Solution to a Problem Found by Calculating the Magnetic Internal Energy Inside Nonextensive Statistical Mechanics

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EJTP 7, No. 24 (2010) 131 138 Electronic Journal of Theoretical Physics Solution to a Problem Found by Calculating the Magnetic Internal Energy Inside Nonextensive Statistical Mechanics Felipe A. Reyes Navarro and Jaime Francisco V. Flores Universidad Nacional Mayor de San Marcos, Ciudad Universitaria, Av. Venezuela Cdra 3. Lima-Peru Received 18 February 2010, Accepted 16 August 2010, Published 10 October 2010 Abstract: Herein, in the context of third version of nonextensive statistical mechanics, a theory that generalizes the Boltzmann- Gibbs-Shannon s statistics, we display a solution for an anomaly found by calculating the internal energy for a composite A + B, of 2 spines 1 2, with additive Hamiltonian H = H A + H B. Specifically, the calculations of the internal energy in the full Hilbert space is different from the calculations done in the Hilbert subspaces, in other words, U total is different to U A + U B. We carry out analytical calculations. The results exactly indicate that the alternative method of matrices E A and E B is suitable for the calculations of the internal energy. Conseuently, the matrix that holds the physical information of the system is ρ. c Electronic Journal of Theoretical Physics. All rights reserved. Keywords: Nonextensive Quantum Statistical Mechanics, Partial Trace, Density Matrix; Thermodynamic Potential PACS (2010): 05.30.-d; 03.70.+k, 11.10.-z; 05.70.-a 1. Introduction The motivation for this work are the researches that have been developed in the last 20 years with the nonextensive statistical mechanics [1], a theory developed by the Brazilian researcher C. Tsallis, which would be an alternative for Boltzmann-Gibbs- Shannon s statistics. The amount of scientific investigations with respect to this theory is large; nowadays there have more than 2000 publications [2].Those studies have included various disciplines from uantum field theories to condensate matter physics. In this ar- farnape@gmail.com jventof@unmsm.edu.pe

132 Electronic Journal of Theoretical Physics 7, No. 24 (2010) 131 138 ticle, we analyzed a problem found in the internal energy (also called -internal energy): the calculations of this thermodynamic potential, in different Hilbert spaces, produce result uneual. Previous researches for magnetization can be seen in the article A Study on Composed Nonextensive Magnetic Systems that contain computer simulations [3]. 2. Theoretical Frame The construction of the nonextensive statistical mechanics begins by postulating the Tsallis entropy S [4]: 1 Tr(ρ ) S = k B, (1) 1 where ρ is the density matrix ρ powered to, with the entropic index and k B the Boltzmann s constant. Tr symbolizes the trace operation over all states from ρ. In the limit, when tends to 1, we recovered the Boltzmann-Gibbs entropy, S 1 = - k B Tr(ρLnρ), (2) next, we apply the maximum entropy method that considers the statistical mechanics as an inference process and was invented by the American Edward T. Jaynes in 1957 [5, 6]. And so, the following step is find out the probability distribution that maximizes the entropy of the E. (1); we want emphasize that the entropy is postulated without knowing the probability distribution a priori. For application of the method, we should take into account the constraints of unitary norm and a redefinition of the internal energy, that is, Tr(ρ) = 1 and U = Tr(ρ H) Tr(ρ ), (3) His the Hamiltonian matrix. After maximization of S, we obtain the following probability distribution p i : p i = ρ ii = [1 (1 )β ε i ] 1 1, (4) Z ρ ii is the matrix element ii of density matrix, β = 1 k B is an energy parameter with T the T temperature, see the works [3, 7, 8]; Z is the partition function, Z = Tr[1 (1 )β ε i ] 1 1, (5) For a compound system A + B, we also postulate the entropy of the complete system: 1 Tr (A,B) [ρ ] S A+B = k B, (6) 1 whereas for the entropies from subsystems A and B, we again postulate them: S A = k B 1 Tr A (ρ A ) 1 1 Tr B (ρ B and S B = k ) B, (7) 1

