Aspects of polaron theory: equilibrium and nonequilibrium by N N Bogolubov, Nickolai N Bogolubov Jr

By N N Bogolubov, Nickolai N Bogolubov Jr

The linear polaron version is a superb instance of an precisely soluble, but nontrivial polaron process. It serves as a tribulation process or zero-level approximation in lots of refined equipment of polaron research. This publication analyzes, specifically, the potential of relief of the complete polaron Hamiltonian to the linear one, and introduces a distinct approach to calculating thermodynamical features according to the calculation of the averages of T-products. This T-product formalism appears to be like a less difficult manner of doing comparable calculations concerning Feynman's course indispensable method.

This publication follows a step by step process, from relatively basic actual principles to a transparent realizing of subtle mathematical instruments of research in sleek polaron physics. The reader is ready to evaluate the actual viewpoint with tools proposed within the ebook, and while seize the underlying arithmetic.

a few familiarity with quantum statistical mechanics is fascinating in analyzing this publication.

Contents: Linear Polaron version; Equilibrium Thermodynamic country of Polaron procedure; Kinetic Equations in Polaron concept.

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Free Energy Calculation 37 We shall calculate explicitly the free energy Fint = −ϑ ln Tr e−βH Tr e−βH = −3ϑ ln , −βHΣ −βHS Tr(S) e Tr(Σ) e Tr(S) e Tr(Σ) e−βHΣ −βHS where the one-dimensional Hamiltonians H, HS and HΣ are given by the following expressions: H= HS = p2 K2 1 + 0 (x − X)2 + P 2, 2m 2 2M p2 , 2m HΣ = P2 K2 + 0 X 2. 2M 2 To diagonalize the one-dimensional Hamiltonian H, we introduce normal coordinates q, Q and corresponding normal momentum variables y, Y: mx + M X , Q = x − X. 58b) ∂ M ∂ ∂ = − , ∂X m + M ∂q ∂Q we put p= m y+Y, m+M P = M y − Y.

51) we see that λ ∂H(λ) ∂λ =3 λ,eq i¯h +∞ 1 mΩ2 − η 2 ∫ 2π −∞ 1 − e−β¯hω mΩ2 − η 2 + λ2 Ω (Ω) ω+i0 ω−i0 dω. Let us note that Ωλ2 (Ω) mΩ2 − η 2 = 1 − , mΩ2 − η 2 + λ2 Ω (Ω) mΩ2 − η 2 + λ2 Ω (Ω) mΩ2 − η 2 mΩ2 − η 2 + λ2 Ω (Ω) ω+i0 ω−i0 =− Ωλ2 (Ω) mΩ2 − η 2 + λ2 Ω (Ω) ω+i0 ω−i0 . 51) gives us ∂H(λ) ∂λ = −3 λ,eq = −3 Ωλ (Ω) i¯h +∞ 1 ∫ 2π −∞ 1 − e−β¯hω mΩ2 − η 2 + λ2 Ω (Ω) λ (Ω) i¯h +∞ Ω ∫ 2π −∞ 1 − e−β¯hΩ mΩ2 − η 2 + λ2 Ω (Ω) ω+i0 ω−i0 ω+i0 ω−i0 dω dω . 54) Using a similar approach for the calculation of the correlation function pα (t)pα (τ ) eq , we find in the limit η → 0, V → ∞ that ∂H(λ) ∂λ =− λ,eq λ ∞ (Ω) Ω 3i¯h +∞ 1 ∫ 2π −∞ 1 − e−β¯hΩ Ω mΩ + λ2 ∞ (Ω) ω+i0 ω−i0 dω .

7. 139) where U (0) = 1 — is the unit operator, and H1 (s) is an operator that can depend explicitly on s. It is easily seen that 1 s U (s) = T e −¯ ∫ {H0 +H1 (σ)} dσ h . 136) U (s) = e− H0 s ¯ h C(s). This ansatz leads to the following equation for C(s) ¯h H0 s H0 s dC(s) = −e ¯h H1 (s)e− ¯h C(s), ds C(0) = 1, which equation can be solved formally as 1 s C(s) = T e −¯ ∫ dσ e h H0 σ h ¯ 0 H1 (σ)e− H0 σ h ¯ . Therefore h 1 β¯ U (β¯h) = T e −¯ ∫ ds {H0 (s)+H1 (s)} h 0 h 1 β¯ =e −H0 β T e ∫ ds e −¯ h 0 H0 s h ¯ H1 (s)e− H0 s h ¯ .

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