Recovering symmetry in coupled oscillators

Question

Consider a pair of LC oscillators, one with capacitance $C_1$ and inductance $L_1$ and the other with capacitance $C_2$ and inductance $L_2$. Suppose they're connected through a capacitor $C_g$. We want to find the normal modes and frequencies.

If we write out Kirchhoff's laws, we find \begin{align} V_1 + \ddot{V}_1 \left(1 + \epsilon_1 \right)/\omega_1^2 - (\epsilon_1/\omega_1^2)\ddot{V}_2 &= 0 \\ V_2 + \ddot{V}_2 \left(1 + \epsilon_2 \right)/\omega_2^2 - (\epsilon_2/\omega_2^2)\ddot{V}_1 &= 0 \\ \end{align} where $\epsilon_i \equiv C_g / C_i$ and $\omega_i^2 \equiv 1/L_i C_i$. These equations can be written in matrix form as $$ \left( \begin{array}{c} V_1 \\ V_2 \end{array} \right) = \left( \begin{array}{cc} (1 + \epsilon_1)/\omega_1^2 & - \epsilon_1 / \omega_1^2 \\ - \epsilon_2 / \omega_2^2 & (1 + \epsilon_2)/\omega_2^2 \\ \end{array} \right) \left( \begin{array}{c} \ddot{V}_1 \\ \ddot{V}_2 \end{array} \right) \tag{$\star$} \, . $$ Now if $L_1 = L_2$ and $C_1 = C_2$ then $\epsilon_1 = \epsilon_2 \equiv \epsilon$ and $\omega_1 = \omega_2 \equiv \omega_0$ and the matrix equation becomes $$ \left( \begin{array}{c} V_1 \\ V_2 \end{array} \right) = \left( \begin{array}{cc} (1 + \epsilon)/\omega_0^2 & - \epsilon / \omega_0^2 \\ - \epsilon / \omega_0^2 & (1 + \epsilon)/\omega_0^2 \\ \end{array} \right) \left( \begin{array}{c} \ddot{V}_1 \\ \ddot{V}_2 \end{array} \right) \, . $$ In this particular case, the matrix can be written in the nice form $$ \frac{1 + \epsilon}{\omega_0^2} \, \mathbb{I} - \frac{\epsilon}{\omega_0^2} \sigma_x \tag{$\star \star$} $$ and it's pretty easy to find the normal modes and normal frequencies.$^{[a]}$

However, when the oscillators aren't identical, e.g. Eq. ($\star$), expressions for the normal modes and frequencies are pretty messy. Is there a transformation we can apply to ($\star$) to bring it into a simple form like ($\star \star$) so that the mode analysis results in simpler equations?

Perhaps another way to ask this would be to ask for a systematic way to rescale the variables so that the matrix in the equations of motion is symmetric or perhaps Hermitian.

[a] The frequencies are $\omega_0$ (even mode) and $\omega_0 / \sqrt{1 + 2 \epsilon}$ (odd mode).

Answer 1

How about this:

I will write your general matrix in the form \begin{align} M=\left(\begin{array}{cc} a&b\\ c&d \end{array}\right) \end{align} so that your system is $$ V=M \ddot{V} $$

Consider \begin{align} U=\left(\begin{array}{cc} e^{\alpha}&0\\ 0&e^{-\alpha} \end{array}\right) \end{align} with $\alpha$ to be determined. The choice of this is closely related to a rotation $e^{-i\alpha \hat L_z}$ that would do the trick if you had a hermitian matrix and wanted to rotate the $\sigma_y$ component away.

Upon conjugation: \begin{align} UMU^{-1}= \left(\begin{array}{cc} a&be^{2\alpha}\\ ce^{-2\alpha}&d \end{array}\right) \end{align} and choose $\alpha$ so that $$ be^{2\alpha}=ce^{-2\alpha}=b’ $$ to bring your original $M$ to the form \begin{align} UMU^{-1}= \left(\begin{array}{cc} a&b’\\ b’&d \end{array}\right) \end{align} which is of the form \begin{align} \frac{1}{2}(a+d)\mathbb{I}+\frac{1}{2}(a-d)\sigma_z+b’\sigma_x \end{align} A further unitary rotation about $y$, generated by $e^{-i\beta\sigma_y}$ can get rid of either the $\sigma_x$ or the $\sigma_z$ term.

