By N N Bogolubov, Nickolai N Bogolubov Jr
The linear polaron version is a superb instance of an precisely soluble, but nontrivial polaron procedure. It serves as an ordeal method or zero-level approximation in lots of refined equipment of polaron research. This e-book analyzes, particularly, the potential of relief of the complete polaron Hamiltonian to the linear one, and introduces a unique approach to calculating thermodynamical features according to the calculation of the averages of T-products. This T-product formalism seems a easier approach of doing related calculations related to Feynman's direction necessary procedure.
This publication follows a step by step procedure, from relatively basic actual principles to a transparent knowing of refined mathematical instruments of research in sleek polaron physics. The reader is ready to evaluate the actual perspective with tools proposed within the e-book, and even as grab the underlying arithmetic.
a few familiarity with quantum statistical mechanics is fascinating in examining this e-book.
Contents: Linear Polaron version; Equilibrium Thermodynamic kingdom of Polaron procedure; Kinetic Equations in Polaron idea.
Read Online or Download Aspects of polaron theory: equilibrium and nonequilibrium problems PDF
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Additional resources for Aspects of polaron theory: equilibrium and nonequilibrium problems
57). We now return to the expression for the free energy Fint in the singlefrequency case and consider the passage to classical mechanics. 56), we get the classical result for the free energy: Fint = 0. 1) in the case K 2 = K02 + η 2 . 65) where 0<ε< 2π . ¯hβ In the classical limit, lim ¯ h→0 1 ¯hΩ = = ϑ. β 1 − e−β¯hΩ Consequently, it is true for the classical mechanics that ∂H(λ) ∂λ =− λ,eq 3iϑ 2π iε+∞ ∫ iε−∞ where F (Ω) = F (Ω) dΩ − −iε+∞ ∫ −iε−∞ λ (Ω) . 67) It should be observed that F (Ω) is a regular analytic function on the halfplane Im (Ω) Thus ∫ ε > 0.
1. , As is left uncoupled, and Aj Γ = 0, because Aj is a linear form composed of operators bα and b†α and Γ is a quadratic form. Then we apply this well-known technique to the calculation of the expression eA Γ , where A is some linear form composed of the above-mentioned Bose operators. We arrive at the following result: eA ∞ Γ = n=0 1 An n! ∞ Γ = k=0 1 A2k (2k)! ∞ Γ =1+ k=1 1 A2k (2k)! Γ. 83) Thanks to the Bloch–Dominicis theorem, A2k Γ = G(k) A2 kΓ , where G(k) is the number of all possible couplings in the expression A1 · · · As Γ .
61) 38 Ch. 1. Linear Polaron Model It is well known that the free energy of the oscillator HΣ with the 1/2 frequency ν0 = K02 /M in the one-dimensional case is g ¯hν0 1 − ϑ ln , 2 1 − e−β¯hν0 FΣ = and because the oscillator Hosc has frequency K02 (M + m) Mm µ= then Fosc is Fosc = 1/2 = ν0 1 + M m 1/2 , ¯hµ 1 − ϑ ln . 2 1 − e−β¯hµ Therefore 3 Tr e−βHin 1 − e−β¯hν0 + ¯h(µ − ν0 ). − 3ϑ ln −βHS −β¯ hµ 2 Tr e 1−e Fint = −3ϑ ln Because the position x — belongs to the interval −L/2 < x < L/2, the corresponding momentum variable p can take only discrete values 2π n¯h, L n = 0, ±1, ±2, ...