A quantum model of feed-forward neural networks with unitary learning algorithms

Abstract

Quantum neural networks (QNNs) are promised to be powerful computing devices that integrate the advantages of artificial neural networks (ANNs) and quantum computing. Due to the different dynamics between ANN and quantum computing, constructing a reasonable QNN model with efficient learning algorithms is still an open challenge. In this paper, we propose a new quantum model for feed-forward neural networks whose learning algorithm applies quantum superposition and parallelism features. This model contains classical feed-forward neural network and the amplitude encoding QNN model as special cases. Moreover, it inherits the advantages and avoids the disadvantages of the amplitude encoding model. We give a quantum-classical hybrid procedure to implement the learning algorithm. The result shows that we can train this QNN by a series of unitary operators efficiently.

Brief review of Hadamard test and swap test

In this appendix, we briefly review two quantum algorithms (Hadamard test and swap test) to extract inner product of quantum states. Each method has its own advantages and disadvantages. First, we consider the following amplitude estimation problem [21]. Let

$$\begin{aligned} |\phi \rangle =U|0\rangle ^{\otimes k}=\sin \theta |0\rangle |u\rangle +\cos \theta |1\rangle |v\rangle \end{aligned}$$

(18)

be a quantum state that can be prepared by a unitary operator U in time $O(T_{\text {in}})$. The problem we plan to solve is to estimate $|\sin \theta |$ and $|\cos \theta |$ (i.e., $\pm \theta$) in quantum computer to accuracy $\epsilon$ with high success probability at least $1-\delta$.

The quantum algorithm to solve this problem is a direct application of quantum phase estimation. Let Z be the two-dimensional Pauli-Z matrix that maps $|0\rangle$ to $|0\rangle$ and $|1\rangle$ to $-|1\rangle$. Denote

$$\begin{aligned} G=(I-2|\phi \rangle \langle \phi |)(Z\otimes I)=U(I-2|0\rangle ^{\otimes k}\langle 0|^{\otimes k})U^\dag (Z\otimes I). \end{aligned}$$

Then,

$$\begin{aligned} G=\left[ \begin{array}{cc} \cos 2\theta &{}\quad \sin 2\theta \\ -\sin 2\theta &{}\quad \cos 2\theta \\ \end{array} \right] \end{aligned}$$

in the space spanned by $\{|0\rangle |u\rangle ,|1\rangle |v\rangle \}$. The eigenvalues of G are $e^{\pm \i 2\theta }$, and the corresponding eigenvectors are

$$\begin{aligned} |w_1\rangle =\frac{1}{\sqrt{2}}\Big (|0\rangle |u\rangle +\i |1\rangle |v\rangle \Big ),~~ |w_2\rangle =\frac{1}{\sqrt{2}}\Big (|0\rangle |u\rangle -\i |1\rangle |v\rangle \Big ) \end{aligned}$$

, respectively. Note that $|\phi \rangle =-\frac{\i }{\sqrt{2}} (e^{\i \theta }|w_1\rangle -e^{-\i \theta }|w_2\rangle ).$ Performing quantum phase estimation to G with initial state $|0\rangle ^n|\phi \rangle$ for some $n=O(\log 1/\delta \epsilon )$ [5]. Then, with probability at least $1-\delta$, we will obtain an approximate of the following state

$$\begin{aligned} -\frac{\i }{\sqrt{2}}\Big (e^{\i \theta }|y\rangle |w_1\rangle -e^{-\i \theta }| -y\rangle |w_2\rangle \Big ), \end{aligned}$$

(19)

where $y\in {\mathbb {Z}}_{2^n}$ satisfies $|\theta -y\pi /2^n|\le \epsilon$. The time complexity of the above procedure is $O(T_{\text {in}}/\epsilon \delta )$. Performing a measurement on (19), we will get an $\epsilon$ approximate of $\theta$ or $-\theta$ with a high probability at least $1-\delta$.

Furthermore, let f be an even function such that there is an efficient oracle $U_f$ to implement f, that is $U_f|j,k\rangle = |j,k\oplus f(j)\rangle$. Then, by Eq. (19), we can get an approximation of

$$\begin{aligned} |\phi \rangle |f(\tilde{\theta })\rangle , \end{aligned}$$

(20)

by adding a register to store $f(\tilde{\theta })$ and undoing the quantum phase estimation, where $|\theta -\tilde{\theta }|\le \epsilon$.

Now let $|x\rangle ,|y\rangle$ be two quantum states that can be prepared in time $O(T_{\text {in}})$. Then, the above method provides us two quantum algorithms to estimate $\langle x|y\rangle$ to accuracy $\epsilon$ in time $O(T_{\text {in}}/\epsilon )$.

First, we discuss the Hadamard test. Assume that there are two unitaries $U_0,U_1$ such that $U_0|0..0\rangle = |x\rangle , U_1|0..0\rangle = |y\rangle$. The Hadamard test can be described by the quantum circuit in Fig. 6.

