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Section 3.3 Coupled harmonic oscillators

In the previous section we argued that quantum mechanics was somewhat easier than classical mechanics, because Schrodinger equation is linear, and hence representation theory naturally plays a role. But there is one particular context in classical mechanics where the equation of motion is linear: it is the case of a system of coupled harmonic oscillators. As Michael Peskin, a famous physicist, said: “Physics is that subset of human experience which can be reduced to coupled harmonic oscillators.” The dynamics of systems of coupled harmonic oscillators is crucial in physics, and in fact forms the basis of much of quantum mechanics and quantum field theory.

In classical mechanics, one can think of a system of coupled harmonic oscillators as a system of masses connected by springs. But in fact what we will do in this section applies more generally to study oscillations of any physical system about a stable equilibrium point, as you may have studied in your classical mechanics course.

Going back to the spring-mass system, the system is governed by Hooke's law, which says that the force on the mass by the spring is proportional to its displacement from equilibrium. For instance, if we consider the very simple system of a mass moving in one dimension, and attached by a spring to a fixed wall, then the equation of motion would be

\begin{equation*} m \frac{d^2}{dt^2} x = - k x, \end{equation*}

where \(m\) is the mass, \(k\) is Hooke's constant, and \(x\) is the displacement of the mass from its equilibrium position. The goal of classical mechanics is to solve the equation of motion to find the displacement \(x\) as a function of time. In this case, we see that the equation of motion is linear. We know that the general solution will take the form \(x(t) = C \cos(\omega t + \delta)\text{,}\) with \(\omega = \sqrt{k/m}\text{.}\)

For a more general system of \(N\) masses moving in \(D\) dimensions, each mass can move in \(D\) independent directions, and hence the displacement of the whole system is given by \(N\) \(D\)-dimensional displacement vectors. All-in-all, this can be packaged in a \(N D\)-dimensional real vector, which we call \(\vec{\eta}(t)\text{.}\) The equation of motions will look like (in vector form):

\begin{equation} \frac{d^2}{dt^2} \vec{\eta} = - H \vec{\eta},\label{equation-coupled-harmonic}\tag{3.3.1} \end{equation}

for some \(N D \times N D\) matrix of coefficients \(H\text{.}\) The key point here is that the equation is still linear. But given a particular spring-mass system, the precise form of the matrix \(H\) may be mess, and in fact not so easy to write down. The goal of this section will be to extract a lot of information about the motion of the system without ever having to write down \(H\) explicitly. As you could guess, what we will use is simply the symmetries of the system, and representation theory.

Subsection 3.3.1 Normal modes

To analyze the behaviour of spring-mass systems in more than one dimension, one needs to understand the concept of normal modes of the system.

Definition 3.3.1. Normal modes.

We say that a system is in a normal mode if all its parts are oscillating with the same frequency.

Let us go back to (3.3.1). To solve this equation, we will assume that normal modes exist, and see what they look like. We take the ansatz

\begin{equation*} \vec{\eta}(t) = \vec{a} \cos(\omega t + \delta), \end{equation*}

for some positive constant \(\omega\) and a vector of coefficients \(\vec{a}\) (for the case \(\omega=0\text{,}\) we would take the ansatz \(\vec{\eta}(t) = \vec{a} t + \vec{b}\)). Plugging back in (3.3.1), we obtain the equation

\begin{equation*} H \vec{a} = \omega^2 \vec{a}, \end{equation*}

which is nothing but an eigenvalue problem for the \(N D \times N D\) matrix of coefficients \(H\text{!}\)

In other words, what we have found is that the normal modes are given by eigenvectors of \(H\text{,}\) with oscillating frequency given by the square root of their eigenvalues. Because the equations of motion are linear, we can conclude that the general motion of the system will be given by a linear combination (superposition) of these \(N D\) normal modes.

A few things to note: for the motion to be oscillatory, all eigenvalues of \(H\) must be non-negative (otherwise we would have an unstable equilibrium point). For a normal mode with positive \(\omega\text{,}\) all masses are oscillating with the same frequency \(\omega\text{.}\) For a normal mode with a zero eigenvalue, the corresponding motion is uniform translation of all masses.

The upshot of this discussion is that to understand the motion of the system, we need to find its normal modes, and to find its normal modes, we need to find the eigenvectors of the \(N D \times N D\) matrix \(H\text{.}\) This is where representation theory comes into play!

Subsection 3.3.2 Representation theory to the rescue

Our goal is now to find the eigenvectors and eigenvalues of \(H\text{.}\) As mentioned previously, for large systems of masses, \(H\) may be a mess. Can we get information about the normal modes without writing \(H\) explicitly?

