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  The integrable spin-boson and 1D inverse-square Ising models: thermal two-point functions

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Consider the spin-boson model. Letting the system Hamiltonian for this model be set to zero, we get an integrable system,

H=Xuλuxu+1HB,

Where HB is the Hamiltonian for a collection of independent harmonic oscillators with displacements given by xu. In particular, the Pauli matrix X is a constant of motion, and so the total Hamiltonian is unitarily equivalent to a direct-sum of two copies of HB via a polaron transformation:

U1HUHBHB

For example, the thermal Green's function of the spin is equal to unity: because [X,eτH]=0,

K(ττ)X(τ)X(τ)β                                          

                                     =1Ztr(eτHXe(ττ)HXe(βτ)H)=1Ztr(XXeβH)=1

However, it is also known that integrating-out the oscillators in the Feynman-Kac representation of the equilibrium partition function (for a detailed derivation, see equations (93)-(102) of Quantum Dissipative Systems by F Bascones et. al.) yields an inverse-square Ising model on a one-dimensional periodic lattice (with lattice spacing equal to the trotter time-step τc) in imaginary time:

tr(eβH)={X(τ)}eβSτc(X(τ)),                                                           

                 βSτc(X(τ))=ττJ(ττ)X(τ)X(τ),     J(ττ)τττcτ2c(ττ)2

Now, I can evaluate the thermal Green's function K(ττ) for my integrable model using the effective Euclidean action above. However, there is no way that an inverse-square Ising model has a two-point function which is identically one across all temperature ranges! What went wrong? 

asked Jun 26, 2017 in Theoretical Physics by David B Roberts (135 points) [ revision history ]
edited Jun 26, 2017 by David B Roberts

1 Answer

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The integrable spin-boson model is not an inverse-square Ising model. For that equivalence to hold, one has to include terms that do not commute with X1.

Let's re-examine the Feynman-Kac representation for the integral kernel of the partition function: when [H,X]=0, Kronecker deltas appear which massively simplify the partition function:

Z[τc]  τc0{X(τ)},{xu(τ)}X(τi1),xu(τi1)|eτcX(τi)uλuxu(τi)eτcHB|X(τi),xu(τi)

       ={X(τ)},{xu(τ)}δ(X(τi1),X(τi))xu(τi1)|eτcX(τi)uλuxu(τi)eτcHB|xu(τi)

To justify the intermediate step, since HB does not affect the system degrees of freedom, X(τi) is just an eigenvalue, and so the bra X(τi1)| meets the ket |X(τi) directly, inducing a prefactor. These Kronecker deltas in the integral kernel are precisely the reason why the thermal Green's function X(τ)X(τ)β for the integrable spin-boson model is identically one (as you correctly deduced by a direct calculation sans path integral techniques).

Of course, if we turned on a nontrivial system Hamiltonian HS (e.g. add a term proportional to Z1), then there would be a possibility to connect different eigenspaces of X at different imaginary times, and these Kronecker-deltas would disappear (this is the starting point of the paper which you cited). But then the integrability of the model would be lost. Either way, the model is either solvable via a polaron transformation, or is equivalent to an inverse-square Ising model; both cannot be true simultaneously.

answered Jun 28, 2017 by David B Roberts (135 points) [ revision history ]
edited Jun 28, 2017 by David B Roberts

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