Exact solution of the two-axis two-spin Hamiltonian
Feng Pan^1,3, Yao-Zhong Zhang²¹¹1The corresponding author.
E-mail address: [email protected], Xiaohan Qi¹, Yue Liang¹, Yuqing Zhang¹, and Jerry P. Draayer³

Quantum squeezing of both Bose and Fermi many-body systems 1 ; 11 ; 12 ; 13 ; 14 ; 15 ; 17 ; 18 ; 31 ; 32 ; 33 ; 34 is effective and useful in quantum metrology and its applications in quantum informatics 222 ; 333 . Two previous models for dynamical generation of spin-squeezed states are the one-axis twisting and two-axis countertwisting Hamiltonians, respectively 1 , of which the latter model gives rise to maximal squeezing with a squeezing angle independent of system size or evolution time. Very recently, two-axis one-spin (2A1S) countertwisting Hamiltonian 1 has been generalized to the two-axis two-spin (2A2S) case 2222 . As shown in 2222 , the 2A2S Hamiltonian produces the spin EPR states, the analog of the two-mode squeezed state for spins, which are able to violate the Bell-CHSH inequality when the quantum numbers of the two spins are finite.

It wss shown in our previous works pan-zhang1 ; pan-zhang2 that the 2A1S Hamiltonian is exactly solvable and its solution can be obtained by using the Bethe ansatz method. As noted in pan-zhang2 , the 2A1S Hamiltonian is equivalent to a special case of the Lipkin-Meshkov-Glick (LMG) model vi1 ; vi2 after an Euler rotation. Similar to other many-spin systems be ; gau ; ric1 ; ric2 , the LMG model can be solved analytically by using the algebraic Bethe ansatz pan ; mori . The same problem can also be solved by using the Dyson boson realization of the SU(2) algebra vi1 ; vi2 ; zhang1 ; zhang2 , of which the solution may be obtained from the Riccati differential equations vi1 ; vi2 . Recently, the Bethe ansatz method has also been applied to generate exact solution of mean-field plus orbit-dependent non-separable pairing model with two non-degenerate $j$-orbits pan2019 and that of the dimer Bose-Hubbard model with multi-body interactions pan2020 . In these two models pan2019 ; pan2020 , there is an additional $S_{2}$ symmetry with respect to the two species of bosons or quasi-spins, which is helpful in constructing operators involved in the corresponding Bethe ansatz states and the related operator algebra calculations.

The purpose of this work is to construct exact and complete solution of the 2A2S Hamiltonian. In Sec. II, the 2A2S Hamiltonian is written in terms of boson operators after the Jordan-Schwinger boson realization of the two related SU(2) algebras, which can then be expressed in terms of generators of two copies of SU(1,1) algebra. Since there are only two sets of SU(1,1) generators involved in the Hamiltonian, the technique used in pan2019 ; pan2020 is helpful in constructing the Bethe ansatz states and the related operator algebra calculations. The derivation of the related extended Heine-Stieltjes polynomials is presented, whose zeros are related to the exact solution of the 2A2S Hamiltonian. Symmetry properties of excited levels of the system and zeros of the related extended Heine-Stieltjes polynomials are also discussed. Some numerical examples of the two equal spin case are presented in Sec. III, which demonstrate the main features of the solution. A brief summary is provided in Sec. IV.

The two-axis two-spin (2A2S) Hamiltonian is given by 2222

where $\chi$ is a constant and $\{S^{\pm}_{i},S^{0}_{i}\}$ ($i=1,2$) are generators of two copies of the SU(2) algebra satisfying the following commutation relations:

Though the Hamiltonian (1) is not commutative with $S^{\nu}_{i}$ ($\nu=+,\,-,\,0;\,i=1,\,2$), it is commutative with the two SU(2) Casimir invariants $C^{(i)}_{2}({\rm SU}(2))=(S^{+}_{i}S^{-}_{i}+S^{-}_{i}S^{+}_{i})/2+(S^{0}_{i})^{2}$ for $i=1,\,2$. Thus, the two spins are good quantum numbers of the system, while the total spin and its projection are not. The generators of the two copies of the SU(2) can be represented using the Jordan-Schwinger boson realization

where $a_{i}$ and $b_{i}$ ($a^{{\dagger}}_{i}$ and $b^{{\dagger}}_{i}$) are the boson annihilation (creation) operators satisfying

By substituting (II) into (1), the Hamiltoanian (1) can be expressed in terms of the generators of two copies of SU(1,1) algebra,

with

which satisfy the commutation relations

As shown in the following, the SU(1,1) type Bethe ansatz states for the Hamiltonian (5) can be written as

where $\eta$ is an additional label needed, $|\lambda_{1},\lambda_{2}\rangle$ is one of the lowest weight states of ${\rm SU}_{1}(1,1)\otimes\rm{SU}_{2}(1,1)$ satisfying

and

with variable $x$ to be determined. The allowed $\lambda_{1}$ and $\lambda_{2}$ values are provided in Table 1.

