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I JUST WANT YOU TO FIND HOW WE GET FROM this: II. QUANTUM PHASE TRANSITION IN THE HEISENBERG CHAIN A. Approach description We begin by explaining the improvements we made in this work to the...

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I JUST WANT YOU TO FIND HOW WE GET FROM this: II. QUANTUM PHASE TRANSITION IN THE HEISENBERG CHAIN A. Approach description We begin by explaining the improvements we made in this work to the zero-field bond-mean-field theory9 BMFT for the Heisenberg chain. Then, we extend the BMFT to the case when an external magnetic field is applied to the chain. 1. Zero-field BMFT The 1D Heisenberg model is H1D = J i,j Si · Sj, 2 where Si is the spin-1/ 2 operator at site i and i, j designates the sum over adjacent sites only. Using the well-known 1D Jordan-Wigner JW transformation Si - = ci eii , i = j=0 i-1 ni, Si z = ni - 1/2, 3 with ni=ci † ci , the Heisenberg chain maps onto a Hamiltonian of 1D interacting spinless fermions; i.e., H1D = J 2 i ci † ci+1 + H.c. + J i ci † ci - 1 2 ci+1 † ci XXXXXXXXXX . 4 In the framework of the BMFT, we consider the bond parameter ci ci+1 †  to decouple the Ising quartic term in Eq. 4. The procedure is as follows. By neglecting the fluctuations in ci ci+1 † , we write ci † ci ci+1 † ci+1  ci ci+1 † ci † ci+1 + ci+1ci † ci+1 † ci + ci ci+1 † ci+1ci †  = Qi ci † ci+1 + Qi * ci+1 † ci + Qi 2, 5 with Qi=ci ci+1 †  the mean-field bond parameter. Note that Si - Si+1 +  = ci ci+1 †  , which means that the parameter Qi accounts for the quantum fluctuations in the Heisenberg Hamiltonian. To recover a result comparable to the des Cloiseaux–Pearson10 spin-excitation spectrum for the single Heisenberg chain, namely, Ek = J 2 sin k , 6 one has to choose Qi=-1 i Q Qeixi , where Q is site independent. Here, xi is an integer that labels the site coordinate along the Heisenberg chain. This yields a phase configuration along the chain of the form xi ...,,0,,0,..., which will in turn give9 Ek = J1 sin k , 7 when the single-fermion term the XY term is also written within the alteranted phase scheme. The hopping amplitude for the JW fermions resulting from this treatment is -1 i J1, with J1=J1+2Q indicating that a renormalization of the single-particle term in Eq. 4 by the Ising interaction took place. Choosing a uniform Qi and neglecting the alternation of the sign of XY-hopping amplitude gives a cos k spectrum that does not agree with the des Cloiseaux–Pearson result. Physically, we can justify the choice of the staggered Qi in the following way. Consider two adjacent bonds 2i, 2i + 12i+1,2i+ 2. Because of the AF correlations, the bonds 2i, 2i+ 12i+1,2i+ 2 may be in the spin configuration ?2i ,?2i+1?2i+1 ,?2i+2. Using the argument of only shortrange and short-lived AF order, one has to view the object ?2i ,?2i+1 as a renormalized spin singlet, rather than a static spin arrangement, because of the important spin fluctuations. 74, 174422  XXXXXXXXXXIn the mean-field treatment based on the JW fermions, any two adjacent bonds can be quantitatively represented by the product c2i † c2i+1c2i+1c2i+2 † , which turns out to be -Q2 , since Q=c2i+1c2i+2 †  and c2i † c2i+1=-c2i+1c2i † =-Q. It is thus justified to use opposite signs for Q on adjacent bonds. We assumed that creating a JW fermion is equivalent to having a spin-up state and destroying a JW fermion corresponds to a spin-down state. With all this and Eq. 5, the mean-field Hamiltonian takes on the form H1D = NJQ2 + k k † H1Dk, 8 where the spinor  is given by k † = ck A† ck B† 9 and the Hamiltonian density by H = 0 ek e* k 0 , 10 with ek=iJ1 sin k. ck is the Fourier transform of ci . =A or B is the label of the two AF sublattices. The eigenenergies are E±k= ±J1+2Q sin k . The ground-state energy per site is EGS = JQ XXXXXXXXXXdk 2 p=± ln 1 + e- Epk , 11 which gives EGST= 0=JQ2- 1 J1+2Q at zero temperature because only the band with negative energies is filled at this temperature. Minimizing EGS with respect to Q gives Q =1/ at T= 0. At nonzero temperature Q is given by Q = - 1 2 dk 2 sin k  p=± pnF Epk . 12 Here, nF Epk =1/ 1+e Epk is the Fermi-Dirac factor. In the high-T limit with kBT J, we get Q J 8kBT 1 1-4kBT/J , a result that rules out any finite-T phase transition, in agreement with the Mermin-Wagner theorem.11 Next, we will extend the analysis we just finished to the case of a Heisenberg chain coupled to an external magnetic field.
Answered Same Day Dec 23, 2021

Solution

Robert answered on Dec 23 2021
122 Votes
Solution
In the above expression we know that,
nF = [EP(k)] = 1 + eβE(k)
Q = -1/2 ∫ ????2?? |????????|∑ ??????[????(??)]??
Q = -1/2 ∫ ????2?? |????????|∑ ??1/1 + e
βE(k)
??
Where β = 1/kbT from Fermi Dirac Hypothesis
Q = -1/2 ∫ ????2?? |????????|∑ ??1/1 + e
βE(k)
??
Here in the above expression we can write
|????????| = k – k3/3! + k5/5! -………………. ~ k
Hence
Q ~ -1/2 ∫ ?? ????2?? ∑ ??1/1 + e
βE(k)
??
Now For...
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