For finite magnetic field B, the exact ground state of transverse Ising model still does not break the →− symmetry In the comments of OP What is spontaneous symmetry breaking in QUANTUM systems?
There is a statement by OP Xiao-Gang Wen saying "the ground state of transverse Ising model $$=−∑__+∑_$$ of  spins. For small , the exact ground state still do not break the →− symmetry. So it is non-trivial to see the $$→−$$ symmetry breaking for small $$."
Also later "A finite  does not break →− symmetry. Also there is Lorentz symmetry."
Can any expert explain why?
I know that $→−$ can be induced by  $$→_x_x^{-1}=−$$
and such operations does not transform any term in $∑_$ since $_x→_x_x_x^{-1}=_x$. But other comments seem subtle. Do we worry the magnitude of $B$?
 A: This Hamiltonian has what is known as a "spin-flip" symmetry. It means that due to the term $\sum S_{z_i}S_{z_j}$, we can simultaneously change the sign of all the $S_{z_i}$ operators and we still have the same Hamiltonian (the operator that commutes with the Hamiltonian is $G=\prod_i S_{x_i}$, which produces a global spin-flip over states in the $z$-basis).
The eigenstates (and the ground state) of this Hamiltonian depend on the magnitude of $B$:
-If $B>>1$, the dominant term of the Hamiltonian is $B\sum_i S_{x_i}$, so the eigenstates are close to product states in the $x$ basis: $|n\rangle\simeq|\leftarrow \rightarrow \rightarrow ...\rangle$. This phase is called the paramagnetic phase.
-If $B<<1$, the dominant term of the Hamiltonian is $-\sum S_{z_i}S_{z_j}$ so the eigenstates are close to cat states of the form $|n_{\pm}\rangle\simeq=\uparrow \downarrow \downarrow ...\rangle\pm |\downarrow \uparrow \uparrow ...\rangle$, oriented in the $z$-axis. Due to the negative sign in $-\sum S_{z_i}S_{z_j}$, this phase would be called antiferromagnetic.
This change of order in the eigenstates (and the ground state) is a phase transition that goes from eigenstates with a defined parity ($B<<1$) to states without a well defined parity $B>>1$. This phase transition is said to have a spontaneous symmetry-brake and I think that in quantum mechanics, although the eigenstates of $B<<1$ do not explicitly break the symmetry (because the cat states are a superposition of the two possibilities of the symmetry-break), the name of "spontaneous..." is kept because of the similarities with the classical case.
Going now to the question itself, when Xiao-Gang Wen says says that small $B$ does not break the symmetry, I think he is meaning respect to the order of the ground state, saying that samll $B$ would not be enough to produce the phase transition.
And when it is said that a finite $B$ does not brake the symmetry neither, I think he is meaning that you cannot explicitly break/destroy the symmetry of the Hamiltonian with the term $B\sum_i S_{x_i}$, i.e., removing the spin-flip property, that is always present in our case. However, you could introduce a term in the $z$ direction like $B_z\sum_i S_{z_i}$ that explicitly breaks the symmetry of the Hamiltonian, and you lose the spin-flip property.
I hope this can help you!