Electronic Journal of Theoretical Physics 7, No. 24 (2010) 131 138 133 Nowadays, there are four versions of nonextensive statistical mechanics that are researched [9]; they all utilize the same entropy S of E. (1). Nevertheless, the third version is the most accepted and widely investigated by the scientific community. This last version was devised by C. Tsallis, A. R. Plastino and R. S. Mendes in 1998 in the paper The Role of Constraints within Generalized Nonextensive Statistics [10]. 3. Procedure We study a magnetic system of two sublattices A and B. The sublattice A has N A ions with spin 1 2, and the sublattice Bhas N B ions also with spin 1 2.We utilize the Heisenberg model in the mean field approach. Hereafter, the Hamiltonian and other physical observables will be represented by operators, that is: with Ĥ A = gμ h B A N i Ĥ A+B = ĤA + ĤB, (8) N ŜA,i z and Ĥ B = gμ h B B ŜB,i, z (9) g is the gyromagnetic factor, μ h is the Bohr magneton, for didactic reasons we changed the usual symbol μ B ; ŜA,i z and Ŝz B,i spin operators, A(B) indicates respective sublattice and the subscripti, the i-th ion; B A and B B are the called effective magnetic fields and they are defined by: B A = B 0 + λ A M A + λ AB M B and B B = B 0 + λ B M B + λ AB M A, (10) B 0 is the external magnetic field; λ A and λ B are the intralattice coupling parameters and λ AB the interlattice coupling parameter. By considering that spines are independent the E. (9) is expressed in this way: Ĥ A = gμ h B A N A Ŝ z A and Ĥ B = gμ h B B N B Ŝ z B. (11) In the following subsections we show calculations to finding the composite internal energies. i 3.1 Calculation of the Internal Energies in Full Hilbert Space 4x4 In this subsection we apply the nonextensive statistical mechanics for researching the magnetic system represented by the Es. (8) and (9). For simplifying the calculations, we allow forn A = N B = N. In full Hilbert space, the internal energies have the general formulas: U A = Tr (A,B)(ˆρ Ĥ A ) Tr (A,B) (ˆρ ) and U B = Tr (A,B)(ˆρ Ĥ A ), (12) Tr (A,B) (ˆρ )

134 Electronic Journal of Theoretical Physics 7, No. 24 (2010) 131 138 next, for our specific system of 2 spines 1 2, in a Hilbert space 4x4, the respective internal energies take the form: Tr (A,B) (ˆρ Ŝ z A ) Tr (A,B) (ˆρ ) and U B = Ngμ h B B Tr (A,B) (ˆρ Ŝ z B ) Tr (A,B) (ˆρ ). (13) In order to calculate these parameters, we will need the density matrix operator, as well as the spin operators. Firstly, we obtain the density matrix operator from the E. (4): which we power to : ˆρ = ˆρ = { 1+(1 ) 2Nμ } 1 1 h K B T (Ŝz AB A + Ŝz BB B ) Z (14) { 1+(1 ) 2Nμ } 1 h K B T (Ŝz AB A + Ŝz BB B ) Z (15) On the other hand, the dimensionless spin operators in the full Hilbert space 4x4 are (the constant h is included in the Bohr magneton): ŜA z = { ++ ++ + + + + + } 2 ŜB z = { ++ ++ + + + + + } 2 (16) where ++ ++, + +, + + and are external operators that are constructed by bras and kets, which represent uantum states of the complete system. The bras are ++, +, + and ; the respective kets are ++, +, + and. Now, we can obtain the density matrix elements of ˆρ, ρ ++ = ++ ˆρ ++ = α 1 Z, and ρ + = + ˆρ + = α 2 Z ρ + = + ˆρ + = α 3 Z and ρ = ˆρ = α 4 Z (17) for practical reasons, we have introduced the following parameters: [ ] 1 [ α 1 = 1+(1 ) 2Nμ h(b A +B B ) 1 k B, α T 2 = 1+(1 ) 2Nμ h(b A B B ) k B T ] 1 1 α 3 = [ 1+(1 ) 2Nμ ] 1 h( B A + B B ) 1 k B T and α 4 = [ 1 (1 ) 2Nμ h(b A + B B ) k B T ] 1 1 ; (18)

