Parity-Dependent Quantum Phase Transition in the Quantum Ising Chain in a Transverse Field

Abstract

Phase transitions—both classical and quantum types—are the perfect playground for appreciating universality at work. Indeed, the fine details become unimportant and a classification in very few universality classes is possible. Very recently, a striking deviation from this picture has been discovered: some antiferromagnetic spin chains with competing interactions show a different set of phase transitions depending on the parity of number of spins in the chain. The aim of this article is to demonstrate that the same behavior also characterizes the most simple quantum spin chain: the Ising model in a transverse field. By means of an exact solution based on a Wigner–Jordan transformation, we show that a first-order quantum phase transition appears at the zero applied field in the odd spin case, while it is not present in the even case. A hint of a possible physical interpretation is given by the combination of two facts: at the point of the phase transition, the degeneracy of the ground state in the even and the odd case substantially differs, being respectively 2 and $2N$, with N being the number of spins; the spin of the most favorable kink shows changes at that point.

1. Introduction

In the non-relativistic microscopic world, the fundamental laws of physics regulate the interactions within pairs of constituents [1]. In the macroscopic scenario, emergent phenomena such as the onset of order parameters, the appearance of phase transitions, and the development of dissipation, are observed [2]. Connecting the two counterparts is a central aim of statistical mechanics. Conceptually, the procedure is the following: one calculates the properties of the system under inspection for an arbitrary number N of constituents, and then performs the thermodynamic limit N to infinity. In such a highly nontrivial limit, the emergent properties show up. Although it is most often not possible to apply this strategy directly, this paradigm has proven extremely useful. Indeed, it led to the description of the different phases of matter and the transitions between them. This statement holds true for both the symmetry-related phases of matter [2], and for the topological phases [3,4,5]. At the same time, it applies to both thermal [2] and quantum [6] phase transitions.

Very recently, a striking discovery was made: there are models in which the result of the limit, at zero temperature, crucially depends on the parity of N [7,8,9,10,11,12,13]. We will here call this phenomenon even–odd criticality. More specifically, what has been shown is that, depending on the parity of N, the system is gapless or gapped. Moreover, again depending on the parity, quantum phase transitions can be present or not. All the models known to show this spectacular behavior are antiferromagnetic spin chains with competing orders and subject to compactifying (periodic or twisted) boundary conditions. The physics behind the phenomenon is the competition between the local antiferromagnetic order and the global constraints posed by the boundary conditions.

In this respect, the main difference between the even and the odd N cases is that only in the odd N case at least one link between spins must be in the energetically unfavorable condition; in other words, at least one kink is present in the ground state. Moreover, the translational invariance implies the formation of a gapless band corresponding to the fact that the kink can be anywhere in the chain, thus altering significantly the spectrum of the system. A schematic is presented in Figure 1.

Since the discovery is extremely recent, not much is known about even–odd criticality. However, its potential is huge. From the theoretical side, it challenges the definition of phases, even thermal phases if classical models with even–odd criticality will be found, and represents a strikingly new playground for the study of the quantum quench dynamics of many-body physics and relaxation [14,15,16,17,18,19,20,21,22,23,24,25,26,27,28]. From a more practical perspective, it paves the way to unprecedented possibilities for quantum technologies: spin chains represent the typical models for quantum buses transferring quantum information. Even–odd critical systems are in principle able to allow or stop the flow of information by switching on or off a single site.

In this work, we show for the first time that even the most simple, yet fruitful and inspiring, spin chain shows even-odd criticality: we analyze the antiferromagnetic Ising spin chain with N sites in a transverse field to show, by means of the exact solution based on the Wigner–Jordan transformation, the presence of a first-order quantum phase transition that is only present in the case of odd N. For first-order quantum phase transition, we here mean a discontinuity of the ground-state energy with respect to some (non-trivial) parameters appearing in the Hamiltonian. Such quantum phase transition is manifested in a discontinuity of the first derivative of the ground state energy with respect to the applied field, calculated at the zero field.

The rest of the article is structured as follows: in Section 2. we outline the Ising model in the antiferromagnetic regime and its solution; in Section 3, we analyze the ground state and its energy, and show that a quantum phase transition is only present in the odd N case. Finally, in Section 3, we discuss the results and we draw our conclusions.

