# For finite magnetic field B, the exact ground state of transverse Ising model still does not break the

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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 

edited Dec 3, 2020

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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.

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