Single mode

Suppose I have two $LC$ oscillators, one with $L_1$ and $C_1$, and the other with $L_2$ and $C_2$. If uncoupled, each oscillator has resonant frequency $\omega \equiv 1/\sqrt{LC}$. Using the flux in the inductor as the coordinate, the equation of motion for each oscillator is

$$\ddot{\Phi} = -\omega^2 \Phi .$$

We can rewrite this in Hamiltonian form like this

$$ \frac{d}{dt} \left[ \begin{array}{c} \Phi \\ Q \end{array} \right] = \left[ \begin{array}{cc} 0 & 1/C \\ -1/L & 0\end{array} \right] \left[ \begin{array}{c} \Phi \\ Q \end{array} \right]. $$

We can clean this up by defining $X \equiv (C/L)^{1/4} \Phi$ and $Y \equiv (L/C)^{1/4} Q$, which leads to

$$ \frac{d}{dt} \left[ \begin{array}{c} X \\ Y \end{array} \right] = \omega^2 \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0\end{array} \right] \left[ \begin{array}{c} X \\ Y \end{array} \right]. $$

Coupled modes

Now suppose the oscillators are coupled through an inductor $L_g$. The coupled equations of motion are

$$ \left[ \begin{array}{c} \ddot{\Phi}_1 \\ \ddot{\Phi}_2 \end{array} \right] = \left[ \begin{array}{cc} -\omega_1^2(1+L_1/L_g) & \omega_1^2(L_1/L_g) \\ \omega_2^2 (L_2/L_g) & -\omega_2^2(1+L_2/L_g) \end{array} \right] \left[ \begin{array}{c} \Phi_1 \\ \Phi_2 \end{array} \right] $$

Note that the weak coupling limit is that where $L_g \gg L_1,L_2$. From here, one can find the normal frequencies by the usual means of finding eigenvalues of a matrix.

One can also work with the Hamiltonian form:

$$ \frac{d}{dt} \left[ \begin{array}{c} X_1 \\ Y_1 \\ X_2 \\ Y_2 \end{array} \right] = \left[ \begin{array}{cccc} 0 & \omega_1 & 0 & 0 \\ -\omega_1' & 0 & -g & 0 \\ 0 & 0 & 0 & \omega_2 \\ -g & 0 & -\omega_2' & 0 \end{array} \right] \left[ \begin{array}{c} X_1 \\ Y_1 \\ X_2 \\ Y_2 \end{array} \right] $$

where $\omega_1' \equiv \omega_1 (1-L_1/L_g)$, $\omega_2' \equiv \omega_2 (1-L_2/L_g)$, and $g\equiv (\sqrt{L_1 L_2}/L_g)\sqrt{\omega_1 \omega_2}$.

Is there any advantage in terms of intuition for the physics or mathematical elegance/simplicity to extract information from the Hamiltonian matrix as opposed to the one which came from the 2$^{\text{nd}}$ order equations?

  • $\begingroup$ It was just pointed out to me that the matrices I've written here are not really Hamiltonians. They're something else, and I'm not sure what they're called. $\endgroup$
    – DanielSank
    Commented Jun 27, 2016 at 22:34

1 Answer 1


I guess this question might spawn a very large spectrum of answers, I'll try to take on the (simple) mathematical side of dealing with a system of first order ODE instead of an higher order one. In this sense (as in many others!) the Hamiltonian formulation of a dynamical problem, like the one you have posed as an example, gives in an easier way plenty of information that the Lagrangian or Newtonian (as $F=ma$) do not.

The reason i'll give you at first sight will look as purely mathematical but as it turns out it gives clear physical insights that the other formulation do not give. I will denote as the n-th derivative of the function $x(t)$ the symbol $x^{(n)}$, Given an ODE written in it's normal form $$\tag 1 x^{(n)}=\mathcal{F}\left(t,x'...x^{(n-1)}\right) \\ \mathcal F:A\subseteq \mathbb R\times \mathbb R^n\longrightarrow \mathbb R^n$$ if we define the variables $x_{k}=x^{(k)}$ so $x=x_1,\ x'=x_2 ,\ x''=x_3,\ ecc \ $ than we can always write it in the form: \begin{align} \tag 2 x &= x_1 \\ x_1' &= x_2 \\ x_2' &= x_3\\ &\ldots \\ x_n' &= \mathcal F(t, x_1,x_2,...., x_n) \end{align} Now, since dynamical problems are of the form $$\tag 3 m \ x''(t)=\mathcal F(t,x,x')$$ the problem can be broken in a system of two first order differential equations: $$\tag 4 \begin{cases}x'_1=x_2 \\ m \ x'_2=\mathcal F(t, x_1, x_2) \end{cases}$$

