# COM2: Superconducting Quantum Computer

The 26th Jyväskylä Summer School (JSS26)

Jyväskylä, Finland, 11.–12.8.2016

Lecturer: Prof. Mikko Möttönen, University of Jyväskylä, Finland

Assistant: Ph.D. Juha Jeronen, University of Jyväskylä, Finland

\label{fig:pendulum}The classical pendulum.

## Introduction: the classical pendulum

During this course, Lagrangian and Hamiltonian mechanics will be used for analyzing quantum computing circuits. As a refresher on Lagrangian mechanics, let us first consider a classical pendulum, with a point mass $$m$$ suspended by a rigid wire of length $$\ell$$, subjected to a uniform gravitational field having acceleration $$g$$. Refer to Figure \ref{fig:pendulum}.

The magnitude of the linear velocity of the bob (point mass) is

$$\label{eq:v-pendulum} \label{eq:v-pendulum}v=\ell\,\dot{\theta}\;,\\$$

where $$\theta$$ is the angle of the pendulum as measured from the vertical.

The gravitational potential energy of the bob is, by geometry,

$$\label{eq:V-pendulum} \label{eq:V-pendulum}V=mgh=mg\ell\,(1-\cos\theta)\approx\frac{1}{2}mg\ell\theta^{2}\;,\\$$

where the zero level for the height has been taken as the stable equilibrium position of the pendulum at $$\theta=0$$, and the last form is an $$O(\theta^{4})$$ Taylor approximation valid near $$\theta=0$$.

The kinetic energy of the bob is

$$\label{eq:T-pendulum} \label{eq:T-pendulum}T=\frac{1}{2}mv^{2}=\frac{1}{2}m\ell^{2}\dot{\theta}^{2}\;,\\$$

where we have used (\ref{eq:v-pendulum}).

Recall that the Lagrangian is defined as the kinetic energy minus the potential energy,

$$\label{eq:L-general} \label{eq:L-general}L\equiv T-V\;.\\$$

We choose the generalized position and momentum variables as

\begin{align} \label{eq:q-pendulum} q & \label{eq:q-pendulum}=\ell\,\theta\;, \\ p & \label{eq:p-pendulum}\equiv\frac{\partial L}{\partial\dot{q}}\approx\frac{\partial}{\partial\dot{q}}\left(\frac{1}{2}m\dot{q}^{2}-\frac{mg}{2\ell}q^{2}\right)=m\dot{q}=m\ell\,\dot{\theta}\;.\\ \end{align}

The Euler–Lagrange equation in Lagrangian mechanics states that

$$\label{eq:Euler-Lagrange-general} \label{eq:Euler-Lagrange-general}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{q}}\right)=\frac{\partial L}{\partial q}\;.\\$$

Inserting (\ref{eq:q-pendulum}) and (\ref{eq:p-pendulum}) into (\ref{eq:Euler-Lagrange-general}), we have

$$\label{eq:p-dot-pendulum-1} \label{eq:p-dot-pendulum-1}\dot{p}=-mg\,\theta\;,\\$$

or without using the Taylor approximation in the potential energy (\ref{eq:V-pendulum}),

$$\label{eq:p-dot-pendulum-exact} \label{eq:p-dot-pendulum-exact}\dot{p}=-mg\sin\theta\;.\\$$

On the other hand, by differentiating (\ref{eq:p-pendulum}) with respect to time, we have

$$\label{eq:p-dot-pendulum-2} \label{eq:p-dot-pendulum-2}\dot{p}=m\ell\,\ddot{\theta}\;.\\$$

Combining (\ref{eq:p-dot-pendulum-1}) and (\ref{eq:p-dot-pendulum-2}), we obtain the linearized equation of motion, describing small vibrations of the pendulum, as

$$m\ell\,\ddot{\theta}+mg\theta=0\;,\nonumber \\$$

i.e.

$$\label{eq:equation-of-motion-pendulum} \label{eq:equation-of-motion-pendulum}\ddot{\theta}+\frac{g}{\ell}\theta=0\;.\\$$

Inserting the standard trial function

$$\label{eq:standard-trial} \label{eq:standard-trial}\theta=\exp(i\omega t)\;,\\$$

where $$i$$ is the imaginary unit and $$\omega$$ the angular frequency, we obtain

$$i^{2}\omega^{2}\exp(i\omega t)+\frac{g}{\ell}\exp(i\omega t)=0\;,\nonumber \\$$

or after regrouping,

$$\label{eq:temp1} \label{eq:temp1}\left(-\omega^{2}+\frac{g}{\ell}\right)\exp(i\omega t)=0\;.\\$$

Equation (\ref{eq:temp1}) holds at all $$t$$ if and only if the expression in the parentheses vanishes:

$$\label{eq:characteristic-polynomial-pendulum} \label{eq:characteristic-polynomial-pendulum}-\omega^{2}+\frac{g}{\ell}=0\;.\\$$

Equation (\ref{eq:characteristic-polynomial-pendulum}) is the characteristic polynomial of the linearized equation of motion (\ref{eq:equation-of-motion-pendulum}). Solving (\ref{eq:characteristic-polynomial-pendulum}) for $$\omega$$, we obtain the natural angular frequency of small vibrations of the pendulum as

$$\label{eq:omega} \label{eq:omega}\omega=\sqrt{g/\ell}\;.\\$$