Electronic Journal of Theoretical Physics 7, No. 24 (2010) 131 138 135 so that we also can get the matrix elements of ˆρ : ρ ++ = ++ ˆρ ++ = α 1 Z, ρ + = + ˆρ + = α2z ρ + = + ˆρ + = Z and ρ = ˆρ = α 4 Z (19) Therefore, by using the external operators, the density matrix operator ˆρ can be rrewritten as follows: ˆρ = {α 1 ++ ++ + α 2 + + + α 3 + + + α 4 }Z, (20) and, analogously, for ˆρ we have: ˆρ = {α 1 ++ ++ + α 2 + + + α 3 + + + α 4 }Z, (21) next, replacing Es. (16) and (21) into the E. (13), we obtain the internal energies in the full Hilbert space 4x4: α 1 + α 2 α 3 α 4 α 1 + α 2 + α 3 + α 4 and so, we also get the total internal energy: α 1 + α 3 α 2 α 4 and U B = Ngμ h B B α 1 + α 2 + α 3 + α, (22) 4 B A {α 1 + α 2 α 3 α U A+B = Ngμ 4} + B B {α 1 + α 3 α 2 α4} h α 1 + α 2 + α 3 + α. (23) 4 In following two subsections, by using two procedures, these results will be compared with calculations in Hilbert subspaces 2x2. 3.2 Calculation of the Internal Energies in Hilbert Subspaces 2x2, by Using the Formulas of the Third Version Now, we calculate U A and U B in Hilbert subspaces 2x2. In principle, the results must be identical with the outcomes obtained in the Hilbert space 4x4. In the Hilbert subspaces 2x2 the internal energies for the system of the Es. (8) and (9) are defined as: Tr A (ˆρ ŝz A ) Tr B (ˆρ A Tr A (ˆρ A ) and U B = Ngμ h B Bŝz B ) B Tr B (ˆρ B ), (24) in order to calculate these thermodynamic parameters, we take the partial trace over ˆρ in the E. (21), so we obtain the respective partial matrices ˆρ A and ˆρ B : now, we elevate to power and obtain ˆρ A = {(α 1 + α 2 ) + + +(α 3 + α 4 ) } Z and ˆρ B = {(α 1 + α 3 ) + + +(α 2 + α 4 ) } Z (25) ˆρ A = {(α 1 + α 2 ) + + +(α 3 + α 4 ) } Z and ˆρ B = {(α 1 + α 3 ) + + +(α 2 + α 4 ) } Z (26)

