Will show you.

Well, we're going to the final ones. I think we'll get started.

So, we finished up last night writing down classical waves of motion for our electrical circuits with mechanically movable compliant capacitors.

And if I remember correctly, that was the simple Lagrangian we wrote down where all the interestingness is really in the fact that the capacitance is a function of some form of x.

And then I'll bring it down again to drop the imposed force and the voltage that it originally had.

And so at this point, we could continue with the classical-aggrated diffusion and solve them in all different limits.

But I think in keeping with the quantum spirit of this school, we'll go ahead and quantize the problem and solve the quantum-aggrated diffusion.

And it will be remarkably simple for the classical ones.

I suppose the primary virtue of doing the quantum version is that it's really kind of relevant to the ambitions of the field,

and it's not natural to treat the quantum noise of the fields that are going to couple to our circuit.

So, to do our standard transformation from Lagrangian written in terms of coordinates and generalized velocities to our Hamiltonian,

it's sort of a function of coordinates and the associated generalized momentum.

And the entirely easy to anticipate outcome of that is just...

We have now what looks like completely like two harmonic oscillators.

And in fact, if there were no position dependence of the capacitance of the problem, it would be too uncoupled from our oscillators.

So at this point, we could carry through writing out...we could carry through with our canonical quantization, but sort of useful at this point just to linearize the problem.

We'll spare some of the notation for the complexity, we'll do that. So we'll just approximate the one over the capacitance for a small displacement of the equilibrium.

And so now, this last term just becomes...

whereby C just being the capacitance and I'll write it as x to the sixth.

So, in the end, we'll now imagine that we put hats on the coordinates and momentum, insist that they will be canonical commutation relations, and voila, we have one Hamiltonian.

The thing which is nice to do is to get rid of all of the dimension equal stuff and convert to ladder operators, annihilation creation operators, and then we'll end up with the equal optomechanical Hamiltonian.

But we can trace all of the various definitions back to the original circuit model. So, we'll do that.

So we're already defining the impedance of the resonator, the associated quantity for the mechanical oscillator.

And then we'll do the power point notion.

In terms of these definitions, we can define the microwave photon annihilation operator.

So...

All right, after a lot of tedious writing on the board, I'll talk to you guys for a moment.

The annihilation operators have this nice dimensional form, and all of the circuit information is contained within the resonance frequency and this sort of impedance.

So in particular, you'll notice that if we have a high impedance resonator, we make this quantity big, then what we mean by a photon is much more charge fluctuations and much less flux and vice versa for a low impedance resonator.

There's the equivalent association with what we mean by one phonon corresponds either to much more displacement or much more momentum, depending on the impedance of the mechanical structure.

And once we've done that, then all of the kind of physics of the circuit and the spring and mass can be used to sort of disappear into these dimensional operators.

Konrad, can you say some more about the mechanical impedance? What's the meaning of this?

You can be a stiff light thing or a heavy floppy thing and resonate with the same frequency. Your zero point motion will be composed of much more displacement and much less momentum in one case versus the other.

Just the ratio between the momentum and the space magnitude.

That's right.

Maybe even, because KS is M omega squared.

Would you like me to write this as M omega squared?

So this thing is M omega squared.

Is that right? This one makes sense to me.

Somehow it's the geometric mean of these two quantities that appear in front of the energy I just erased.

I was just going to point out, maybe another way of seeing it is, I'm sure you were going to go into this later, but you just write that mechanics is an equivalent circuit and it looks like it has a serious RL speed.

So if I make the equivalent electrical circuit of the mechanical structure, I actually won't do that, but I think I can do that.

So there would be a kind of motionless inductance and motionless capacitance.

I'm going to do that now.

So I'm going to take all of those substitutions and the associated commutation relations.

I'm going to take the termination coupling.

And then the piece that will drop.

So the termination coupling is zero and the dimension for coupling is zero.

Obviously the familiar uncouple harmonic oscillator and Hamiltonian is for the resonator mechanics.

The now fully anticipated optomechanical interaction piece.