Learning Symmetric Hamiltonian

The specific form of HL can be reduced to determine the unknowns $\vec{a}$ in a local Hamiltonian

where $h_{n}$ is the Pauli operator that acts on $k$ continuously site.

In general, the algorithms of HL from steady state often can be written in the forms of homogeneous linear equations

where $\vec{x}$ is the unknown vector for $(\vec{a},E)$, and $E$ is the eigenvalue of corresponding eigenstate, $Q$ is constraint matrix for unknowns $\vec{x}$. Eq. (2) extracts information from measurements and can be varied in different algorithms. For the EEE and the OSE method, $Q$ is deduced from energy eigen-value equations. By calculating the number of LIE in Eq. (2), we can evaluate the degree to which a steady state characterizes a Hamiltonian and thus determine the recoverability of the Hamiltonian from its steady state.

We note that the rank of matrix $Q$ equals the number of LIE for $\vec{x}$, which we denote as $K$. The number of LIE for $\vec{a}$, denoted as $M$, is equal to $K-1$. In short, we have $\operatorname{Rank}Q=K=M+1$. Eq. (2) represents a system of homogeneous linear equations, such that the parameter of Hamiltonian can only be determined up to a linear factor $\alpha$, where $\vec{a}_{recover}=\alpha\vec{a}_{true}$. As a consequence, when the value of $M$ reaches $M=N-1$, the Hamiltonian can be uniquely determined.

In this paper, we propose the first algorithm to calculate the number of LIE obtained from an eigenstate of symmetric Hamiltonian.

The degeneracy of Hamiltonian’s energy level is intimately linked to its symmetry group. Supposing that Hamiltonian $H$ is invariant under the transformation operator $S$, such that

If $H$ is invariant under transformation $S$ and $S^{\prime}$, then it follows that $H$ also remains invariant under their composition $S\cdot S^{\prime}$. Therefore, all transformations satisfying (3) constitute the symmetry group of $H$. Eq. (3) can also be written as

Now we consider an eigenstate of Hamiltonian $H$, $|\psi\rangle$ with eigenvalue $E$. Applying the transformation operator $S$ to $|\psi\rangle$, the resulting state remains an eigenstate of $H$,

Similarly, repeatedly applying $S$ on eigenstate $S|\psi\rangle$, the resulting state $S^{2}|\psi\rangle$ also remains an eigenstate of $H$.Consequently, all states $S^{n}|\psi\rangle$, where $n\in\mathbb{N}_{0}$ are eigenstates correspond to the same eigenvalue $E$.

Representation theory plays a vital role in group theory. Here, we introduce two fundamental theorems concerning representations, which reveal the relevance between properties of Hamiltonian and its symmetry group representation[30, 31].

Through the theorems above, it can be inferred that the degeneracy of the Hamiltonian levels equals the dimension of the IR of the symmetry group $G$. This insight inspires a useful concept: If we determine the symmetry group of an unknown Hamiltonian, we can uncover information about the degeneracy of its energy levels and the transformations between degenerate eigenstates.

Now we ask: Is it possible to uncover Hamiltonian’s symmetry group without knowing its specific form? The answer is yes. In fact, we can determine under which transformations the Hamiltonian ansatz specified in Eq. (1) is invariant. As long as all terms $h_{n}$ are invariant under certain transformations, the Hamiltonian $H$ will also be invariant under those transformations. In other words, the Hamiltonians under the same ansatz in Eq. (1) share the same symmetry group, resulting in consistent energy level degeneracy. This degeneracy, caused by symmetry, is referred to as intrinsic degeneracy. Mathematically, any symmetric transformation in this group is defined by

We call the group as the symmetry group $G$ on Hamiltonian learning.

Next, we will explore the impact of this symmetry on Hamiltonian Learning.

Group theory can be utilized in simplifying the solution of Schodinger equations. In this subsection, we will introduce the application of group theory in the diagonalization of secular determination.

Supposing that Hamiltonian $H$ having group $G$ as its symmetry group. Starting with the Schodinger equation

Expanding $H$ in the basis $\{h_{n}\}$, $H=\sum_{n}a_{n}h_{n}$,

Choosing a set of symmetrized basis $\{|\phi_{im}^{p}\rangle\}$ for eigenstate $|\psi\rangle$, whose transformation is in accordance with IR of group $G$. $|\phi_{im}^{p}\rangle$ is the $m$-th base of the $p$-th IR of group $G$, and $i$ is the index of the repeating IR. Expanding the eigenstate $|\psi\rangle$ in the basis $\{|\phi_{im}^{p}\rangle\}$, and Eq. (2) can be written as

Taking the inner product of $|\phi_{jl}^{q}\rangle$ with the equation above

To solve the tedious equations above, we introduce the theorem of matrix elements for the invariant operator[32]:

