3.3.3 Linearization of the NSO

All terms have to be evaluated in such a way, that the total expression rests of order Oð Þ h . For expressions (25) and (26), we find after integration by parts

$$\rho\_{\epsilon}(t) = \rho\_{\text{rel}}(t) - \int\_{-\infty}^{t} \text{d}t\_{1} \mathbf{e}^{c(t\_{1}-t)} \mathbf{U}(t, t\_{1}) \left\{ \frac{\mathbf{i}}{\hbar} \left[ (\mathbf{H}\_{\text{S}} + \mathbf{H}\_{\text{F}}^{t}), \rho\_{\text{rel}}(t\_{1}) \right] + \frac{\partial}{\partial t\_{1}} \rho\_{\text{rel}}(t\_{1}) \right\} \mathbf{U}^{\dagger}(t, t\_{1}) . \tag{156}$$

Since HS commutes with ρeq (equilibrium!), the curly bracket is of order Oð Þ h . In particular, we have for the first term the time derivative in the Heisenberg picture,

$$\frac{1}{\hbar} \left[ \mathbf{H}\_{\rm S}, \beta \int \mathbf{d}\lambda \sum\_{n} F\_{n}(t\_{1}) \mathbf{B}\_{n}(\mathbf{i}\lambda\beta\hbar) \rho\_{\rm eq} \right] = \beta \int \mathbf{d}\lambda \sum\_{n} F\_{n}(t\_{i}) \dot{\mathbf{B}}\_{n}(i\lambda\beta\hbar) \rho\_{\rm eq}.\tag{157}$$

For the second term of the integral in Eq. (156), we use Kubo's identity

$$\mathbb{E}\left[\mathbf{B},\mathbf{e}^{\mathbf{A}}\right] = \int\_{0}^{1} \mathbf{d}\lambda \,\mathbf{e}^{\lambda\mathbf{A}}[\mathbf{B},\mathbf{A}] \,\mathbf{e}^{(1-\lambda)\mathbf{A}}.\tag{158}$$

so that

$$\frac{\dot{\mathbf{i}}}{\hbar} \left[ \mathbf{H}\_F^{t\_1}, \rho\_{\mathbf{eq}} \right] = -\beta e^{-i\alpha t\_1} \int\_0^1 \mathbf{d}\lambda \sum\_j h\_j \dot{\mathbf{A}}\_j(\mathbf{i}\lambda\beta\hbar) \rho\_{\mathbf{eq}}.\tag{159}$$

that in the classical limit the relations become more simple because the variables commute, and we have not additional integrals expanding the exponential. We can solve this linear system of equations (162) using Cramers rule. The response parameters Fn are found to be proportional to the external fields hj with coefficients that are ratios of two determinants. The matrix elements are given by equilibrium correlation functions. This way, the self-consistency conditions are solved, and the Lagrange multipliers can be eliminated. The nonequilibrium problem is formally solved. The second problem, the evaluation of equilibrium correlation functions, can be solved by different methods such as numerical simulations, quantum statistical perturbation theories such as thermodynamic Green functions and Feynman diagrams, path integral methods, etc. Using partial integration, we

Then, the generalized linear response equations (162) can be rewritten in the

<sup>þ</sup> <sup>B</sup>\_ <sup>m</sup>; <sup>A</sup>\_ <sup>j</sup>

Having the response parameters Fn to our disposal, we can evaluate averages of

Eliminating Fm, these average fluctuations Bh i<sup>n</sup> <sup>t</sup> are proportional to the fields

As an example for the generalized linear response theory, we calculate the conductivity of a plasma of charged particles (electrons and ions) that is exposed to

HF ¼ �eEX, X ¼ ∑

Instead of hj, we have only one constant external field E. For the treatment of arbitrary ω to obtain the dynamical (optical) conductivity see [11, 13, 16, 17]. The conjugated variable A from Eq. (149) that couples the system to the external field is

For simplicity, the ions are considered here as fixed in space because of the large mass ratio (adiabatic approximation). Then, the transport of charge is owing to the

Ne i

that can be interpreted as generalized transition rates (collision integral, left-hand side) and the influence of external forces (drift term, right-hand side of

Fmei<sup>ω</sup><sup>t</sup>

<sup>ω</sup>þi<sup>ϵ</sup> � <sup>i</sup>ωð Þ� <sup>B</sup>mjB<sup>n</sup> <sup>i</sup><sup>ω</sup> <sup>B</sup>\_ <sup>m</sup>; <sup>B</sup><sup>n</sup>

<sup>z</sup> <sup>¼</sup> ð Þ� <sup>A</sup>j<sup>B</sup> <sup>A</sup>; <sup>B</sup>\_

ωþiϵ

Nmn, Nmn ¼ ð Þ BmjB<sup>n</sup> : (169)

z

ωþiϵ

: (168)

xi: (170)

<sup>i</sup> px,i denoting the

: (166)

, (167)

�izh i <sup>A</sup>; <sup>B</sup> <sup>z</sup> <sup>¼</sup> ð Þþ <sup>A</sup>j<sup>B</sup> <sup>A</sup>\_ ; <sup>B</sup>

rel ¼ �β ∑ m

3.3.4 Force-force correlation function and static (dc) conductivity

a static homogeneous electric field in x-direction: ω ¼ 0, E ¼ Eex,

<sup>A</sup> <sup>¼</sup> <sup>e</sup>X. The time derivative follows as <sup>A</sup>\_ <sup>¼</sup> ð Þ <sup>e</sup>=<sup>m</sup> P, with P <sup>¼</sup> <sup>∑</sup>Ne

Dmj <sup>¼</sup> <sup>B</sup>mjA\_ <sup>j</sup>

short form (165) with the matrix elements

the relevant observables, see Eq. (151),

h i <sup>B</sup><sup>n</sup> <sup>t</sup> <sup>¼</sup> h i <sup>B</sup><sup>n</sup> <sup>t</sup>

total momentum in x direction.

