Clifford Algebras, Quantum Neural Networks and Generalized Quantum Fourier Transform

Abstract

We propose models of quantum perceptrons and quantum neural networks based on Clifford algebras. These models are capable to capture geometric features of classical and quantum data as well as producing data entanglement. Due to their representations in terms of Pauli matrices, the Clifford algebras seem to be a natural framework for multidimensional data analysis in a quantum setting. In this context, the implementation of activation functions, and unitary learning rules are discussed. In this scheme, we also provide an algebraic generalization of the quantum Fourier transform containing additional parameters that allow performing quantum machine learning based on variational algorithms. Furthermore, some interesting properties of the generalized quantum Fourier transform have been proved.

Appendices

Appendix

A Basic Elements of Clifford Algebras

In this appendix, we review some basic concepts about Clifford algebras [24, 32].

Given a vector space V, the Clifford algebra can be defined as quotient $Cl(V,Q)=\frac{T(V)}{I}$, where $I_{Q}$ is a two sided ideal generated by elements

$$\begin{aligned} v \otimes v - Q(v)1, \end{aligned}$$

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for all $v \in V$; Q is the quadratic form and T(V) is the tensor algebra. Alternatively, let V be a space vector over $\mathbb {R}$ equipped with a symmetric bilinear form g, A an associative algebra with unit $1_{A}$, and $\gamma$ a linear application $\gamma :V\rightarrow A$. The pair $(A, \gamma )$ is a Clifford algebra for the quadratic space (V, g) if A is generated as an algebra by $\{\gamma (v); v \in V\}$, $\{a 1_{A}; a \in \mathbb {R}\}$ and satisfies

$$\begin{aligned} \gamma (v)\gamma (u)+\gamma (u)\gamma (v)=2g(v,u)1_{A}, \end{aligned}$$

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for all $v,u \in V$. Let V be a vector space $\mathbb {R}^{n}$ and g a symmetric bilinear form in $\mathbb {R}^{n}$ of signature (p, q) with $p+q=n$. We will denote by $Cl(p,q) \equiv Cl_{p,q} \equiv Cl(\mathbb {R}^{p,q})$ the Clifford algebra associated with the quadratic space $\mathbb {R}^{p,q}$. In addition, we denote $Cl(2n,0)\equiv Cl(2n)$. The even subalgebra is defined by:

$$\begin{aligned} Cl^{+}_{p,q}=\{\Gamma \in Cl_{p,q} ; \Gamma = \widehat{\Gamma }\}, \end{aligned}$$

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where $\widehat{\ \cdot \ }$ denotes graded involution, which keeps the sign of the elements belonging to even subspaces. Note that the involution of a k-vector $\Gamma _{[k]}$ is given by $\widehat{\Gamma }_{[k]}=(-1)^{k}\Gamma _{[k]}$. The groups spin(p, q) and $spin_{+}(p,q)$ are given by:

$$\begin{aligned} spin(p,q)=\{a \in Cl_{p,q}^{+}; N(a)=\pm 1 \} \end{aligned}$$

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and

$$\begin{aligned} spin_{+}(p,q)=\{a \in Cl_{p,q}^{+}; N(a)= 1 \}, \end{aligned}$$

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respectively, where $N(a)=\vert a \vert ^{2}=\langle \widetilde{a} a \rangle _{0}$ is related to the norm of elements of Clifford algebra, and $\widetilde{\ \cdot \ }$ represents the reversion operator defined by $\widetilde{\Gamma }_{[k]}=(-1)^{k(k-1)/2}\Gamma _{[k]};$ $\langle \ \ \rangle _{k}:Cl_{p,q}\rightarrow \bigwedge _{k}(\mathbb {R}_{p,q})$, with $\bigwedge _{k}(\mathbb {R}_{p,q})$ denoting the exterior algebra of vector space $\mathbb {R}_{p,q}$.

The Lie Algebra of $spin_{+}(p,q)$ is the space of bivectors $\bigwedge ^{2} \mathbb {R}^{p,q}$. Let $B_{1}$ and $B_{2}$ be two bivectors. Then the commutator

$$\begin{aligned}{}[B_{1},B_{2}] \in {\bigwedge \nolimits ^{\!\!2}} \mathbb {R}^{p,q} \end{aligned}$$

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is a bivector.

