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  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 $$

asked Sep 28, 2020 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]
edited Dec 3, 2020

1 Answer

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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.

I hope this can help you!

This post imported from StackExchange Physics at 2020-12-03 17:31 (UTC), posted by SE-user RMPsp
answered Oct 4, 2020 by RMPsp (20 points) [ no revision ]

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