Ok, let's define the differentiable curve ${x}(t)\in C'(I\subset \mathbb R,\mathbb R^n)$ and the matrix $\mathrm A \in GL(\mathbb R^n)$ such that we can write the cauchy problem $$ \tag 5 \begin{cases} x'(t)=\mathrm A \cdot x(t)\\ x(t_0)=\xi_0 \end{cases}$$ long story short you can write the solution as $$\tag 6 x(t)=e^{\mathrm A(t-t_0)}\cdot\xi_0$$ (There is a simple theory on how you can diagonalize and write the matrices $\mathrm A$ and $e^{\mathrm A t}$ depending on the nature of it's eigenvalues) Once you know the eigenvalues of $\mathrm A$ you can determine the asymptotic behavior of the system: like it's stability, if the solution is limited and in some cases even the presence of attractors just by knowing the eigenvalues of $\mathrm A$. (the Hamiltonian matrix you mention)

The Hamiltonian formulation of mechanics is the most natural way to write Newton's law, a second order differential equation, into a system of two first order equations. Thus for the linearized theory and even for the case $x'(t)=\mathrm A\cdot x(t)+g(t,x)$ where $g(t,x)$ is "small" respect to $x(t)$ (and under different assumptions, for more general and more complicated cases too!). You can write the solution as $$\tag 7 x(t)=e^{\mathrm A(t-t_0)}\cdot\xi_0+\int_{t_0}^t e^{\mathrm A(t-s)}\cdot g(s,x(s)) \ ds$$ and extract alot of information from the matrix $\mathrm A$ eigenvalues, as I already emphasized.

The morale of the story is that, in mechanics the natural framework to carry on these very fast analysis (and more sophisticated ones too!) on the nature of the solutions is the Hamiltonian formulation.

For example:

When you write the matrix for $$\tag 8 \Phi''=-\omega^2 \Phi$$ and get $$\tag 9 \frac{d}{dt} \left[ \begin{array}{c} X \\ Y \end{array} \right] = \omega^2 \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0\end{array} \right] \left[ \begin{array}{c} X \\ Y \end{array} \right]. $$ it's easy to compute the eigenvalues which are pure imaginary $\lambda_{\pm}=\pm i \omega$ the theory tells you right away that the solutions are limited for $t \rightarrow \infty$ and now (knowing a cuple of theorems) you can write just by knowing the eigenvalues: \begin{align} \tag {10} \mathcal U(t) & \equiv e^{\mathrm A t} = \left[ \begin{array}{cc} \cos(\omega t) & \sin(\omega t) \\ -\sin(\omega t) & \cos(\omega t)\end{array} \right] \\ S(t) &\equiv \left[ \begin{array}{c} X \\ Y \end{array} \right](t)= \mathcal U(t) \left[ \begin{array}{c} X \\ Y \end{array} \right](0) \end{align} so $$\tag{11}||S(t)||=\sqrt{X^2(0)+Y^2(0)} = \text{constant}$$ i.e. the solutions are circular orbits! In less trivial examples those kinds of analysis really saves you a lot of time in my personal experience.

  • $\begingroup$ I'm not seeing how the eigenvalues of the matrix in equation (9) can be $\pm i \omega$ since there's an $\omega^2$ in front. $\endgroup$
    – DanielSank
    Commented Jun 27, 2016 at 22:36
  • $\begingroup$ $$M = \left[ \begin{array}{cc} 0 &\omega^2\\-\omega^2 & 0\end{array} \right]$$ the eigenvalue problem is $det(M-\lambda \mathcal{I}) = 0$ where $\mathcal{I}$ is the identity. So the characteristic polynomial is: $$\lambda^2+\omega^2=0$$ therefore the eigenvalues are $\lambda_{\pm} = \pm i \omega$ $\endgroup$
    – Fra
    Commented Jul 10, 2016 at 9:12

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