Master equation on spontaneous emission.

I will explain how to handle the simplest noise in quantum computers, natural radiation, in theory. Spontaneous emission is the process of state transition by emitting photons in any quantum system. It is the simplest of the states that can be solved by the master equation. To do so, we solve the master equation as shown below [1].

\begin{equation} \frac{d}{dt}=[H,\rho]\frac{1}{i\hbar}+\gamma(\bar{n}+1)(\sigma^+ \rho \sigma^- -\frac{1}{2}\sigma^- \sigma^+ \rho -\frac{1}{2}\rho \sigma^- \sigma^+)+\gamma\bar{n}(\sigma^- \rho \sigma^+ - \frac{1}{2}\sigma^+ \sigma^-\rho -\frac{1}{2}\rho \sigma^+ \sigma^-) -(1) \end{equation}

where $\gamma$ and $\bar{n}$ are the decoherence factor and the number of photons in Bose-Einstein statistics, respectively. The Hamiltonian is as follows \begin{equation} H=\frac{1}{2}\omega_0 \hbar \sigma^z -(2) \end{equation}

Solve for this and you get

\begin{eqnarray} \frac{d}{dt}\rho = \left( \begin{array}{cc} -\gamma\bar{n}\rho_{00}+\gamma(\bar{n}+1)\rho_{11}&-0.5(\omega_0 i + 2\gamma\bar{n}-\gamma)\rho_{01} \ 0.5(\omega_0 i -2\gamma\bar{n}-\gamma)\rho_{10}&\gamma\bar{n}\rho_{00}-\gamma(\bar{n}+1)\rho_{11} \ \end{array} \right)-(3) \end{eqnarray} The diagonal term is \begin{eqnarray} \frac{d}{dt}{\rho_d}=\left( \begin{array}{cc} -\gamma\bar{n}&\gamma(\bar{n}+1)\ \gamma\bar{n}&-\gamma(\bar{n}+1)\ \end{array}\right) {\rho_d}-(4) \end{eqnarray} We can do this as In the same way as solving a simultaneous differential equation, we have the following \begin{equation} {\rho_d}=\left( \begin{array}{c} Ae^{-(2\bar{n}+1)\gamma t} + c_0 \ -Ae^{-(2\bar{n}+1)\gamma t} - c_1 \ \end{array} \right)-(5) \end{equation} From equations (4) and (5), the coefficients look like this

\begin{equation} c_0=\frac{\bar{n}+1}{2\bar{n}+1}, c_1=\frac{-\bar{n}}{2\bar{n}+1}-(6) \end{equation} Thus, each element of the density operator looks like this \begin{eqnarray} \rho_{00}(t)&=&Ae^{-(2\bar{n}+1)\gamma t}+\frac{\bar{n}+1}{2\bar{n}+1} \ \rho_{11}(t)&=&Ae^{-(2\bar{n}+1)\gamma t} + \frac{\bar{n}}{2\bar{n}+1} \ \rho_{10}(t)&=&A_1e^{-\frac{1}{2}(2\bar{n}+1)\gamma t}e^{\frac{1}{2}\omega_0 it} \ \rho_{01}(t)&=&A_2e^{-\frac{1}{2}(2\bar{n}+1)\gamma t}e^{-\frac{1}{2}\omega_0 it} \ \end{eqnarray}-(7)

We can see that the damping term in the non-diagonal term is exactly half of the damping term in the diagonal term. This is the inverse of what is called the transverse and longitudinal relaxation times, respectively. This means that the superposition in the quantum state is stronger than in the classical state. Since this is a calculation of natural radiation, I used a lot of equations, but even this is a small number for a noise calculation. The calculation is quite trivial, but I hope you will be able to see for yourself because it is simple.

[1]Giuliano Benenti, Giuliano Casati, and Giuliano Strini. Principles of Quantum Computation and Information Volume I: Basic Concepts, Splinger, 2004