Fermi Machine
— Quantum Many-Body Solver Derived from Correspondence
between Noninteracting and Strongly Correlated Fermions

Fractionalization of electrons in strongly correlated electron systems has been proposed in several different phenomena. One dimensional itinerant electron systems exhibit the separation of spin and charge degrees of freedom [1, 2]. Models of polyacetylene accommodate spin and charge solitons with fractionalized charge as elementary excitations [3]. Fractional quantum Hall states under the stong magnetic fields [4] as well as under the emergent chiral symmetry breaking [5] show the fractionalization of electronic charge and fractional quantizations. Slave bosons and fermions were proposed to approximately describe the 2D systems as models of cuprate high-$T_{c}$ superconductors [6, 7]. Recently, the fractionalization of an electron into two or multiple fermion components has been examined and has successfully accounted for otherwise puzzling spectroscopic experimental results of the cuprates [8, 9, 10, 11, 12, 13].

Meanwhile, numerical method to construct ground state wave functions of strongly interacting quantum many-body systems is one of the grand challenges in physics. Various algorithms such as a variety of auxiliary-field quantum Monte Carlo [14, 15, 16], variational Monte Carlo [17, 18, 19, 20, 21, 22], density matrix renormalization group [23], tensor network [24, 25, 26], density matrix embedding [27] and dynamical mean-field theory [28, 29] have been proposed. Recently, neural network with Boltzmann machine has established an accurate way to approximate ground states of quantum spin systems [30] as well as itinerant fermion systems [31]. Vision transformer, another architecture of the neural network, has also shown the state-of-the-art accuracy for quantum spin models [32, 33]. We note that these neural networks are constructed by introducing hidden variables that are described by classical degrees of freedom.

In this paper, we discuss mapping of the Hubbard model (or in more general, interacting lattice fermions) to noninteracting multi-component fermion models in the ground states that materializes the fractionalization, from which we propose a quantum machine learning algorithm aiming at an efficient quantum many-body solver. Here, we introduce a fermion system as the hidden part coupled to the physical system, instead of the classical one such as the Ising spins in the previous neural networks.

In Secs. I and II, the equivalence between the Hubbard model and the two-component fermion model (TCFM) is shown for the atomic limit and for the 2-site systems, respectively. Based on this equivalence, we propose a quantum machine learning algorithm in Sec. III, where the Hubbard model is approximated by a deep-layer representation of noninteracting multi-component fermion models. The couplings between the layers are represented by the hybridization of fermions belonging to neighboring layers. This method is viewed as an extension of restricted Boltzmann-machine representation of the neural quantum states proposed by Carleo and Troyer [30] to a quantum neural network, where the Ising hidden variable in the Boltzmann machine is replaced by the fermionic operators that allows hybridization to generate quantum entanglement. Furthermore, it is also regarded as an embodiment of the electron fractionalization [9, 10], which has succeeded in solving a number of experimental puzzles in the cuprate high-$T_{c}$ superconductors [8, 12, 10, 13]. This quantum neural network can also be viewed as a tractable realization of the ground states by a systematic expansion to take account of correlation effects similarly to the formulation by the equation of motion method or equivalently the continued fraction expansion of the Green’s function for the excited states [34, 35, 36]. The present method does not seem to have difficulty of exponentially increasing terms encountered in the conventional equation of motion method.

In this section, the result shown in Appendix A of Ref. \citenImada2019 is briefly summarized for the self-contained presentation. We consider the Hubbard $U$ term

$\displaystyle{\cal H}_{U}$

$\displaystyle=$

$\displaystyle Un_{\uparrow}n_{\downarrow},$

(1)

with $n_{\sigma}=c^{\dagger}_{\sigma}c_{\sigma}$ for the creation (annihilation) operators $c_{\sigma}^{\dagger}$ ($c_{\sigma}$) of the spin $\sigma$, and introduce an auxiliary fermion

$\displaystyle\tilde{d}_{\sigma}=c_{\sigma}(1-2n_{-\sigma}),$

(2)

together with

$\displaystyle\tilde{c}_{\sigma}=c_{\sigma}.$

(3)

Equation (1) can then be rewritten in a form of the TCFM as

	$\displaystyle{\cal H}_{\rm TCFM}$	$\displaystyle=$	$\displaystyle{\cal H}^{(\tilde{c})}+{\cal H}^{(\tilde{d})}+{\cal H}^{(\tilde{c}\tilde{d})},$		(4)
	$\displaystyle{\cal H}^{(\tilde{c})}$	$\displaystyle=$	$\displaystyle\mu_{\tilde{c}}\tilde{c}_{\sigma}^{\dagger}\tilde{c}_{\sigma},$		(5)
	$\displaystyle{\cal H}^{(\tilde{d})}$	$\displaystyle=$	$\displaystyle\mu_{\tilde{d}}\tilde{d}_{\sigma}^{\dagger}\tilde{d}_{\sigma},$		(6)
	$\displaystyle{\cal H}^{(\tilde{c}\tilde{d})}$	$\displaystyle=$	$\displaystyle\Lambda(\tilde{c}_{\sigma}^{\dagger}\tilde{d}_{\sigma}+{\rm H.c}),$		(7)

with the mapping

$\displaystyle\mu_{\tilde{c}}=\mu_{\tilde{d}}=-\Lambda=\frac{U}{2}.$

(8)

For the derivation of Eq.(4) from Eq. (1), see below.

