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?

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?

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

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

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

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

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:

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

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?

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?

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

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

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

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

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:

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

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?

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

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

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

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

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:

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

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?