# Learning Symmetric Hamiltonian

Jing Zhou  Institute of Physics, Beijing National Laboratory for Condensed
Matter Physics,
Chinese Academy of Sciences, Beijing 100190, China  School of Physical
Sciences, University of Chinese Academy of Sciences, Beijing 100049, China  D.
L. Zhou<EMAIL_ADDRESS>Institute of Physics, Beijing National Laboratory
for Condensed Matter Physics,
Chinese Academy of Sciences, Beijing 100190, China  School of Physical
Sciences, University of Chinese Academy of Sciences, Beijing 100049, China

###### Abstract

Hamiltonian Learning is a process of recovering system Hamiltonian from
measurements, which is a fundamental problem in quantum information
processing. In this study, we investigate the problem of learning the
symmetric Hamiltonian from its eigenstate. Inspired by the application of
group theory in block diagonal secular determination, we have derived a method
to determine the number of linearly independent equations about the
Hamiltonian unknowns obtained from an eigenstate. This number corresponds to
the degeneracy of the associated irreducible representation of the Hamiltonian
symmetry group. To illustrate our approach, we examine the XXX Hamiltonian and
the XXZ Hamiltonian. We first determine the Hamiltonian symmetry group, then
work out the decomposition of irreducible representation, which serves as
foundation for analyzing the uniqueness of recovered Hamiltonian. Our
numerical findings consistently align with our theoretical analysis.

††preprint: APS/123-QED

##  I introduction

Hamiltonian Learning (HL) refers to recovering the Hamiltonian of a quantum
system, typically from observations of its steady state or dynamics. HL plays
a pivotal role in quantum physics. In recent years, there has been a rapid
development in quantum simulators and related quantum computing devices,
including controllable trapped ion [1, 2, 3, 4] and superconducting quantum
circuits [5, 6]. The exponential growth of noise with the size of quantum
hardware has made HL an essential tool for verification and benchmarking [7,
8, 9, 10, 11, 12]. In the filed of condensed matter physics, HL aids in
understanding the behavior of quantum material and may be utilized to explore
new material systems [13, 14, 15, 16, 17, 18, 19].

One of the challenge in HL lies in determining when we can uniquely recover a
Hamiltonian from its steady state. This question was initially raised and
partially addressed by Qi and Ranard , and they present that the most of the
generic local Hamiltonian can be recovered from measurements of a single
eigenstate [20]. From the perspective of dimension counting, the space of
$k$-local Hamiltonian scales linearly with the system size $n$, whereas the
space of states scales exponentially with $n$. As a result, it becomes
feasible to recover local Hamiltonian from its single eigenstate. Qi and
Ranard introduced the concept of “correlation matrix”, which successfully
recovered local Hamiltonian from a single eigenstate.

Following Qi’s question, various methods for recovering the Hamiltonian from a
steady state have been proposed [21, 22, 23]. Most of these methods aim to
simplify the process of measuring the steady state, while ignoring the case
when recovering the Hamiltonian from its steady state is not feasible. In
fact, there exists critical chain length $L_{c}$ in spin system with length
$L$ for generic local Hamiltonian. When $L$ is smaller than $L_{c}$, we cannot
recover Hamiltonian form its steady state. The value of $L_{c}$ depends on the
rank of the steady state and Hamiltonian model. In previous work, we utilize
the energy eigen-value equations (EEE) to analytically determine the value of
$L_{c}$, successfully applying this approach to models of generic local
Hamiltonians [24]. Subsequently, we extended this investigation to degenerate
steady states using the orthogonal space equations (OSE) method [25].

So far, we focused on the recoverability of a generic local Hamiltonian from
its steady state, while many Hamiltonians arouse interests exhibit symmetries
[26, 27]. Such as the Heisenberg spin chain models [28]. Symmetric
Hamiltonians inherently entail fewer unknowns compared to generic ones.
However, it doesn’t necessarily make them easier to recover. According to
Noether’s theorem, there is a direct relationship between the Hamiltonian
symmetry and the conserved quantities in the associated physical system. It
directly reduces the degrees of freedom of associated eigenstates [29]. In
other words, the space dimension of eigenstate is reduced, which decreases the
number of linearly independent equations (LIE) that can be derived from an
eigenstate.

