Linear Response, Fluctuation-Dissipation Theorem for $L(t)$

We first check that

$$L(t) = \langle x(t) x \rangle_0,$$ (186)

the (van-Hove) position correlation function of the harmonic oscillator with co-ordinate $x$ in thermal equilibrium: write

$$x = \sqrt{\frac{1}{2M\Omega}} \left( a+a^{\dagger}\right),\quad x(t) = \sqrt{\frac{1}{2M\Omega}} \left( ae^{-i\Omega t}+a^{\dagger}e^{i\Omega t}\right)$$

$$L(t) = \langle x(t) x \rangle_0= \frac{1}{2M\Omega} \langle a a^{\dagger... ...= \frac{1}{2M\Omega} \left\{ (1+n_B) e^{-i\Omega t} + n_B e^{i\Omega t}\right\}$$

$$= \frac{1}{2M\Omega} \left\{ (1+2 n_B) \cos \Omega t -i \sin \Omega... ...} \left\{ \coth \frac{\beta \Omega}{2} \cos \Omega t -i \sin \Omega t \right\},$$ (187)

where we again have used the relation

$$1+2n_B = 1+\frac{2}{e^{\beta\Omega}-1} = \frac{e^{\beta\Omega}+1 }{e^{\beta\Omega}-1} = \coth \frac{\beta \Omega}{2}.$$ (188)

Now let us have another look at this function. Consider the Hamiltonian

$$H_B[x] + H_{SB}[xq]\equiv H(t) = \frac{p^2}{2M}+ \frac{1}{2}M\Omega^2 x^2 + f(t) x,$$ (189)

where we consider the function $f[q_t]=f(t)$ for a fixed path $q_t$ as an external classical force acting on the oscillator. The density matrix $\rho_B(t)$ of the oscillator in the interaction picture fulfills, cf Eq.(7.9),

$$\tilde{\rho}_B(t) = {\rho}_0 -i \int_0^t dt'f(t')[\tilde{x}(t'),\tilde{\rho}_B(t')]$$ (190)

$$\approx {\rho}_0 -i \int_0^t dt'f(t')[\tilde{x}(t'),{\rho}_0] ,$$

where $\rho_0={\rho}_B(t=0)$ is assumed to be the thermal equilibrium density matrix. The expectation value of the position is then

$$\langle x \rangle_t \equiv {\rm Tr} {\rho}_B(t) x = {\rm Tr} \tilde{\rho}_B(t) \tilde{x}(t)$$

$$= \langle x\rangle_0 - i \int_0^t dt' f(t') {\rm Tr} [\tilde{x}(t')... ...\rangle_0 - i \int_0^t dt' f(t') {\rm Tr} {\rho}_0 [\tilde{x}(t),\tilde{x}(t')]$$

$$= \langle x\rangle_0 - i \int_0^t dt' f(t') \langle [\tilde{x}(t),\tilde{x}(t')] \rangle_0, .$$

We check that

$$\langle [\tilde{x}(t),\tilde{x}(t')] \rangle_0 = \langle [\tilde{x}(t-t'),\tilde{x}(0)] \rangle_0$$ (191)

(definition of $\rho_0$ !) and define the linear susceptibility

$$\chi_{xx}(t-t') \equiv i \theta(t-t') \langle [\tilde{x}(t-t'),\tilde{x}(0)] \rangle_0 ,$$ (192)

so that we can write

$$\langle x \rangle_t= \langle x\rangle_0 - i \int_0^t dt' \chi_{xx}(t-t') f(t').$$ (193)

The theta function in $\chi_{xx}(t-t')$ guarantees causality: the response of $x$ at time $t$ is determined by the system at earlier times $t'\le t$ only.

Define additional functions and their symmetric and antisymmetric (in time) linear combinations,

$$C^+(t) \equiv \langle \tilde{x}(t) x \rangle_0,\quad C^-(t) \equiv \langle \tilde{x}(-t) x \rangle_0 = \langle {x} \tilde{x}(t) \rangle_0$$

$$C^{\pm}(t) \equiv S(t) \pm i A(t)$$ (194)

$$S(t) = S(-t) = \frac{1}{2}\langle \tilde{x}(t) x + x \tilde{x}(t) \rangl... ...A(t) = -A(-t) = \frac{1}{2i} \langle \tilde{x}(t) x - x \tilde{x}(t) \rangle_0.$$

We thus have

$$\chi_{xx}(t)= -2 \theta(t) A(t)$$ (195)