Note that my $U$ is not a unitary transformation: your $M$ isn’t hermitian either so something’s gotta give. $U$ is a rescaling of the original basis vectors, stretching one and compressing the other. The transformed basis vector remain orthogonal but no longer have length 1. The transformed $M$ is hermitian, as requested.

This kind of “diagonalization” of non-hermitian operator using a non-unitary transformation is explored in

Rashid MA. The intelligent states. I. Group‐theoretic study and the computation of matrix elements. Journal of Mathematical Physics. 1978 Jun;19(6):1391-6.

Intelligent states are states that saturate the uncertainly relations; they are eigenstates of a non-hermitian operator.

Answer 2

We wish to find a basis in which

$$ \left( \begin{array}{c} V_1 \\ V_2 \end{array} \right) = \underbrace{\left( \begin{array}{cc} (1 + \epsilon_1)/\omega_1^2 & - \epsilon_1 / \omega_1^2 \\ - \epsilon_2 / \omega_2^2 & (1 + \epsilon_2)/\omega_2^2 \\ \end{array} \right)}_{:= M} \left( \begin{array}{c} \ddot{V}_1 \\ \ddot{V}_2 \end{array} \right) \, $$

is diagonal. This can be done only if the eigenvalues of $M$ are distinct.

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Outline: We want to diagonalize $M$, but we first need to find if this is possible. It is possible if $M$ has distinct eigenvalues. Using the facts that

$$ tr(M) = \lambda_1 + \lambda_2 = \frac{1+\epsilon_1}{\omega_1^2}+ \frac{1+\epsilon_2}{\omega_2^2} $$

$$\det(M) = \lambda_1\lambda_2 = \frac{(1+\epsilon_1)(1+\epsilon_2)}{\omega_1^2\omega_2^2}{} + \frac{\epsilon_1\epsilon_2}{\omega_1^2\omega_2^2}$$

which gives us the eigenvalues

$$ \{\lambda_1, \lambda_2\}= \\ \left\{\frac{-\sqrt{(-\epsilon_2 \omega_1 -\epsilon_1 \omega_2-\omega_2-\omega_1 )^2-4 (\epsilon_2 \omega_2 \omega_1 +\epsilon_1 \omega_2 \omega_1 +\omega_2 \omega_1 )}+\epsilon_2 \omega_1 +\epsilon_1 \omega_2+\omega_2+\omega_1 }{2 \omega_2 \omega_1 },\\ \frac{\sqrt{(-\epsilon_2 \omega_1 -\epsilon_1 \omega_2-\omega_2-\omega_1 )^2-4 (\epsilon_2 \omega_2 \omega_1 +\epsilon_1 \omega_2 \omega_1 +\omega_2 \omega_1 )}+\epsilon_2 \omega_1 +\epsilon_1 \omega_2+\omega_2+\omega_1 }{2 \omega_2 \omega_1 }\right\} $$

which are in general distinct. So the point is that we can find a suitable transformation matrix $S$ such that $M$ is diagonal (The columns of $S$ are the eigenvectors of $M$).

Now label the states in the new basis with primes. Then we go to the new basis $$ S \left( \begin{array}{c} V_1 \\ V_2 \end{array} \right) = S\left( \begin{array}{cc} (1 + \epsilon_1)/\omega_1^2 & - \epsilon_1 / \omega_1^2 \\ - \epsilon_2 / \omega_2^2 & (1 + \epsilon_2)/\omega_2^2 \\ \end{array} \right)S^{-1}S \left( \begin{array}{c} \ddot{V}_1 \\ \ddot{V}_2 \end{array} \right) \, $$

becomes

$$ \left( \begin{array}{c} V'_1 \\ V'_2 \end{array} \right) = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \left( \begin{array}{c} \ddot{V'}_1 \\ \ddot{V'}_2 \end{array} \right) $$

and so our equations are decoupled in this basis.

Answer 3

Is there a transformation we can apply to $(*)$ to bring it into a simple form like $(**)$ so that the mode analysis results in simpler equations?

Actually there is, but it isn't a simple rescaling. Maybe the easiest way to see what to do is to proceed in two steps. First, a substitution $V_2=k\,V_2'$ leaves unaltered diagonal elements but makes off-diagonal terms equal to each other for some $k$.

Now the matrix is a linear combination of $\Bbb I$, $\sigma_1$, $\sigma_3$ and it shouldn't be difficult to find eigenvalues and eigenvectors,

Hope this also answers your question.