Improving the system tunnelling probability via an ancillary system

Info

This is a detailed version of calculations originally done in Machine Learning Catalysis of quantum tunneling. Thank you to Emmanuele Cinnirella for sharing the calculations below

Setup

We have an Hamiltonian given by $H=H_S+H_A+H_{int}$, where $\{\begin{array}{rl}{H}_A& =-{\sigma }_z-\gamma {\sigma }_x\\ H_S& =-J_z\\ H_{int}& =\alpha {\sigma }_z\otimes J_z\end{array}$ , and:

• ${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ is a Pauli Matrix
• ${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ is a Pauli Matrix
• $J_z$ is the Angular Momentum Operator along the $z$ axis.

Goal

Given the initial state ${\rho }^{(S)}\otimes {\rho }^{(A)}$, where ${\rho }^{(S)}=|1⟩⟨1|$ and ${\rho }^{(A)}=|{\psi }_A⟩⟨{\psi }_A|$, we want to calculate the Quantum tunnelling probability ${\mathbb{P}}_{L\to R}(t):=⟨0|{\rho }^{(S)}(t)|0⟩$.

Solution

$$\begin{array}{rll}{\rho }^{(S)}(t)& ={\mathrm{Tr}}_A\left[e^{-itH}{\rho }^{(S)}\otimes {\rho }^{(A)}{e}^{itH}\right]& \text{By definition of trace}\\ & ={\sum }_{k=0}^N⟨k|e^{-itH}{\rho }^{(S)}\otimes {\rho }^{(A)}{e}^{itH}|k⟩& \text{Substituting our Hamiltonian}\\ & ={\sum }_{k=0}^N⟨k|e^{-it(-{\sigma }_z-\gamma {\sigma }_x-J_z+\alpha {\sigma }_z\otimes J_z)}{\rho }^{(S)}\otimes {\rho }^{(A)}{e}^{it(-{\sigma }_z-\gamma {\sigma }_x-J_z+\alpha {\sigma }_z\otimes J_z)}|k⟩& \text{Since Jz∣k⟩=(N−2k)∣k⟩}\\ & ={\sum }_{k=0}^N⟨k|e^{-it(-{\sigma }_z-\gamma {\sigma }_x-(N-2k)+\alpha (N-2k){\sigma }_z)}{\rho }^{(S)}\otimes {\rho }^{(A)}{e}^{it(-{\sigma }_z-\gamma {\sigma }_x-(N-2k)+\alpha (N-2k){\sigma }_z)}|k⟩& \text{Cancelling the e±(N−2k) constant}\\ & ={\sum }_{k=0}^N⟨k|e^{-it(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)}{\rho }^{(S)}\otimes {\rho }^{(A)}{e}^{it(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)}|k⟩& \text{Taking out the ⟨k∣ψA⟩ factor}\\ & ={\sum }_{k=0}^N|⟨k|{\psi }_A⟩{|}^2{e}^{-it(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)}{\rho }^{(S)}{e}^{it(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)}& \text{Since eiθn⋅σ=cos(θ)12+isin(θ)n⋅σ whenever ∣n∣=1}\\ & ={\sum }_{k=0}^N|⟨k|{\psi }_A⟩{|}^2\left(\cos ({\omega }_{k}t)+i\frac{\sin ({\omega }_{k}t)}{{\omega }_k}(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)\right){\rho }^{(S)}\left(\cos ({\omega }_{k}t)+i\frac{\sin ({\omega }_{k}t)}{{\omega }_k}(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)\right)& {\omega }_k=|\overrightarrow{n}|=\sqrt{{\gamma }^2+(1-\alpha (N-2k){)}^2}\end{array}$$

The value of interest is ${\mathbb{P}}_{L\to R}(t):=⟨0|{\rho }^{(S)}(t)|0⟩$. Since ${\rho }^{(S)}(0)=|1⟩⟨1|$ we can ignore all terms in $\cos ({\omega }_{k}t)\pm i\frac{\sin ({\omega }_{k}t)}{{\omega }_k}(-{\sigma }_z-\gamma {\sigma }_x+\alpha (N-2k){\sigma }_z)$ that do not involve ${\sigma }_x$. This means we only care about the $\pm i\frac{\sin ({\omega }_{k}t)}{{\omega }_k}\gamma {\sigma }_x$ term, getting the result

$$\begin{array}{rl}{\mathbb{P}}_{L\to R}(t)& =⟨0|{\rho }^{(S)}(t)|0⟩\\ & ={\sum }_{k=0}^N|⟨k|{\psi }_A⟩{|}^2|i\frac{\sin ({\omega }_{k}t)}{{\omega }_k}\gamma {|}^2\\ & ={\sum }_{k=0}^N|⟨k|{\psi }_A⟩{|}^2\frac{{\gamma }^2}{{\omega }_k^2}{\sin }^2({\omega }_{k}t)\end{array}$$

We maximise the Quantum tunnelling probability when $|{\psi }_A⟩$ = $|k⟩\exists k\in \{0,1,...,N\}$ and, for the given $k$, we have

$$\begin{array}{rl}1& =\frac{{\gamma }^2}{{\omega }_k^2}\\ {\gamma }^2& ={\omega }_k^2\\ {\gamma }^2& =(1-\alpha \frac{N-2k}{2}{)}^2+{\gamma }^2\\ 0& =(1-\alpha \frac{N-2k}{2})\\ \alpha & =\frac{2}{N-2k}\end{array}$$