Note first that $\tilde{d}$ and $\tilde{c}$ satisfy the exact anticommutation relation in the ground state average,

	$\displaystyle\langle\tilde{c}_{\sigma}\tilde{c}_{\sigma}^{\dagger}+\tilde{c}_{\sigma}^{\dagger}\tilde{c}_{\sigma}\rangle=1,$		(9)
	$\displaystyle\langle\tilde{d}_{\sigma}\tilde{d}_{\sigma}^{\dagger}+\tilde{d}_{\sigma}^{\dagger}\tilde{d}_{\sigma}\rangle=1,$		(10)
	$\displaystyle\langle\tilde{c}_{\sigma}\tilde{d}_{\sigma}^{\dagger}+\tilde{d}_{\sigma}^{\dagger}\tilde{c}_{\sigma}\rangle=0,$		(11)

where $\langle\cdots\rangle$ is defined by

$\displaystyle\langle\cdots\rangle=\frac{\langle\Phi_{0\uparrow}^{(1)}|\cdots|\Phi_{0\uparrow}^{(1)}\rangle+\langle\Phi_{0\downarrow}^{(1)}|\cdots|\Phi_{0\downarrow}^{(1)}\rangle}{\langle\Phi_{0\uparrow}^{(1)}|\Phi_{0\uparrow}^{(1)}\rangle+\langle\Phi_{0\downarrow}^{(1)}|\Phi_{0\downarrow}^{(1)}\rangle},$

(12)

because of the ground state degeneracy of Kramers doublet, when the one-particle ground state is described as we prove later as

$$|\Phi_{0\sigma}^{(1)}\rangle=(\alpha_{c}\tilde{c}_{\sigma}^{\dagger}+\alpha_{d}\tilde{d}_{\sigma}^{\dagger})|0\rangle$$

(13)

with $\sigma$ being either $\uparrow$ or $\downarrow$. (Note that the one-particle state must be degenerate between $|\Phi_{0\uparrow}^{(1)}\rangle$ and $|\Phi_{0\downarrow}^{(1)}\rangle$.) In this sense, $\tilde{c}$ and $\tilde{d}$ behave as orthogonal fermions for single-particle excitations from the ground state, which is the degenerate ensemble of $|\Phi_{0\uparrow}^{(1)}\rangle$ and $|\Phi_{0\downarrow}^{(1)}\rangle$.

By diagonalizing Eq.(4) for the case $\mu_{\tilde{c}}=\mu_{\tilde{d}}$, one obtains for the spin $\sigma$ part

	$\displaystyle{\cal H}_{\rm DTCFM}$	$\displaystyle=$	$\displaystyle\mu_{a}a_{\sigma}^{\dagger}a_{\sigma}+\mu_{b}b_{\sigma}^{\dagger}b_{\sigma},$		(14)
	$\displaystyle\mu_{b}$	$\displaystyle=$	$\displaystyle\frac{1}{2}(\mu_{\tilde{c}}+\mu_{\tilde{d}})+\Lambda,$		(15)
	$\displaystyle\mu_{a}$	$\displaystyle=$	$\displaystyle\frac{1}{2}(\mu_{\tilde{c}}+\mu_{\tilde{d}})-\Lambda$		(16)

with

	$\displaystyle b_{\sigma}$	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{2}}(\tilde{c}_{\sigma}+\tilde{d}_{\sigma})=\sqrt{2}c_{\sigma}(1-n_{-\sigma}),$		(17)
	$\displaystyle a_{\sigma}$	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{2}}(\tilde{c}_{\sigma}-\tilde{d}_{\sigma})=\sqrt{2}c_{\sigma}n_{-\sigma}.$		(18)

The bonding and antibonding states are given by Eqs. (17) and (18), respectively, which are nothing but the operators for the lower and upper Hubbard levels, respectively, in the Hubbard model whose averaged energies are given by $E=0$ and $U$. Namely, aside from the normalization factor, ${b}_{\sigma}^{\dagger}$ creates an electron with the spin $\sigma$ when it is not occupied by the opposite spin electron (namely it fills the lower Hubbard state in the corresponding Hubbard model) as is apparent from the last equation of Eq.(17). On the other hand, $a_{\sigma}^{\dagger}$, creates an electron with the spin $\sigma$, when the opposite spin electron already exists, namely it creates a doublon at the upper Hubbard state in the Hubbard model. It helps an intuitive interpretation of the correspondence between the Hubbard model and the TCFM. With the choice Eq.(8), $\mu_{a}=0$ and $\mu_{b}=U$ are obtained and the Hubbard gap is reproduced. In addition, the ground state $b_{\sigma}^{\dagger}|0\rangle$ indeed satisfies the form of Eq.(13). Then the mapping between Eqs.(1) and (4) becomes exact. In this correspondence, the Mott gap in the Hubbard model is interpreted as the hybridization gap in the TCFM.

In general, one can show that the single-particle Green’s function for the TCFM is given by

	$\displaystyle G_{\tilde{c}_{\sigma},\tilde{c}_{\sigma}^{\dagger}}(\omega)$	$\displaystyle=$	$\displaystyle\frac{1}{\omega-\mu_{\tilde{c}}-\frac{\Lambda^{2}}{\omega-\mu_{\tilde{d}}}},$
	$\displaystyle G_{\tilde{c}_{\sigma},\tilde{d}_{\sigma}^{\dagger}}(\omega)$	$\displaystyle=$	$\displaystyle G_{\tilde{d}_{\sigma},\tilde{c}_{\sigma}^{\dagger}}(\omega)=\frac{-\Lambda}{(\omega-\mu_{\tilde{c}})(\omega-\mu_{\tilde{d}})-\Lambda^{2}},$
	$\displaystyle G_{\tilde{d}_{\sigma},\tilde{d}_{\sigma}^{\dagger}}(\omega)$	$\displaystyle=$	$\displaystyle\frac{1}{\omega-\mu_{\tilde{d}}-\frac{\Lambda^{2}}{\omega-\mu_{\tilde{c}}}}.$		(19)

Then from Eq.(8), we obtain

	$\displaystyle G_{\tilde{c}_{\sigma},\tilde{c}_{\sigma}^{\dagger}}(\omega)$	$\displaystyle=$	$\displaystyle\frac{1}{\omega-\frac{U}{2}-\frac{\frac{U^{2}}{4}}{\omega-\frac{U}{2}}}$		(20)
		$\displaystyle=$	$\displaystyle\frac{1}{2}\left[\frac{1}{\omega}+\frac{1}{\omega-U}\right].$		(21)

This is equivalent to the Green’s function of the atomic Hubbard, Eq.(1). Then the self-energy has the correct form as well:

$\displaystyle{\Sigma}(\omega)$

$\displaystyle=$

$\displaystyle\frac{\frac{U^{2}}{4}}{\omega-\frac{U}{2}}.$

(22)

In this way, exact correspondence is established between the TCFM (4) and the Hubbard model (1) in the atomic limit for the half-filled ground state as well as for single-particle excitations from it. Namely, the full Hilbert space of the Hubbard model in the atomic limit is equivalent to that of the TCFM.