The single- and double-commutators of the Hamiltonian (5) with the operator (10) can be expressed as

while other higher order commutators of the Hamiltonian (5) with the operator (10) vanish. Once these commutators are obtained, the corresponding polynomials $G_{1}(x)$ and $G_{2}(x,y)$ in $\Lambda^{+}_{1}$ and $\Lambda^{+}_{2}$ on the lowest weight states of ${\rm SU}_{1}(1,1)\otimes{\rm SU}_{2}(1,1)$, defined as pan2020

can be expressed in the form

Here $g$ is a free parameter whose allowed values will be determined later and

It can be observed that the single- and double-commutators of the Hamiltonian (5) with the operator (10) are quite similar to the corresponding ones appearing in the Richardosn-Gaudin type models pan2019 ; pan2020 . Therefore, the SU(1,1) type Bethe ansatz (8) works for this case.

Similar to what is shown in pan2019 , using the commutation relations (11), (12), and the expressions shown in (14) and (15), we can directly check that

It is clear that the second and the fourth terms in (18) are proportional to the Bethe ansatz state (8), which is assumed to be the eigenstate of the system. Therefore, the eigen-energy of the 2A2S Hamiltonian is given by

as long as the first and the third terms proportional to $\Lambda^{+}(g)\prod_{\rho\,(\neq i)}^{k}\Lambda^{+}(x^{(\eta)}_{\rho})|\lambda_{1},\lambda_{2}\rangle$ in (18) for given $i$ are cancelled out, which leads to the corresponding equations in determining the $k$ variables $\{x_{1}^{(\eta)},\,x_{2}^{(\eta)},\cdots,\,x_{k}^{(\eta)}\}$:

Using $\alpha(x^{(\eta)}_{i})$, $\beta(x^{(\eta)}_{i})$, $a(x_{i^{\prime}}^{(\eta)},x_{i}^{(\eta)})$ and $c(x_{i}^{(\eta)},x_{i^{\prime}}^{(\eta)})$ in (16) and (II), the eigen-energy can be simplified to

where the $k$ variables $\{x_{i}^{(\eta)}\}$ should satisfy

under the condition that $g\neq x^{(\eta)}_{i}~{}\forall~{}\eta,\,i$. It is obvious that $g=x^{(\eta)}_{i}$ for any $\eta$ and $i$ is the singular point of (21) and should be avoided. It can be verified that root components $\{x^{(\eta)}_{i}\}$ of (22), which are always real and unequal one another, lie in the two open intervals $(-\infty,0)\bigcup(0,+\infty)$.

By using (22), (21) can be expressed as

It is obvious that the first part $1/(x^{(\eta)}_{i}-x^{(\eta)}_{i^{\prime}})$ within the double sum over $i$ and $i^{\prime}$ in (23) is antisymmetric, while the last part is symmetric with respect to the permutation $i\rightleftharpoons i^{\prime}$, which ensures that the second term in (23) is zero. Hence, the eigen-energies (21) are independent of $g$ as long as $g\neq x^{(\eta)}_{i}~{}\forall~{}\eta,\,i$, and can be further simplified to the form

It is now clear that $\eta$ labels the $\eta$-th solution of (22). In addition, though the eigenstates provided in (8) are not normalized, they are always orthogonal with

where ${\cal N}(\eta,k;\lambda_{1},\lambda_{2})$ is the corresponding normalization constant.