B Swap Test

The swap test consists of the decomposition [34]

$$\begin{aligned} (\mathbb {C}^{d})^{\otimes 2}=Sym^{2}(\mathbb {C}^{d})\oplus \Lambda ^{2} (\mathbb {C}^{d}), \end{aligned}$$

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where $Sym^{2}(\mathbb {C}^{d})$ and $\Lambda ^{2} (\mathbb {C}^{d})$ are symmetric and antisymmetric spaces, respectively. Consider the system in the state $\vert 0, \phi , \psi \rangle$. The Hadamard gate transforms this state in

$$\begin{aligned} \frac{1}{\sqrt{2}}(\vert 0, \phi , \psi \rangle + \vert 1, \phi , \psi \rangle ). \end{aligned}$$

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Then, the controlled swap gate produces

$$\begin{aligned} \frac{1}{\sqrt{2}}(\vert 0, \psi , \phi \rangle + \vert 1, \phi , \psi \rangle ). \end{aligned}$$

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After a second application of Hadamard gate, we obtain

$$\begin{aligned} \frac{1}{2} \vert 0 \rangle (\vert \psi , \phi \rangle + \vert \phi , \psi \rangle )+\frac{1}{2} \vert 1 \rangle (\vert \psi , \phi \rangle - \vert \phi , \psi \rangle ). \end{aligned}$$

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If we perform a measurement on the first qubit, we would get

$$\begin{aligned} Pr(outcome =0)=\frac{1}{2}\left( 1+\vert \langle \psi \vert \phi \rangle \vert ^{2}\right) . \end{aligned}$$

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By repeating N times this procedure, we obtain

$$\begin{aligned} \vert \langle \psi \vert \phi \rangle \vert =\left( 1-2 \frac{\#\{outcome=0\}}{N}\right) ^{1/2}. \end{aligned}$$

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This equation allows the calculation of quantum fidelity in a quantum computer through measurements on the first qubit.

C A Simple Example of Decomposition of Quantum Gates

Unitary operators can be obtained through exponentials of anti-Hermitian operators. We will show how a simple circuit can be built from exponentials of elements of representations of Clifford algebras following the prescription obtained in the reference [25]. Thus we will build a circuit related to the unitary transformation

$$\begin{aligned} U(\theta _{1}, \theta _{2})= & {} \exp [i(\theta _{1}(\sigma _{x} \otimes \sigma _{y}) + \theta _{2} (\sigma _{y}\otimes \sigma _{x})] \\= & {} U( \theta _{1}) U( \theta _{2}), \end{aligned}$$

where

$$\begin{aligned} U(\theta _{1})= \left( \begin{array}{cccc} \cos (\theta _{1})&{} 0 &{} 0 &{} \sin (\theta _{1}) \\ 0 &{} \cos (\theta _{1}) &{} \sin (\theta _{1}) &{} 0 \\ 0 &{} -\sin (\theta _{1}) &{} \cos (\theta _{1}) &{} 0 \\ -\sin (\theta _{1}) &{} 0 &{} 0 &{} \cos (\theta _{1}) \\ \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} U(\theta _{2})= \left( \begin{array}{cccc} \cos (\theta _{2})&{} 0 &{} 0 &{} \sin (\theta _{2}) \\ 0 &{} \cos (\theta _{2}) &{} \sin (\theta _{2}) &{} 0 \\ 0 &{} -\sin (\theta _{2}) &{} \cos (\theta _{2}) &{} 0 \\ -\sin (\theta _{2}) &{} 0 &{} 0 &{} \cos (\theta _{2}) \\ \end{array} \right) . \end{aligned}$$

$U(\theta _{1})$ can be expressed as a product of two-level unitary gates

$$\begin{aligned} U(\theta _{1})=U_{1}(\theta _{1})U_{2}(\theta _{1})U_{3}(\theta _{1}), \end{aligned}$$

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where

$$\begin{aligned}{} & {} U_{1}(\theta _{1})= \left( \begin{array}{cccc} \cos (\theta _{1})&{} 0 &{} 0 &{} -\sin (\theta _{1}) \\ 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 \\ -\sin (\theta _{1}) &{} 0 &{} 0 &{} -\cos (\theta _{1}) \\ \end{array} \right) , \\{} & {} U_{2}(\theta _{2})= \left( \begin{array}{cccc} 1&{} 0 &{} 0 &{} 0 \\ 0 &{} \cos (\theta _{2}) &{} \sin (\theta _{2}) &{} 0 \\ 0 &{} \sin (\theta _{2}) &{} -\cos (\theta _{2}) &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \\ \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} U_{3}(\theta _{2})= \left( \begin{array}{cccc} 1&{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} 0 &{} -1 \\ \end{array} \right) . \end{aligned}$$