In this section, the mapping of the atomic limit already established and summarized in the last section is extended and a mapping between the 2-site Hubbard model and the 2-site TCFM is shown.

The 2-site Hubbard Hamiltonian reads

	$\displaystyle{\cal H}$	$\displaystyle=$	$\displaystyle{\cal H}_{t}+{\cal H}_{U},$		(23)
	$\displaystyle{\cal H}_{t}$	$\displaystyle=$	$\displaystyle\sum_{\sigma}[-t(c_{1\sigma}^{\dagger}c_{2\sigma}+c_{2\sigma}^{\dagger}c_{1\sigma})+\mu\sum_{i=1,2}n_{i,\sigma}],$		(24)
	$\displaystyle{\cal H}_{U}$	$\displaystyle=$	$\displaystyle U\sum_{i=1,2}n_{i\uparrow}n_{i\downarrow}.$		(25)

with $n_{i,\sigma}=c^{\dagger}_{i,\sigma}c_{i,\sigma}$. We take $\mu=0$, because in the canonical ensemble, spatially uniform chemical potential does not change physical properties except for a trivial energy shift.

We first analyze the half-filled case with one $\uparrow$ and one $\downarrow$ electron and use the basis of the full Hilbert space expanded by $|\uparrow,\downarrow\rangle,|\downarrow,\uparrow\rangle,|\uparrow\downarrow,0\rangle,|0,\uparrow\downarrow\rangle$ in the notation $|n_{1\uparrow}n_{1\downarrow},n_{2\uparrow}n_{2\downarrow}\rangle$, where $n_{i\sigma}$ is the number of spin $\sigma$ fermion at the site $i$ and is denoted as $\uparrow$ if $n_{i\uparrow}=1$ and $\downarrow$ if $n_{i\downarrow}=1$ while $0$ if $n_{i\uparrow}=n_{i\downarrow}=0$ at the site $i$. For instance, $|\uparrow,\downarrow\rangle$ represents $c_{1\uparrow}^{\dagger}c_{2\downarrow}^{\dagger}|0\rangle$ and $|\downarrow,\uparrow\rangle=c_{2\uparrow}^{\dagger}c_{1\downarrow}^{\dagger}|0\rangle$ such that up-spin creation operators are ordered in the left-hand side and spatial sites are ordered in the site-number order from 1 to 2 for the same spin. Here $|0\rangle$ is the vacuum. In this notation, the Hamiltonian matrix is written as

$\displaystyle{\cal H}=\left(\begin{array}[]{c|cccc}&|\uparrow,\downarrow\rangle&|\downarrow,\uparrow\rangle&|\uparrow\downarrow,0\rangle&|0,\uparrow\downarrow\rangle\\ \hline\cr\langle\uparrow,\downarrow|&0&0&-t&-t\\ \langle\downarrow,\uparrow|&0&0&-t&-t\\ \langle\uparrow\downarrow,0|&-t&-t&U&0\\ \langle 0,\uparrow\downarrow|&-t&-t&0&U\end{array}\right).$

(31)

Here and hereafter, we have explicitly written the choice of the basis as the row and column vectors to make the definition of the components of the $4\times 4$ matrix clearer.

The eigenvalues are

	$\displaystyle E_{0}$	$\displaystyle=$	$\displaystyle\frac{U}{2}Q,$
	$\displaystyle E_{1}$	$\displaystyle=$	$\displaystyle 0,$
	$\displaystyle E_{2}$	$\displaystyle=$	$\displaystyle U,$
	$\displaystyle E_{3}$	$\displaystyle=$	$\displaystyle\frac{U}{2}P,$		(32)

where $P=1+\sqrt{1+R^{2}}$, $Q=1-\sqrt{1+R^{2}}$ and $R=4t/U$. The normalized eigenfunction of the ground state with the energy $E_{0}$ is given by

$$(R,R,-Q,-Q)/\sqrt{2(R^{2}+Q^{2})},$$

(33)

which is the coefficients in the basis of $(|\uparrow,\downarrow\rangle,|\downarrow,\uparrow\rangle,|\uparrow\downarrow,0\rangle,|0,\uparrow\downarrow\rangle)$. Namely, the normalized ground state is

$$\frac{1}{\sqrt{2(R^{2}+Q^{2})}}(R|\uparrow,\downarrow\rangle+R|\downarrow,\uparrow\rangle-Q|\uparrow\downarrow,0\rangle-Q|0,\uparrow\downarrow\rangle).$$

(34)

Now we introduce the same auxiliary fermions Eqs.(2) and (3) as the atomic limit. By using $\tilde{c}$ and $\tilde{d}$, the Hubbard Hamiltonian (23) can be mapped to a noninteracting 2-site TCFM as