To determine the number of the LIE obtained from an eigenstate of symmetric
Hamiltonian, we introduce the block diagonalization of secular determination,
which is a commonly used technique in group theory. We expand the EEE in a
symmetrized basis, and analyze the repeating structure of linearly equations.
We find that the number of LIE equals the degeneracy of the associated
irreducible representation (IR) of the Hamiltonian symmetry group. In other
words, if we know the symmetry group of a Hamiltonian, we can infer how much
information is contained in its eigenstate. To verify our theoretical
derivation, we perform simulations on the XXX and XXZ Hamiltonians. Given only
the Hamiltonian’s symmetry group and its IRs, we are able to successfully
predict the number of LIEs obtained from the eigenstate.

##  II framework

###  II.1 Problem setting

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

$H=\sum_{n=1}^{N}a_{n}h_{n},$ (1)

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

$Q\vec{x}=0,$ (2)

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.

###  II.2 The symmetry group of 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

$SHS^{-1}=H.$ (3)

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

$HS=HS.$ (4)

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 $H$,

$H(S|\psi\rangle)=S(H|\psi\rangle)=E(S|\psi\rangle).$ (5)

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].

######  Theorem 1.

The eigenfunctions of the Hamiltonian $H$ with the same eigenvalues constitute
a set of basis functions for the symmetry group $G$ of the Hamiltonian.

######  Theorem 2.

In the absence of accidental degeneracy, a group of eigenstates that are
transformed according to an IR of group $G$, belong to the same energy level.

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

$Sh_{n}S^{-1}=h_{n},\;\forall n.$ (6)

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

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

###  II.3 The diagonalization of the Schodinger equation

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

$H|\psi\rangle=E|\psi\rangle$ (7)

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

$\sum_{n}a_{n}h_{n}|\psi\rangle=E|\psi\rangle.$ (8)

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

$\sum_{n}a_{n}h_{n}\sum_{i}\sum_{p}\sum_{m}c_{im}^{p}|\phi_{im}^{p}\rangle=E\sum_{i}\sum_{p}\sum_{m}c_{im}^{p}|\phi_{im}^{p}\rangle.$
(9)

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

$\displaystyle\sum_{n}a_{n}\sum_{i}\sum_{p}\sum_{m}c_{im}^{p}\langle\phi_{jl}^{q}|h_{n}|\phi_{im}^{p}\rangle$
(10)
$\displaystyle-E\sum_{i}\sum_{p}\sum_{m}c_{im}^{p}\langle\phi_{jl}^{q}|\phi_{im}^{p}\rangle=0.$

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

######  Theorem 3 ( Theorem of matrix elements for the invariant operator).

If $G$ is the symmetry group of Hamiltonian $H=\sum_{n}a_{n}h_{n}$, and
$\\{\phi_{jl}^{q}\\}$ is the basis of IR for $G$, then the matrix elements of
$h_{n}$ satisfy that

$\langle\phi_{jl}^{q}|h_{n}|\phi_{im}^{p}\rangle=\delta_{pq}\delta_{lm}\langle\phi_{j\mu}^{p}|h_{n}|\phi_{i\mu}^{p}\rangle,$
(11)

where $\langle\phi_{j\mu}^{p}|h_{n}|\phi_{i\mu}^{p}\rangle$ is independent of
$\mu$.

According to Theorem 3, Eq. (10) becomes

$\sum_{n}a_{n}\sum_{i}c_{il}^{q}\langle\phi_{jl}^{q}|h_{n}|\phi_{il}^{q}\rangle-
Ec_{jl}^{q}=0,$ (12)

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

$\sum_{i}\frac{c_{il}^{q}}{c_{jl}^{q}}\langle\phi_{jl}^{q}|H|\phi_{il}^{q}\rangle-E=0,$
(13)

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

$\sum_{n}a_{n}\sum_{i}\alpha_{ij}\langle\phi_{jl}^{q}|h_{n}|\phi_{il}^{q}\rangle-E=0,$
(14)

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.

Figure 1: Structure of system of linear equations for basis
$\\{|\phi_{jl}^{q}\rangle\\}$ with 8 base functions, containing a
$4$-dimensional IR $D^{1}$ and two degenerate $2$-dimensional IR $D^{2}$.

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].

######  Theorem 4.

Eigen-states with the same eigen-value furnish an IR of Hamiltonian symmetry
group.