136 Electronic Journal of Theoretical Physics 7, No. 24 (2010) 131 138 For our calculation, we also know that the spin operators in the Hilbert subspaces 2x2 are: ŝ z A = { + + } 2 and ŝ z B = { + + } 2 (27) in this manner, by replacing Es. (26) and (27) into E. (24), we get the following result in the respective Hilbert spaces 2x2: (α 1 + α 2 ) (α 3 + α 4 ) (α 1 + α 2 ) +(α 3 + α 4 ) and (α 1 + α 3 ) (α 2 + α 4 ) U B = Nμ h B A (α 1 + α 2 ) +(α 3 + α 4 ), (28) so that the total internal energy is: U A+B = Ngμ h B A {(α 1 + α 2 ) (α 3 + α 4 ) } + B B {(α 1 + α 3 ) (α 2 + α 4 ) } (α 1 + α 2 ) +(α 3 + α 4 ), (29) we notice that these outcomes are different from the results got in the full Hilbert space 4x4, Es. (22) and (23). This is interpreted to be an anomalous result but we solve it in the next subsection. 3.3 The Matrices E A and E B for the Calculation of the Internal Energies, in the Hilbert Subspaces 2x2 For the magnetic system represented by the Es. (8) and (9), we introduce a procedure that was known in [7]. Herein, for didactic reasons, we called E A and E B to the matrices ρ A, and ρ B, of the reference [7]. Therefore, in this context, we define the internal energies as: Tr A (ÊAŝ z A ) Tr A (ÊA) and U B = Ngμ h B B Tr B (ÊBŝ z B ) Tr B (ÊB) (30) where partial matrices operators ÊA and ÊB are defined as: Ê A =Tr B (ˆρ ) and Ê B =Tr A (ˆρ ), (31) then, we carry out the partial trace over ˆρ in the E. (21): Ê A =(α 1 + α2) + + +(α 3 + α4) Z and Ê B =(α 1 + α 3) + + +(α 2 + α 4) Z (31) we must emphasize that these matrices no longer need to be powered. Therefore, by replacing Es. (27) and (31) into E. (30), we found that the employment of the matrices E A and E B produces the following energies, in the Hilbert subspaces 2x2: α 1 + α 2 α 3 α 4 α 1 + α 2 + α 3 + α 4 α 1 + α 2 α 3 α 4 and U B = Ngμ h B B α 1 + α 2 + α 3 + α. (32) 4

Electronic Journal of Theoretical Physics 7, No. 24 (2010) 131 138 137 Hence, it follows that the total energy of system A + B is: B A {α 1 + α 2 α 3 α U A+B = Ngμ 4} + B B {α 1 + α 3 α 2 α4} h α 1 + α 2 + α 3 + α ; (33) 4 These last results agree exactly with the Es. (22) and (23) that were calculated in the full Hilbert space 4x4. This fact shows clearly that partial matrices E A and E B are the adeuate ones for the calculation of the internal energy. Conclusions By using a magnetic system of 2 spines 1 2 as a specific example, we have shown explicitly that happen an anomaly by calculating the internal energy of a composite A + B, when the formulas from third version of nonextensive statistical mechanics are utilized, in the Hilbert subspaces. Also, we demonstrated that the solution to that problem is introducing matrices E A and E B, these matrices contain the physical information of the subsystems. Therefore, the operation of partial trace should be taken on the matrix ρ but not on the matrixρ. Finally, we want stress that the proposal of the method with matrices E A and E B is confirmed by an exact analytical way. References [1] http:tsallis.cat.cbpf.brbiblio.htm [2] C. Tsallis (2004), Phys. D 193, 3 cond-mat0403012. [3] F. A. R. Navarro, M. S. Reis, E. K. Lenzi and I. S. Oliveira (2004), Phys. A 343, 499. [4] C. Tsallis (1988), Possible generalization of Boltzmann-Gibbs statistics, Jour. of Stat. Phys., 52, 479. [5] E. T. Jaynes (1957), Phys. Rev., 106, 620. E. T. Jaynes (1963), Lectures in Theoretical Physics 3: Statistical Physics, 181.Ed. K.W. Ford, Brandeis University. E. T. Jaynes (2003), Probability Theory: The Logic of Science, Cambridge. Ed. G.L. Bretthorst. [6] Robert K. Niven (2009), ArXiv:0908.0990. [7] M. S. Reis, J. P. Araujo and V. S. Amaral, E. K. Lenzi and I. S. Oliveira (2002), Phys. Rev. B 66, 134417, cond-mat0207245. [8] F. A. R. Navarro, M. S. Reis, E. K Lenzi and I. S. Olivera (2007); cond-mat0701495. [9] G. L. Ferri, S. Martínez and A. Plastino (2005), Euivalence of the four versions of Tsallis s statistics, J. of Stat. Mech.: Theory and Experiment, P04009. [10] C. Tsallis, R. S. Mendes and R. Plastino (1998), Physica A 261, 534.