2 level system tunnelling probability

Info

This is an expansion of the namesake section in Machine Learning Catalysis of quantum tunneling

Setup

We have a 2-level system described by

• the Hamiltonian $H=-\delta {\sigma }_z-\gamma {\sigma }_x$
• ${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ is a Pauli Matrix
• ${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ is a Pauli Matrix
• Our system starts in the Quantum state $|0⟩=(10{)}^T$

Goal

We want to explicitly calculate the probability $\mathbb{P}[{\psi }_t=|1⟩]$, given the initial state ${\psi }_0=|0⟩$.

Solution

The time evolution of ${\psi }_t$ is given by the formal solution ${\psi }_t=e^{-\frac{i}{\hslash }Ht}$. We will solve this problem by:

1. Using the Eigendecomposition ${\sum }_i{\lambda }_i|i⟩⟨i|$ of $H$.
2. Writing $e^{-\frac{i}{\hslash }Ht}$ as ${\sum }_i\exp (-\frac{i}{\hslash }{\lambda }_{i}t)|i⟩⟨i|$
3. Calculating $\mathbb{P}[{\psi }_t=|1⟩]$ = $\Vert ⟨1|e^{-\frac{i}{\hslash }Ht}|0⟩{\Vert }^2$.

We have $H=-\delta {\sigma }_z-\gamma {\sigma }_x=\left(\begin{array}{cc}-\Delta & -\gamma \\ -\gamma & \Delta \end{array}\right)$. We solve $0=\det (H-\lambda 1_2)$ to get the roots $\pm \sqrt{{\Delta }^2+{\gamma }^2}$ . Let $\lambda :=\sqrt{{\Delta }^2+{\gamma }^2}$, with the two roots of $H$ being $\pm \lambda$. By solving the 2 systems of equations $H\psi =\lambda \psi$ and $H\psi =-\lambda \psi$, for example using Gaussian elimination, we obtain the 2 Eigenstates $|+⟩:=\left(\begin{array}{c}-\gamma \\ \Delta +\lambda \end{array}\right)$ and $|-⟩:=\left(\begin{array}{c}\Delta +\lambda \\ \gamma \end{array}\right)$, with respective eigenvalues $\lambda$ and $-\lambda$. After normalisation, the Eigenstates are $|+⟩=\sqrt{\frac{1}{2\lambda (\Delta +\lambda )}}\left(\begin{array}{c}-\gamma \\ \Delta +\lambda \end{array}\right)$ and $|-⟩:=\sqrt{\frac{1}{2\lambda (\Delta +\lambda )}}\left(\begin{array}{c}\Delta +\lambda \\ \gamma \end{array}\right)$. Hence

$$\{\begin{array}{rcrcr}H& =& \lambda |+⟩⟨+|& -& \lambda |-⟩⟨-|\\ e^{-\frac{i}{\hslash }Ht}& =& \exp (-\frac{i\lambda }{\hslash }t)|+⟩⟨+|& +& \exp (\frac{i\lambda }{\hslash }t)|-⟩⟨-|\end{array}$$

We preemptively calculate the Inner Products

$$\{\begin{array}{rl}⟨+|0⟩& =-\gamma \sqrt{\frac{1}{2\lambda (\Delta +\lambda )}}\\ ⟨+|1⟩& =\sqrt{\frac{\Delta +\lambda }{2\lambda }}\\ ⟨-|0⟩& =\sqrt{\frac{\Delta +\lambda }{2\lambda }}\\ ⟨-|1⟩& =\gamma \sqrt{\frac{1}{2\lambda (\Delta +\lambda )}}\end{array}$$

to make the calculation

$$\begin{array}{rll}\sqrt{\mathbb{P}[{\psi }_t=|1⟩]}& =⟨1|\exp (\frac{i\lambda }{\hslash }Ht)|0⟩& \text{ As exp(ℏiλHt)=exp(−ℏiλt)∣+⟩⟨+∣+exp(ℏiλt)∣−⟩⟨−∣}\\ & =⟨1|\exp (-\frac{i\lambda }{\hslash }t)|+⟩⟨+|+\exp (\frac{i\lambda }{\hslash }t)|-⟩⟨-||0⟩& \text{Splitting the braket}\\ & =\exp (-\frac{i\lambda }{\hslash }t)⟨1|+⟩⟨+|0⟩+\exp (\frac{i\lambda }{\hslash }t)⟨1|+⟩⟨+|0⟩& \text{Using the innter products above}\\ & =\exp (-\frac{i\lambda }{\hslash }t)(-\frac{\gamma }{2\lambda })+\exp (\frac{i\lambda }{\hslash }t)(\frac{\gamma }{2\lambda })& \\ & =\frac{\gamma }{\lambda }\frac{1}{2}\left[\exp (\frac{i\lambda }{\hslash }t)-\exp (-\frac{i\lambda }{\hslash }t)\right]& \\ & =\frac{\gamma }{\lambda }\sin (\frac{i\lambda t}{\hslash })& \\ & \Downarrow & \\ \mathbb{P}[{\psi }_t=|1⟩]& =\frac{{\gamma }^2}{{\lambda }^2}{\sin }^2(\frac{\lambda t}{\hslash })& \text{Setting ℏω=λ}\\ & =\frac{{\gamma }^2}{{\hslash }^2{\omega }^2}{\sin }^2(\omega t)& \end{array}$$

Generalisation

The calculations above have been generalised by introducing an ancillary system, with the goal of reducing the asymmetry $\Delta$ of the system. This has been done on Improving the system tunnelling probability via an ancillary system