	$\displaystyle{\cal H}_{\rm TCFM}$	$\displaystyle=$	$\displaystyle{\cal H}^{(\tilde{c})}+{\cal H}^{(\tilde{d})}+{\cal H}^{(\tilde{c}\tilde{d})},$		(35)
	$\displaystyle{\cal H}^{(\tilde{c})}$	$\displaystyle=$	$\displaystyle\sum_{\sigma}[-t_{\tilde{c}}(\tilde{c}_{1,\sigma}^{\dagger}\tilde{c}_{2,\sigma}+{\rm H.c.})+\mu_{\tilde{c}}\sum_{i=1,2}n_{\tilde{c},i,\sigma}],\ \ \ \$		(36)
	$\displaystyle{\cal H}^{(\tilde{d})}$	$\displaystyle=$	$\displaystyle\sum_{\sigma}[-t_{\tilde{d}}(\tilde{d}_{1,\sigma}^{\dagger}\tilde{d}_{2,\sigma}+{\rm H.c.})+\mu_{\tilde{d}}\sum_{i=1,2}n_{\tilde{d},i,\sigma}],\ \ \ \$		(37)
	$\displaystyle{\cal H}^{(\tilde{c}\tilde{d})}$	$\displaystyle=$	$\displaystyle\Lambda\sum_{\sigma}\sum_{i=1,2}(\tilde{c}_{i,\sigma}^{\dagger}\tilde{d}_{i,\sigma}+{\rm H.c}),$		(38)

where $n_{\tilde{c},i,\sigma}=\tilde{c}_{i,\sigma}^{\dagger}\tilde{c}_{i,\sigma}$ and $n_{\tilde{d},i,\sigma}=\tilde{d}_{i,\sigma}^{\dagger}\tilde{d}_{i,\sigma}$ again with the same mapping Eq. (8) together with

$\displaystyle\tilde{t}_{\tilde{c}}=t,\tilde{t}_{\tilde{d}}=-t.$

(39)

The Hilbert space of this TCFM wave function may be expanded in the basis of $(\tilde{c}_{1,\sigma}^{\dagger}+\tilde{d}_{1,\sigma}^{\dagger})/\sqrt{2},(\tilde{c}_{2,\sigma}^{\dagger}+\tilde{d}_{2,\sigma}^{\dagger})/\sqrt{2},(\tilde{c}_{1,\sigma}^{\dagger}-\tilde{d}_{1,\sigma}^{\dagger})/\sqrt{2},(\tilde{c}_{2,\sigma}^{\dagger}-\tilde{d}_{2,\sigma}^{\dagger})/\sqrt{2}))|0\rangle,$ where $|0\rangle$ is the vacuum of $\tilde{c}$ and $\tilde{d}$. Since the spin degeneracy exists and the correlation between opposite spin fermions does not exist, we use an abbreviated notation of this basis as $\tilde{b}_{i}=(\tilde{c}_{i}+\tilde{d}_{i})/\sqrt{2},\tilde{a}_{i}=(\tilde{c}_{i}-\tilde{d}_{i})/\sqrt{2}$ for $i=1,2$. The Hamiltonian matrix in this representation is

$\displaystyle{\cal H}_{\rm TCFM}=\left(\begin{array}[]{c|cccc}&\tilde{b}_{1}&\tilde{b}_{2}&\tilde{a}_{1}&\tilde{a}_{2}\\ \hline\cr\tilde{b}_{1}^{\dagger}&0&0&-t&-t\\ \tilde{b}_{2}^{\dagger}&0&0&-t&-t\\ \tilde{a}_{1}^{\dagger}&-t&-t&U&0\\ \tilde{a}_{2}^{\dagger}&-t&-t&0&U\end{array}\right),$

(45)

for both of up and down spin sectors, which is the same as Eq.(31). Then the eigenvalues and eigenfunctions are of course the same as Eqs.(32) and (34), respectively. The ground-state wave function is given by

	$\displaystyle|\Phi_{0}^{\rm TCFM}\rangle$	$\displaystyle=$	$\displaystyle\frac{1}{2(R^{2}+Q^{2})}(R(\tilde{b}_{1,\uparrow}^{\dagger}+\tilde{b}_{2,\uparrow}^{\dagger})+Q(\tilde{a}_{1,\uparrow}^{\dagger}+\tilde{a}_{2,\uparrow}^{\dagger}))$		(46)
		$\displaystyle\times$	$\displaystyle(R(\tilde{b}_{1,\downarrow}^{\dagger}+\tilde{b}_{2,\downarrow}^{\dagger})+Q(\tilde{a}_{1,\downarrow}^{\dagger}+\tilde{a}_{2,\downarrow}^{\dagger}))|0\rangle,$

which has the symmetries of spin singlet and spatial parity even. The first excited state is given by

	$\displaystyle|\Phi_{1}^{\rm TCFM}\rangle$	$\displaystyle=$	$\displaystyle\frac{\sqrt{2}}{4(R^{2}+Q^{2})}$
		$\displaystyle\times$	$\displaystyle((R(\tilde{b}_{1,\uparrow}^{\dagger}-\tilde{b}_{2,\uparrow}^{\dagger})+Q(\tilde{a}_{1,\uparrow}^{\dagger}-\tilde{a}_{2,\uparrow}^{\dagger}))$
			$\displaystyle\times(R(\tilde{b}_{1,\downarrow}^{\dagger}+\tilde{b}_{2,\downarrow}^{\dagger})+Q(\tilde{a}_{1,\uparrow}^{\dagger}+\tilde{a}_{2,\uparrow}^{\dagger}))$
		$\displaystyle-$	$\displaystyle(R(\tilde{b}_{1,\uparrow}^{\dagger}+\tilde{b}_{2,\uparrow}^{\dagger})+Q(\tilde{a}_{1,\uparrow}^{\dagger}+\tilde{a}_{2,\uparrow}^{\dagger}))$
			$\displaystyle\times(R(\tilde{b}_{1,\downarrow}^{\dagger}-\tilde{b}_{2,\downarrow}^{\dagger})+Q(\tilde{a}_{1,\downarrow}^{\dagger}-\tilde{a}_{2,\downarrow}^{\dagger})))|0\rangle,$

which is spin triplet and spatial parity odd. Other two higher excited states can be given similarly.