2. Model

The model under inspection is the antiferromagnetic quantum Ising chain in a transverse field [6]. Explicitly, we consider the Hamiltonian

$$H=\frac{J}{2}\sum _{j=1}^N\left[{\sigma }_j^x{\sigma }_{j+1}^x+h{\sigma }_j^z\right].$$

(1)

Here, we impose periodic boundary conditions ${\sigma }_{N+1}^{\alpha }={\sigma }_1^{\alpha }$. Moreover, in the Hamiltonian, $J>0$ parametrizes the antiferromagnetic coupling and sets the energy scale, h parametrizes the transverse magnetic field (we will only consider $\left|h\right|<1$), j is an index running over the lattice sites, N is the number of lattice sites, and ${\sigma }_j^{\alpha }$ with $\alpha =x,y,z$ are the three Pauli matrices defined on the j-th site.

The Hamiltonian can be exactly diagonalized by means of a Wigner–Jordan transformation to free fermions [29,30]. To do so, previously we defined

$${\sigma }_j^{\pm }\equiv \frac{{\sigma }_j^x\pm i{\sigma }_j^y}{2}.$$

(2)

Then, the transformation to free fermions is introduced, namely

$$\begin{array}{ccc}\hfill {\sigma }_j^{+}& \equiv & e^{i\pi {\sum }_{l=1}^{j-1}{\psi }_l^{†}{\psi }_l}{\psi }_j,\hfill \end{array}$$

(3)

$$\begin{array}{ccc}\hfill {\sigma }_j^{-}& =& e^{-i\pi {\sum }_{l=1}^{j-1}{\psi }_l^{†}{\psi }_l}{\psi }_j^{†},\hfill \end{array}$$

(4)

$$\begin{array}{ccc}\hfill {\sigma }_j^z& =& 1-2{\psi }_j^{†}{\psi }_j,\hfill \end{array}$$

(5)

where $\{{\psi }_j^{†},{\psi }_l\}={\delta }_{j,l}$ and $\{{\psi }_j^{†},{\psi }_l^{†}\}=\{{\psi }_j,{\psi }_l\}=0$. Here $\{·,·\}$ is the anticommutator and ${\psi }_j$ is the spinless fermionic operator associated to a particle on the site j. From the interpretation point of view, it is fruitful to note that the number $n_j={\psi }_j^{†}{\psi }_j$ of spinless femions on the j-th site equals one or zero if the z projection of the spin on that site j is 1 or −1, respectively. The advantage brought by this transformation is that the Hamiltonian, written in terms of the fermions, can be diagonalized by simple Bogoliubov transformations. Indeed, one finds

$$H=\mathcal{P}{H}^{(+)}\mathcal{P}+\mathcal{Q}{H}^{(-)}\mathcal{Q}.$$

(6)

Here, $\mathcal{P}$ is the projector onto the even parity sector of the fermionic Fock space, given by

$$\mathcal{P}\equiv \frac{1+{\prod }_{j=1}^N\left(1-2{\psi }_j^{†}{\psi }_j\right)}{2},$$

(7)

that satisfies $\left[\mathcal{P},H\right]=0$, with $\left[·,·\right]$ the commutator. Furthermore, $\mathcal{Q}\equiv 1-\mathcal{P}$. The Hamiltonians $H^{(\pm )}$ are given by

$$H^{(\pm )}=\frac{J}{2}\left\{\sum _{j=1}^{N-1}\left[{\psi }_j^{†}{\psi }_{j+1}^{}+{\psi }_j^{†}{\psi }_{j+1}^{†}-h{\psi }_j^{†}{\psi }_j^{}+\frac{h}{2}\right]\mp {\psi }_N^{†}{\psi }_1^{}\mp {\psi }_N^{†}{\psi }_1^{†}\right\}+\mathrm{h}.\mathrm{c}.$$

(8)

The problem is hence highly non-linear due to the presence of the projectors. However, as mentioned, such projectors commute with the Hamiltonian. The best strategy to diagonalize H is hence to separately diagonalize $H^{(+)}$ and $H^{(-)}$ and observe that only the eigenstates with an even and odd number of fermions, respectively, are also eigenstates of H.

