Thermodynamics of the classical spin triangle

Heinz-Jürgen Schmidt; Christian Schröder

doi:10.1515/zna-2022-0034

Article

Thermodynamics of the classical spin triangle

Heinz-Jürgen Schmidt and Christian Schröder

Published/Copyright: April 13, 2022

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From the journal Zeitschrift für Naturforschung A Volume 77 Issue 11

Abstract

The classical spin system consisting of three spins with Heisenberg interaction is an example of a completely integrable mechanical system. In this paper, we explicitly calculate thermodynamic quantities such as density of states, specific heat, susceptibility and spin autocorrelation functions. These calculations are performed (semi-)analytically and shown to agree with corresponding Monte Carlo simulations. It is shown that the thermodynamic functions depend qualitatively on the character of the system in terms of its frustration, especially w. r. t. their low temperature limit. For the long-time autocorrelation function, we find, for certain values of the coupling constants, a decay to constant values in the form of an 1/t damped harmonic oscillation and propose a theoretical explanation.

Keywords: autocorrelation function; integrable systems; thermodynamics

Corresponding author: Heinz-Jürgen Schmidt, Department of Physics, Osnabrück University, Osnabrück D-49069, Germany, E-mail: hschmidt@uos.de

Author contributions: All the authors have accepted responsibility for the entire content of this submitted manuscript and approved submission.
Research funding: None declared.
Conflict of interest statement: The authors declare no conflicts of interest regarding this article.

Appendix A: Details of the explicit time evolution

Instead of u we will use the variable x given by

(A1) x = x 0 + g u ,

where the constants x₀ and g will be determined later such that the Weierstrass differential equation [[30], 23.3.10] is obtained. Also v and w can be linearly expressed in terms of the variable x in the form:

(A2) v = J 3 − J 1 J 2 − J 3 x g + v 0 ,

(A3) w = J 1 − J 2 J 2 − J 3 x g + w 0 .

We consider the time derivative of x:

(A4) x ̇ = ( A 1 ) g u ̇

(A5) = ( 22 ) ± g J 3 − J 2 1 − u 2 − v 2 − w 2 + 2 u v w = ( 14 ) g J 3 − J 2 δ .

By substituting (A2), (A3) and (A1), the square of (A5) can be written as a 3rd order polynomial Π(x). g and x₀ will be chosen such that the cubic term of Π(x) reads 4x³ and the quadratic term of Π(x) vanishes and hence

(A6) d x d t 2 = x ̇ 2 = g 2 J 3 − J 2 2 δ 2 = Π ( x ) = 4 x 3 − g 2 x − g 3 .

This is achieved by setting

(A7) g = − 1 2 J 1 − J 2 J 1 − J 3 ,

and

(A8) x 0 = 1 6 − J 1 2 − J 2 2 − J 3 2 + J 1 J 3 + J 2 J 3 + J 1 J 2 + 2 J 1 − J 2 − J 3 ε + 2 J 2 J 3 − J 1 J 2 + J 3 σ .

The explicit form of the coefficients g₂ and g₃ is more complicated:

(A9) g 2 = 1 3 J 1 2 J 2 2 ( σ + 1 ) ( σ + 3 ) + J 3 2 ( σ + 1 ) ( σ + 3 ) − J 3 ( σ + 3 ) ϵ − J 2 J 3 σ ( σ + 2 ) + ( σ + 3 ) ϵ + ϵ 2 − J 1 J 2 3 ( σ + 2 ) + J 2 J 3 2 σ ( σ + 2 ) − 6 J 3 ( σ + 2 ) ϵ + ϵ 2 + J 2 2 J 3 σ ( σ + 2 ) + ( σ + 3 ) ϵ + J 3 J 3 + ϵ J 3 ( σ + 2 ) + ϵ + J 2 2 J 3 2 ( σ + 1 ) ( σ + 3 ) − J 3 ( σ + 3 ) ϵ + ϵ 2 − J 1 3 J 2 ( σ + 2 ) + J 3 ( σ + 2 ) − 2 ϵ − J 2 3 J 3 ( σ + 2 ) − 2 ϵ − J 2 J 3 J 3 + ϵ J 3 ( σ + 2 ) + ϵ + J 3 2 J 3 + ϵ 2 + J 1 4 + J 2 4 ,

and

(A10) g 3 = 1 108 ∑ i = 1 8 g 3 ( i ) ,

where

(A11) g 3 ( 1 ) = 4 J 1 6 + 4 J 2 6 + 4 J 3 3 J 3 + ϵ 3 − 6 J 2 5 J 3 ( σ + 2 ) − 2 ϵ − 6 J 2 J 3 2 J 3 + ϵ 2 J 3 ( σ + 2 ) + ϵ ,