With this correspondence, the TCFM ground state (Eq.(46)) is exactly mapped to the Hubbard terminology as

	$\displaystyle|\Phi_{0}^{\rm Hub}\rangle$	$\displaystyle=$	$\displaystyle\frac{1}{2(R^{2}+Q^{2})}(R(\tilde{b}_{1,\uparrow}^{\dagger}\tilde{b}_{2,\downarrow}^{\dagger}+\tilde{b}_{2,\uparrow}^{\dagger}\tilde{b}_{1,\downarrow}^{\dagger})$		(48)
			$\displaystyle+Q(\tilde{a}_{1,\uparrow}^{\dagger}\tilde{b}_{1,\downarrow}^{\dagger}+\tilde{a}_{2,\uparrow}^{\dagger}\tilde{b}_{2,\downarrow}^{\dagger}))|0\rangle$
		$\displaystyle\Leftrightarrow$	$\displaystyle{\rm Eq.(29)}.$

Here, in the Hubbard model, we used the relations $(\tilde{c}_{i,\sigma}^{\dagger}-\tilde{d}_{i,\sigma}^{\dagger})|0\rangle=0$, $(\tilde{c}_{i,\sigma}^{\dagger}+\tilde{d}_{i,\sigma}^{\dagger})(\tilde{c}_{i,-\sigma}^{\dagger}+\tilde{d}_{i,-\sigma}^{\dagger})|0\rangle=0$ and $(\tilde{c}_{1i,\sigma}^{\dagger}-\tilde{d}_{i,\sigma}^{\dagger})(\tilde{c}_{j,-\sigma}^{\dagger}+\tilde{d}_{j,-\sigma}^{\dagger})|0\rangle=0$ for $i\neq j$, because $\tilde{c}_{i,\sigma}^{\dagger}-\tilde{d}_{i,\sigma}^{\dagger}$ creates a doublon at the $i$th site singly occupied by a spin $-\sigma$ electron and $\tilde{c}_{i,\sigma}^{\dagger}+\tilde{d}_{i,\sigma}^{\dagger}$ creates an electron at the empty $i$th site. Mapping for the excited states is similarly shown. Therefore, the exact mapping of not only the ground state but also the whole structure of the spectra between the TCFM and Hubbard model is proven for the two-site system as well.

Here, we remark on the spin degeneracy in the TCFM. Because of the spin degeneracy, the total ground state of the TCFM with one up and one down spin sector is given by the product state $|\Phi_{0}\rangle=|\Phi_{0\uparrow}\rangle|\Phi_{0\downarrow}\rangle$. Then the total ground-state energy with one up and one down spin fermions is twice of the ground state energy $E_{0}$ in Eq.(32). However, as in the single-site problem, the ground state average in the TCFM $\langle\cdots\rangle$ should be taken as Eq.(12) and the factor 2 is canceled.

For the filling doped with one hole or one electron from the half filling has also trivially the same mapping. For instance, the energy of a hole doped ground state is

$\displaystyle|\Phi_{0}^{\rm TCFM}\rangle$

$\displaystyle=$

$\displaystyle\frac{1}{2}(\tilde{b}_{1,\uparrow}^{\dagger}+\tilde{b}_{2,\uparrow}^{\dagger})|0\rangle,$

(49)

which has the energy $E_{0}=0$ and is the same as the one-hole doped Hubbard model.

We have shown that the Hubbard model has an exact mapping to the TCFM for one- and two-site problems and the ground-state wave function of the Hubbard model can be constructed from the ground state of the TCFM. It demonstrates that the TCFM is able to describe the energy scales of the Mott gap separating the lower and upper Hubbard bands scaled by $E_{2}-E_{0}\propto U$ and the singlet-triplet gap ($E_{1}-E_{0}$ and $E_{3}-E_{2}$) associated with the superexchange interaction $J$, which scales to $4t^{2}/U$ in the strong coupling limit. Up to the 2 sites, the exact ground and excited states of the Hubbard model are represented by the corresponding ground and excited states of the TCFM, respectively by optimizing the variational parameters given by Eqs.(8) and (39), namely, $\mu_{\tilde{c}},\mu_{\tilde{d}},\Lambda,t_{\tilde{c}}$ and $t_{\tilde{d}}$.

This suggests a potential to represent $N_{s}$-site Hubbard models defined by

	$\displaystyle{\cal H}$	$\displaystyle=$	$\displaystyle{\cal H}_{t}+{\cal H}_{U},$		(50)
	$\displaystyle{\cal H}_{t}$	$\displaystyle=$	$\displaystyle\sum_{\langle i,j\rangle,\sigma}[-t(c_{i\sigma}^{\dagger}c_{j\sigma}+{\rm H.c.}+\mu\sum_{i}^{N_{s}}n_{i,\sigma}],$		(51)
	$\displaystyle{\cal H}_{U}$	$\displaystyle=$	$\displaystyle U\sum_{i}^{N_{s}}n_{i\uparrow}n_{i\downarrow}$		(52)

for any system size $N_{s}$ by a quantum neural network along this line. We will discuss the representability of the wave function later.

In principle, by introducing the self-energies of the visible electron $c$ through the effect of the hidden fermions $d$, it was shown that the spectral function and energy spectra of any interacting system can be represented formally but exactly by the hierarchy of the self-energy structure, which represents the energy eigenvalues of the Hamiltonian by poles and zeros of Green’s function through the continued fraction expansion [34, 35, 36] as

	$\displaystyle G(q,\omega)$	$\displaystyle=$	$\displaystyle\frac{1}{\omega-\epsilon_{c}(q)-\Sigma_{1}(q,\omega)},$		(53)
	$\displaystyle\Sigma_{1}(q,\omega)$	$\displaystyle=$	$\displaystyle\frac{\eta_{1}(q)}{\omega-\epsilon_{1}(q)-\Sigma_{2}(q,\omega)},$
	$\displaystyle\Sigma_{2}(q,\omega)$	$\displaystyle=$	$\displaystyle\frac{\eta_{2}(q)}{\omega-\epsilon_{2}(q)-\Sigma_{3}(q,\omega)}.$
	$\displaystyle\cdots$

Here we assume the translational symmetry of the Hamiltonian parameters to allow the momentum representation. This can be achieved by formally tracing the equation of motion for the $c$ operator in the Heisenberg representation as

$\displaystyle i\hbar\frac{dc(t)}{dt}$

$\displaystyle=$

$\displaystyle[c,{\cal H}],$

(54)

for any Hamiltonian $\cal H$ and $[A,B]\equiv AB-BA$.