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

$M=m_{q}-1,$ (15)

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

##  III examples

###  III.1 The XXX Spin Chain

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

$H_{XXX}=\sum_{l=1}^{L-1}J_{l}(\sigma_{l}^{x}\sigma_{l+1}^{x}+\sigma_{l}^{y}\sigma_{l+1}^{y}+\sigma_{l}^{z}\sigma_{l+1}^{z}),$
(16)

where

$\sigma_{l}^{i}=\mathbb{I}\otimes\cdots\otimes\mathbb{I}\otimes\sigma^{i}\otimes\mathbb{I}\cdots\otimes\mathbb{I},$
(17)

$\sigma^{i}$ 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

$M=\\{R\\}+\\{TR\\},$ (18)

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.

$L$ | Decomposition of representation
---|---
2 | $\mathbf{2}^{\otimes 2}=\mathbf{3}\oplus\mathbf{1}$
3 | $\mathbf{2}^{\otimes 3}=\mathbf{4}\oplus(2\times\mathbf{2})$
4 | $\mathbf{2}^{\otimes 4}=\mathbf{5}\oplus(3\times\mathbf{3})\oplus(2\times\mathbf{1})$
5 | $\mathbf{2}^{\otimes 5}=\mathbf{6}\oplus(4\times\mathbf{4})\oplus(5\times\mathbf{2})$
6 | $\mathbf{2}^{\otimes 6}=\mathbf{7}\oplus(5\times\mathbf{5})\oplus(9\times\mathbf{3})\oplus(5\times\mathbf{1})$
7 | $\mathbf{2}^{\otimes 7}=\mathbf{8}\oplus(6\times\mathbf{6})\oplus(14\times\mathbf{4})\oplus(14\times\mathbf{2})$
Table 1: Dimensions of SU(2) IRs of $L$-site XXX Hamiltonian.

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:

$S^{\pm}=S^{x}\pm iS^{y}.$ (19)

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

$S^{\pm}|\pm j\rangle=0.$ (20)

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 ($M$).

eigenstate | $|\psi_{0}\rangle$ | $|\psi_{1}\rangle$ | $|\psi_{2}\rangle$ | $|\psi_{3}\rangle$ | $|\psi_{4}\rangle$ | $|\psi_{5}\rangle$ | $|\psi_{6}\rangle$ | $|\psi_{7}\rangle$
---|---|---|---|---|---|---|---|---
$g$ | 2 | 2 | 2 | 2 | 4 | 4 | 4 | 4
$l$ | 2 | 2 | 2 | 2 | 4 | 4 | 4 | 4
$m$ | 2 | 2 | 2 | 2 | 1 | 1 | 1 | 1
$M$ | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0
success? | O | O | O | O | X | X | X | X
Table 2: Recovery of 3-site XXX Hamiltonian from its eigenstate. eigenstate |  $g$ |  $l$ |  $m$ |  $M$ |  success?
---|---|---|---|---|---
$|\psi_{0}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{1}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{2}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{3}\rangle$ | 1 | 1 | 2 | 1 | X
$|\psi_{4}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{5}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{6}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{7}\rangle$ | 5 | 5 | 1 | 0 | X
$|\psi_{8}\rangle$ | 5 | 5 | 1 | 0 | X
$|\psi_{9}\rangle$ | 5 | 5 | 1 | 0 | X
$|\psi_{10}\rangle$ | 5 | 5 | 1 | 0 | X
$|\psi_{11}\rangle$ | 5 | 5 | 1 | 0 | X
$|\psi_{12}\rangle$ | 1 | 1 | 2 | 1 | X
$|\psi_{13}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{14}\rangle$ | 3 | 3 | 3 | 2 | O
$|\psi_{15}\rangle$ | 3 | 3 | 3 | 2 | O
Table 3:  Recovery of 4-site XXX Hamiltonian from its eigenstate.

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:

$H_{XXX}=\sum_{l=1}^{L-1}J_{l}(\sigma_{l}^{z}\sigma_{l+1}^{z}+\sigma_{l}^{+}\sigma_{l}^{-}+\sigma_{l}^{-}\sigma_{l}^{+}).$
(21)

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 $H_{XXX}$.