(A12) g 3 ( 2 ) = − 6 J 1 5 J 2 ( σ + 2 ) + J 3 ( σ + 2 ) − 2 ϵ + 3 J 2 4 J 3 2 ( σ ( 7 σ + 10 ) − 1 ) − 2 J 3 ( 2 σ + 5 ) ϵ + 4 ϵ 2 ,

(A13) g 3 ( 3 ) = 3 J 2 2 J 3 J 3 3 ( σ ( 7 σ + 10 ) − 1 ) + 4 J 3 ( 2 σ + 3 ) ϵ 2 − 2 J 3 2 σ 2 − 2 ϵ − 2 ϵ 3 ,

(A14) g 3 ( 4 ) = 3 J 1 4 J 2 2 ( σ ( 7 σ + 10 ) − 1 ) + J 3 2 ( σ ( 7 σ + 10 ) − 1 ) − 2 J 2 J 3 5 σ 2 − 11 + ( 2 σ + 5 ) ϵ − 2 J 3 ( 2 σ + 5 ) ϵ + 4 ϵ 2 ,

(A15) g 3 ( 5 ) = 2 J 2 3 J 3 3 ( σ ( σ ( 2 σ − 15 ) − 18 ) + 13 ) − 3 J 3 ( σ + 4 ) ϵ 2 − 3 J 3 2 σ 2 − 2 ϵ + 2 ϵ 3

(A16) g 3 ( 6 ) = 2 J 1 3 J 2 3 ( σ ( σ ( 2 σ − 15 ) − 18 ) + 13 ) + J 3 3 ( σ ( σ ( 2 σ − 15 ) − 18 ) + 13 ) − 3 J 3 ( σ + 4 ) ϵ 2 − 3 J 2 J 3 2 ( σ ( ( σ − 1 ) σ + 4 ) + 11 ) − 4 J 3 ( σ + 2 ) 2 ϵ + ( σ + 4 ) ϵ 2 − 3 J 3 2 σ 2 − 2 ϵ − 3 J 2 2 J 3 ( σ ( ( σ − 1 ) σ + 4 ) + 11 ) + σ 2 − 2 ϵ + 2 ϵ 3 ,

(A17) g 3 ( 7 ) = − 6 J 1 J 2 5 ( σ + 2 ) + J 2 3 J 3 2 ( σ ( ( σ − 1 ) σ + 4 ) + 11 ) − 4 J 3 ( σ + 2 ) 2 ϵ + ( σ + 4 ) ϵ 2 + J 3 J 2 J 3 3 5 σ 2 − 11 + 2 J 3 σ ϵ 2 − 4 J 3 2 ( σ + 2 ) 2 ϵ − 4 ϵ 3 + J 2 2 J 3 3 ( σ ( ( σ − 1 ) σ + 4 ) + 11 ) + 2 J 3 σ ϵ 2 + 2 J 3 2 ( σ ( σ + 6 ) + 6 ) ϵ + ϵ 3 + J 2 4 J 3 5 σ 2 − 11 + ( 2 σ + 5 ) ϵ + J 3 2 J 3 + ϵ 2 J 3 ( σ + 2 ) + ϵ ,

(A18) g 3 ( 8 ) = 3 J 1 2 J 2 4 ( σ ( 7 σ + 10 ) − 1 ) + 2 J 2 2 J 3 2 ( σ ( σ ( 4 σ + 3 ) + 18 ) + 33 ) − 2 J 3 ( σ ( σ + 6 ) + 6 ) ϵ + 2 ( 2 σ + 3 ) ϵ 2 + J 3 J 3 3 ( σ ( 7 σ + 10 ) − 1 ) + 4 J 3 ( 2 σ + 3 ) ϵ 2 − 2 J 3 2 σ 2 − 2 ϵ − 2 ϵ 3 − 2 J 2 J 3 3 ( σ ( ( σ − 1 ) σ + 4 ) + 11 ) + 2 J 3 σ ϵ 2 + 2 J 3 2 ( σ ( σ + 6 ) + 6 ) ϵ + ϵ 3 − 2 J 2 3 J 3 ( σ ( ( σ − 1 ) σ + 4 ) + 11 ) + σ 2 − 2 ϵ .

For statistical considerations parts of the phase space with zero measure can be neglected (but note that the dos may diverge for states with aperiodic motion according to (48)). Hence we can restrict ourselves to the “generic case” where certain exceptions are excluded, see [11]. In this generic case the polynomial Π(x) will have three real simple roots x₁ < x₂ < x₃ satisfying x₁ + x₂ + x₃ = 0 and Π(x) > 0 for x₁ < x < x₂. The explicit form of the roots is known but of overwhelming complexity if expressed in terms of the physical parameters ɛ, σ, J₁, J₂, J₃.

