Intermediate qutrit-assisted Toffoli gate decomposition with quantum error correction

Abstract

Introducing a few intermediate qutrits for efficient decomposition of 3-qubit unitary gates has been proposed recently to obtain an exponential reduction in the depth of the decomposed circuit. An intermediate qutrit implies that a qubit is operated as a qutrit in a particular execution cycle. This method, primarily for the NISQ era, treats a qubit as a qutrit only for the duration when it requires access to the state $\left| {2}\right\rangle$ during the computation. In this article, we study the challenges of extending this decomposition to the error-corrected regime. We first we show that if a qubit has to be in state $\left| {2}\right\rangle$ at any point of time, then it must be encoded using a qutrit quantum error correcting code (QECC), thus resulting in a circuit with both qubits and qutrits. Qutrits being noisier than qubits, the former are expected to require higher levels of concatenation to achieve a particular accuracy than that for qubit-only decomposition. We derive analytically a relation between the levels of concatenation required for qubit-only and that for qubit–qutrit decomposition to achieve the same level of accuracy. Finally, we estimate (i) the degree of concatenation for both qubit–qutrit and qubit-only decompositions as a function of the probability of error and (ii) the criterion for which qubit–qutrit decomposition leads to a lower gate count than that for qubit-only decomposition.

Appendices

Appendix

1.1 Proof of Theorem 3

Proof

Let the accuracy obtained using only qubit and qubit–qutrit decomposition after k levels of concatenations be $\epsilon _2$ and $\epsilon _3$, respectively, where $\epsilon _3 = \delta \cdot \epsilon _2$. Let $\frac{1}{c_3}$ and $\frac{1}{c_2}$ be the thresholds of the ternary and binary QECCs used for encoding. Then,

$$\begin{aligned} \frac{1}{c_3}(c_3.p_{2,3})^{2^k}= & {} \frac{\delta }{c_2}(c_2.p_2)^{2^k} \nonumber \\ \Rightarrow \frac{c_2}{c_3}(c_3.p_{2,3})^{2^k}= & {} \delta (c_2.p_2)^{2^k} \nonumber \\ \Rightarrow c(c_3.p_{2,3})^{2^k}= & {} \delta (c_2.p_2)^{2^k} ~\text {where} ~c=\frac{c_2}{c_3} \nonumber \\ \Rightarrow log(c) + 2^k log(c_3.p_{2,3})= & {} log(\delta ) + 2^k.log(c_2.p_2) \nonumber \\ \Rightarrow log(\delta )-log(c)= & {} 2^k[log(c_3.p_{2,3})-log(c_2.p_2)] \nonumber \\ \Rightarrow log(\delta )-log(c)= & {} 2^k log\frac{p_{2,3}}{c.p_2} \nonumber \\ \Rightarrow log(\delta )= & {} 2^k log\frac{p_{2,3}}{c.p_2} + log(c) \end{aligned}$$

(8)

$\square$

Proof of Theorem 4

Proof

The accuracy obtained after k levels of concatenation with a QECC having threshold $\frac{1}{c}$ is $\frac{1}{c}(c.p)^{2^k}$, where p is the probability of error. In the current setting, both types of decomposition are attaining the same accuracy after $k_2$ and $k_3$ levels of concatenations. If $\frac{1}{c_3}$ and $\frac{1}{c_2}$ be the thresholds for ternary and binary QECCs used, then,

