Auxiliary-qubit-assisted holonomic quantum gates on superconducting circuits

Abstract

Embraced with the built-in noise resilience feature for quantum evolutions, holonomic quantum computation provides high fidelity quantum gate operations, and thus has attracted much attention from researchers for stepping forward the development of robust quantum computation. Here in this work, we propose a physically feasible scheme to perform scalable non-adiabatic non-Abelian holonomic quantum computations in a physical system whose Hamiltonian is described by a quantum spin model with tunable coupling strengths. Our scheme is realizable from experimentally achievable Hamiltonian in superconducting-qubit systems and therefore, it is of practical importance in implementing protected quantum computation.

Appendix: The two-qubit gate

As presented in the main text, a holonomic two-qubit gate $U'$ (c.f. Eq. (18)) can be realized with assistance of the auxiliary qubit. It is well-known that concurrence can be used to evaluate the entanglement of two-qubit states. For any pure state $|{\psi }{{\rangle }_{12}}$ on a pair of quantum systems 1 and 2, the concurrence is defined as [48]

$$\begin{aligned} C({{|\psi \rangle }_{12}})= |{}_{12} \langle \psi | {\tilde{\psi }} \rangle _{12}|, \end{aligned}$$

(21)

with ${{| {\tilde{\psi }} \rangle }_{12}}=\sigma _{y}^{1}\otimes \sigma _{y}^{2}{{\left| {{\psi }^{*}} \right\rangle }_{12}}$ and ${{\left| {{\psi }^{*}} \right\rangle }_{12}}$ is the complex conjugation of the state ${|\psi \rangle }_{12}$. Concurrence ranges from 0 to 1, with $C =0$ for product state, and $C=1$ for maximum entangled state. In the following, by calculating the concurrence of the two-qubit state before and after the two-qubit gate $U'$, we prove that the gate $U'$ performs an entangling operation. Suppose the initial state of the two-qubit system is an arbitrary product state,

$$\begin{aligned} |\psi {{\rangle }_{12}}={{(a|-\rangle }_{1}}+b|+{{\rangle }_{1}})\otimes {{(c|-\rangle }_{2}}+d|+{{\rangle }_{2}}), \end{aligned}$$

(22)

where the coefficients a, b, c, d are complex values. It can be easily checked that $C({{|\psi \rangle }_{12}})=0$ for the initial product state ${{|\psi \rangle }_{12}}$. Applying the gate $U'$ onto the state $|\psi {{\rangle }_{12}}$, the two-qubit state becomes

$$\begin{aligned} |{\psi }'{{\rangle }_{12}}= & {} {U}'|\psi {{\rangle }_{12}}, \nonumber \\= & {} -\left( ac{{\left| -- \right\rangle }_{12}}+ad{{\left| +- \right\rangle }_{12}}+bc{{\left| -+ \right\rangle }_{12}}-bd{{\left| ++ \right\rangle }_{12}} \right) . \end{aligned}$$

(23)

For our transformed state $|{\psi }'{{\rangle }_{12}}$, the concurrence is calculated to be

$$\begin{aligned} C{{(|{\psi }'\rangle }_{12}})=4\left| abcd \right| . \end{aligned}$$

(24)

Therefore, as long as the condition of $abcd\ne 0$ is satisfied, $C{{(|{\psi }'\rangle }_{12}}) \ne 0$, which indicates that $|{\psi }'{{\rangle }_{12}}$ is an entangled state and the two-qubit gate ${U}'$ is an entangled gate. For example, suppose the initial product state reads

$$\begin{aligned} |\psi {{\rangle }_{12}}= & {} \frac{1}{\sqrt{2}}\left( {{\left| - \right\rangle }_{1}}+{{\left| + \right\rangle }_{1}} \right) \otimes \frac{1}{\sqrt{2}}\left( {{\left| - \right\rangle }_{2}}+{{\left| + \right\rangle }_{2}} \right) , \nonumber \\= & {} \frac{1}{2}\left( {{\left| -- \right\rangle }_{12}}+{{\left| -+ \right\rangle }_{12}}+{{\left| +- \right\rangle }_{12}}+{{\left| ++ \right\rangle }_{12}} \right) , \end{aligned}$$

(25)

The state after applying the gate $U'$ is then given by

$$\begin{aligned} |{\psi }'{{\rangle }_{12}}=-\frac{1}{2}\left( {{\left| -- \right\rangle }_{12}}+{{\left| -+ \right\rangle }_{12}}+{{\left| +- \right\rangle }_{12}}-{{\left| ++ \right\rangle }_{12}} \right) . \end{aligned}$$

(26)

It is clear that $|{\psi }'{{\rangle }_{12}}$ becomes an entangled state.