It follows [11] that in the generic case (A5) has the solution

(A19) x ( t ) = ℘ t + t 0 ; g 2 , g 3 ,

with the above-mentioned parameters g₂, g₃ and the imaginary parameter t₀ can be expressed through an elliptical integral:

(A20) t 0 ≔ i ∫ x 2 x 3 d x ′ 4 x ′ 3 − g 2 x ′ − g 3 = i ∫ − ∞ x 1 d x ′ 4 x ′ 3 − g 2 x ′ − g 3 = i x 3 − x 1 K x 3 − x 2 x 3 − x 1 ,

see [[30], 23.6.34-35] and [[40], 17.4.61 ff]. Moreover, this solution will be T -periodic where

(A21) T 2 = ∫ x 1 x 2 d x ′ 4 x ′ 3 − g 2 x ′ − g 3 = ∫ x 3 ∞ d x ′ 4 x ′ 3 − g 2 x ′ − g 3 = 1 x 3 − x 1 K x 2 − x 1 x 3 − x 1 .

Hence, for given coupling constants J₁, J₂, J₃, the period T can be viewed as a function T ( σ , ε ) although the explicit form of this function is too complicated to be reproduced here.

Appendix B: Distribution of random variables with a saddle point

We will prove the following proposition which is tailored to its application in Section 7.2.2 and not formulated as general as possible:

Proposition 1

Let Z ⊂ R 2 be an open bounded domain and φ : Z → R + a continuous probability distribution. Further, let f : Z → R be a smooth function (“random variable”) with a saddle point z 0 ∈ Z such that φ(z₀) > 0. Let ρ f : R → R + be the corresponding probability distribution, i.e., satisfying

(B1) ∫ u 1 u 2 ρ f ( u ) d u = ∫ Z ( f ; u 1 , u 2 ) φ ( z ) d z

for all u 1 < u 2 ∈ R and

(B2) Z ( f ; u 1 , u 2 ) ≔ { z ∈ Z | u 1 ≤ f ( z ) ≤ u 2 } .

Then ρ^f has a logarithmic singuarity at u = u 0 ≔ f z 0 .

Proof

Without loss of generality we may assume u 0 = f z 0 = 0 . Then there exist local coordinates x, y in a neighborhood of z₀ such that z₀ has the coordinates (0, 0) and f(x, y) = x² − y² or, after a rotation with π/4, f(x, y) = x y for, say, |x|, |y| ≤ R and some R > 0. Consider an arbitrary ϵ > 0 and choose u₂ = −u₁ = ϵ such that

(B3) p ( ϵ ) ≔ ∫ − ϵ ϵ ρ f ( u ) d u = ∫ Z ( f ; − ϵ , ϵ ) φ ( z ) d z ≥ ∫ H ( ϵ ) φ ( x , y ) d x d y ,

where H(ϵ) is the hyperbolic region

(B4) H ( ϵ ) ≔ { ( x , y ) ∈ R 2 | x | , | y | ≤ R , and x y ≤ ϵ } ,

see Figure 16. By assumption, φ(0, 0) > 0 and, since φ is continuous, we may choose R > 0 so small such that

(B5) φ ( x , y ) ≥ c > 0 for all ( x , y ) ∈ H ( ϵ ) .

This implies

(B6) p ( ϵ ) ≥ ( B 3 ) ∫ H ( ϵ ) φ ( x , y ) d x d y ≥ c H ( ϵ ) ,

where H ( ϵ ) denotes the area of H(ϵ) given by

(B7) H ( ϵ ) = 4 ϵ R R + ∫ ϵ / R R ϵ x d x = 4 ϵ − ϵ log ϵ + 2 ϵ log R ,

see Figure 16. If u↦ρ^f(u) would be continuous in a neighborhood of u = 0 then it would follow that

(B8) ρ f ( 0 ) = ( B 3 ) lim ϵ → 0 p ( ϵ ) 2 ϵ ≥ ( B 6 , B 7 ) lim ϵ → 0 2 c 1 − log ϵ + 2 log R ,

which is a contradiction due to the divergence of −log ϵ. Hence ρ^f(0) is divergent and the singularity is, at least, of logarithmic order.

The singularity is exactly of logarithmic order since the contribution to ρ^f(0) from other possible zeroes f z ν = 0 except the considered saddle point at z₀ would be of order O(1) for regular zeroes, or of order O(ϵ) for local maxima or minima, or of order O(log ϵ) for other saddle points.□

$Figure 16: Plot of the hyperbolic domain H(ϵ) given by |x y| ≤ ϵ and |x|, |y| ≤ R. The area of the intersection of H(ϵ) with the positive quadrant is given by ϵ R R + ∫ ϵ / R R ϵ x d x $\frac{{\epsilon}}{R}R+{\int }_{{\epsilon}/R}^{R}\frac{{\epsilon}}{x}\,\mathrm{d}x$ .$

Figure 16:

Plot of the hyperbolic domain H(ϵ) given by |x y| ≤ ϵ and |x|, |y| ≤ R. The area of the intersection of H(ϵ) with the positive quadrant is given by ϵ R R + ∫ ϵ / R R ϵ x d x .

