What class of materials are closest to realizing the tunable coupling Hamiltonian?

Question

From a physics point of view, there is an effective (approximation to second-order coupling Jaynes-Cummings) Hamiltonian of the form [1] \begin{equation} H=\sum_j\omega_j(t)\sigma_j^z+\sum_{\langle i,j\rangle}J_{ij}(t)(\sigma_i^+\sigma_j^- + \text{h.c.}) \end{equation} where $J_{ij}(t)$ is tunable, meaning that it can always be tuned continuously w.r.t. a parameter (such as capacitance in superconducting circuits). $\langle i,j\rangle$ denotes the sum over adjacent spin-1/2's that are coupled to each other. We may assumed ferromagnetic or anti-ferromagnetic configuration. But my focus shall be in just the functional form of the Hamiltonian. This coupling Hamiltonian actually resembles the $XY$ model \begin{equation} H_{XY}=\sum_{\langle i,j\rangle}2J_{ij}(\sigma_i^+\sigma_j^- + \text{h.c.})=\sum_{\langle i,j\rangle}J_{ij}(\sigma_i^x\sigma_j^x +\sigma_i^y\sigma_j^y) \end{equation} where $J_{ij}$ are fixed in experiment, not tunable. My aim is to see the purely material implementation for the same where we have some tunable parameter to control and we can selectively and dynamically choose each $J_{ij}$ for the coupling for each pair. Please suggest such materials or theoretical models for such materials if they exist or are proposed.

[1] Fei Yan, Philip Krantz, Youngkyu Sung, Morten Kjaergaard, Daniel L. Campbell, Terry P. Orlando, Simon Gustavsson, and William D. Oliver. Tunable Coupling Scheme for Implementing High-Fidelity Two-Qubit Gates. Phys. Rev. Applied, 10:054062, Nov 2018.

Answer 1

Realizing this Hamiltonian in a natural material

• I cannot imagine a material in which all nearest-neighbor spin-spin interactions can be adjusted arbitrarily at the same time. Spin-spin couplings that are stronger when the spins are closer together, and weaker when the spins are farther apart, can be adjusted by moving the spins relative to each other; so if we have a 2D sheet of spins and we pull it from all four corners, all the spin-spin distances will get larger, and in that sense we might have $J_{ij}(t) = J_{ij}/t$, meaning that the coupling potential decreases linear as we do the stretching. But not only will the couplings of the nearest neighbors be affected, but all couplings will, since all distances change. There would also not be much control over the functions $J_{ij}(t)$. Perhaps we would have to change the couplings without changing relative distances, but in some other way, such as with a laser, but I don't know how this would happen (an NMR expert might).
• Any Hamiltonian with only Pauli operators is quite an approximation of real-world materials existing in nature. The electronic Hamiltonian of a material does not really resemble this Hamiltonian, but the nuclear Hamiltonian used in NMR studies might resemble it more closely (if spin-orbit coupling and other types of terms can be ignored, and if the nuclear spins indeed are $\pm 1/2$ like in the case of hydrogen atoms). Hopefully an NMR expert can give a better answer here.

Realizing this Hamiltonian in an artificial material

Apart from the example in the paper you gave, the closest example I know is from a paper published fewer than 2 months before you asked this question. The Hamiltonian is given in the image I posted in this question, where the time-dependent "driver" part is shown in a bit more general form in Eq. 3 of the paper I mentioned (whether viewing on arXiv or Physical Review, it's still Eq. 3). I will repeat the driver and "problem" Hamiltonians here for the case where the coupling in the driver and the problem Hamiltonians are chosen to be the same (which needs to be perfectly doable anyway):

\begin{align} \tag{1} H_{\rm{driver}} &= \sum_{i}\omega_i \sigma^x_i + \sum_{ij}J_{ij}\sigma_i^y\sigma_j^y \\ \tag{2} H_{\rm{problem}} &= \sum_{ij}J_{ij}\sigma_i^z\sigma_j^z\\ \end{align}

Now assuming that the normal AQC driving is done, we get:

\begin{align} \tag{3} H(t) & = (1-t)H_{\rm{driver}} + tH_{\rm{problem}} \\ \tag{4} & = \sum_{i}(1-t)\omega_i \sigma^x_i + \sum_{ij}(1-t)J_{ij}\sigma_i^y\sigma_j^y + \sum_{ij}tJ_{ij}\sigma_i^z\sigma_j^z\\ \tag{5}\label{similar} & =\sum_{i}\omega_i(t) \sigma^x_i + \sum_{ij}J_{ij}(t)\sigma_i^y\sigma_j^y + \sum_{ij}K_{ij}(t)\sigma_i^z\sigma_j^z,\\ \end{align}

where I have defined:

\begin{align} \tag{6} \label{defineTimeDependence} \omega_i(t) \equiv (1-t)\omega, & ~~~~~~J_{ij}(t)\equiv (1-t)J_{ij}, & K_{ij}(t)\equiv tJ_{ij}. \end{align}

When $t=1/2$, we get:

\begin{align} \tag{7} H(t) & =\sum_{i}\omega_i(t) \sigma^x_i + \sum_{ij}J_{ij}(t)\left(\sigma_i^y\sigma_j^y + \sigma_i^z\sigma_j^z\right).\\ \end{align}

Now let's apply the unitary transormation:

\begin{align} \tag{8} H(t) & =\sum_{i}\omega_i(t) \sigma^x_i + \sum_{ij}J_{ij}(t)\left(\sigma_i^y\sigma_j^y + \sigma_i^z\sigma_j^z\right),\\ \end{align}

which is strikingly close to your Hamiltonian:

\begin{align} \tag{9} H(t) & =\sum_{i}\omega_i(t) \sigma^z_i + \sum_{ij}J_{ij}(t)\left(\sigma_i^y\sigma_j^y + \sigma_i^x\sigma_j^x\right).\\ \end{align}

So the closest I could get was the same Hamiltonian except it's only precisely the same as yours when:

• $t=1/2$ holds true in my case (Eq. $\eqref{similar}$ is quite similar for other values of $t$ though).
• Eq. $\eqref{defineTimeDependence}$ holds true in your case.
• $X$ is interchanged with $Z$.

In the 2-spin case, if a material needs to behave like Eq. $\eqref{similar}$, it can work if it contains the following superconducting circuit with Josephson junctions represented by the x's (it looks only slightly different from the circuit here maybe because this circuit is actually reported to be able to have linear $Z$ terms in addition to the linear $X$ terms, but I've set those field strengths to 0 in my analysis, which is a perfectly fine thing to do):