Therefore, one can expect that the ground state of the Hubbard model (or any interacting lattice fermions) can be mapped to an optimized non-interacting multi-component fermion model (MCFM) represented by

	$\displaystyle{\cal H}_{\rm MCFM}$	$\displaystyle=$	$\displaystyle{\cal H}^{(\tilde{c})}+{\cal H}^{(\tilde{c}\tilde{d})}+{\cal H}^{(\tilde{d})},$		(55)
	$\displaystyle{\cal H}^{(\tilde{c})}$	$\displaystyle=$	$\displaystyle\sum_{\langle i,j\rangle,\sigma}(-t_{\tilde{c},i,j})(\tilde{c}_{i,\sigma}^{\dagger}\tilde{c}_{j,\sigma}+{\rm H.c.})+\sum_{i,\sigma}\mu_{\tilde{c},i}n_{\tilde{c},i,\sigma},$
	$\displaystyle{\cal H}^{(\tilde{c}\tilde{d})}$	$\displaystyle=$	$\displaystyle\Lambda\sum_{\sigma}\sum_{i}(\tilde{c}_{i,\sigma}^{\dagger}\tilde{d}_{i,\sigma}^{(1)}+{\rm H.c}),$		(57)
	$\displaystyle{\cal H}^{(\tilde{d})}$	$\displaystyle=$	$\displaystyle\sum_{m=1}^{M}[\sum_{\langle i,j\rangle\sigma}(-t_{\tilde{d},i,j}^{(m)})(\tilde{d}_{i,\sigma}^{(m)\dagger}\tilde{d}_{j,\sigma}^{(m)}+{\rm H.c.})$		(58)
			$\displaystyle+\sum_{i,\sigma}\mu_{\tilde{d},i}^{(m)}\tilde{d}_{i,\sigma}^{(m)\dagger}\tilde{d}_{i,\sigma}^{(m)}]$
		$\displaystyle+$	$\displaystyle\sum_{m=1}^{M-1}\Lambda^{(m)}\sum_{i,\sigma}(\tilde{d}_{i,\sigma}^{(m)\dagger}\tilde{d}_{i,\sigma}^{(m+1)}+{\rm H.c.}),$

as is illustrated in Fig. 1, because the Green’s function of Eq. (55) is given by the hierarchical structure given by Eq.(53) with the correspondence

	$\displaystyle\epsilon_{c}(q)$	$\displaystyle=$	$\displaystyle\frac{1}{N_{s}}\sum_{\langle\ell,j\rangle}e^{iq(\ell-j)}(-t_{\tilde{c},\ell,j})+\mu_{\tilde{c}},$
	$\displaystyle\eta_{1}(q)$	$\displaystyle=$	$\displaystyle\Lambda,$
	$\displaystyle\epsilon_{m}(q)$	$\displaystyle=$	$\displaystyle\frac{1}{N_{s}}\sum_{\langle\ell,j\rangle}e^{iq\ell-j)}(-t_{\tilde{d},\ell,j}^{(m)})+\mu_{\tilde{d},i}^{(m)},$
	$\displaystyle\eta_{m}(q)$	$\displaystyle=$	$\displaystyle\Lambda^{(m)},$		(59)

if the chemical potential is spatially uniform.

The algorithm of the present quantum neural network to represent the ground state of the Hubbard model is the following:

• •

  We diagonalize the noninteracting MCFM Hamiltonian Eq. (55) and obtain the ground state by filling the low-energy fermions up to the Fermi level for a given set of parameters in Eqs. (LABEL:eq:MCFM_Hc)-(58).

• •

  Next we replace the hidden fermions $\tilde{d}$ by using the rule Eq.(2) together with similar correspondence such as $\tilde{d}_{i,\sigma}^{(m)}=\tilde{d}_{i,\sigma}^{(m-1)}(1-2n_{\tilde{d}^{(m-1)},i,-\sigma})$ where $n_{\tilde{d}^{(m)},i,-\sigma}=\tilde{d}_{i,-\sigma}^{(m)\dagger}\tilde{d}_{i,-\sigma}^{(m)}$.

• •

  Then by using the relations such as those below Eq. (48) used to derive Eq. (48) and its extension to deep-layer hidden operators, one can represent the wave function for the Hubbard model $|\Phi_{0}^{\rm Hub}\rangle$ as in the derivation of Eq. (48), from which one can easily calculate the matrix elements of the Hubbard model and the energy expectation value of the Hubbard model by

  $\displaystyle\langle E_{0}\rangle=\langle{\cal H}\rangle=\frac{\langle\Phi_{0}^{\rm Hub}|{\cal H}|\Phi_{0}^{\rm Hub}\rangle}{\langle\Phi_{0}^{\rm Hub}|\Phi_{0}^{\rm Hub}\rangle}$

  (60)

  using Monte Carlo sampling of the real space configuration of $c$ fermion occupation. Here $\cal H$ is given by Eqs. (50)-(52). Note that ${\cal H}_{t}|\Phi_{0}^{\rm Hub}\rangle$ is not a single Slater determinant even when $|\Phi_{0}^{\rm Hub}\rangle$ is so. An easy way to obtain the average is

  $\displaystyle\langle E_{0t}\rangle=\langle{\cal H}_{t}\rangle=\lim_{\tau\rightarrow 0}\frac{1}{\tau}\log\left[\frac{\langle\Phi_{0}^{\rm Hub}|\exp[\tau{\cal H}_{t}]|\Phi_{0}^{\rm Hub}\rangle}{\langle\Phi_{0}^{\rm Hub}|\Phi_{0}^{\rm Hub}\rangle}\right]$

  (61)

  because $\exp[\tau{\cal H}_{t}]|\Psi\rangle$ remains a single Slater determinant when $|\Psi\rangle$ is a single Slater determinant [16].

• •

  Finally, by using the variational principle, the parameters in the MCFM Hamiltonian Eqs. (55)-(58) are regarded as the variational parameters and are optimized to lower the energy Eq. (60) by following the variational principle.

