Optimal quantum state transformations based on machine learning

Abstract

It is well known that quantum algorithms may solve problems efficiently that are intractable using conventional algorithms. Quantum algorithms can be designed with a set of universal quantum gates that transform input states into desired output states. However, designing quantum algorithms that transform states in desired ways is challenging due to its complexity. In this paper, we propose a machine learning framework for the transformation of unknown states into their corresponding target states. Specifically, a parameterized quantum circuit learns a given task by tuning its parameters. After the learning is done, the circuit is competent for the quantum task. This allows us to circumvent cumbersome circuit design based on universal quantum gates. If perfect transformation is forbidden by quantum theory, an optimal transformation can be obtained in terms of fidelity. This provides a research method to study various quantum no-go theorems that characterize the intrinsic gap between quantum and classical information. As examples, quantum state rotation and quantum state cloning are studied using numerical simulations. We also show the good robustness of our machine learning framework to corrupted training data, which is a very nice property for physical implementation on near-term noisy intermediate-scale quantum devices.

Appendix

We here give a derivation of the probabilities of measurement in Eqs. (12) and (15).

For QSR, the quantum circuit is shown in Fig. 4. Suppose that an input state $|\psi _\mathrm{in}^i\rangle$ and the corresponding teaching state $|\psi _{t}^i\rangle$ are fed into the quantum circuit. The $|\psi _\mathrm{in}^i\rangle$ is sent through the learning circuit to give

$$\begin{aligned} |\psi _\mathrm{out}^i \rangle = U(\vec {\theta })|\psi _\mathrm{in}^i\rangle . \end{aligned}$$

(22)

The state before evaluation by the teaching circuit is

$$\begin{aligned} |\psi _\mathrm{out}^i \rangle |\psi _{t}^i\rangle |0\rangle . \end{aligned}$$

(23)

The first Hadamard gate on the last qubit gives us

$$\begin{aligned} \frac{1}{\sqrt{2}}\left( |\psi _\mathrm{out}^i \rangle |\psi _{t}^i\rangle |0\rangle +|\psi _\mathrm{out}^i \rangle |\psi _{t}^i\rangle |1\rangle \right) . \end{aligned}$$

(24)

Then, a controlled-SWAP gate is applied, which gives

$$\begin{aligned} \frac{1}{\sqrt{2}}\left( |\psi _\mathrm{out}^i \rangle |\psi _{t}^i\rangle |0\rangle +|\psi _{t}^i \rangle |\psi _\mathrm{out}^i\rangle |1\rangle \right) . \end{aligned}$$

(25)

Next, a Hadamard gate is applied again on the last qubit, giving

$$\begin{aligned} |\Psi \rangle =\frac{1}{2} \left[ \left( |\psi _\mathrm{out}^i \rangle |\psi _{t}^i\rangle +|\psi _{t}^i \rangle |\psi _\mathrm{out}^i\rangle \right) |0\rangle +\left( |\psi _\mathrm{out}^i \rangle |\psi _{t}^i\rangle -|\psi _{t}^i \rangle |\psi _\mathrm{out}^i\rangle \right) |1\rangle \right] .\qquad \end{aligned}$$

(26)

Finally, by measuring the last qubit, we obtain a result of 0 with probability

$$\begin{aligned} \begin{aligned} p_{0}^i =&\langle \Psi |(I\otimes I \otimes |0\rangle \langle 0|)|\Psi \rangle \\ =&\frac{1+{\mathcal {F}}^i}{2}. \end{aligned} \end{aligned}$$

(27)

Similarly, we obtain a result of 1 with probability

$$\begin{aligned} p_{1}^i =\frac{1-{\mathcal {F}}^i}{2}. \end{aligned}$$

(28)

For $1-2$ QSC, the quantum circuit is shown in Fig. 6. Suppose that an input state $|\psi _\mathrm{in}^i\rangle$ and the corresponding teaching state $|\psi _{t}^i\rangle =|\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle$ are fed into the quantum circuit. The $|\psi _\mathrm{in}^i\rangle$ together with auxiliary qubits are sent through the learning circuit to give

$$\begin{aligned} |\psi _\mathrm{out}^i\rangle = U(\vec {\theta })|\psi _\mathrm{in}^i\rangle |0\rangle |0\rangle . \end{aligned}$$

(29)

Suppose that $|\psi _\mathrm{out}^i\rangle =\sum _{l,m,n=0}^{1} \alpha _{lmn}^{i}|l\rangle |m\rangle |n\rangle$. The state before evaluation by the teaching circuit is

$$\begin{aligned} \sum _{l,m,n=0}^{1} \alpha _{lmn}^{i}|l\rangle |m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |0\rangle |0\rangle . \end{aligned}$$

(30)

The first two Hadamard gates on the last two qubits give us

$$\begin{aligned} \frac{1}{2}\sum _{l,m,n=0}^1\alpha _{lmn}^{i}|l\rangle |m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle (|0\rangle |0\rangle +|0\rangle |1\rangle +|1\rangle |0\rangle +|1\rangle |1\rangle ). \end{aligned}$$

(31)

Then, two controlled-SWAP gates are applied, which gives

$$\begin{aligned}&\frac{1}{2}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |0\rangle |0\rangle +\frac{1}{2}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |m\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |n\rangle |0\rangle |1\rangle \nonumber \\ +&\frac{1}{2}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |\psi _\mathrm{in}^i\rangle |n\rangle |m\rangle |\psi _\mathrm{in}^i\rangle |1\rangle |0\rangle +\frac{1}{2}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |m\rangle |n\rangle |1\rangle |1\rangle . \end{aligned}$$

(32)