The idea is to use symmetries of the system. We consider the group \(G\) of symmetries that leave the system invariant. Concretely, this means that it leaves \(H\) invariant. As before, this means that \(G\) acts on the space of solutions as an \(N D\)-dimensional representation that commutes with \(H\text{.}\) This representation is in general reducible, but one can choose a basis for the space of states such that it decomposes as a direct sum of irreducible representations. By Schur's lemma, then we know that in this basis, the matrix \(H\) becomes diagonal. Thus the basis vectors are eigenvectors of \(H\text{.}\) Moreover, the space of states also decomposes as a direct sum of subspaces that transform according to the irreducible representations in the decomposition, and any two states in the same subspace share the same eigenvalue.

So given a particular spring-mass system, to understand the different normal modes, or eigenstates of the system, one proceeds as follows:

  • We find the symmetry group \(G\) of the system, and how it acts on the space of states as an \(N D\)-dimensional (reducible) representation (in fact we only need its characters).
  • We use Theorem 2.9.2 to calculate how it decomposes as a direct sum of irreducible representations.
  • We know that in this basis, the matrix \(H\) becomes diagonal, with eigenvalues appearing with degeneracy equal to the dimension of the corresponding irreducible representation. The different eigenvalues give us the possible different normal modes of the system.
  • We use physical insight or further calculations to determine what these normal modes are.

Subsection 3.3.3 Examples

In this section we consider two examples. The first one is very simple, and is studied simply to set the stage; the second one is more involved.

We start with the simple example of two masses connected by a spring and restricted to move in one dimension. We want to understand the normal modes of the system. In this case we can do everything explicitly, so let us start by doing that. But the goal of this example is to show how representation theory can be used to gain information on the system without solving explicitly.

Let \(\eta_1(t)\) and \(\eta_2(t)\) be the displacements of the two masses. The equations of motion are (setting the constants \(m_1=m_2=k=1\) for simplicity):

\begin{equation*} \frac{d^2}{dt^2} \eta_1 = -(\eta_1-\eta_2), \qquad \frac{d^2}{dt^2}\eta_2 = -(\eta_2-\eta_1). \end{equation*}

In matrix form, those reads:

\begin{equation*} \frac{d^2}{dt^2} \vec{\eta} = - \begin{pmatrix} 1 \amp -1 \\ -1 \amp 1 \end{pmatrix} \vec{\eta}. \end{equation*}

We could of course solve this equation explicitly here. But let us use representation theory instead. To get the normal modes we want to find eigenvectors for the matrix

\begin{equation} H = \begin{pmatrix} 1 \amp -1 \\ -1 \amp 1 \end{pmatrix}.\label{equation-H}\tag{3.3.2} \end{equation}

We want to do so using symmetries of the system.

The symmetry group \(G\) here is rather simple. It has two elements: the identity, and the exchange of the two masses. Thus \(G \cong \mathbb{Z}_2\text{.}\) It acts on the two-dimensional space of states in a two-dimensional representation \(T\text{.}\) We can write down the two-dimensional representation explicitly (we will do that below for completeness), but to find its decomposition in terms of irreducible representations we only need its characters. The identity element has character \(\chi(e) = 2\text{,}\) while the non-trivial element does not leave any mass fixed, hence it must have character \(\chi(a) = 0\text{.}\) For completeness, the two-dimensional representation is given by the matrices:

\begin{equation*} T(e) = \begin{pmatrix} 1 \amp 0 \\ 0 \amp 1 \end{pmatrix}, \qquad T(a) = \begin{pmatrix} 0 \amp 1 \\ 1 \amp 0 \end{pmatrix}. \end{equation*}

Now let us decompose \(T\) as a direct sum of irreducible representations. Using the character table for \(\mathbb{Z}_2\text{,}\) and the decomposition theorem Theorem 2.9.2, we can find the decomposition of \(T\text{.}\) If we denote by \(T^{(1)}\) the trivial representation, and \(T^{(2)}\) the parity representation (the other irreducible representation of \(\mathbb{Z}_2\)), then we can write \(T = m_1 T^{(1)} \oplus m_2 T^{(2)}\text{.}\) From Theorem 2.9.2, we find the coefficients to be:

\begin{gather*} m_1 = \frac{1}{2} ( (2)(1) + (0)(1) ) = 1,\\ m_2 = \frac{1}{2} ( (2)(1) + (0)(-1) ) = 1. \end{gather*}

Thus \(T = T^{(1)} \oplus T^{(2)}\text{.}\) In this basis the representation is given by diagonal matrices.

By Schur's lemma, we know that in this basis \(H\) is also diagonal, i.e. the basis vectors are eigenvectors of \(H\text{,}\) and hence give normal modes of the system. The eigenvalues here are non-degenerate since the irreducible representations are one-dimensional. We do not know from symmetry alone however what the diagonal entries of \(H\) should be; those correspond to the eigenvalues (of frequencies of the normal modes).

But since here we know the specific form of \(H\) (see (3.3.2)), and see that the trace is 2 and the determinant is 0, the diagonal form of \(H\text{,}\) must be

\begin{equation*} H = \begin{pmatrix} 0 \amp 0 \\ 0 \amp 2 \end{pmatrix}. \end{equation*}

So there is one normal mode with zero frequency, and one normal mode with a non-zero frequency. It is easy to identify these normal modes physically. The first mode corresponds to linear translation of the whole system (the spring is not stretched at all). The second mode corresponds to the masses oscillating with the same frequency in opposite directions (the “breathing mode”).

Of course this example was rather trivial. We could have solved the equations of motion explicitly. But in more complicated systems, \(H\) may be difficult to write down, and studying symmetries provides valuable information on the normal modes of the system without having to solve any equation of motion.

We now consider the more interesting example of 3 masses in two dimensions connected by three springs in the form of an equilateral triangle. Each mass has two displacement coordinates (horizontal and vertical directions), and hence the displacement vector is \(3 \times 2 = 6\)-dimensional.

Writing down the matrix \(H\) for this system would be pretty annoying. So let us not do that. We will use symmetry instead to gain information on the normal modes of the system.

The symmetry group here is the dihedral group \(D_3\) which is isomorphic to the symmetric group \(S_3\text{.}\) It acts on the space of states in the form of a \(6\)-dimensional representation \(T\text{.}\) We will not write down the representation explicitly, but simply deduce its characters on the three conjugacy classes of \(S_3\text{.}\) We will use the notation in Section 2.10.

In the conjugacy class \(C_1\) containing the identity element, the character of \(T\) is \(\chi_1 = 6\text{.}\) In the conjugacy class \(C_2\) corresponding to the transpositions, the action should leave one mass fixed, and hence its two displacement coordinates invariant. Therefore, the character must be \(\chi_2 = 2\text{.}\) For the class \(C_3\) corresponding to the cyclic permutations, nothing is left invariant, and hence \(\chi_3 = 0\text{.}\)

From this we can get the decomposition of \(T\text{.}\) Using Theorem 2.9.2 and the character table Table 2.10.2 (with the notation that \(T^{(1)}\) is the trivial irrep, \(T^{(2)}\) is the parity one-dimensional irrep, and \(T^{(3)}\) is the two-dimensional irrep) we get the coefficients:

\begin{gather*} m_1 = \frac{1}{6} ( 1 (6)(1) + 3 (2)(1) + 2 (0)(1) ) = 2,\\ m_2 = \frac{1}{6} ( 1 (6)(1) + 3 (2)(-1) + 2 (0) (1) ) = 0,\\ m_3 = \frac{1}{6} ( 1 (6)(2) + 3 (2)(0) + 2 (0)(-1) ) = 2. \end{gather*}

Thus \(T = T^{(1)} \oplus T^{(1)} \oplus T^{(3)} \oplus T^{(3)}\text{.}\) So in this basis the matrix \(H\) takes the diagonal form:

\begin{equation*} H = \text{diag}(\omega_1, \omega_2, \omega_3, \omega_3, \omega_4, \omega_4), \end{equation*}

for some eigenvalues \(\omega_1, \omega_2, \omega_3, \omega_4\text{.}\) Thus there should be four types of normal modes, two of which would have a two-dimensional eigenspace. At least that's what we get from symmetry; there could be additional accidental degeneracies (i.e. some of the \(\omega_i\) could be identical).

In fact one can identify the zero modes physically. Two of them correspond to horizontal and vertical translations of the whole system; those have zero frequency. This corresponds to one of the two-dimensional irreps. There is in fact an accidental degeneracy here; there is another mode with zero frequency, corresponding to cyclic rotation of the whole system. This corresponds to one of the one-dimensional irreps. So there are three zero modes for this system.

We still have one one-dimensonal irrep and one two-dimensional irrep. There is one “breathing mode”, where all masses oscillate inwards and outwards simultaneously. This corresponds to the remaining one-dimensional irrep. The two other normal modes are not so easy to guess, but one could calculate the eigenvectors from linear algebra, since they should be orthogonal to the four eigenvectors that we have already determined, and then see how they act on the three masses. See for instance Section III.2 in Zee's book.

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