There exist a few useful relations to simplify the MCFM state:

	$\displaystyle\tilde{d}_{i,\sigma}^{(m)}$	$\displaystyle=$	$\displaystyle\tilde{d}_{i,\sigma}^{(m-1)}(1-2n_{\tilde{d}^{(m-1)},i,-\sigma})=\cdots$		(62)
		$\displaystyle=$	$\displaystyle c_{i,\sigma}(1-2n_{c_{i,-\sigma}})^{m},$
	$\displaystyle n_{\tilde{d}_{i,\sigma}^{(m)}}$	$\displaystyle=$	$\displaystyle n_{\tilde{d}^{(m-1)}_{i,\sigma}}=\cdots=n_{c_{i,\sigma}}$		(63)

The ground state of the MCFM can be represented generally as

	$\displaystyle|\Psi_{0}^{\rm MCFM}\rangle$	$\displaystyle=$	$\displaystyle\begin{pmatrix}|\Psi_{0\ \uparrow}^{\rm MCFM}\rangle&0\\ 0&|\Psi_{0\ \downarrow}^{\rm MCFM}\rangle\end{pmatrix},$		(64)
	$\displaystyle|\Psi_{0\ \sigma}^{\rm MCFM}\rangle$	$\displaystyle=$	$\displaystyle\prod_{q}^{k_{\rm F}}|\Psi_{\sigma}^{\rm MCFM}(q)\rangle,$		(65)
	$\displaystyle|\Psi_{\sigma}^{\rm MCFM}(q)\rangle$	$\displaystyle=$	$\displaystyle[\alpha_{0,q}\tilde{c}_{q,\sigma}^{\dagger}+\sum_{m=1}^{M}\alpha_{m,q}\tilde{d}_{m,q,\sigma}^{\dagger}]|0\rangle,$		(66)

where $q$ is the momentum. Here, in the ground state of the MCFM, fermions occupy up to the fermi momentum $k_{\rm F}$. Since the MCFM is diagonal in the momentum space, the ground-state wave function in the $k$ space is trivial if the parameters in Eqs. (LABEL:eq:MCFM_Hc)-(58) are given. The coefficient $\alpha_{m,q}$ ($m=0,\cdots M$) is determined from the eigenvector of the ground state, which can be easily calculated numerically.

This MCFM ground state is mapped to the variational form of the ground state of the Hubbard model as

$\displaystyle|\Psi_{0}^{\rm Hub}\rangle$

$\displaystyle=$

$\displaystyle\begin{pmatrix}|\Psi_{0\ \uparrow}^{\rm Hub}\rangle&0\\ 0&|\Psi_{0\ \downarrow}^{\rm Hub}\rangle\end{pmatrix},$

(67)

$$|\Psi_{0\ \sigma}^{\rm Hub}\rangle=\prod_{q}^{k_{\rm F}}[\sum_{m=0}^{M}\alpha_{m,q}\sum_{\ell}\frac{e^{iq\ell}}{\sqrt{N_{s}}}c_{\ell,\sigma}^{\dagger}(1-2n_{\ell,-\sigma})^{m}]|0\rangle.$$

(68)

It can be rewritten as

$\displaystyle|\Psi_{0\ \sigma}^{\rm Hub}\rangle$

$\displaystyle=$

$\displaystyle\prod_{q}^{k_{\rm F}}|\Psi_{\sigma}^{\rm Hub}(q)\rangle,$

(69)

$$|\Psi_{\sigma}^{\rm Hub}(q)\rangle=[\sum_{m=0}^{M}\alpha_{m,q}\sum_{\ell}\frac{e^{iq\ell}}{\sqrt{N_{s}}}c_{\ell,\sigma}^{\dagger}(-)^{mn_{\ell,-\sigma}}]|0\rangle.$$

(70)

It should be noted that $|\Psi_{\sigma}^{\rm Hub}(q)\rangle$ contains electron operators $c^{\dagger}_{p,\sigma}$ with not only the momentum $q$ but also all the momentum $p$ in the Brillouin zone in the momentum representation, which is a consequence of the interference with the spin $-\sigma$ electron at the $\ell$-th site in Eq.(70) reflecting the scattering of $\sigma$ and $-\sigma$ electrons at the $\ell$-th site in the original Hubbard model. Indeed, intuitively, the minus factor for the odd $m$ in Eq. (70) represents the reduction of the weight of $c_{\ell,\sigma}^{\dagger}$ electron if $-\sigma$ electron exists at the $\ell$-th site to punish the double occupation leading to the entanglement of spin up and down electrons. For instance, in the large $U$ limit, the corresponding $\Lambda\rightarrow-\infty$ makes the perfect exclusion of the double occupation through Eq. (70). The representability of this wave function is discussed in Appendix.

To numerically optimize the variational parameters in the MCFM (namely the parameters in Eqs. (55)-(58)), we need to estimate the ground-state energy using the obtained wave function through Eq. (60), by the average of the inserted sample $|x\rangle$ as

	$\displaystyle E_{0}$	$\displaystyle=$	$\displaystyle\langle{\cal H}\rangle=\sum_{x}\rho(x)F(x),$		(71)
	$\displaystyle\rho(x)$	$\displaystyle=$	$\displaystyle|\langle x|\Psi_{0}^{\rm Hub}\rangle|^{2}/\langle\Psi_{0}^{\rm Hub}|\Psi_{0}^{\rm Hub}\rangle,$		(72)
	$\displaystyle F(x)$	$\displaystyle=$	$\displaystyle\sum_{x^{\prime}}\frac{\langle\Psi_{0}^{\rm Hub}|x^{\prime}\rangle}{\langle\Psi_{0}^{\rm Hub}|x\rangle}\langle x^{\prime}|{\cal H}|x\rangle.$		(73)

Here, $\langle x^{\prime}|{\cal H}_{t}|x\rangle$ can be calculated in the same way as Eq. (61).