It is obvious that the Hamiltonian (1) is invariant under the permutations of the two spin operators with $S^{\nu}_{1}\leftrightarrow S^{\nu}_{2}$ for $\nu=0,\,+,\,-$, which corresponds to the permutation of $a$-bosons with $b$-bosons. The ${\rm SU}(1,1)$ operators $\Lambda^{+}(x_{i})$ used in (10) are invariant under the permutation of $a$-bosons with $b$-bosons. Therefore, the SU(1,1) lowest weight states $|\lambda_{1},\lambda_{2}\rangle$ should be invariant under the permutation of $a$-bosons with $b$-bosons. As clearly shown in Table 1, the first two and the last two two-spin intrinsic states $\left|\begin{array}[]{l}~{}~{}S_{1}^{\rm in}\\ \ M_{1}^{\rm in}\end{array}\begin{array}[]{l}S_{2}^{\rm in}\\ M_{2}^{\rm in}\end{array}\right\rangle$ and $\left|\begin{array}[]{l}~{}~{}S_{2}^{\rm in}\\ \ M_{2}^{\rm in}\end{array}\begin{array}[]{l}S_{1}^{\rm in}\\ M_{1}^{\rm in}\end{array}\right\rangle$ are indeed the same SU(1,1) lowest weight state $|\lambda_{1},\lambda_{2}\rangle$ when both $\mu_{1}$ and $\mu_{2}\neq 0$. The two-spin intrinsic state is unique with $\left|\begin{array}[]{l}~{}~{}S_{1}^{\rm in}\\ \ M_{1}^{\rm in}\end{array}\begin{array}[]{l}S_{2}^{\rm in}\\ M_{2}^{\rm in}\end{array}\right\rangle=\left|\begin{array}[]{l}~{}0\\ \ 0\end{array}\begin{array}[]{l}0\\ 0\end{array}\right\rangle$ only when $\mu_{1}=\mu_{2}=0$. It is also obvious that the two-spin intrinsic states with the same $M^{\rm in}=M_{1}^{\rm in}+M_{2}^{\rm in}$ value are the same SU(1,1) lowest weight state $|\lambda_{1},\lambda_{2}\rangle$.

Once the Bethe ansatz equations (22) are solved, the eigenstates (8), up to the two-spin permutation and a normalization constant, can be expressed in terms of uncoupled two-spin states as

where

are symmetric functions of $\{x^{(\eta)}_{1},\cdots,\,x^{(\eta)}_{k}\}$.

Moreover, it can be verified directly that (22) and the corresponding eigen-energy (21) are invariant under the simultaneous interchanges $\lambda_{1}\leftrightarrow\lambda_{2}$ and $x^{(\eta)}_{i}\leftrightarrow 1/x^{(\eta)}_{i}$, which corresponds to the permutation of the two copies of the SU(1,1) generators, $\Lambda^{\mu}_{1}\leftrightarrow\Lambda^{\mu}_{2}$ for $\mu=+,-,0$. It is obvious that the 2A2S Hamiltonian (5) is also invariant under the permutation $\Lambda^{\mu}_{1}\leftrightarrow\Lambda^{\mu}_{2}$. Therefore, when $\lambda_{1}\neq\lambda_{2}$, the eigenvalue of the 2A2S Hamiltonian with $\{x_{i}^{(\eta)}\}$ built on the SU(1,1) lowest weight state $|\lambda_{1},\lambda_{2}\rangle$ and that with $\{(x_{i}^{(\eta)})^{-1}\}$ built on the $|\lambda_{2},\lambda_{1}\rangle$ are the same. Thus, when $\lambda_{1}=\lambda_{2}$, if $\{x_{i}^{(\eta)}\}$ is a solution, $\{(x_{i}^{(\eta)})^{-1}\}$ gives the same solution. Namely, for fixed $i$, if $x_{i}^{(\eta)}$ is one of the root component, $x^{(\eta)}_{i^{\prime}}=(x_{i}^{(\eta)})^{-1}$ is also a root component of the same root. We can also verify that the roots of (22) have the mirror symmetry. Namely, if $\{x_{i}^{(\eta)}\}$ is a solution, $\{-x_{i}^{(\eta)}\}$ is also a solution. Thus, if the eigen-energy is nonzero, the sign of the eigen-energy with $\{x_{i}^{(\eta)}\}$ is opposite to that with $\{-x_{i}^{(\eta)}\}$.

According to the Heine-Stieltjes correspondence pan2019 ; 3 ; 4 ; 5 , the second-order Fuchsian equation of the extended Heine-Stieltjes polynomials whose zeros are the roots of (22) can be established. By using the identity

it can be verified that the related extended Heine-Stieltjes polynomials $y^{{(\eta)}}_{k}(x)$ of degree $k$ should satisfy

where the Van Vleck polynomial $V^{{(\eta)}}(x)$ is simply a binomial to be determined by (29). Write $y^{(\eta)}_{k}(x)=\sum_{n=0}^{k}f^{(\eta)}_{n}x^{n}$ and $V^{{(\eta)}}(x)=v^{(\eta)}_{0}+v^{(\eta)}_{1}x$, where $\eta$ labels the $\eta$-th polynomial. It can be verified directly that the expansion coefficients $f^{(\eta)}_{n}$ ($n=0,\cdots,k$) should satisfy the following three-term recurrence relations:

with

which is independent of $\eta$. Instead of solving the three-term recurrence relations (30) for the expansion coefficients $f^{(\eta)}_{n}$ ($n=0,\cdots,k$) and $v^{(\eta)}_{0}$, we can construct the corresponding $(k+1)\times(k+1)$ bidiagonal matrix $A$ with entries