Appendix C: Asymptotic expansions of Fourier integrals involving logarithmic singularities

We consider the case of a distribution function ρ(ω) with a logarithmic singularity at ω = ω₀ and will investigate the decay of the corresponding Fourier transform ρ ̂ ( t ) for t → ∞. More specifically, we assume that ρ(ω) is of the form

(C1) ρ ( ω ) = ϕ ( ω ) log ω − ω 0 ,

where ϕ(ω) is N times continuously differentiable for γ₁ < ω < γ₂ and γ₁ < ω₀ < γ₂. For our purposes we may assume that ϕ(ω) is a real function. Then we consider the Fourier integral

(C2) ρ ̂ ( t ) = ∫ γ 1 γ 2 ϕ ( ω ) log ω − ω 0 exp ( i ω t ) d ω ,

and the asymptotic expansion of ρ ̂ ( t ) for t → ∞. This problem has been solved in [[41], Th. 4] for the one-sided Fourier integral

(C3) A = ∫ ω 0 γ 2 ϕ ( ω ) log ω − ω 0 exp ( i ω t ) d ω .

We will utilize this solution to obtain the asymptotic expansion for the two-sided Fourier integral (C2). For this purpose, we will quote the corresponding theorem 4 of [41] in full detail, with slight modifications according to our notation.

Proposition 2

(Erdélyi). Under the preceding assumptions on ϕ(ω) we have

(C4) A = ∑ n = 0 N − 1 i n + 1 ϕ ( n ) ( ω 0 ) ψ ( n + 1 ) − log t + i π 2 t − n − 1 × exp i ω 0 t + o t − N ,

for t → +∞, where ψ(z) denotes the logarithmic derivative of Γ(z).

Let us denote the complementary integral of (C3) by

(C5) A ̃ = ∫ γ 1 ω 0 ϕ ( ω ) log ω − ω 0 exp ( i ω t ) d ω ,

such that ρ ̂ ( t ) = A + A ̃ and denote complex conjugation by an overline. Then

(C6) A ̃ ̄ = ∫ γ 1 ω 0 ϕ ( ω ) log ω − ω 0 exp ( i ω t ) d ω ̄

(C7) = − ∫ − γ 1 − ω 0 ϕ ( − ω ) log ω − ω 0 exp ( − i ω t ) d ω ̄

(C8) = ∫ − ω 0 − γ 1 ϕ ( − ω ) log ω − ω 0 exp ( i ω t ) d ω ,

where we have used that ϕ(ω) is real. This form of A ̃ ̄ is suited for the application of Proposition 2. Using the abbreviation ϕ ( − ω ) = ϕ ̃ ( ω ) which yields

(C9) ϕ ̃ ( n ) − ω 0 = − 1 n ϕ ( n ) ω 0 d ω

we thus obtain from (C4) and the replacement ω₀↦ −ω₀

(C10) A ̃ ̄ = ∑ n = 0 N − 1 i n + 1 ϕ ̃ ( n ) ( − ω 0 ) ψ ( n + 1 ) − log t + i π 2 t − n − 1 × exp − i ω 0 t + o t − N .

This entails

(C11) A ̃ = ( C 9 ) ∑ n = 0 N − 1 − i n + 1 ( − 1 ) n ϕ ( n ) ( ω 0 ) × ψ ( n + 1 ) − log t − i π 2 t − n − 1 exp i ω 0 t + o t − N

(C12) = − ∑ n = 0 N − 1 i n + 1 ϕ ( n ) ( ω 0 ) ψ ( n + 1 ) − log t − i π 2 t − n − 1 × exp i ω 0 t + o t − N ,

and, finally,

(C13) ρ ̂ ( t ) = A + A ̃ = − π ∑ n = 0 N − 1 i n ϕ ( n ) ω 0 t − n − 1 × exp i ω 0 t + o t − N .

We note that the terms containing log t cancel and the leading term corresponding to n = 0 in the asymptotic expansion (C13) is proportional to 1/t.

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Received: 2022-02-04

Revised: 2022-03-21

Accepted: 2022-03-29

Published Online: 2022-04-13

Published in Print: 2022-11-25

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https://doi.org/10.1515/zna-2022-0034

Keywords for this article

autocorrelation function; integrable systems; thermodynamics

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