write

$$\omega_- = \omega_0 \sqrt{1+\frac{2\kappa}{m\omega_0^2}}.$$

Using $\sqrt{1+\epsilon}\approx1+\epsilon/2$ gives

$$\omega_- \approx \omega_0 \left( 1+\frac{\kappa}{m\omega_0^2} \right).$$

Thus

$$\omega_- \approx \omega_0+\frac{\kappa}{m\omega_0}.$$

Where This Is Used

• Quantum Harmonic Oscillator gives the single-mode model.
• Ladder-Operator Solution: First Encounter supplies the creation and annihilation operators used for each mode.
• Number States explains occupation-number language.
• Oscillator as a Universal Local Model explains why quadratic normal-mode approximations arise near stable equilibria.
• Normal Modes of Polyatomics applies the matrix pattern to molecular masses, Cartesian Hessians, rigid-motion projection, and spectroscopy.
• Harmonic Chain Model Dossier extends the same normal-coordinate logic to a periodic many-body chain while treating the acoustic zero mode explicitly.
• Harmonic Oscillator to Fields records the bridge from oscillator modes to free fields.
• Eigenvalues and Eigenvectors supplies the linear-algebra background for normal modes.

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Saunders, 1976.