Week 8: Physical Implementation of Qubits: Harmonic Oscillators and Superconductors

Coursework focused on A Quantum Engineer’s Guide to Superconducting Qubits.

Exercise 1

The Schrödinger equation is given as: $\Psi_{E_{n}}(x) = C_{n}H_{n}\left(\sqrt{\frac{m\omega_{0}}{\hbar}}x\right)e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}$

For the lowest energy level state $(n=0)$ , write the solution of $\Psi(x)$ . Then substitute it back in the Schrödinger equation and show it’s a solution. Use the time independent Schrödinger equation:

$\left[\frac{-\hbar^{2}}{2m}\frac{\delta^{2}}{\deltax^2}+\frac{kx^{2}}{2}\right]\Psi_{E_{n}}(x) = E_{n} \Psi_{En}^{x}$

$\Psi_{E_{n}}(x) = C_{n}H_{n}\left(\sqrt{\frac{m\omega_{0}}{\hbar}}x\right)e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}$

Which is equivalent to,

$\frac{d^{2}H_n(u)}{du^{2}} - 2u\frac{dH_{n}(u)}{du} + 2\left(\frac{E_n}{\hbar\omega_{0}} - \frac{1}{2}\right)H_{n}(u) = 0$

If $n = 0$ then we know, $H_n(u) = 1$ ; because $H_n(u)$ is a Hermite Polynomial

Therefore,

$\Psi_{0}(x) = C_{0}(1)\left(e^{\frac{-m\omega}{2\hbar}x^{2}}\right)$

$\Psi_{0}(x) = C_{0}e^{\frac{-m\omega}{2\hbar}x^{2}}$

Plugged back into the time independent Schrödinger equation, we have:

$\left(-\frac{\hbar^{2}}{2m}\frac{\delta^{2}}{\deltax^{2}} + \frac{kx^{2}}{2}\right)\Psi_{E_{n}}(x) = E_{n}\Psi_{n}(x)$

Where, $n = 0$ :

$= -\frac{\hbar C_{0}}{2m}\left[\frac{\delta}{\delta x}\left(-\frac{2m\omega_{0}x}{2\hbar}e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}\right)\right] + \frac{C_{0}k}{2}x^{2}e^{-\frac{m_\omega_{0}}{2\hbar}x^{2}}$

$= \frac{2m\omega_{0} \hbar^{2} C_{0}}{4m\hbar}\left[\frac{\delta}{\delta x}\left(xe^{-\frac{m\omega_{0}}{2\hbar}x^{2}}\right)\right] + \frac{C_{0}k}{2}x^{2}e^{-\frac{m_\omega_{0}}{2\hbar}x^{2}}$

$= \frac{C_{0}\omega{0}\hbar}{2}\left[x\left(-\frac{2m\omega_{0}x}{2\hbar}\right)e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}\right] + \frac{C_{0}kx^{2}}{2}e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}$

$= \left[-\frac{m\omega_{0}^{2}}{2}C_{0}x^{2} + \frac{\hbar\omega_{0}C_{0}}{2} + \frac{C_{0}kx^{2}}{2}\right]e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}$

$= \left[-\frac{m\omega_{0}^{2}}{2}x^{2} + \frac{\hbar\omega_{0}}{2} + \frac{kx^{2}}{2}\right]C_{0}e^{-\frac{m\omega_{0}}{2\hbar}x^{2}}$

$=\left(\frac{\hbar\omega_{0}}{2} + \frac{kx^{2}}{2} - \frac{m\omega_{0}^{2}x^{2}}{2}\right)\Psi_{0}(x)$

Therefore,

$\Psi_{0}(x) = C_{0}e^{-\frac{m\omega_0x^{2}}{2\hbar}}$

Exercise 2

The expectation value of a variable $x$ is given by $\int_{\infty}^{-\infty}\Psi^{*}x\Psi dx$ . Show for the lowest energy state of the quantum harmonic oscillator that the expected value of the position $x$ is zero.

$<x> = \int_{\infty}^{-\infty}\Psi_{0^{n}}n \Psi_{0^{n}}dn$

$= C_{0}^{2} \int_{\infty}^{-\infty}e^{\frac{m\omega}{\hbar}n^{2}}dn$

$f(n) = ne^{\frac{m\omega}{\hbar}n^{2}}$ where $<x> = 0$

Therefore,

$\int{\infty}{-\infty}f(n)dn = 0$

Exercise 3

What is the Josephson oscillation frequency for a $0.5V$ potential drop across a junction? Is this frequency change detectable by standard electronics? What voltage drop would give a Josephson oscillation frequency of $10GHz$ ?