$S^{-}|\uparrow\uparrow\uparrow\rangle=\frac{1}{\sqrt{3}}(|\downarrow\uparrow\uparrow\rangle+|\uparrow\downarrow\uparrow\rangle+|\uparrow\uparrow\downarrow\rangle),$
(22)

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

$(S^{-})^{2}|\uparrow\uparrow\uparrow\rangle=\frac{1}{\sqrt{3}}(|\downarrow\downarrow\uparrow\rangle+|\downarrow\uparrow\downarrow\rangle+|\uparrow\downarrow\downarrow\rangle),$
(23a)

$(S^{-})^{3}|\uparrow\uparrow\uparrow\rangle=\frac{1}{\sqrt{3}}(|\downarrow\downarrow\downarrow\rangle+|\downarrow\downarrow\downarrow\rangle+|\downarrow\downarrow\downarrow\rangle),$
(23b)

$(S^{-})^{4}|\uparrow\uparrow\uparrow\rangle=0.$ (23c)

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

$M=\min\\{m-1,N-1\\}.$ (24)

###  III.2 The XXZ Spin Chain

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

$H_{XXZ}=\sum_{l=1}^{L-1}J_{l}^{z}\sigma_{l}^{z}\sigma_{l+1}^{z}+J_{l}^{xy}(\sigma_{l}^{x}\sigma_{l+1}^{x}+\sigma_{l}^{y}\sigma_{l+1}^{y}).$
(25)

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

$P=\\{U\\}+\\{TU\\},$ (26)

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.

$L$ | Decomposition of representation
---|---
2 | $\mathbf{2}^{\otimes 2}=\mathbf{1}^{0,1}\oplus\mathbf{1}^{0,-1}\oplus\mathbf{2}^{\pm 1}$
3 | $\mathbf{2}^{\otimes 3}=\mathbf{2}^{\pm\frac{3}{2}}\oplus(3\times\mathbf{2}^{\pm\frac{1}{2}})$
4 | $\mathbf{2}^{\otimes 4}=(3\times\mathbf{1}^{0,1})\oplus(3\times\mathbf{1}^{0,-1})\oplus(4\times\mathbf{2}^{\pm 1})\oplus\mathbf{2}^{\pm 2}$
5 | $\mathbf{2}^{\otimes 5}=\mathbf{2}^{\pm\frac{5}{2}}\oplus(5\times\mathbf{2}^{\pm\frac{3}{2}})\oplus(10\times\mathbf{2}^{\pm\frac{1}{2}})$
6 | $\mathbf{2}^{\otimes 6}=(10\times\mathbf{1}^{0,1})\oplus(10\times\mathbf{1}^{0,-1})\oplus(15\times\mathbf{2}^{\pm 1})\oplus 6\times\mathbf{2}^{\pm 2}\oplus\mathbf{2}^{\pm 3}$
7 | $\mathbf{2}^{\otimes 7}=\mathbf{2}^{\pm\frac{7}{2}}\oplus(7\times\mathbf{2}^{\pm\frac{5}{2}})\oplus(21\times\mathbf{2}^{\pm\frac{3}{2}})\oplus(35\times\mathbf{2}^{\pm\frac{1}{2}})$
Table 4:  Decomposition of the U(1)+T symmetry of the $L$-site XXZ Heisenberg
model into a series of fundamental, IRs.

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.

$L$ $S^{z}$ | 0 | $\pm\frac{1}{2}$ | $\pm 1$ | $\pm\frac{3}{2}$ | $\pm 2$ | $\pm\frac{5}{2}$ | $\pm 3$ | $\pm\frac{7}{2}$ | N
---|---|---|---|---|---|---|---|---|---
2 | 0 | / | 0 | / | / | / | / | / | 2
3 | / | 2 | / | 0 | / | / | / | / | 4
4 | 2 | / | 3 | / | 0 | / | / | / | 6
5 | / | 7 | / | 4 | / | 0 | / | / | 8
6 | 9 | / | 9 | / | 5 | / | 0 | / | 10
7 | / | 11 | / | 11 | / | 6 | / | 0 | 12
Table 5:  Recovery of XXZ Hamiltonian from its eigenstate. The number of LIE
as a function of $S^{z}$ for chain lengths ranging from 2 to 7.

