【Read between lines】Hal Tasaki "Physics and mathematics of quantum many body systems" p. 496 (Problem 2.4.a: uniqueness in each M sector of ferromagnetic ground state via Perron-Frobenius theorem)

Key Words

• Perron-Frobenius theorem

• A complete description of ground states for ferromagnetic system

Relevant section

It suffices to show the uniqueness of ground state in each $${\mathscr{H}_M}$$ . This is easily done by applying the Perron–Frobenius theorem to the matrix representation $${\bra{Ψ^σ}\hat{H}\ket{\Psi^\sigma}}$$ of the Hamiltonian.

Solution

Perron-Frobenius theorem says,

Let $${M=(m_{i,j})_{i,j=1, …, N}}$$ be an $${N\times N}$$ real symmetric matrix with the properties that (i) $$m_{ij}\leq0}$$ for any $${i\neq j}$$, and (ii) All $${i\neq j}$$ are connected via nonvanishing matrix elements of $${M}$$, or, more precisely, for any $${i\neq j}$$, we can take a sequence $${(i_1, …, i_K)}$$ such that $${i_1=i, i_K=j}$$, and $${m_{i_k, i_{k+1}}\neq0}$$ for all $${k=1, 2, …, K-1}$$. Then the eigenvalue of $${M}$$ is nondegenerate and the corresponding eigenvector $${\bm{v}=(v_i)_{i=1, …, N}}$$ can be taken to satisfy $${v_i>0}$$ for all i.

p. 475

In the sector $${M}$$, we have to check whether
(0) The matrix representation $${\bra{Ψ^σ}\hat{H}\ket{\Psi^{\sigma'}}}$$ is real and symmetric,
(1) $${\bra{Ψ^σ}\hat{H}\ket{\Psi^{\sigma'}}\leq0}$$ for all $${\sigma\neq\sigma'}$$, and
(2) All off-diagonal elements are connected.
If all of them are satisfied, the minimum eigenvalue of this matrix does not degenerate, i.e. the ground state is unique.

(0) Are the matrix elements real and symmetric?

Every term in the Hamiltonian is written as $${\hat{\bm{S}}_x\cdot\hat{\bm{S}}_y=(\hat{S}_x^+\hat{S}_y^-+\hat{S}_x^-\hat{S}_y^-)/2+\hat{S}_x^{(3)}\hat{S}_y^{(3)}}$$. Sandwiching this operator with $${\bra{\Psi^\sigma}}$$ and $${\ket{\Psi^{\sigma'}}}$$, 

$$
\bra{\Psi^\sigma}(\hat{S}_x^+\hat{S}_y^-+\hat{S}_x^-\hat{S}_y^+)/2+\hat{S}_x^{(3)}\hat{S}_y^{(3)}\ket{\Psi^{\sigma'}}\\=\mathrm{factor}\times\delta_{\sigma_x,\sigma'_x+1}\delta_{\sigma_x,\sigma'_x-1}+\mathrm{factor}\times\delta_{\sigma_x,\sigma'_x-1}\delta_{\sigma_y,\sigma'_y+1}+\mathrm{factor}\times\delta_{\sigma_x,\sigma'_x}\delta_{\sigma_y,\sigma'_y}
$$

All the factors are real and positive. Since $${\hat{H}}$$ is hermitian,

$$
\bra{\Psi^\sigma}\hat{H}\ket{\Psi^{\sigma'}}=\bra{\hat{\Psi^{\sigma'}}}\hat{H}\ket{\Psi^\sigma}^\ast=\bra{\hat{\Psi^{\sigma'}}}\hat{H}\ket{\Psi^\sigma}
$$

thus the matrix elements are real and symmetric.

(1) Are off-diagonal elements are zero or negative?

When $${\sigma\neq\sigma'}$$, 

$$
\bra{\Psi^\sigma}\hat{H}\ket{\Psi^{\sigma'}}=\left(\bigotimes_{j\in\Lambda}\bra{\psi_j^{\sigma_j}}\right)-\sum_{\{x,y\}\in\mathscr{B}}\left(\frac{\hat{S}_x^+\hat{S}_y^-+\hat{S}_x^-\hat{S}_y^+}{2}+\hat{S}_x^{(3)}\hat{S}_y^{(3)}\right)\left(\bigotimes_{j\in\Lambda}\ket{\psi_j^{\sigma'_j}}\right)\\=-\sum_{\{x,y\}\in\mathscr{B}}\left(\prod_{j\in\mathscr{B}\smallsetminus\{x,y\}}\delta_{\sigma_j,\sigma'_j}\right)\bra{\psi_x^{\sigma_x}}\bra{\psi_y^{\sigma_y}}\left[\mathrm{positive}\ket{\psi_x^{\sigma'_x+1}}\ket{\psi_y^{\sigma'_y-1}}+\mathrm{positive}\ket{\psi_x^{\sigma'_x-1}}\ket{\psi_y^{\sigma'_y+1}}+\sigma_x'\sigma_y'\ket{\psi_x^{\sigma'_x}}\ket{\psi_y^{\sigma'_y}}\right]
$$

The third term in R.H.S. vanishes if $${\sigma=\sigma'}$$. Therefore off-diagonal elements are zero or negative.

(2) Are off-diagonal elements are connected via non-vanishing elements?

This is verified in the Proposition in 2.5.