$$\begin{aligned} \frac{1}{c_3}(c_3.p_{2,3})^{2^{k_3}}= & {} \frac{1}{c_2}(c_2.p_2)^{2^{k_2}} \\ \Rightarrow \frac{c_2}{c_3}(c_3.p_{2,3})^{2^{k_3}}= & {} \delta (c_2.p_2)^{2^{k_2}} \\ \Rightarrow c(c_3.p_{2,3})^{2^{k_3}}= & {} \delta (c_2.p_2)^{2^{k_2}} ~\text {where} ~c=\frac{c_2}{c_3} \\ \Rightarrow log(c) + 2^{k_3}log(c_3.p_{2,3})= & {} 2^{k_2}log(c_2.p_2) \\ \Rightarrow 2^{k_3}log(c_3.p_{2,3})= & {} 2^{k_2}log(c_2.p_2) - log(c) \\ \Rightarrow 2^{k_3 - k_2}log(c_3.p_{2,3})= & {} log(c_2.p_2) - \frac{1}{2^{k_2}}log(c) \\ \Rightarrow 2^{k_3 - k_2}log(c_3.p_{2,3})= & {} log(c_2.p_2) - log(c)^{\frac{1}{2^{k_2}}} \\ \Rightarrow 2^{k_3 - k_2}= & {} \frac{log(c_2.p_2) - log(c)^{\frac{1}{2^{k_2}}}}{log(c_3.p_{2,3})} \\ \Rightarrow k_3 - k_2= & {} log\left( \frac{log(c_2.p_2) - log(c)^{\frac{1}{2^{k_2}}}}{log(c_3.p_{2,3})}\right) \\ \Rightarrow k_3= & {} k_2 + log\left( \frac{log(c_2.p_2) - \frac{1}{2^{k_2}}log(\frac{c_2}{c_3})}{log(\delta ) + log(c_3.p_{2})}\right) ~\text {where} ~p_{2,3} = \delta .p_2 \end{aligned}$$

$\square$

Proof of Theorem 5

Proof

We require the overall gate count of the qubit–qutrit decomposition to be lower than that of the qubit decomposition. In other words, our requirement is

$$\begin{aligned} \left( \sum _{g \in \mathcal {G}_{2,3}} \kappa _g n_g\right) ^{k_3} \le \left( \sum _{g \in \mathcal {G}_{2}} \kappa _g n_g\right) ^{k_2} \end{aligned}$$

Now,

$$\begin{aligned} \left( \sum _{g \in \mathcal {G}_{2,3}} \kappa _g n_g\right) ^{k_3}\le & {} \left( \sum _{g \in \mathcal {G}_{2}} \kappa _g n_g\right) ^{k_2} \nonumber \\ \Rightarrow k_3 log\left( \sum _{g \in \mathcal {G}_{2,3}} \kappa _g n_g \right)\le & {} k_2 log\left( \sum _{g \in \mathcal {G}_{2}} \kappa _g n_g\right) \nonumber \\ \Rightarrow \frac{k_3}{k_2}\le & {} \frac{log\left( \sum _{g \in \mathcal {G}_{2}} \kappa _g n_g\right) }{log\left( \sum _{g \in \mathcal {G}_{2,3}}\kappa _g n_g\right) } \nonumber \\ \Rightarrow \frac{k_3}{k_2} - 1\le & {} \frac{log\left( \sum _{g \in \mathcal {G}_{2}} \kappa _g n_g\right) }{log\left( \sum _{g \in \mathcal {G}_{2,3}}\kappa _g n_g\right) } - 1 \nonumber \\ \Rightarrow k_3 - k_2 \le k_2 \cdot \frac{log\left( \frac{\sum _{g \in \mathcal {G}_{2}} \kappa _g n_g}{\sum _{g \in \mathcal {G}_{2,3}}\kappa _g n_g}\right) }{log\left( \sum _{g \in \mathcal {G}_{2,3}}\kappa _g n_g\right) } \end{aligned}$$

(9)

By Theorem 4, we have

$k_3 = k_2 + log\left( \frac{log(c_2.p_2) - \frac{1}{2^{k_2}}log(\frac{c_2}{c_3})}{log(\delta ) + log(c_3.p_{2})}\right)$. Substituting this in Eq. (9), we have

$$\begin{aligned} log\left( \frac{log(c_2.p_2) - \frac{1}{2^{k_2}}log(\frac{c_2}{c_3})}{log(\delta ) + log(c_3.p_{2})}\right) \le k_2 \cdot \frac{log\frac{\sum _{g \in \mathcal {G}_{2}} (\kappa _g n_g)}{\sum _{g \in \mathcal {G}_{2,3}}(\kappa _g n_g)}}{log\sum _{g \in \mathcal {G}_{2,3}}(\kappa _g n_g)} \end{aligned}$$

$\square$