Abstract
In this chapter, we start by motivating the problem studied throughout Part I, and then discuss simple problems with discrete symmetry. In Sect. 3.1, we briefly discuss what is expected and known in the antiferromagnetic Heisenberg model, and introduce the essential idea of “long-range order (LRO) without spontaneous symmetry breaking (SSB)”. In Sect. 3.2, we discuss a similar problem in equilibrium states of the classical Ising model, which is simpler and easier-to-understand. Then, in Sect. 3.3, we study the simplest quantum many-body system that exhibits “LRO without SSB”, namely, the quantum Ising model, and see how the above mentioned “paradox” is solved. At the end of this section, we make an important remark about the essential role played by symmetry in the notion of phases. In the final Sect. 3.4, we describe general rigorous theories of LRO and SSB when the relevant symmetry is Ising-like.
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Notes
- 1.
See [3] for a recent introduction to the concept of SSB from a different perspective.
- 2.
But the situation is different in systems exhibiting Bose–Einstein condensation or superconductivity. We will discuss this interesting topic in Chap. 5.
- 3.
The same model can be described as a quantum spin system with Hamiltonian \(\hat{H}_0=-\sum _{\{x,y\}\in \mathcal{B}_L}\hat{S}^{(3)}_x\hat{S}^{(3)}_y\). The thermal expectation value of an operator \(\hat{A}\) is defined as \(\langle \hat{A}\rangle _{\beta ,L}:={\text {Tr}}[\hat{A}e^{-\beta \hat{H}_0}]/{\text {Tr}}[e^{-\beta \hat{H}_0}]\).
- 4.
To be rigorous, the existence of the limit in (3.2.3) has been proved only for \(\beta \le \beta _\mathrm {c}\) when periodic boundary conditions are used. For \(\beta >\beta _\mathrm {c}\) one should replace \(\lim \) by \(\limsup \) or \(\liminf \), or understand that a subsequence is taken when necessary. Then the following statements hold rigorously. See, e.g., [9].
- 5.
More precisely there is power law correction to the exponential decay. The asymptotic decay of the correlation function is (believed to be) given by \(\langle \sigma _x\sigma _y\rangle _{\beta ,\infty }\simeq (\text {constant})|x-y|^{-(d-1)/2}\times \)
\(\exp [-|x-y|/\xi (\beta )]\), which is known as the Ornstein-Zernike form. See, e.g., [5].
- 6.
The other strategy is to impose boundary conditions which break the symmetry. The most sophisticated method is to define equilibrium states in the infinite volume, and apply abstract decomposition theories. See Sect. 4.3.
- 7.
\(\{\hat{1},\hat{U}^{(1)}_\pi \}\) can be regarded as (a representation of) the discrete group \(\mathbb {Z}_2\). See Appendix A.5.
- 8.
More generally any nonzero operator \(\hat{\mathcal{O}}_L'\) such that \((\hat{U}^{(1)}_\pi )^\dagger \hat{\mathcal{O}}_L'\hat{U}^{(1)}_\pi =-\hat{\mathcal{O}}_L'\) has a chance to play the role of the order operator.
- 9.
Proof We abbreviate \(\hat{U}^{(1)}_\pi \) as \(\hat{U}\). Since the invariance of the Hamiltonian reads \(\hat{H}\hat{U}=\hat{U}\hat{H}\), we see \(\hat{H}(\hat{U}|\varPhi _\mathrm {GS}\rangle )=\hat{U}\hat{H}|\varPhi _\mathrm {GS}\rangle =E_\mathrm{GS}\hat{U}|\varPhi _\mathrm {GS}\rangle \), which means that \(\hat{U}|\varPhi _\mathrm {GS}\rangle \) is also a ground state. The uniqueness then implies \(\hat{U}|\varPhi _\mathrm {GS}\rangle =\alpha |\varPhi _\mathrm {GS}\rangle \) with some \(\alpha \in \mathbb {C}\) with \(|\alpha |=1\). (Note that this conclusion is valid for any nondegenerate energy eigenstate.) For the unique ground state \(|\varPhi _\mathrm {GS}\rangle \), (3.3.7) with \(c_{\varvec{\sigma }}>0\), along with (2.1.26) or (S.4), implies \(\alpha =(-i)^L\).
- 10.