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:

$M=\min\\{m-1,N-1\\}.$ (27)

##  IV conclusions

In this paper, we investigate the problem of when a local Hamiltonian can be
uniquely recovered from its eigenstate, which extends the previous work to the
realm of symmetric Hamiltonian. Our work shows that the number of LIEs derived
from an eigenstate corresponds to the degeneracy of IR of Hamiltonian symmetry
group. An open question remains regarding the potential extension of our
results to a broader range of Hamiltonians, such as those featuring all-to-all
two-body interactions, or those of indistinguishable particles. Additionally,
we ask whether our findings are applicable to systems possessing spatial
symmetries, such as crystallographic point groups that are intricately linked
to the symmetries of crystals. Addressing these inquiries is pivotal for the
practical implementation of HL in the exploration of novel quantum materials.

##  V acknowledgement

This work is supported by National Key Research and Development Program of
China (Grant No. 2021YFA0718302 and No. 2021YFA1402104), National Natural
Science Foundation of China (Grants No. 12075310).

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##  Appendix A Co-representation of group containing anti-unitary elements

Considering a non-unitary group $Q$ that composes half of the anti-unitary
elements

$Q=\\{S\\}+\\{AS\\},$ (28)

where $\\{S\\}$ is a unitary subgroup and $A$ is an anti-unitary operator. The
representation of such group is called co-representation. The co-
representation can be educed from an IR $\mathbf{\Gamma}_{S}$ of unitary
subgroup $\\{S\\}$

$\mathbf{D}(S)=\begin{pmatrix}\mathbf{\Gamma}(S)&0\\\
0&\mathbf{\Gamma}(A^{-1}SA)^{*}\end{pmatrix},S\in\\{S\\},$ (29a)
$\mathbf{D}(B)=\begin{pmatrix}0&\mathbf{\Gamma}(BA)\\\
\mathbf{\Gamma}(A^{-1}B)^{*}&0\end{pmatrix},B\in\\{AS\\}.$ (29b)

This co-representation is denoted as $D\Gamma$, whose representation matrix
$\mathbf{D}$ is unitary. Since $D\Gamma$ is educed from IR
$\mathbf{\Gamma}_{S}$, its reducibility is directly related to
$\mathbf{\Gamma}_{S}$ and can be divided into three conditions:

  1. (1)

If
$\mathbf{\Gamma}(S)=\mathbf{C}\mathbf{\Gamma}(A^{-1}SA)^{*}\mathbf{C}^{-1}$,
and $\mathbf{C}\mathbf{C}^{*}=\mathbf{\Gamma}(A^{2})$, then $D\Gamma$ is
reducible and the degeneracy will not be doubled,

  2. (2)

If
$\mathbf{\Gamma}(S)=\mathbf{C}\mathbf{\Gamma}(A^{-1}SA)^{*}\mathbf{C}^{-1}$,
and $\mathbf{C}\mathbf{C}^{*}=-\mathbf{\Gamma}(A^{2})$, then $D\Gamma$ is
irreducible and the degeneracy will be doubled,

  3. (3)

If two IR $\mathbf{\Gamma}_{S}$ and $\mathbf{\Gamma}_{S}^{*}$ of subgroup
$\\{S\\}$ are inequivalent, then $D\Gamma$ is irreducible and the degeneracy
will be doubled.

Character of representation $\mathbf{\Gamma}_{S}$ can be utilized to
distinguish these three cases. Supposing that $\chi$ is the character of
representation $\mathbf{\Gamma}_{S}$, then

$\sum_{B\in\\{AS\\}}\chi(B^{2})=\left\\{\begin{aligned}
=&+g_{s},&\mathbf{\Gamma}_{S}\text{ in case (1)}\\\
=&-g_{s},&\mathbf{\Gamma}_{S}\text{ in case (2) }\\\ =&\ \
0,&\mathbf{\Gamma}_{S}\text{ in case (3) }\end{aligned}\right.$ (30)

where $g_{s}$ is the rank of subgroup $\\{S\\}$. Through co-representation of
Hamiltonian symmetry group, we can educe the irreducible subspace of the
system’s Hilbert space and thus the energy level degeneracy.

