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7:00 PM

@EmilioPisanty I got this far: $$H = \frac{\omega_1}{2}(X_1^2 + Y_1^2) + \frac{\omega_2}{2}(X_2^2 + Y_2^2) + \frac{Y_1 Y_2}{\sqrt{Z_1 Z_2}} \frac{1}{C_g''}$$ where the $Z$'s are constants and $C_g''$ has to do with the coupling strength.

The $X$'s and $Y$'s are conjugate.

You said to make the flux part symmetric, so I guess now I need to rescale $X_1$ and $Y_1$ to make the prefactor of the first term the same as the prefactor of the second term. Working...

Oh my goodness I'm so glad I wrote things down in grad school.

PSA: write things down.

Emilio Pisanty

7:21 PM

-2

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@DanielSank looks about right, but I'm on mobile and sans mathjax

@DanielSank yes, looks like that.

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DanielSank

@EmilioPisanty Now I have something a bit more interesting. Rescaling the variables, and defining $\omega_0 \equiv \sqrt{\omega_1 \omega_2}$ and $\rho \equiv \sqrt{\omega_1 / \omega_2}$, we get $$H = \frac{\omega_0}{2} X_1^2 + \frac{\omega_0}{2} X_2^2 + \frac{1}{2} \rho \omega_1 Y_1^2 + \frac{1}{2} \frac{1}{\rho} \omega_2 Y_2^2 + \frac{Y_1 Y_2}{\sqrt{Z_1 Z_2} C_g''} \, .$$

Now I'm not sure what to do.

Emilio Pisanty

7:40 PM

@DanielSank rotate the Ys

DanielSank

Ok, so what you're saying is to write the $Y$ part as $$(Y_1 \, Y_2) \left( \begin{array}{cc} \rho \omega_1 / 2 & 1 / 2\sqrt{Z_1 Z_2} C_g'' \\ 1 / 2 \sqrt{Z_1 Z_2} C_g'' & \omega_2 / 2 \rho \end{array} \right) \left( \begin{array}{c} Y_1 \\ Y_2 \end{array} \right)$$

and invert the matrix.

@EmilioPisanty it's interesting that the matrix for the $Y$ part is now symmetric.

Let's define $Z_0 \equiv \sqrt{Z_1 Z_2}$ so that the $Y$ matrix is $$\left( \begin{array}{cc} \rho \omega_1 / 2 & 1 / 2 Z_0 C_g'' \\ 1 / 2 Z_0 C_g'' & \omega_2 / 2 \rho \end{array} \right) \, .$$

The inverse of the matrix is $$\frac{4}{\omega_1 \omega_2 \left(1 - C_1 C_2/C_g''^2 \right)} \left( \begin{array}{cc} \omega_2 / 2 \rho & -1/2 Z_0 C_g'' \\ -1 / 2Z_0 C_g'' & \rho \omega_1 /2 \end{array} \right) \, .$$

The original matrix can be written $$\frac{1}{4} \left( \rho \omega_1 + \frac{\omega_2}{\rho} \right) \mathbb{I} + \frac{1}{4} \left( \rho \omega_1 - \frac{\omega_2}{\rho} \right) \sigma_z + \frac{1}{2 Z_0 C_g''} \sigma_x \, .$$

DanielSank

8:18 PM

I'm going to continue this later. I think there's a nice way to simplify by defining $$\Delta \equiv \omega_2 - \omega_1$$ to simplify the $\sigma_z$ term and defining $$g \equiv \frac{\omega_0 \sqrt{C_1 C_2}}{2 C_g''}$$ to simplify the $\sigma_x$ term.

DanielSank

8:36 PM

Actually we should just rename the coefficient of $\sigma_z$ to be $\Delta / 4$ and the coefficient of $\sigma_x$ to be $g$.