According to Theorem 3, Eq. (10) becomes

which holds for any $\{q,j,l\}$. Noting that coefficient $c_{il}^{q}/c_{jl}^{q}$ is independent of $l$. Eq. (12) can be written as

which we see as a linear equation for the unknowns $\{c_{il}^{q}/c_{jl}^{q}\}_{i}$. Since$\langle\phi_{jl}^{q}|H|\phi_{il}^{q}\rangle$ is independent of $l$, $\{c_{il}^{q}/c_{jl}^{q}\}=\alpha_{ij}$ is also independent of $l$. Consequently, Eq. (12) is rewritten as

which reveals that once $q,j$ are fixed, Eqs. (12) are equivalent for all $l$. Therefore Eqs. (12) can be divided into series of subsystem of linear equations, which are composed of repeating unit of equations. Noting that Eqs. (14) are complex equations for generic Hamiltonian. Here we consider a symmetric Hamiltonian whose basis $\{h_{n}\}$ can be written as a real matrix, making Eqs. (14) real equations.

The number of LIE in each unit depends on the degeneracy of IR, and repeating times of units depends on the dimension of IR. An example of such structure of linear equations with $8$ base functions which contains one 4-dimensional IR and two degenerate 2-dimensional IR is shown in Fig. 1.

So far, we have shown the repeating structure of system of linear equations and given the possible number of LIE obtained from a eigenstate of Hamiltonian. However, while considering a specific eigenstate, some of the equations in Eqs. (12) could be trivial [33].

Eigenstate $|\psi\rangle$ lies in a subspace spanned by the basis of a IR of $G$, that is to say, every eigenstate has a corresponding $q$. Eqs. (12) are nontrivial only when the value of $q$ is conform to $|\psi\rangle$. Consequently, the number of LIE obtained from an eigenstate is

where $m_{q}$ is the degeneracy of IR $q$ in the Hilbert space.

Considering an 1-dimensional spin chain of $L$ sites, with a spin $\frac{1}{2}$ particle at each site. The Hamiltonian of this model can be written as

where

$\sigma^{i}$ is the $i$’th Pauli matrix, and $\mathbb{I}$ is the $2\times 2$ identity matrix.

XXX Hamiltonian has time reversal and SU(2) symmetry. Its symmetry group is composed of

where $T$ is the time reversal operator, and $SU(2)=\{R\}$ and $\{TR\}$ is a set of anti-unitary operators. The representation of non-unitary group $M$ is denoted as co-representation.

To determine whether time-reversal symmetry will cause a doubling of energy degeneracy, it’s instructive to study the co-representation $D\Gamma$ of the symmetry group $M$ [34]. $D\Gamma$ can be derived from an IR $\mathbf{\Gamma}_{R}$ of subgroup $\{R\}$, whose specific forms are illustrated in appendix. A. The dimension of $D\Gamma$ is doubled compared to the basis of $\mathbf{\Gamma}_{R}$. Without the time reversal symmetry, the energy level degeneracy of Hamiltonian is determined by the dimension of the IR of the group $\{R\}$. If the co-representation $D\Gamma$ is reducible, the degeneracy will not be doubled. However, if $D\Gamma$ is irreducible, then the degeneracy will be doubled.

The co-representation $D\Gamma$ of group $M$ is reducible and the degeneracy will not be doubled (see appendix. B). Therefore the dimension of IR of $M$ and SU(2) group is consistent. TABLE. 1 shows the decomposition of the SU(2) symmetry of the $L$-site XXX Hamiltonian into a series of IRs.

Up to now, we have calculated the dimensions of IR of the symmetry group $M$. Next, we will leverage the properties of these IRs to determine how much information can be extracted from an eigenstate of the XXX Hamiltonian. In other words, we will explore the number of linearly independent equations that can be deduced from a single eigenstate.

The number of LIE obtained from an eigenstate corresponds to the degeneracy of its associated irreducible subspace. For example, when the chain length $L=3$, the Hilbert space is segmented into one subspace of dimension 4, and two subspaces of dimension 2, resulting in energy level degeneracies of 4, 2 and 2, respectively. For eigenstates lie in the 4-dimensional subspace, the degeneracy of the corresponding IR is 1. Thus we can only derive a single LIE from it. Conversely, eigenstates situated in the 2-dimensional subspaces have a degeneracy of 2, allowing us to obtain two LIEs. This elucidates that determining which subspace an eigenstate lies in enables us to ascertain how much information about the Hamiltonian it encodes.

To determine the subspace occupancy of an eigenstate within the formalism of angular momentum, one often utilizes ladder operators:

Upon applying $S^{\pm}$ to an eigenstate $|m\rangle$ of $S^{z}$ with eigenvalue $m$, the resulting state $S^{\pm}|m\rangle$ remains an eigenstate of $S^{z}$ with eigenvalue $m\pm 1$.

In the context of a finite spin chain, the irreducible subspace is finite. Starting from an eigenstate of the XXX Hamiltonian, one can apply the ladder operators a finite number of times until it reaches 0. Within the irreducible subspace labeled by $j$, the maximum eigenvalue of $S^{z}$ is $+j$, the minimum is $-j$. This can be expressed as

The space that the ladder operators acting on, is exactly an irreducible subspace with dimension $2j+1$. All vectors within this subspace can be connected through the action of the ladder operators. Therefore, when given an eigenstate, by applying ladder operators, we can traverse its subspace and thereby determine where it lies in.