##  Appendix B The XXX spin chain

We will first show that the XXX Hamiltonian is invariant under the time
reversal transformation. The time reversal is represented by an antiunitary
operator

$T=UK,$ (31)

where $K$ is the complex conjugate operator and $U$ is an unitary operator.
The spin angular momentum is transformed in the way

$T\sigma^{x}T^{-1}=-\sigma^{x},\ T\sigma^{y}T^{-1}=-\sigma^{y},\
T\sigma^{z}T^{-1}=-\sigma^{z}.$ (32)

We take the definition of $T$ for a single spin as

$T=i\sigma^{y}K.$ (33)

For a system contains $L$ spins, we choose

$T=(i\sigma^{y})^{\otimes L}K.$ (34)

It can be verified that $H_{XXX}$ is invariant under the rime reversal
transformation

$TH_{XXX}T^{-1}=H_{XXX}.$ (35)

We now show that the XXX Hamiltonian has SU(2) symmetry by confirming that it
commutes with the total spin operator

$S^{i}=\sum_{l=1}^{L}\sigma_{l}^{i},\ i=x,y,z.$ (36)

It can be checked that

$[H,S^{x}]=[H,S^{y}]=[H,S^{z}]=0.$ (37)

The spin operators form a SU(2) algebra and consequently the XXX spin chain
has SU(2) as a symmetry group.

Each Lie group has a corresponding Lie algebra, the elements of the group can
be constructed using the exponential mapping from the algebra elements.
Therefore the elements of the SU(2) group can be written as

$R(\hat{\mathbf{n}},\omega)=e^{-i\omega\hat{\mathbf{n}}\cdot\vec{\mathbf{S}}}=e^{-i\omega(n_{x}S^{x}+n_{y}S^{y}+n_{z}S^{z})},$
(38)

which describes the state transformation corresponding to the rotation with
angle $\omega$ around the axis $\hat{n}$.

So far, we have shown that the XXX Hamiltonian has time reversal and SU(2)
symmetry. Thus its symmetry group is composed of

$M=\\{R\\}+\\{TR\\},$ (39)

where $T$ is the time reversal operator, and $SU(2)=\\{R\\}$. Note that
$\\{TR\\}$ is a set of anti-unitary operators. Time reversal symmetry may
cause the doubling of energy degeneracy, which is distinguished by Eq. (30).
We first calculate the left-hand-side of Eq. (30),

$\displaystyle\sum_{B\in\\{TR\\}}\chi(B^{2})=$
$\displaystyle\sum_{B\in\\{TR\\}}\chi(TRTR)$ (40) $\displaystyle=$
$\displaystyle\sum_{R\in\\{R\\}}\chi(T^{2})\chi(T^{-1}RTR)$ $\displaystyle=$
$\displaystyle\chi(T^{2})\sum_{R\in\\{R\\}}\chi(R^{2}).$

In a finite group, the group function averages over the group elements. When
this group function is extended to a Lie group, it becomes an integral of the
group function over the group elements. Eq. (40) becomes

$\sum_{B\in\\{TR\\}}\chi(B^{2})=\chi(T^{2})\int_{R\in\\{R\\}}\chi(R^{2})dR.$
(41)

As is known that the IRs of SU(2), $\mathbf{\Gamma}^{j}$ are labeled by $j$,
where $j=0,1/2,1,3/2,\cdots$, whose dimension is $2j+1$. The characters of the
representation $\mathbf{\Gamma}^{j}$ of an arbitrary group element
$R(\hat{\mathbf{n}},\omega)$ is

$\chi^{j}(R)=\frac{\sin[(j+1/2)\omega]}{\sin(\omega/2)}.$ (42)

We have $[R(\hat{\mathbf{n}},\omega)]^{2}=R(\hat{\mathbf{n}},2\omega)$, such
that

$\chi^{j}(R^{2})=\frac{\sin[(2j+1)\omega]}{\sin(\omega)}.$ (43)

The integration over SU(2) is thus [35]

$\displaystyle\sum_{B\in\\{TR\\}}\chi(B^{2})=$
$\displaystyle\chi(T^{2})\cdot\frac{1}{\pi}\int_{0}^{2\pi}\frac{\sin[(2j+1)\omega]}{\sin(\omega)}\sin^{2}(\frac{\omega}{2})d\omega.$
(44) $\displaystyle=$ $\displaystyle(-1)^{L}\cdot\frac{1}{\pi}(-1)^{2j}\pi$
$\displaystyle=$ $\displaystyle 1.$

Now it’s time to figure the rank of subgroup SU(2). The rank of the Lie
algebra is known as the number of generators that can be simultaneously
diagonalized [33]. The rank of SU(2) is thus 1: we can only diagonalize
$S_{z}$. We can now conclude that

$\sum_{B\in\\{TR\\}}\chi(B^{2})=+g_{s},\ \mathbf{\Gamma}_{R}\text{ in case
(1)},$ (45)

which means that the co-representation $D\Gamma$ is reducible and the
degeneracy will not be doubled. Therefore the dimension of IR of $M$ and
$SU(2)$ group is the same.