Fig. 6

Quantum circuit of Hadamard test

The final quantum state is

$$\begin{aligned} |\phi \rangle =\frac{1}{2}(|0\rangle (|x\rangle +|y\rangle )+|1\rangle (|x\rangle -|y\rangle )). \end{aligned}$$

The probability of $|0\rangle$ is $(1+\mathrm{Re}\langle x|y\rangle )/2$, and the probability of $|1\rangle$ is $(1-\mathrm{Re}\langle x|y\rangle )/2$. So we can set $\sin \theta =\sqrt{(1+\mathrm{Re}\langle x|y\rangle )/2}$ and $\cos \theta =\sqrt{(1-\mathrm{Re}\langle x|y\rangle )/2}$. The quantum state $|\phi \rangle$ can be rewritten in the form of (18) with $|u\rangle ,|v\rangle$ corresponding to the normalization of $|x\rangle +|y\rangle ,|x\rangle -|y\rangle$. Therefore, $\mathrm{Re}\langle x|y\rangle$ can be evaluated in time $O(T_{\text {in}}/\epsilon )$ with accuracy $\epsilon$. However, if considering

$$\begin{aligned} |\psi \rangle =\frac{1}{2}(|0\rangle (|x\rangle +\i |y\rangle )+|1\rangle (|x\rangle -\i |y\rangle )), \end{aligned}$$

then we can estimate $\mathrm{Im}\langle x|y\rangle$ in time $O(T_{\text {in}}/\epsilon )$ with accuracy $\epsilon$. Moreover, we actually can obtain the following quantum state

$$\begin{aligned} \frac{1}{\sqrt{2}}(|+\rangle |x\rangle +|-\rangle |y\rangle )\otimes |g(s)\rangle , \end{aligned}$$

(21)

for any function g with an efficient circuit to implement $U_g:|j,k\rangle = |j,k\oplus g(j)\rangle$, where $|s-\langle x|y\rangle | \le \epsilon$. More precisely, we can first generate $|\phi \rangle |\mathrm{Re}(\langle x|w\rangle )\rangle$ by taking $f(\theta ) = 1- 2 \cos ^2\theta$ in Eq. (20). Since $|\psi \rangle$ can be obtained from $|\phi \rangle$, we can further generate $|\psi \rangle |\mathrm{Re}(\langle x|w\rangle )\rangle |\mathrm{Im}(\langle x|w\rangle )\rangle$. Now based on the oracle $U_g$ and the unitary to perform $|j,k,0\rangle \rightarrow |j,k,j+\i k\rangle$, we can get $|\psi \rangle |\mathrm{Re}(\langle x|w\rangle )\rangle |\mathrm{Im}(\langle x|w\rangle )\rangle |g(\langle x|w\rangle )\rangle$. Finally, undo the first two steps to yield state (21). Concluding the above analysis, we have the following proposition.

Proposition 1

Let $|x\rangle ,|y\rangle$ be two quantum states that can be prepared by two unitaries in time $O(T_{in })$. Let g be a univariate function over ${\mathbb {C}}$ with an efficient circuit to implement $U_g:|x,y\rangle = |x,y\oplus g(x)\rangle$. Then, the following transformation

$$\begin{aligned} \frac{1}{\sqrt{2}}(|0\rangle |x\rangle +|1\rangle |y\rangle ) |0\rangle \mapsto \frac{1}{\sqrt{2}}(|0\rangle |x\rangle +|1\rangle |y\rangle )|g(s)\rangle \end{aligned}$$

(22)

can be achieved in time $O(T_{in }/\epsilon )$, where $|\langle x|y\rangle -s|\le \epsilon$.

To apply Hadamard test, we need to know the unitaries $U_0,U_1$ to prepare $|x\rangle ,|y\rangle$ in advance. Otherwise, we cannot use it to extract $\mathrm{Re}\langle x|y\rangle$ and $\mathrm{Im}\langle x|y\rangle$. On the other hand, swap test does not need such an assumption [24]. However, swap test can only estimate the norm of $\langle x|y\rangle$. The quantum circuit to implement swap test is shown in Fig. 7.

Fig. 7

Quantum circuit of swap test

The final state of Fig. 7 has the following form

$$\begin{aligned} \frac{1}{2}|0\rangle (|x\rangle |y\rangle + |y\rangle |x\rangle ) +\frac{1}{2}|1\rangle (|x\rangle |y\rangle - |y\rangle |x\rangle ). \end{aligned}$$

(23)

The probability of $|0\rangle$ is $\frac{1}{2} ( 1+ |\langle x|y\rangle |^2)$. Consequently, we can get an approximation of $|\langle x|y\rangle |$. The complexity to prepare state (23) is $O(T_\mathrm{in})$; thus, the complexity to get an $\epsilon$-approximation of $|\langle x|y\rangle |$ still equals $O(T_\mathrm{in}/\epsilon )$ by the method discussed in the beginning of this appendix. We can also obtain a similar result as Proposition 1 by swap test; that is, s in Proposition 1 becomes an approximation of $|\langle x|y\rangle |$.

However, there is a simple technique to extract $\langle x|y\rangle$ from swap test if we are given two unit real vectors x, y. More precisely, denote ${\tilde{x}}=(1,x)$ and ${\tilde{y}}=(1,y)$; then, it is easy to verify that $|\langle {\tilde{x}}|{\tilde{y}}\rangle |=(1+\langle x|y\rangle )/2$. Thus, to compute $\langle x|y\rangle$, we can apply swap test to $|{\tilde{x}}\rangle , |{\tilde{y}}\rangle$. With this simple technique, Proposition 1 can also be obtained.