The same model with periodic boundary conditions can be exactly solved by mapping it to a free fermion system. See [16] or Appendix A of [6]. (Note that the version of [6] on the arXiv is more recent and accurate.) The above conclusions about low-lying energy levels for \(0<\lambda \ll 1\) can be confirmed by the exact solution. The calculation becomes very hard if one uses open boundary conditions as in (3.3.1).
- 11.
We will encounter a similar correspondence between bulk properties and edge modes in \(S=1\) antiferromagnetic chains in Sect. 8.2.
- 12.
We note however that a naive convergence estimate (as we learn in elementary quantum mechanics or linear algebra) is not enough for a macroscopic system as the present one. The estimate works for sufficiently small \(\lambda \), but how small it should be depends on the system size L. In order to get a convergence estimate which is uniform in L, one needs to use an argument which takes into account the locality of the interactions, such as the machinery called the cluster expansion. See, e.g., [13] and references therein.
- 13.
The quotation marks indicate that they are not ground states in the standard definition in quantum mechanics.
- 14.
This definition does not assume that the two ranges of parameters belong to distinct phases. In fact we do observe a “phase transition” within a single phase. A notable example is the transition between gas and liquid, which are both parts of the same fluid phase. See Fig. 3.5. See also Footnote 18 for the notion of phase. We should say that the term “phase transition” is confusing.
- 15.
To be rigorous \(\lim \) should be replaced by \(\liminf \) for \(0<\lambda <\lambda _\mathrm {c}\).
- 16.
To be more precise, there exists a constant \({\Delta E}>0\) independent of L, and the first excited energy \(E_\mathrm{1st}\) satisfies \(E_\mathrm{1st}-E_\mathrm{GS}\ge {\Delta E}\) for any L.
- 17.
There appear many low-energy excited states but the unique ground state does not break the symmetry.
- 18.
To be precise a phase is a region in the parameter space that is separated from the rest by phase transition points. This definition clearly depends on the choice of the parameter space.
- 19.
This is true for any phase transitions in classical physics. In quantum many-body systems there can be phases with topological order, which need not be protected by any symmetry. See Sect. 8.4.
- 20.
The theory can readily be extended to more general Hamiltonian with short-range interactions. See Problem 3.4.a below.
- 21.
Suppose that the relevant symmetry is described by a unitary operator \(\hat{U}\) such that \(\hat{U}^\dagger \hat{H}\hat{U}=\hat{H}\) and \(\hat{U}^\dagger \hat{\mathcal{O}}_L\hat{U}=-\hat{\mathcal{O}}_L\). This implies \(\hat{U}^\dagger (\hat{\mathcal{O}}_L)^n\hat{U}=-(\hat{\mathcal{O}}_L)^n\) for any odd n. Suppose that the ground state \(|\varPhi _\mathrm {GS}\rangle \) is invariant under \(\hat{U}\), i.e., \(\hat{U}|\varPhi _\mathrm {GS}\rangle =c|\varPhi _\mathrm {GS}\rangle \) with \(|c|=1\). (This is always the case when the ground state is unique. When the ground states are degenerate, one can always find a ground state which is invariant under \(\hat{U}\).) Then, for odd n, we see that \(\langle \varPhi _\mathrm {GS}|(\hat{\mathcal{O}}_L)^n|\varPhi _\mathrm {GS}\rangle =\langle \varPhi _\mathrm {GS}|\hat{U}^\dagger (\hat{\mathcal{O}}_L)^n\hat{U}|\varPhi _\mathrm {GS}\rangle =-\langle \varPhi _\mathrm {GS}|(\hat{\mathcal{O}}_L)^n|\varPhi _\mathrm {GS}\rangle \), and hence \(\langle \varPhi _\mathrm {GS}|(\hat{\mathcal{O}}_L)^n|\varPhi _\mathrm {GS}\rangle =0\).
- 22.
In the classical Ising model the equivalence is rigorously known. See Sect. 3.2.
- 23.
- 24.
To be rigorous \(\lim \) should be replaced by \(\liminf \).
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Tasaki, H. (2020). Long-Range Order and Spontaneous Symmetry Breaking in the Classical and Quantum Ising Models. In: Physics and Mathematics of Quantum Many-Body Systems. Graduate Texts in Physics. Springer, Cham. https://doi.org/10.1007/978-3-030-41265-4_3
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