It is here important to underline that the number of fermions, and hence their parity, is not related to the number of sites N, but rather to the magnetization in the z direction.

To proceed with the diagonalization of $H^{(+)}$ and $H^{(-)}$, we switch to the appropriate Fourier space, implicitly defining the ${\tilde{\psi }}_q^{(\pm )}$ operators via the Fourier series

$${\psi }_j^{}\equiv \frac{e^{i\frac{\pi }{4}}}{\sqrt{N}}\sum _{q\in {\Gamma }^{(\pm )}}{e}^{iqj}{\tilde{\psi }}_q^{(\pm )},$$

(9)

where ${\Gamma }^{(+)}\equiv \left\{\frac{\pi }{N}\left(2k+1\right)\right\}$ and ${\Gamma }^{(-)}\equiv \left\{\frac{2\pi }{N}k\right\}$, with $k=0,\dots ,N-1$. The Ensembles ${\Gamma }^{(+)}$ and ${\Gamma }^{(-)}$ for even and odd N are shown for clarity in Figure 2, since they will be of fundamental importance in the following.

The usefulness in the definition of two different Fourier series for the two parity sectors is associated with the different boundary conditions that need to be imposed on the fermionic operators in order to achieve compact forms for $H^{(+)}$ and $H^{(-)}$. Such point of view is discussed in detail in Ref. [29] and more briefly in Appendix A.

By substituting the Fourier expansion, we can now find the diagonal form of $H^{(\pm )}$ in terms of the operators ${\tilde{\psi }}_q^{(\pm )}$ respectively. We find

$$H^{(\pm )}=-J{\sum }_{q\in {\Gamma }^{(\pm )}}^{}\left[\left(h-cos\left(q\right)\right){\tilde{\psi }}_q^{(\pm )†}{\tilde{\psi }}_q^{(\pm )}+\frac{1}{2}sin\left(q\right)\left({\tilde{\psi }}_{-q}^{(\pm )†}{\tilde{\psi }}_q^{(\pm )†}+{\tilde{\psi }}_q^{(\pm )}{\tilde{\psi }}_{-q}^{(\pm )}\right)\right]+\frac{JhN}{2}.$$

(10)

It is here crucial to notice that two values of q need to be treated with particular care: $q=0$ and $q=\pi$. In those cases, no superconducting-like coupling is present and they need to be treated separately. However, this fact only generates subtleties for $q=0$. Subsequently, we rotate the fields, introducing

$$\begin{array}{ccc}\hfill {\tilde{\psi }}_q^{(+)}& \equiv & cos\left({\theta }_q\right){\chi }_q^{(+)}+sin\left({\theta }_q\right){\chi }_{2\pi -q}^{(+)†},\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill {\tilde{\psi }}_{q\ne 0}^{(-)}& \equiv & cos\left({\theta }_q\right){\chi }_q^{(-)}+sin\left({\theta }_q\right){\chi }_{2\pi -q}^{(-)†},\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill {\tilde{\psi }}_0^{(-)}& \equiv & {\chi }_0^{(-)},\hfill \end{array}$$

(13)

with ${\theta }_q$ satisfying

$$tan2{\theta }_q=\frac{sin\left(q\right)}{h-cos\left(q\right)}.$$

(14)

We thus find

$$H^{(+)}=-J\sum _{q\in {\Gamma }^{(+)}}\left[ϵ\left(q\right)\left({\chi }_q^{(+)†}{\chi }_q^{(+)}-\frac{1}{2}\right)\right]$$

(15)

and

$$H^{(-)}=-J\sum _{q\in {\Gamma }^{(-)},q\ne 0}\left[ϵ\left(q\right)\left({\chi }_q^{(-)†}{\chi }_q^{(-)}-\frac{1}{2}\right)\right]+Jϵ\left(0\right)\left({\chi }_0^{(-)†}{\chi }_0^{(-)}-\frac{1}{2}\right),$$

(16)

with

$$ϵ\left(q\right)\equiv \sqrt{{(h-cos\left(q\right))}^2+{sin}^2\left(q\right)}.$$