Let ${\bf F}^{(\eta)}=(f^{(\eta)}_{0},\,f^{(\eta)}_{1},\,\cdots,\,f^{(\eta)}_{k})^{\bf T}$, where ${\bf T}$ stands for the transpose operation. The $k+1$ dimensional vector ${\bf F}^{(\eta)}$ is the $\eta$-th eigenvector of the bidiagonal matrix $A$ with

where $-v^{(\eta)}_{0}$ is the corresponding eigenvalue, which clearly shows that there are $k+1$ sets of solutions of (22) for the eigen-energies (24) and the corresponding eigenstates (26) with $\eta=1,\,2,\,\cdots,\,k+1$. It is also obvious that not only the construction of the matrix $A$, but also its diagonalization is easier with smaller size of the matrix and thus more efficient than the direct diagonalization of the 2A2S Hamiltonian (1) in the original uncoupled two-spin basis, especially when $k$ is large. Once the $\eta$-th set of the expansion coefficients $f^{(\eta)}_{n}$ ($n=0,\cdots,k$) are known from the eigen-equation (33), the $k$ zeros $\{x_{i}^{(\eta)}\}$ ($i=0,\cdots,k$) of the polynomial $y^{(\eta)}_{k}(x)/{f}^{(\eta)}_{k}=\sum_{n=0}^{k}\bar{f}^{(\eta)}_{n}x^{n}$, where, up to an overall factor, $\bar{f}^{(\eta)}_{n}={f}^{(\eta)}_{n}/{f}^{(\eta)}_{k}$ ($n=1,\cdots,k$), can easily be calculated due to the fact that $y^{(\eta)}_{k}(x)$ is one-variable polynomial and ${f}^{(\eta)}_{k}$ is always nonzero. In addition, $y^{(\eta)}_{k}(x)/{f}^{(\eta)}_{k}$ can also be expressed in terms of the zeros $\{x^{(\eta)}_{j}\}$ ($j=1,\cdots,k$) as

where ${S}_{q}^{(k,\eta)}$ is the symmetric function defined in (27). Comparing (34) with $y^{(\eta)}_{k}(x)/{f}^{(\eta)}_{k}=\sum_{n=0}^{k}\bar{f}^{(\eta)}_{n}x^{n}$, we get

which can be used to avoid unnecessary computation of $S^{(k,\eta)}_{q}$ from $\{x^{(\eta)}_{1},\cdots,~{}x^{(\eta)}_{k}\}$ needed in the eigenstates (26). Furthermore, using the three-term recurrence relations (30), we have

where the relation (35) is used for the second equality, which shows that the constant term $v_{0}^{(\eta)}$ in the Van Vleck polynomial equals exactly to the corresponding eigen-energy $E^{(\eta)}_{k,\mu_{1},\mu_{2}}/\chi$ of the 2A2S Hamiltonian.

