3
$\begingroup$

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).

$\endgroup$
6
  • $\begingroup$ Have you tried working in a basis where the matrix is diagonal? $\endgroup$ – InertialObserver Mar 16 '19 at 18:56
  • $\begingroup$ @InertialObserver Finding that basis in a systematic way is exactly the point of this question. $\endgroup$ – DanielSank Mar 16 '19 at 18:59
  • $\begingroup$ I don’t understand.. this isn’t a problem about coupled oscillators then.. do you know about matrix diagonalization? $\endgroup$ – InertialObserver Mar 16 '19 at 19:00
  • $\begingroup$ @InertialObserver Yes I know about matrix diagonalization, and yes this is a question about coupled oscillators. The equation ($\star$) are the equations of two coupled electrical harmonic oscillators. The question is how to diagonalize the matrix (i.e. decouple the equations of motion) in a systematic way when the oscillators are not identical. $\endgroup$ – DanielSank Mar 16 '19 at 19:17
  • $\begingroup$ I suppose you could write the matrix in general as $a_0 I + \vec{a} \cdot \vec{\sigma}$ and then apply a rotation (by conjugating with $e^{- i \theta \hat{n} \cdot \vec{\sigma}}$) to align $\vec{a}$ with $\hat{x}$. However, doing this explicitly might be more complicated than doing it the normal way. $\endgroup$ – knzhou Mar 16 '19 at 23:06
0
$\begingroup$

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.


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.

$\endgroup$
1
  • $\begingroup$ This doesn't answer the question. The question asks if there's a systematic way to handle the extra complexity that comes in when the oscillators aren't identical. This answer states the eigenvalues but it doesn't make an attempt to simplify them, e.g. by defining useful quantities shared between the two expressions. Neither does this answer explain how to formally resolve the missing symmetry. $\endgroup$ – DanielSank Mar 17 '19 at 0:12
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.