Next, two Hadamard gates are applied again on the last two qubits, giving

$$\begin{aligned}&\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle (|0\rangle |0\rangle +|0\rangle |1\rangle +|1\rangle |0\rangle +|1\rangle |1\rangle )\nonumber \\&+\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |m\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |n\rangle (|0\rangle |0\rangle -|0\rangle |1\rangle +|1\rangle |0\rangle -|1\rangle |1\rangle )\nonumber \\&+\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |\psi _\mathrm{in}^i\rangle |n\rangle |m\rangle |\psi _\mathrm{in}^i\rangle (|0\rangle |0\rangle +|0\rangle |1\rangle -|1\rangle |0\rangle -|1\rangle |1\rangle )\nonumber \\&+\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |m\rangle |n\rangle (|0\rangle |0\rangle -|0\rangle |1\rangle -|1\rangle |0\rangle +|1\rangle |1\rangle ). \end{aligned}$$

(33)

It can be rewritten as

$$\begin{aligned}&\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle \Big (|m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle +|m\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |n\rangle +|\psi _\mathrm{in}^i\rangle |n\rangle |m\rangle |\psi _\mathrm{in}^i\rangle \nonumber \\&+|\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |m\rangle |n\rangle \Big )|0\rangle |0\rangle \nonumber \\&+\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle \Big (|m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle -|m\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |n\rangle +|\psi _\mathrm{in}^i\rangle |n\rangle |m\rangle |\psi _\mathrm{in}^i\rangle \nonumber \\&-|\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |m\rangle |n\rangle \Big )|0\rangle |1\rangle \nonumber \\&+\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle \Big (|m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle +|m\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |n\rangle -|\psi _\mathrm{in}^i\rangle |n\rangle |m\rangle |\psi _\mathrm{in}^i\rangle \nonumber \\&-|\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |m\rangle |n\rangle \Big )|1\rangle |0\rangle \nonumber \\&+\frac{1}{4}\sum _{l,m,n}\alpha _{lmn}^{i}|l\rangle \Big (|m\rangle |n\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle -|m\rangle |\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |n\rangle -|\psi _\mathrm{in}^i\rangle |n\rangle |m\rangle |\psi _\mathrm{in}^i\rangle \nonumber \\&+|\psi _\mathrm{in}^i\rangle |\psi _\mathrm{in}^i\rangle |m\rangle |n\rangle \Big )|1\rangle |1\rangle . \end{aligned}$$

(34)

Thus, by measuring the last two qubits, we obtain a result of 00 with probability

$$\begin{aligned} p^{i}_{00}=&\frac{1}{4}\Big [1+\sum _{l,m,n,n'}\alpha _{lmn}^{i}\alpha _{lmn'}^{*i}\Big \langle \psi _\mathrm{in}^i|n\Big \rangle \Big \langle n'|\psi _\mathrm{in}^i\Big \rangle +\sum _{l,m,m',n}\alpha _{lmn}^{i}\alpha _{lm'n}^{*i}\Big \langle \psi _\mathrm{in}^i|m\Big \rangle \Big \langle m'|\psi _\mathrm{in}^i\Big \rangle \nonumber \\&+\sum _{l,m,m',n,n'}\alpha _{lmn}^{i}\alpha _{lm'n'}^{*i}\Big \langle \psi _\mathrm{in}^i|m\Big \rangle \Big \langle \psi _\mathrm{in}^i|n\Big \rangle \Big \langle m'|\psi _\mathrm{in}^i\Big \rangle \Big \langle n'|\psi _\mathrm{in}^i\Big \rangle \Big ]. \end{aligned}$$

(35)

Notice that $\sum _{l,m,m',n}\alpha _{lmn}^{i}\alpha _{lm'n}^{*i}|m\rangle \langle m'|$ is the reduced density matrix of copy one, denoted as $\rho _1$, $\sum _{l,m,n,n'}\alpha _{lmn}^{i}\alpha _{lmn'}^{*i}|n\rangle \langle n'|$ is the reduced density matrix of copy two, denoted as $\rho _2$. $\sum _{l,m,m',n,n'}\alpha _{lmn}^{i}\alpha _{lm'n'}^{*i}|m\rangle |n\rangle \langle m'|\langle n'|$ is the reduced density matrix of copy one and two, denoted as $\rho _{12}$. Then, $p^{i}_{00}$ can be written as

$$\begin{aligned} p^{i}_{00}&=\frac{1}{4}\left( 1+\left\langle \psi _{in}^i|\rho _1|\psi _{in}^i\right\rangle +\left\langle \psi _{in}^i|\rho _2|\psi _{in}^i\right\rangle +\left\langle \psi _{in}^i\left|^{\otimes 2}\rho _{12}\right|\psi _{in}^i\right\rangle ^{\otimes 2}\right) \nonumber \\&=\frac{1}{4}\left( 1+{\mathcal {F}}^{i}_{1}+{\mathcal {F}}^{i}_{2}+{\mathcal {F}}^{i}_{12}\right) . \end{aligned}$$

(36)

Similarly, we can obtain

$$\begin{aligned} p^{i}_{01}&=\frac{1}{4}\left( 1+{\mathcal {F}}^{i}_{1}-{\mathcal {F}}^{i}_{2}-{\mathcal {F}}^{i}_{12}\right) ,\nonumber \\ p^{i}_{10}&=\frac{1}{4}\left( 1-{\mathcal {F}}^{i}_{1}+{\mathcal {F}}^{i}_{2}-{\mathcal {F}}^{i}_{12}\right) ,\nonumber \\ p^{i}_{11}&=\frac{1}{4}\left( 1-{\mathcal {F}}^{i}_{1}-{\mathcal {F}}^{i}_{2}+{\mathcal {F}}^{i}_{12}\right) . \end{aligned}$$

(37)