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.

Helpful background. Quantum Harmonic Oscillator to recognize oscillator energies, lengths, and eigenfunctions.

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:

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

Assume $m>0$ and $\omega_0>0$. 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 $M$ and $K$ are real symmetric positive-definite mass and stiffness matrices. A normal displacement $\mathbf v$ satisfies $K\mathbf v=\omega^2M\mathbf v$: the frequencies depend on the kinetic energy as well as the potential. Mass weighting converts this generalized eigenvalue problem into a symmetric one,

$$D=M^{-1/2}KM^{-1/2} =O\operatorname{diag}(\omega_\alpha^2)O^T, \qquad O^TO=I.$$

The canonical variables $\mathbf Q=O^TM^{1/2}\mathbf x$ and $\mathbf P=O^TM^{-1/2}\mathbf p$ give unit mode masses. The inverse mass factors in the momentum transformation are essential:

$$[Q_\alpha,P_\beta] =i\hbar\bigl(O^TM^{1/2}M^{-1/2}O\bigr)_{\alpha\beta} =i\hbar\delta_{\alpha\beta}.$$

Diagonalizing $K$ alone generally leaves a coupled kinetic energy when the masses differ; exercise 2 illustrates this. If $K$ instead has a zero eigenvalue, that mode is a free coordinate, not an oscillator with a normalizable Gaussian vacuum; a negative eigenvalue signals an instability.

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. Two masses $m$ and $2m$ are each attached to a fixed support by a spring of stiffness $k>0$ and joined by another spring of stiffness $k$. Their mass and stiffness matrices are

$$M=m\begin{pmatrix}1&0\\0&2\end{pmatrix}, \qquad K=k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.$$

Find the normal-mode frequencies and displacement ratios $x_2/x_1$. Why do the equal-mass combinations $(x_1\pm x_2)/\sqrt2$ no longer decouple the Hamiltonian?

Solution

Mass weighting gives

$$D=\frac{k}{m} \begin{pmatrix}2&-1/\sqrt2\\-1/\sqrt2&1\end{pmatrix}.$$

Writing $\lambda=m\omega^2/k$, its characteristic equation is $\lambda^2-3\lambda+3/2=0$. Hence

$$\omega_{\mathrm{low}}^2=\frac{k}{2m}(3-\sqrt3), \qquad \omega_{\mathrm{high}}^2=\frac{k}{2m}(3+\sqrt3).$$

In the original coordinates, the first row of $(K-\omega^2M)\mathbf v=0$ gives $x_2/x_1=2-\lambda$. The two ratios are therefore

$$\left(\frac{x_2}{x_1}\right)_{\mathrm{low}}=\frac{1+\sqrt3}{2}, \qquad \left(\frac{x_2}{x_1}\right)_{\mathrm{high}}=\frac{1-\sqrt3}{2}.$$

The low mode is in phase and the high mode is out of phase, but their displacement magnitudes are unequal. The equal-mass combinations still diagonalize $K$; they do not diagonalize the kinetic energy $p_1^2/(2m)+p_2^2/(4m)$. As a check, the two displacement vectors $\mathbf v_{\mathrm{low}}=(1,(1+\sqrt3)/2)^T$ and $\mathbf v_{\mathrm{high}}=(1,(1-\sqrt3)/2)^T$ obey

$$\mathbf v_{\mathrm{low}}^TM\mathbf v_{\mathrm{high}} =m\left[1+2\frac{1+\sqrt3}{2}\frac{1-\sqrt3}{2}\right]=0.$$

Their orthogonality uses the mass matrix, not the ordinary Euclidean dot product.

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

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

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Saunders, 1976.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• Massachusetts Institute of Technology, “Coupled Linear Oscillators”, 8.03 at ESG Supplemental Notes, undated, pp. 1–3. The notes define their force matrix with the opposite sign to the positive stiffness matrix $K$ used here.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.