(17)

Importantly, $ϵ\left(q\right)$ is flat for $h=0$, while it has its minimum for $q=0$ ($q=\pi$) for $h>0$ ($h<0$). This fact is shown in Figure 3. Note that we did not have to separately treat the $q=\pi$ case since, in the parameter range we inspect, $h+1=ϵ\left(\pi \right)$. This was not the case for $q=0$. Indeed, ${lim}_{q\to 0}$ does not converge to the energy contribution related to the occupation of the fermion ${\chi }_0^{(-)}$, but rather to its opposite. Finally, it is worth stressing that the parity of the fermions $\psi$ is the same as the parity of the fermions $\chi$.

3. Results

In order to demonstrate the presence of the first-order quantum phase transition, we need to study the ground-state energy as a function of h and of the parity of N. To do so, we have to address the lowest energy state of both $H^{(+)}$ and $H^{(-)}$ that is compatible with the parity requirements and can hence contribute to the eigenstates of H. We will hence analyze the eight energies $E_{g/l,e/o}^{(\pm )}$ of the lowest energy eigenstates of H emerging from $H^{(+)}$ and $H^{(-)}$, for even (e) and odd (o) N, and for $h>0$ (g) and $h<0$ (l). We now proceed to the assessment of the eight cases.

(1)

For N even and $h>0$, the lowest energy state of $H^{(+)}$ with an even number of fermions is simply obtained by occupying all the energy levels, so

$$E_{g,e}^{(+)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(+)}}ϵ\left(q\right).$$

(18)

(2)

For N even and $h>0$, the lowest energy state of $H^{(-)}$ with an odd number of fermions is the one obtained by occupying all the fermionic states, except for one. Since ${\Gamma }^{(-)}$ contains $q=0$, and $q=0$ is the only state with positive energy, it is indeed favorable to keep it empty. By doing so, and by noticing that the energy contribution of keeping the $q=0$ empty is $Jϵ\left(0\right)$, the energy becomes

$$E_{g,e}^{(-)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(-)}}ϵ\left(q\right).$$

(19)

(3)

For N even and $h<0$, the lowest energy state of $H^{(+)}$ with an even number of fermions is simply obtained by occupying all the energy levels, so

$$E_{l,e}^{(+)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(+)}}ϵ\left(q\right).$$

(20)

(4)

For N even and $h<0$, the lowest energy state of $H^{(-)}$ with an odd number of fermions is the one obtained by occupying all the fermionic states, except for one. Since ${\Gamma }^{(-)}$ contains $q=0$, and $q=0$ is the only state with positive energy, it is indeed favorable to keep it empty. By doing so, and by noticing that the energy contribution of keeping the $q=0$ empty is $Jϵ\left(0\right)$, the energy becomes

$$E_{l,e}^{(-)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(-)}}ϵ\left(q\right).$$

(21)

(5)

For N odd and $h>0$, the lowest energy state of $H^{(+)}$ with an even number of fermions is the one obtained by filling all the states, except for the one with the smallest energy in the modulus (note that all energies are negative). For $h>0$, such minimum is for $q=\frac{\pi }{N},2\pi -\frac{\pi }{N}$. We hence obtain

$$E_{g,o}^{(+)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(+)}}ϵ\left(q\right)+Jϵ\left(\frac{\pi }{N}\right).$$

(22)

(6)

For N odd and $h>0$, the lowest energy state of $H^{(-)}$ with an even number of fermions is the one obtained by filling all the states. We obtain

$$E_{g,o}^{(-)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(-)}}ϵ\left(q\right)+Jϵ\left(0\right).$$

(23)

(7)

For N odd and $h<0$, the lowest energy state of $H^{(+)}$ with an even number of fermions is the one obtained by filling all the states except for the one with the smallest energy in modulus (note that all energies are negative), given by $q=\pi$. We hence find

$$E_{l,o}^{(+)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(+)}}ϵ\left(q\right)+Jϵ\left(\pi \right).$$

(24)

(8)

For N odd and $h<0$, the lowest energy state of $H^{(-)}$ is the one obtained by leaving both $q=0$ and $q=p$ empty, where p is the nearest to $\pi$ element in ${\Gamma }^{(-)}$. We hence obtain

$$E_{l,o}^{(-)}=-\frac{J}{2}\sum _{q\in {\Gamma }^{(-)}}ϵ\left(q\right)+Jϵ\left(p\right).$$

(25)

In the thermodynamic limit, $N\to \infty$, $E_{g,e}^{(+)}$ and $E_{g,e}^{(-)}$ are degenerate and constitute the ground-state manifold for N even and $h>0$. The same holds for $E_{l,e}^{(+)}$ and $E_{l,e}^{(-)}$ for $h<0$. It is clear that the ground-state energy is smooth in $h=0$.

As far as the odd N is concerned, and again considering the thermodynamic limit, $E_{l,o}^{(+)}$ and $E_{l,o}^{(-)}$ are degenerate, as well as $E_{g,o}^{(+)}$ and $E_{g,o}^{(-)}$. However, here a peculiar effect happens: the ground-state energy is not smoothly connected between $h<0$ and $h>0$. Indeed we find, and this is the central result of the present study, that

$$\underset{N\to \infty }{lim}\left(\frac{\partial E_{g,o}^{(+)}}{\partial h}{|}_{h\to 0^{+}}-\frac{\partial E_{l,o}^{(+)}}{\partial h}{|}_{h\to 0^{-}}\right)=-2J.$$

(26)

We have indeed found that a first-order quantum phase transition appears for odd N only in the quantum Ising chain in a transverse field.

It is here worth pointing out that, even though the full analysis of the ground state energy for both $h>0$ and $h<0$ can represent a direct approach and a useful check, restricting to $h>0$ would also have been possible by observing that the energy spectrum needs to be symmetric under the exchange $h\leftrightarrow -h$, and hence its derivative antisymmetric.

4. Discussion and Conclusions

In this work, we have analytically demonstrated that a first-order quantum phase transition (a discontinuity of the derivative of the ground state energy) is present as a function of the applied field in the case of an odd number of spins, while this is not the case for even N.

The crucial mathematical point is that in the odd case only; for $h=0$, the degeneration of the ground state is remarkably different from the $h\ne 0$ case. Indeed, while in both cases the energy dispersion $ϵ\left(q\right)$ becomes flat for $h=0$, only in the odd N case does that fact influence the ground-state degeneracy due to the parity constraints posed by the projectors $\mathcal{P}$ and $\mathcal{Q}$ appearing in the fermionic description of H. A further ingredient that is crucial for the existence of the effect is that the minimum of $ϵ\left(q\right)$ switches from $q=0$ to $q=\pi$ as h goes from positive to negative.

From the physical point of view, such mathematical statements translate into the following considerations: in the even N case for every h, the ground state is doubly degenerate, gapped, antiferromagnetic and no features appear. On the other hand, in the odd N case, the spectrum is not gapped since at least one kink must be present. Moreover, the kink states form a metallic band. This is due to the fact that not all the antiferromagnetic nearest neighbour interactions can be fulfilled. In the classical $h=0$ case, this reflects into a degeneracy $2N$ of the ground state, since such a kink can be equivalently positioned anywhere in the chain and a full spin flip does not change the energy. For finite h, the degeneracy is lifted by hybridization. In switching from $h<0$ to $h>0$, the most favorable majority spin orientation changes, and this reorganization of the ground state, allowed by the degeneracy at $h=0$, has the first-order quantum phase transition as a signature.

The implications and opportunities for further inspections that are opened by our discovery are, in our opinion, remarkable. Firstly, the effect of the first-order quantum phase transition on the observables needs to be clarified. Then, the finite temperature effects of the phenomenon, the quantum-quench-related thermodynamics of the model, the search for even–odd criticality without classical point and its physical interpretation and the quest for even–odd thermal phase transitions are all possible follow ups of our finding in the prototypical quantum Ising chain. Finally, an intriguing effect should also be inspected in a more general framework: both in our case and in the cases reported earlier in literature, the jump in the ground state energy associated to even–odd criticality is not extensive in the number of particles, but rather it is of order 1. While the mathematical origin is clear in the Wigner–Jordan transformation approach, the physical implications and the generality of this property need to be further inspected.