Finally, we prove that the solutions of the Bethe ansatz equations (22) are complete. When the two spins are unequal with $S_{1}>S_{2}$, there are three sets of uncoupled two-spin states used in the eigenstates (26): Case 1 with $S_{1}={1\over{2}}(k+\mu_{1}+\mu_{2})$ and $S_{2}={1\over{2}}k$ corresponding to the upper one in (26); Case 2 with $S_{1}={1\over{2}}(k+\mu_{1})$ and $S_{2}={1\over{2}}(k+\mu_{2})$ corresponding to the lower one in (26); and Case 3 with $S_{1}={1\over{2}}(k+\mu_{2})$ and $S_{2}={1\over{2}}(k+\mu_{1})$ corresponding to the two-spin permutation $S_{1}\rightleftharpoons S_{2}$ of the lower one in (26). For Case 1, $S_{1}={1\over{2}}(k+\mu_{1}+\mu_{2})>S_{2}={1\over{2}}k$ with $\mu_{1}=2S_{1}-2S_{2}-\mu_{2}\geq 0$, where $\mu_{2}$ can be taken as $0,1,2\cdots,2S_{1}-2S_{2}$. The number of solutions provided by (22) for a fixed $\mu_{2}$ is $k+1=2S_{2}+1$. Thus, the total number of solutions for $S_{1}={1\over{2}}(k+\mu_{1}+\mu_{2})$ and $S_{2}={1\over{2}}k$ case shown by (26) with the upper uncoupled two-spin states is $(2S_{2}+1)(2S_{1}-2S_{2}+1)$, which includes $\mu_{2}=0$ case. For Case 2, $S_{1}={1\over{2}}(k+\mu_{1})>S_{2}={1\over{2}}(k+\mu_{2})$ with $\mu_{1}=\mu_{2}+2S_{1}-2S_{2}$ and $\mu_{2}>0$ because $\mu_{2}=0$ case is already considered in Case 1, the number of solutions provided by (22) is $k+1=2S_{2}-\mu_{2}+1$ for a fixed $\mu_{2}$, where $\mu_{2}$ can be taken as $1,2,\cdots,2S_{2}$. Hence, the total number of solutions for this case shown by (26) with the lower uncoupled two-spin states is $\sum_{\mu_{2}=1}^{2S_{2}}(2S_{2}-\mu_{2}+1)=(2S_{2}+1)S_{2}$, which excludes the $\mu_{2}=0$ case. For Case 3, $S_{1}={1\over{2}}(k+\mu_{2})>S_{2}={1\over{2}}(k+\mu_{1})$ corresponding to the two-spin permutation $S_{1}\rightleftharpoons S_{2}$ of the lower one in (26), where $\mu_{1}=\mu_{2}-2S_{1}+2S_{2}\geq 0$ and $2S_{1}\geq\mu_{2}\geq 2S_{1}-2S_{2}$, the number of solutions provided by (22) is $k+1=2S_{1}-\mu_{2}+1$ for a fixed $\mu_{2}$. Since $\mu_{2}=2S_{1}-2S_{2}$ with $\mu_{1}=0$ case is already considered in Case 2, the total number of solutions for this case is $\sum_{\mu_{2}=2S_{1}-2S_{2}+1}^{2S_{1}}(2S_{1}-\mu_{2}+1)=(2S_{2}+1)S_{2}$, which excludes the $\mu_{2}=2S_{1}-2S_{2}$ case. It is now obvious that the total number of the solutions provided by the three cases equals exactly to $(2S_{1}+1)(2S_{2}+1)$, which is the dimension of the uncoupled two-spin states for $S_{1}>S_{2}$. This conclusion also applies to $S_{1}<S_{2}$ case by the permutation $S_{1}\rightleftharpoons S_{2}$.

For the two equal spin case with $S_{1}=S_{2}=k/2$, which is exemplified in the next section, the upper and lower uncoupled two-spin states shown in (26) are the same. According to (26), the two-spin symmetric or anti-symmetric eigenstates of the 2A2S Hamiltonian in this case can be written uniformly as

for $\mu=0,\,1,\,\cdots,\,k$, where

are the symmetric and anti-symmetric two-spin states, respectively. The number of solutions of (22) with the replacement: $k\rightarrow k-\mu$ with a fixed $\mu$ for (II) is $k-\mu+1$. Due to the two-fold degeneracy of the $\mu\neq 0$ states, the total number of solutions provided by (II) is $2\sum_{\mu=1}^{k}(k-\mu+1)+k+1=(k+1)^{2}$.

To demonstrate the method and results presented above, in this section, we consider the two equal spin case with $S_{1}=S_{2}=k/2$ as studied in 2222 . The two-spin symmetric or anti-symmetric eigenstates of the 2A2S Hamiltonian are given by (II). It is obvious that the $\mu=0$ eigenstates of the 2A2S Hamiltonian are always symmetric, while both symmetric and anti-symmetric states are possible for $\mu\neq 0$. Moreover, the corresponding eigen-energies $E^{(\eta)}_{k-\mu,\mu,\mu}/\chi$ ($\eta=1,2,\cdots,k-\mu+1$) of both two-spin symmetric and anti-symmetric cases are symmetric with respect to the the sign change. Namely, if the $k-\mu+1$ level energies are arranged as $E^{(1)}_{k-\mu,\mu,\mu}/\chi<E^{(2)}_{k-\mu,\mu,\mu}/\chi<\cdots<E^{(k-\mu+1)}_{k-\mu,\mu,\mu}/\chi$, then