Analogously, $U(\theta _{2})$ can be expressed as a product of two-level unitary gates

$$\begin{aligned} U(\theta _{2})=U_{1}(\theta _{2})U_{2}(\theta _{2})U_{3}(\theta _{2}), \end{aligned}$$

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where

$$\begin{aligned}{} & {} U_{1}(\theta _{2})= \left( \begin{array}{cccc} \cos (\theta _{1})&{} 0 &{} 0 &{} -\sin (\theta _{1}) \\ 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 \\ -\sin (\theta _{1}) &{} 0 &{} 0 &{} -\cos (\theta _{1}) \\ \end{array} \right) , \\{} & {} U_{2}(\theta _{1})= \left( \begin{array}{cccc} 1&{} 0 &{} 0 &{} 0 \\ 0 &{} \cos (\theta _{1}) &{} -\sin (\theta _{1}) &{} 0 \\ 0 &{} -\sin (\theta _{1}) &{} -\cos (\theta ) &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \\ \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} U_{3}(\theta _{1})= \left( \begin{array}{cccc} 1&{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} 0 &{} -1 \\ \end{array} \right) , \end{aligned}$$

so that

$$\begin{aligned} {U(\theta _{1},\theta _{2})}=U(\theta _{1})U(\theta _{2})=U_{1}(\theta _{1})U_{2}(\theta _{1})U_{3}(\theta _{1})U_{1}(\theta _{2})U_{2}(\theta _{2})U_{3}(\theta _{2}). \end{aligned}$$

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The circuit is illustrated in the figure below.

where

$$\begin{aligned} \widetilde{U}_{1}(\theta _{1})= & {} \left( \begin{array}{cc} \cos (\theta _{1}) &{} -\sin (\theta _{1}) \\ -\sin (\theta _{1}) &{} -\cos (\theta _{1}) \\ \end{array} \right) , \widetilde{U}_{2}(\theta _{1})= \left( \begin{array}{cc} \cos (\theta _{1}) &{} \sin (\theta _{1}) \\ \sin (\theta _{1}) &{} -\cos (\theta _{1}) \\ \end{array} \right) ,\\ \widetilde{U}_{3}(\theta _{1})= & {} \left( \begin{array}{cc} -1 &{} 0 \\ 0 &{} -1 \\ \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} \widetilde{U}_{1}(\theta _{2})= & {} \left( \begin{array}{cc} \cos (\theta _{2}) &{} -\sin (\theta _{2}) \\ -\sin (\theta _{2} &{} -\cos (\theta _{2}) \\ \end{array} \right) , \widetilde{U}_{2}(\theta _{2})= \left( \begin{array}{cc} \cos (\theta _{2}) &{} -\sin (\theta _{2}) \\ -\sin (\theta _{2}) &{} -\cos (\theta _{2}) \\ \end{array} \right) ,\\ \widetilde{U}_{3}(\theta _{2})= & {} \left( \begin{array}{cc} -1 &{} 0 \\ 0 &{} -1 \\ \end{array} \right) . \end{aligned}$$

Notice that

$$\begin{aligned} U(\theta _{1})U(\theta _{2})\vert 00 \rangle= & {} [\cos (\theta _{1})\cos (\theta _{2})-\sin (\theta _{1})\sin (\theta _{2})] \vert 00 \rangle \nonumber \\{} & {} + [\sin (\theta _{1})\cos (\theta _{2})-\sin {\theta _{2}}\cos {\theta _{1}}] \vert 11 \rangle , \end{aligned}$$

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which is generally an entangled state. We will build states with the following notation:

$$\begin{aligned} \vert x; \theta \rangle= & {} U(\theta )U(x)\vert 00 \rangle \\= & {} [\cos (\theta )\cos (x)-\sin (\theta )\sin (x)] \vert 00 \rangle + [\sin (\theta )\cos (x)-\sin (x)\cos (\theta )] \vert 11 \rangle , \end{aligned}$$

where $\theta _{1}=\theta$ and $\theta _{2}=x$ and

$$\begin{aligned} \vert y \rangle = U(y) \vert 00 \rangle = \cos (y) \vert 00 \rangle - \sin (y) \vert 11 \rangle . \end{aligned}$$