##  Appendix C the XXZ spin chain

Noting that $H_{XXZ}$ commutes with total $z$-spin operator $S^{z}$

$[H_{XXZ},S^{z}]=0.$ (46)

Additionally, the XXZ Hamiltonian is invariant under time reversal $T$,

$TH_{XXZ}T^{-1}=H_{XXZ}.$ (47)

Therefore, system of XXZ Hamiltonian possesses U(1) symmetry associated with
the conservation of total $z$-spin, as well as time reversal symmetry.

The $U(1)$ group consists of all the one dimensional unitary operator

$U(\theta)=e^{i\theta},\theta\in[0,2\pi).$ (48)

The U(1) group is a compact abelian group, and its elements commute with each
other. According to Schur’s theorem, all the IRs of a compact abelian group
are one-dimensional. Therefore, the IR of U(1) symmetry is one-dimensional.
Geometrically, the elements of the U(1) group can be visualized as points on a
unit circle in the complex plane. In the context of the spin chain system, the
U(1) symmetry can be expressed as

$U(\theta)=e^{i\theta S^{z}},$ (49)

where $\theta$ is the continuous parameter, and $S^{z}$ is the total $z$-spin
operator.

Thus the symmetry group $P$ of the XXZ Hamiltonian can be written as

$P=\\{U\\}+\\{TU\\},$ (50)

where $\\{U\\}$ represents U(1) group, $T$ represents time reversal operator.
Similar to the XXX Hamiltonian, co-representation can be utilized to
characterize the representation of group $P$. Next, we will investigate the
reducibility of co-representation of group $P$.

Recalling that there are three cases of criterion for reducibility of co-
representation $D\Gamma$. To determine which case it belongs, we first
calculate $T^{-1}U(\theta)T$:

$T^{-1}U(\theta)T=U(\theta).$ (51)

The co-representation $D\Gamma$ is deduced from IR of subgroup $\Gamma(U)$,
such that the reducibility of $D\Gamma$ is decided by the properties of
$\Gamma$. We first calculate $\Gamma(U)$ and $\Gamma(T^{-1}UT)^{*}$ in the
context of $L$-site spin chain:

$\Gamma(U)=\Gamma(e^{i\theta S^{z}}),$ (52a)
$\Gamma(T^{-1}UT)^{*}=\Gamma(U)^{*}=\Gamma(e^{-i\theta S^{z}}).$ (52b)

Noting that the equivalence between $\Gamma(U)$ and $\Gamma(T^{-1}UT)^{*}$
depends on the value of $S^{z}$. Therefore, we partition Hilbert space into
subspaces by the value of $S^{z}$. This immediately leading to two cases:

  1. 1.

When $S^{z}=0$ (only exist for $L$ is even),
$\Gamma(U)=\Gamma(T^{-1}UT)^{*}=1$, which is a 1-dim identity representation.
The transformation operator is $C=1$, with $CC^{*}=\Gamma(T^{2})$. As a
result, $D\Gamma$ is reducible, and the degeneracy of the corresponding energy
level will not be doubled.

  2. 2.

When $S^{z}\neq 0$ (exist for $L$ is even or odd). $\Gamma(U)$is not
equivalent to $\Gamma(T^{-1}UT)^{*}$. Hence $D\Gamma$ is irreducible, and the
degeneracy of the corresponding energy level will be doubled.

Through the discussion above, we have discovered that the time reversal
symmetry connects certain subspaces with opposite values of $S^{z}$. In the
absence of time reversal symmetry, when only the U(1) symmetry is present, all
IRs are one-dimensional. However, with the introduction of time reversal
symmetry, some of the subspaces become connected, forming two-dimensional IRs.
In other words, when the time reversal operator $T$ acts on a vector in a
subspace with $S^{z}=m$ (where $m\neq 0$), it results in a vector in the
subspace with $S^{z}=-m$. On the other hand, when $T$ acts on a vector in the
subspace with $S^{z}=0$, it leaves the vector unchanged within the same
subspace.

Figure 2:  $T$ operator acting on different space labeled by $S^{z}$.