0
$\begingroup$

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 $$𝐻=−∑𝑆𝑧_𝑖𝑆𝑧_𝑗+𝐵∑𝑆𝑥_𝑖$$ of 𝑁 spins. For small 𝐵, the exact ground state still do not break the 𝑆𝑧→−𝑆𝑧 symmetry. So it is non-trivial to see the $$𝑆𝑧→−𝑆𝑧$$ symmetry breaking for small $𝐵$."

Also later "A finite 𝐵 does not break 𝑆𝑧→−𝑆𝑧 symmetry. Also there is Lorentz symmetry."

Can any expert explain why?

I know that $𝑆𝑧→−𝑆𝑧$ can be induced by $$𝑆𝑧→𝑆_x𝑆𝑧𝑆_x^{-1}=−𝑆𝑧$$ and such operations does not transform any term in $𝐵∑𝑆𝑥_𝑖$ since $𝑆_x→𝑆_x𝑆_x𝑆_x^{-1}=𝑆_x$. But other comments seem subtle. Do we worry the magnitude of $B$?

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.