We perform simulations on recovering the Hamiltonian $H_{XXX}$ from a single eigenstate. The results for chain length $L=3,4$ are shown in TABLE. 2 and 3, respectively. We present the recovery result for all eigenstate, where the symbol ‘O’ indicates the success of HL and ‘X’ signifies a failure. We also record the energy level degeneracy ($g$), the dimension of IR ($l$), the degeneracy of IR ($m$), and the number of LIEs for unknowns $\vec{a}$ of the Hamiltonian ($M$).

When the chain length $L=3$, the Hamiltonian has two unknown parameters. Its Hilbert space is partitioned into one subspace with dimension 4 and two subspaces with dimension 2. When (1) $g=2$, $m=2$, $M=1$, the Hamiltonian can be uniquely recovered, (2) $g=4$, $m=1$, $M=0$, and the Hamiltonian cannot be uniquely recovered. These results are consistent with Equation (15), which relates the number of LIEs $M$ to the IR degeneracy $m$.

When $L=3$ and $g=4$, the dimension of IR is 4, labeled by quantum number $j=\frac{3}{2}$. We can express the XXX Hamiltonian $H_{XXX}$ using the ladder operators $\sigma_{l}^{\pm}=\sigma_{l}^{x}\pm i\sigma_{l}^{y}$ acting on the $l$-th site:

From this expression, we can see that the states $|\uparrow\uparrow\uparrow\rangle$ and $|\downarrow\downarrow\downarrow\rangle$ are eigenstates of $H_{XXX}$, with all spins either up or down in the $z$-direction. Additionally, since the operator $S^{-}$ commutes with $H_{XXX}$, i.e., $[S^{-},H_{XXX}]=0$, applying $S^{-}$ to the eigenstate $|\uparrow\uparrow\uparrow\rangle$ results in another eigenstate of $H_{XXX}$.

repeatedly acting $S^{-}$, until it reaches 0

From this example, we note that for any 3-site $H_{XXX}$ Hamiltonian, the following states are degenerate eigenstates, regardless of the specific model parameters: $|\uparrow\uparrow\uparrow\rangle$, $|\downarrow\downarrow\downarrow\rangle$, $\frac{1}{\sqrt{3}}(|\downarrow\downarrow\uparrow\rangle+|\downarrow\uparrow\downarrow\rangle+|\uparrow\downarrow\downarrow\rangle)$ and $\frac{1}{\sqrt{3}}(|\downarrow\downarrow\downarrow\rangle+|\downarrow\downarrow\downarrow\rangle+|\downarrow\downarrow\downarrow\rangle)$. These four of degenerate eigenstates can be connected through the action of the ladder operators, but they do not provide any information about the Hamiltonian parameters. Therefore, the number of LIE obtained from these eigenstate is 0.

When $L=4$, for all eigenstates, the relationship between the degeneracy of the IR $m$ and number of LIE $M$ is given by $M=m-1$, as shown in TABLE. 3. When $M=N-1$, the Hamiltonian $H_{XXX}$ can be uniquely recovered. Here, we point out that, under the case when the value of $M$ exceeds the value of Hamiltonian unknowns $N$. The number of LIE is given by

Now we consider 1-dimensional spin-$\frac{1}{2}$ chain with XXZ Hamiltonian give by

The system described by XXZ Hamiltonian has U(1) and time reversal symmetry, whose symmetry group can be written as

where $\{U\}$ and $T$ represent U(1) group and time reversal operator, respectively.

The representation $D\Gamma$ of the non-unitary group $P$ is denoted as co-representation. The degeneracy of Hamiltonian energy level is determined by the reducibility of $D\Gamma$. In appendix. C, we show that when $S^{z}=0$, $D\Gamma$ is reducible, and the corresponding energy level will not be doubled; when $S^{z}\neq 0$, $D\Gamma$ is irreducible, and the corresponding energy level will be doubled.

TABLE.  4 shows the decomposition of the symmetry group of the XXZ Hamiltonian into a series of IRs, whose degeneracy is dependent on the value of $|S^{z}|$.This means that for a given eigenstate of the XXZ Hamiltonian, the number of linearly independent eigenstates (LIE) is determined by the value of $|S^{z}|$. Simulations of HL are performed on $H_{XXZ}$, the number of LIE as a function of $S^{z}$ for chain lengths ranging from 2 to 7 are presented in Table 5.

When the degeneracy $m$ of the IR is smaller than the unknowns $N$, the number of LIE $M$ equals $m-1$. In this case, the Hamiltonian cannot be fully recovered. When $m$ is equal to or larger than $N$, $M$ equals to $N-1$, it succeed to recover Hamiltonian. To sum up, the number of LIE is determined by the degeneracy of IR of the corresponding eigenstate or the number of unknowns for the first and the second situation: