Transverse Ising Chain (Pure System)

Abstract

The one-dimensional pure Ising model in a transverse field is the simplest solvable model that shows a quantum phase transition. The properties of this model are investigated in detail in Chap. 2. First, the quantum critical point is identified by employing duality argument. The Hamiltonian of the transverse Ising chain is then diagonalised and its ground-state properties are investigated by using the Jordan-Wigner transformation, which maps the system to a free fermion system. Next, approximate methods such as the exact diagonalisation method with finite size scaling, which can be also applied to interacting fermion systems, and the real-space renormalisation group method are introduced. The finite temperature property and an experimental study of the transverse Ising chain are mentioned in the last part of this chapter. Details of the Jordan-Wiger transformation, diagonalisation of a general Hamiltonian quadratic in fermion operators, and the calculation of correlation functions of the transverse Ising chain are included in appendices.

Appendix 2.A

2.1.1 2.A.1 Jordan-Wigner Fermions

To check that the operators c _i and $c_{i}^{\dag}$ satisfy fermionic anti-commutation relations

$$ \bigl[c_{i}, c_{i}^{\dag}\bigr]_{+} = \delta_{ik}, \qquad [c_{i},c_{k}]_{+} = 0 \quad \mbox{and} \quad \bigl[c_{i}^{\dag},c_{k}^{\dag} \bigr]_{+} = 0, $$

(2.A.1)

one uses the simple relations obeyed by Pauli spin operators

$$ S_{i}^{-}S_{i}^{+} = \frac{1}{2}\bigl(1-S_{i}^{z}\bigr); \quad S_{i}^{+}S_{i}^{-} = \frac{1}{2}\bigl(1+S_{i}^{z}\bigr); \quad \exp \biggl( \frac{i\pi}{2}S_{i}^{z} \biggr) = iS_{i}^{z}. $$

(2.A.2)

So that, the Jordan-Wigner transformation (2.2.2a), (2.2.2b) can be rewritten as

(2.A.3a)

(2.A.3b)

Using (2.A.3a) and (2.A.3b), one can immediately check that

$$ c_{i}c_{i}^{\dag} + c_{i}^{\dag}c_{i} = S_{i}^{-}S_{i}^{+} + S_i^+ S_i^- = 1. $$

(2.A.4)

To prove $[c_{i},c_{k}^{\dag} ]_{+} = 0$, for k≠i, without loss of generality we assume k<i. Now

$$ c_{k}c_{i}^{\dag} + c_{i}^{\dag}c_{k} = S_{k}^{-} \prod_{j=k}^{i-1} \bigl[-S_{j}^{z}\bigr]S_{i}^{+}+ S_{i}^{+} \prod_{j=k}^{i-1} \bigl[-S_{j}^{z}\bigr]S_{k}^{-}. $$

(2.A.5)

If one now uses the fact

$$S_{m}^{-}S_{m}^{z} = -S_{m}^{z}S_{m}^{-} $$

the right hand side of (2.A.5) vanishes identically. Similarly, one can derive the other anti-commutation relations.

To express the spin Hamiltonian in terms of the fermion operators, we use the relation, which can be checked using (2.A.3a) and (2.A.3b). Now from (2.A.2), we get

$$ S_{i}^{z}= 2S_{i}^{+}S_{i}^{-}-1=2c_{i}^{\dag}c_{i}-1, $$

(2.A.6)

which can be easily derived using (2.A.3a), (2.A.3b). The coupling term

$$ S_{i}^{x}S_{i+1}^{x}= \bigl[S_{i}^{+}+S_{i}^{-}\bigr] \bigl[S_{i+1}^{+}+S_{i+1}^{-}\bigr] $$

(2.A.7)

is now to be rewritten in terms of fermions. Consider the product

$$ c_{i}^{\dag} c_{i+1} = S_{i}^{+} \bigl[-S_{i}^{z}\bigr]S_{i+1}^{-} \quad \bigl(\mbox{from (2.A.3a), (2.A.3b)}\bigr). $$

(2.A.8)

But $S_{i}^{+}S_{i}^{z}=-S_{i}^{+}$, so we get

$$ c_{i}^{\dag}c_{i+1} = S_{i}^{+}S_{i+1}^{-}. $$

(2.A.9a)

Similarly one can derive the following relations

(2.A.9b)

(2.A.9c)

(2.A.9d)

The coupling term can be written in terms of fermion operators which are also only coupled to nearest neighbours. This result is a consequence of the Ising character $S_{i}^{2}= + 1$ of the original degrees of freedom. Collecting all the terms in (2.A.5) and (2.A.9a), (2.A.9b), (2.A.9c), (2.A.9d), one can arrive at the Hamiltonian (2.2.6). One should note here that if the original spin Hamiltonian incorporates next-nearest neighbour interaction, the resulting fermion Hamiltonian includes four-fermion term.

2.1.2 2.A.2 To Diagonalise a General Hamiltonian Quadratic in Fermions

We wish to diagonalise a general quadratic Hamiltonian of the form

$$ H = \sum_{ij} c_{i}^{\dag} A_{ij} c_{j} + \frac{1}{2} \sum _{ij} c_{i}^{\dag} B_{ij} c_{j}^{\dag}+ h.c. $$

(2.A.10)

where c _i and $c_{i}^{\dag}$ are fermion annihilation and creation operators respectively. For system size N, A and B are both N×N matrices. Hermiticity of H demands A to be a Hermitian matrix and anti-commutation of fermion operators demands B to be antisymmetric. Both can be chosen to be real.

Equation (2.A.10) is most effectively diagonalised by Bogoliubov transformation employed for fermion operators by Lieb et al. [245]. In particular, the technique is directly applicable for quite general nearest neighbour interactions, including for example quasiperiodic and random ones [77]. One wants to rewrite the Hamiltonian in a diagonal form using variables η _q and $\eta_{q}^{\dag}$ as

$$ H = \sum_{q}\omega_{q} \eta_{q}^{\dag} \eta_{q} + \mbox{const}, $$

(2.A.11)

ω _q are one fermion energies. One makes a linear transformation of the form [245]

(2.A.12a)

(2.A.12b)

where g _qi and h _qi can be chosen to be real. For η _q ’s to satisfy fermionic anti-commutation relations we require

(2.A.13a)

(2.A.13b)

If (2.A.11) holds, then we must have

$$ [\eta_{q},H]_{+} -\omega_{q} \eta_{q} =0. $$

(2.A.14)

Using (2.A.12a), (2.A.12b) in (2.A.14) one finds

(2.A.15a)

(2.A.15b)

The above coupled equations can be written in the following form

(2.A.16a)

(2.A.16b)

where the components of the 2N vectors Φ and Ψ are obtained from the matrices g and h as

(2.A.17a)

(2.A.17b)

One fermion energies ω _q are obtained from the eigenvalues of a N×N matrix

(2.A.18a)

(2.A.18b)

For ω _q ≠0, either (2.A.18a) or (2.A.18b) is solved for Φ _q or Ψ _q and the other vector is obtained from (2.A.17a) or (2.A.17b). For ω _q =0 these vectors are determined by (2.A.18a), (2.A.18b) or more simply by (2.A.17a), (2.A.17b), their relative sign being arbitrary.

The achievement of the method is obvious. The problem of diagonalising a 2N×2N matrix has been reduced to the eigenvalue problem of a N×N matrix [77, 245]. Even when the eigenvalues of M=(A+B)(A−B)=(A−B)(A+B)^T cannot be found analytically, numerical studies can be made for large systems, for example, in one dimensional nearest-neighbour model with random transverse field (see Sect. 5.3). The price to pay is that one only obtains the one particle energies directly. Multiparticle states have to be built up by linear superposition of the one particle ones. Since A is symmetric and B is antisymmetric both (A−B)(A+B) and (A+B)(A−B) are symmetric and at least positive semi-definite. Thus all the λ _q ’s are real and all the Φ _q ’s and Ψ _q ’s can be chosen to be real as well as orthogonal. If Φ _q ’s are normalised, Ψ _q ’s are automatically normalised. To evaluate the constant in (2.A.11) we use the trace invariance of the Hamiltonian under the canonical transformation to the variables η. From (2.A.10)

$$\mathrm{Tr} H = 2^{N-1} \sum_{i} A_{ii}. $$

Again

$$\mathrm{Tr} H = 2^{N-1}\sum_{q} \omega_{q}+2^{N} \times\mbox{const}, $$

from (2.A.11). The constant is thus

$$ \frac{1}{2} \biggl[\sum_{i}A_{ii}- \sum_{q}\omega_{q}\biggr]. $$

(2.A.19)

The complete diagonalised form of the Hamiltonian (2.A.10) is therefore written as

$$ H = \sum_{q}\omega_{q} \eta_{q}^{\dag} \eta_{q} + \frac{1}{2} \biggl(\sum_{i}A_{ii}-\sum _{q}\omega_{q}\biggr). $$

(2.A.20)

We illustrate the above diagonalisation procedure using the example of the one dimensional anisotropic XY chain in a transverse field described by the Hamiltonian (see Sect. 10.1.2)

$$ H = -\frac{J}{2} \sum_{i} \bigl[ (1+ \gamma)S_{i}^{x}S_{i+1}^{x} + (1-\gamma) S_{i}^{y} S_{i+1}^{y} \bigr] - \varGamma\sum_{i}S_{i}^{z} $$

(2.A.21)

where γ is the measure of anisotropy. γ=0 corresponds to the isotropic XY chain in a transverse field, γ=1 corresponds to the transverse Ising chain. The Hamiltonian (2.A.21) is written in terms of Jordan-Wigner fermions as

$$ H = -2 \Biggl( \sum_{i}\biggl(c_{i}^{\dag}c_{i}- \frac{1}{2}\biggr) + \frac{1}{2} \overline{\lambda} \sum _{i=1}^{N} \bigl(c_{i}^{\dag} c_{i+1}+ c_{i}c_{i+1}^{\dag} +\gamma c_{i}^{\dag}c_{i}^{\dag} +\gamma c_{i}c_{i+1}\bigr) \Biggr), $$

(2.A.22)

where $\overline{\lambda}= J/\varGamma$. Equation (2.A.22) can be put in the general form

$$ H = \biggl( \sum_{ij} c_{i}^{\dag} A_{ij} c_{j} + \frac{1}{2} \sum _{ij} \bigl(c_{i}^{\dag} B_{ij} c_{j}^{\dag}+h.c. \bigr) \biggr) + N, $$

(2.A.23)

where

(2.A.24)

The eigenvalue problem (2.A.18a), (2.A.18b) is solved considering an ansatz wave function exp(i q⋅R _j ) and the excitation spectrum is given as

$$ \omega_{q} = \sqrt{ (\gamma\overline{\lambda})^{2} \sin^{2}q + (1-\overline{\lambda}\cos q)^{2}}, $$

(2.A.25)

so that the diagonalised form of the Hamiltonian (with ∑ _i A _ii =−N) is given by

$$ H = 2\sum_{q} \sqrt{ (\gamma\overline{\lambda} \sin q)^{2}+ (1-\overline{\lambda} \cos q)^{2} } \eta_{q}^{\dag} \eta_{q} - \sum _{q}\omega_{q} $$

(2.A.26)

which reduces to the transverse Ising Hamiltonian (2.2.18) for γ=1.

2.1.3 2.A.3 Calculation of Correlation Functions

Longitudinal spin-spin correlation function is defined as

(2.A.27)

In terms of Jordan-Wigner fermions (with j>i)

$$ C_{ij}^{x} = \langle\psi_{0}| \bigl(c_{i}^{\dag}+c_{i}\bigr) \exp \Biggl( -i \pi \sum_{i=0}^{j-1} c_{l}^{\dag} c_{l} \Biggr) \bigl(c_{j}^{\dag} +c_{j}\bigr)| \psi_{0} \rangle $$

(2.A.28)

where the averages are calculated over the ground state. One can now verify using a representation in which $c_{l}^{\dag}c_{l}$ is diagonal, that

Defining $A_{l} = c_{l}^{\dag}+c_{l}$ and $B_{l} = c_{l}^{\dag}-c_{l}$ and noting that $A_{l}^{2}= 1$, we have

(2.A.29)

The complicated expression can be simplified using Wick’s theorem, and following relations

$$ \langle \psi_{0}|A_{i}A_{j}| \psi_{0} \rangle = \langle\psi_{0}| \bigl( \delta_{ij}-c_{j}^{\dag}c_{i} + c_{i}^{\dag}c_{j}\bigr)|\psi_{0} \rangle= \delta_{ij}, $$

(2.A.30)

and

(2.A.31)

Only nonzero contractions are 〈A _j B _i 〉 and 〈B _i A _j 〉, since 〈A _i A _j 〉 and 〈B _i B _j 〉 never occur. Defining 〈B _i A _i+r 〉=−〈A _i+r B _i 〉=G _i i+r =G _r =G _−r , the correlation function is given by a determinant

(2.A.32)

which is of size r. Similarly one can evaluate the transverse correlation function defined as

(2.A.33)

One can now check that,

(2.A.34)

so that

$$ C_{i\,i+r}^{z} = -G_{i\,i+r}G_{i+r\,i}. $$

(2.A.35)

To evaluate G _i i+r we consider the inverse Fourier-Bogoliubov transformation (see (2.2.7a), (2.2.7b) and (2.2.13)) given by

$$ c_i^{\dag} = \sqrt{\frac{1}{N}}\sum _q e^{i q R_i} \bigl(u_q \eta_q^{\dag} + i v_q \eta_{-q} \bigr) $$

(2.A.36)

so that we get

$$ c_i^{\dag} + c_i = \sqrt{\frac{1}{N}} \sum_q \bigl(e^{i q R_i}u_q \eta_q^{\dag} + e^{-i q R_i}u_q \eta_q -i e^{-iq R_i}v_q\eta_{-q}^{\dag} + i e^{i q R_i}v_q\eta_{-q} \bigr) $$

(2.A.37a)

and

(2.A.37b)

Note that u _q and v _q are given by (2.2.17). In the ground state, G _r =G _i i+r =〈ψ ₀|B _i A _i+r |ψ ₀〉. At a finite temperature T, G _r =G _i i+r =〈B _i A _i+r 〉 _β , where 〈B _i A _j 〉 _β denotes an average over the canonical ensemble at temperature k _B T=1/β. Thus,

(2.A.38)

The average fermion occupation at temperature T

$$\bigl\langle\eta_{q}^{\dag} \eta_{q} \bigr \rangle_{\beta} = \bigl\langle\eta_{-q}^{\dag} \eta_{-q} \bigr\rangle_{\beta} = \bigl(\exp (\beta\cdot2 \omega_{q})+1\bigr)^{-1} $$

so that,

(2.A.39)

For ground state tanh(βω _q /2)=1, we have

$$ G_{r} = \int_{-\pi}^{\pi} \frac{dq}{2\pi} e^{i q r} \biggl(\frac{1 + \overline{\lambda}e^{iq}}{1 + \overline{\lambda}e^{-iq}} \biggr)^{1/2}. $$

(2.A.40)

One can now evaluate the following values G _r for some special values of $\overline{\lambda}$, e.g.,

etc.

In order to evaluate the determinant (2.A.32), let us define

(2.A.41)

In terms of this, (2.A.32) is rewritten as

$$ C_{i\,i+r}^{x} = \det \left ( \begin{array}{l@{\quad}l@{\quad}l@{\quad}l} D_{0} & D_{-1} & \cdots& D_{-r+1} \\ D_{1} & D_{0} & \cdots& D_{-r+2} \\ \cdots& \cdots& \cdots& \cdots\\ D_{r-1} & D_{r-2} & \cdots& D_{0} \end{array} \right ). $$

(2.A.42)

This Toeplitz determinant can be evaluated by using the Szegö’s theorem [393] (see also Ref. [144]), which stated as follows. We refer to Ref. [268] for the proof.

Szegö’s Theorem

Assume that $\hat{D}(e^{iq})$ and $\ln\hat{D}(e^{iq})$ are continuous cyclic functions of q for q∈[−π,π]. Let

$$D_p = \int_{-\pi}^{\pi} \frac{dq}{2\pi} \hat{D}\bigl(e^{iq}\bigr) e^{-i q p}, \qquad d_p = \int_{-\pi}^{\pi} \frac{dq}{2\pi} \ln\hat{D}\bigl(e^{iq}\bigr) e^{-i q p}. $$

The Toeplitz determinant

$$\varDelta_r = \det \left ( \begin{array}{l@{\quad}l@{\quad}l@{\quad}l} D_{0} & D_{-1} & \cdots& D_{-r+1} \\ D_{1} & D_{0} & \cdots& D_{-r+2} \\ \cdots& \cdots& \cdots& \cdots\\ D_{r-1} & D_{r-2} & \cdots& D_{0} \end{array} \right ) $$

is given by

$$ \lim_{r\to\infty} \frac{\varDelta_r}{e^{r d_0}} = \exp \Biggl[\sum _{p=1}^{\infty} p d_{-p} d_p \Biggr] , $$

(2.A.43)

whenever the sum in (2.A.43) converges.

Applying this theorem to (2.A.41) and (2.A.42), one finds that the longitudinal correlation function behaves as $C_{i\,i+r}^{x}\approx A e^{-r/\xi}$ for r≫1 and the correlation length ξ is given by

(2.A.44)

(Note that, when $\overline{\lambda} < 1$, one needs additional algebras because $\ln\hat{D}(e^{iq})$ is not cyclic with respect to q. See Ref. [265].) Substituting (2.2.16) for ω _q and recovering the lattice constant a and scale of energy Γ, one can write (2.A.44) as

$$ \frac{1}{\xi} = \left\{ \begin{array}{l@{\quad}l} \int_{0}^{\pi/a}\frac{dq}{\pi} \ln\coth\beta\varGamma\sqrt{1 + 2\overline{\lambda}\cos qa + \overline {\lambda}^2} & \mbox{for }\overline{\lambda} > 1 \\ [3pt] \int_{0}^{\pi/a}\frac{dq}{\pi} \ln\coth\beta\varGamma\sqrt{1 + 2\overline{\lambda}\cos qa + \overline {\lambda}^2} - \frac{1}{a}\ln\overline{\lambda} & \mbox{for }\overline{\lambda} < 1 \end{array} \right.. $$

(2.A.45)

Let us now define $\mathrm{\varDelta} = |1 - \overline{\lambda}|/2\varGamma a^{2}$ and c=2Γa. By making a→0 with fixing β, $\mathrm{\varDelta}$ and c, and defining a new variable of integral y=βcq, we reach an expression for the continuous limit [338]

$$ \frac{1}{\xi} = \frac{1}{\beta c} f\bigl(\beta\mathrm{ \varDelta}c^2\bigr), $$

(2.A.46)

where

$$ f(x) = \left\{ \begin{array}{l@{\quad}l} \int_{0}^{\infty}\frac{dy}{\pi} \ln\coth\frac{\sqrt{y^2 + x^2}}{2} & \mbox{for }x > 0 \\ [3pt] \int_{0}^{\infty}\frac{dy}{\pi} \ln\coth\frac{\sqrt{y^2 + x^2}}{2} - x & \mbox{for }x < 0 \end{array} \right.. $$

(2.A.47)

One can easily show that the function f(x) is not only continuous but also smooth at x=0. This fact shows that the transverse Ising chain has no singularity at finite temperatures. In special cases, f(x) takes following forms:

$$ f(x) = \left\{ \begin{array}{l@{\quad}l} \sqrt{\frac{2 x}{\pi}}e^{-x} & (x \to\infty) \\ [4pt] \frac{\pi}{4} & (x = 0) \\ [3pt] -x + \sqrt{2|x|}{\pi}e^{-|s|} & (x \to-\infty) \end{array} \right.. $$

(2.A.48)

Substituting this in (2.A.46) and recovering the original parameters, one obtains

$$ \frac{1}{\xi} = \left\{ \begin{array}{l@{\quad}l} \frac{1}{a} \sqrt{\frac{T}{\pi}(\overline{\lambda} - 1)}e^{-2(\overline{\lambda} - 1)/T} & (\overline{\lambda} > 1,\ T \to0) \\[4pt] \frac{\pi}{4}\frac{T}{2a} & (\overline{\lambda} = 1) \\[3pt] \frac{1-\overline{\lambda}}{a} + \frac{1}{a}\sqrt{\frac{T}{\pi}(1-\overline{\lambda})} e^{-2(1 - \overline{\lambda})/T} & (\overline{\lambda} < 1,\ T \to0) \end{array} \right.. $$

(2.A.49)

At the zero temperature, one finds $\xi= a/(1-\overline{\lambda})$ for $\overline{\lambda} < 1$, so that the exponent ν turns out to be 1.

The result (2.A.49) implies that the low temperature phase of the transverse Ising chain is divided into three phases by crossover lines $T\approx|\overline{\lambda} - 1|$. We show a schematic phase diagram in Fig. 2.10.