<sup>þ</sup> <sup>B</sup>\_ <sup>m</sup>; <sup>B</sup>\_ <sup>n</sup>

Pmn <sup>¼</sup> <sup>B</sup>mjB\_ <sup>n</sup>

Nonequilibrium Statistical Operator

DOI: http://dx.doi.org/10.5772/intechopen.84707

show the relation

Eq. (165)).

hje�iω<sup>t</sup> .

37

The last term in the curly bracket can be rewritten as

$$\frac{\partial}{\partial t\_1} \rho\_{\rm rel} = \beta \int\_0^1 \mathbf{d}\lambda \sum\_n \dot{F}\_n(t\_1) \mathbf{B}\_n(i\lambda \beta \hbar) \rho\_{\rm eq}. \tag{160}$$

Because we restrict ourselves to the order Oð Þ h , for the time evolution operator we have Uð Þ <sup>t</sup>; <sup>t</sup><sup>1</sup> <sup>≃</sup> <sup>e</sup>�iHSð Þ <sup>t</sup>�t<sup>1</sup> <sup>=</sup>ℏ:

After linearization with respect to the external fields hj and the response parameters Fn, finally we have

$$\begin{aligned} \rho\_c(t) &= \rho\_{\rm rel}(t) - \beta \mathbf{e}^{-\rm{i}\rm{at}} \int\_{-\infty}^{0} \mathbf{d}t\_1 \mathbf{e}^{-\rm{i}\rm{it}\_1} \int\_{0}^{1} \mathbf{d}\lambda \left[ -\sum\_{j} h\_j \dot{\mathbf{A}}\_j (\dot{\mathbf{i}}\lambda\beta\hbar + t\_1) \rho\_{\rm eq} \\\\ &+ \sum\_{n} \left( F\_n \dot{\mathbf{B}}\_n (\dot{\mathbf{i}}\lambda\beta\hbar + t\_1) \rho\_{\rm eq} - \dot{\mathbf{i}}\alpha F\_n \mathbf{B}\_n (\dot{\mathbf{i}}\lambda\beta\hbar + t\_1) \rho\_{\rm eq} \right) \right] \end{aligned} \tag{161}$$

(z ¼ ω þ iϵ). Here, we used that hjð Þt and Fnð Þt , Eq. (154), are proportional to e�iω<sup>t</sup> .

We multiply this equation by Bm, take the trace and use the self-consistency relation (151). We obtain a set of linear equations for the thermodynamically conjugated parameters Fn (response parameters):

$$\sum\_{n} \left\{ \left< \mathbf{B}\_{m}; \dot{\mathbf{B}}\_{n} \right>\_{x} - \mathrm{i}\boldsymbol{\alpha} \langle \mathbf{B}\_{m}; \mathbf{B}\_{n} \rangle\_{x} \right\} \mathbf{F}\_{n} = \sum\_{j} \left< \mathbf{B}\_{m}; \dot{\mathbf{A}}\_{j} \right>\_{x} \mathrm{h}\_{j} \tag{162}$$

with the Kubo scalar product (the particle number commutes with the observables)

$$\mathbf{A}(\mathbf{A}|\mathbf{B}) = \int\_0^1 \mathbf{d}\lambda \text{Tr}\left\{ \mathbf{A} \mathbf{e}^{-\mathbf{i}\beta\hbar} \mathbf{B} \mathbf{e}^{\mathbf{j}\beta\hbar} \rho\_{\text{eq}} \right\} = \int\_0^1 \mathbf{d}\lambda \text{Tr}\left\{ \mathbf{A} \mathbf{B}(\mathbf{i}\lambda\beta\hbar) \rho\_{\text{eq}} \right\},\tag{163}$$

and its Laplace transform, the thermodynamic correlation function

$$\langle \mathbf{A}; \mathbf{B} \rangle\_x = \int\_{-\infty}^0 \mathbf{d}t \,\mathbf{e}^{-\text{i}\text{r}t} (\mathbf{A} \, | \, \mathbf{B}(t)) = \int\_0^\infty \text{d}t \,\mathbf{e}^{\text{i}\text{r}t} (\mathbf{A}(t) \, | \, \mathbf{B}). \tag{164}$$

The linear system of equations (162) has the form

$$\sum\_{n} P\_{mn} F\_n = \sum\_{j} D\_{m\vec{j}} h\_{\vec{j}} \tag{165}$$

to determine the response parameters Fn, the number of equations coincides with the number of variables to be determined. The coefficients of this linear system of equations are given by equilibrium correlation functions. We emphasize