Hereafter, let us consider a $N_{s}$-site system with $N/2$ up spin and down spin electrons each. (Extension to the case of an arbitrary number of up and down spin electrons, $N_{\uparrow}$, and $N_{\downarrow}$, respectively, is easy and straightforward.) The Monte Carlo sampling can be performed by taking $\rho(x)$ as the probability to generate samples by following the importance sampling and the simple average of $F(x)$ over sampling gives us the estimate of $E_{0}$. Here, $|x\rangle$ is a sample of real space configuration (simple product state), such as $(0,\uparrow,\uparrow,\uparrow\downarrow,\downarrow,0,\cdots)$, namely

$\displaystyle|x\rangle$

$\displaystyle=$

$\displaystyle\prod_{\ell}^{N/2}c_{\ell,\uparrow}^{\dagger}\prod_{\ell^{\prime}}^{N/2}c_{\ell^{\prime},\downarrow}^{\dagger}|0\rangle,$

(74)

where the sets of the real space coordinate $\{\ell\}$ and $\{\ell^{\prime}\}$ specify the positions of the up and down spin electrons, respectively in the given sample. Since $\cal H$ is a sparse matrix by assuming the system with the short-ranged hoppings and interactions, and has nonzero matrix element only in their ranges, the summation over $|x^{\prime}\rangle$ can be explicitly taken for each $|x\rangle$.

To enhance the representability of the wave function, one can optionally introduce the “Fermi distribution factor” $f$ on top of Eq.(68) (or Eqs.(69) and (70)) similarly to the finite-temperature Boltzmann factor in the classical Boltzmann machine. Namely, in this scheme, $|\Psi_{0\ \sigma}^{\rm Hub}\rangle$ is modified and replaced by

	$\displaystyle|\Psi_{0\ \sigma}^{\rm Hub}\rangle$	$\displaystyle=$	$\displaystyle\prod_{q}f(\beta,q)|\Psi_{\sigma}^{\rm Hub}(q)\rangle,$		(75)
	$\displaystyle f(\beta,q)$	$\displaystyle=$	$\displaystyle\frac{1}{1+e^{\beta{\cal E}(q)}},$		(76)
	$\displaystyle{\cal E}(q)$	$\displaystyle=$	$\displaystyle\sum_{\sigma}\frac{\langle\Psi_{\sigma}^{\rm MCFM}(q)|{\cal H}^{\rm MCFM}|\Psi_{\sigma}^{\rm MCFM}(q)\rangle}{\langle\Psi_{\sigma}^{\rm MCFM}(q)|\Psi_{\sigma}^{\rm MCFM}(q)\rangle},$
	$\displaystyle|\Psi_{\sigma}^{\rm MCFM}(q)\rangle$	$\displaystyle=$	$\displaystyle\prod_{\sigma}[\alpha_{0,q}\tilde{c}_{q,\sigma}^{\dagger}+\sum_{m=1}^{M}\alpha_{m,q}\tilde{d}_{m,q,\sigma}^{\dagger}]|0\rangle,$		(77)

where $\beta$ is an additional variational parameter, $|\Psi_{\sigma}^{\rm MCFM}(q)\rangle$ is the eigenstate of Eq.(55) with the momentum $q$ and $|\Psi_{\sigma}^{\rm Hub}(q)\rangle$ is the same as Eq.(70). The product with respect to $q$ is over the full Brillouin zone for Eq. (75). If we take $\beta\rightarrow\infty$, it is reduced to the original algorithm using Eqs.(69) and (70). Therefore this Fermi machine extension certainly enhances the representability with a strategy conceptually similar to the Boltzmann machine [30, 31] but by using a substantially different quantum algorithm.

In the practical calculation, it is useful to represent in the matrix form: $|\Psi_{0\ \sigma}^{\rm Hub}\rangle$ is represented by a $N_{s}\times N/2$ matrix ${\cal P}_{\sigma}$, where from Eq.(70), the matrix element is given by

$\displaystyle({\cal P}_{\sigma})_{\ell,k}$

$\displaystyle=$

$\displaystyle\sum_{m=0}^{M}\alpha_{m,q(k,\sigma)}e^{iq(k,\sigma)\ell}(-)^{mn_{\ell,-\sigma}}.$

(78)

Here, $q(k,\sigma)$ is the momentum of the $k$th occupied electron with spin $\sigma$. The matrix representation of the sample $|x\rangle\rightarrow{\cal X}$ is

$\displaystyle({\cal X}_{\sigma})_{\ell,\ell^{\prime}}=\sum_{n}^{N_{s}}\sum_{i}^{N/2}\delta_{\ell,n}\delta_{\ell^{\prime},i}$

(79)

if the $i$th spin-${\sigma}$ electron is at the site $n$ in the given sample.

Since $\langle x^{\prime}|{\cal H}|x\rangle$ is easily calculated in Eq.(73), here we discuss how one can calculate $\langle\Psi_{0}^{\rm Hub}|x^{\prime}\rangle$ and $\langle x|\Psi_{0}^{\rm Hub}\rangle$ in the matrix representation. By defining $2N_{s}\times N$ matrices

$\displaystyle{\cal P}$

$\displaystyle=$

$\displaystyle\begin{pmatrix}{\cal P}_{\uparrow}&0\\ 0&{\cal P}_{\downarrow}\end{pmatrix}$

(80)

and

$\displaystyle{\cal X}$

$\displaystyle=$

$\displaystyle\begin{pmatrix}{\cal X}_{\uparrow}&0\\ 0&{\cal X}_{\downarrow}\end{pmatrix},$

(81)

we can calculate $\langle x|\Psi_{0}^{\rm Hub}\rangle$ from the determinant of the matrix product ${}^{T}{\cal X}{\cal P}$ represented by a $N\times N$ matrix, where ${}^{T}{\cal X}$ is the transpose of $\cal X$. In Eq.(78), we need to be careful about the dependence on $n_{\ell,-\sigma}$. Since $n_{\ell,-\sigma}=\sum_{k}({\cal X}_{-\sigma})_{\ell,k}$, ${\cal P}_{\sigma}$ depends on ${\cal X}_{-\sigma}$, but it can be easily numerically calculated, because practically, $|\Psi_{0}^{\rm Hub}\rangle$ appears only in the inner product with $\langle x|$.