We define the Fourier transforms,

$$\hat{C}^{\pm}(\omega) \equiv \int_{-\infty}^{\infty} dt C^{\pm}(t... ...a t},\quad \hat{A}(\omega) \equiv \int_{-\infty}^{\infty} dt A(t) e^{i\omega t}$$

$$\hat{\chi}(\omega) \equiv \int_{0}^{\infty} dt \chi(t) e^{i\omega t}$$ (196)

and use

$${\rm Tr} (e^{-\beta H_B} x \tilde{x}(t) ) = {\rm Tr} (e^{-\beta H_B} x e^{\beta H_B} e^{-\beta H_B} \tilde{x}(t) ) = {\rm Tr} (\tilde{x}(i\beta) e^{-\beta H_B} \tilde{x}(t) )$$

$$= {\rm Tr} ( e^{-\beta H_B} \tilde{x}(t)\tilde{x}(i\beta) ) = {\rm Tr} ( e^{-\beta H_B} \tilde{x}(t-i\beta) x )$$

$$\leadsto C^-(t) = C^+(t-i\beta),$$ (197)

and therefore in the Fourier transform

$$C^-(\omega) = C^+(\omega) e^{-\beta\omega} .$$ (198)

We now define real and imaginary part of the Fourier transform of the susceptibility,

$$\hat{\chi}_{xx}(\omega) \equiv \hat{\chi}_{xx}'(\omega)+i \hat{\chi}_{xx}''(\omega).$$ (199)

Then,

$$\hat{\chi}''(\omega) = {\rm Im} \int_{0}^{\infty} dt \chi(t) e^{i\omega t} = -2 {\rm Im}... ...} \int_{0}^{\infty} dt \left ( A(t) e^{i\omega t} - A(t) e^{-i\omega t} \right)$$

$$= i \int_{0}^{\infty} dt \left ( A(t) e^{i\omega t} - \int_{-\infty... ...ga t} \right) = [A(t)=-A(-t)] = i \int_{-\infty}^{\infty} dt A(t) e^{i\omega t}$$

$$= i \hat{A}(\omega) = i\frac{1}{2i}\left( \hat{C}^+(\omega) - \hat{... ...\omega)\right) = \frac{1}{2}\left(1 - e^{-\beta\omega}\right) \hat{C}^+(\omega)$$ (200)

The relation

$$\hat{\chi}''(\omega)= \frac{1}{2}\left(1 - e^{-\beta\omega}\right) \hat{C}^+(\omega)$$ (201)

is called Fluctuation-Dissipation Theorem (FDT) (Callen, Welton 1951) and can be re-written, using

$$\hat{S}(\omega) = \frac{1}{2}(\hat{C}^+(\omega) + \hat{C}^-(\omega)) =\frac{1}{2}\left( 1+ e^{-\beta \omega} \right) \hat{C}^+(\omega),$$ (202)

leading to

$$\fbox{$ \begin{array}{rcl} \displaystyle \hat{S}(\omega) = \hat{\chi}''(\omega)\coth \frac{\beta\omega}{2}. \end{array}$\ }$$ (203)

Example- harmonic oscillator: we have

$$\chi(t) = i \theta(t) \langle [x(t),x]\rangle_0 = \frac {i \theta(t)}{2M\Omega} \left(e^{-i\Omega t} - e^{i\Omega t} \right)$$

$$\leadsto { \hat{\chi}_{xx}(\omega)} = {\rm Im} \frac {i }{2M\Omega}\int_{0}^{\infty} dt \left(e^{-i\Omega t} - e^{i\Omega t} \right) e^{i\omega t}$$

$$= \frac {1 }{2M\Omega}\int_{0}^{\infty} dt \left \{ \cos (\omega-\Omega) t - \cos (\omega+\Omega) t \right\}$$

$$= \frac {1 }{2M\Omega} \frac{1}{2} \int_{-\infty}^{\infty} dt \left \{ \cos (\omega-\Omega) t - \cos (\omega+\Omega) t \right\},$$ (204)

therefore

$$\hat{\chi}_{xx}(\omega) = { \frac {1 }{2M\Omega} \frac{2\pi }{2} \left \{ \delta(\omega-\Omega)-\delta(\omega+\Omega) \right\}}.$$ (205)

On the other hand,

$$C(t) \equiv L(t) = \langle \tilde{x}(t) x\rangle_0 = \frac{1}{2M\Omega} \left\{ \coth \frac{\beta \Omega}{2} \cos \Omega t -i \sin \Omega t \right\}$$

$$\leadsto S(t) = \frac{1}{2M\Omega}\coth \frac{\beta \Omega}{2} \cos \Omega t$$

$$\leadsto S(\omega) = \frac{1}{2M\Omega} \pi \left\{ \delta(\omega+\Omega)+\delta(\omega-\Omega) \right\}\coth \frac{\beta \Omega}{2}$$

$$= \frac{1}{2M\Omega} \pi \left\{ -\delta(\omega+\Omega)+\delta(\ome... ... \frac{\beta \omega}{2} = \hat{\chi}_{xx}(\omega) \coth \frac{\beta \omega}{2},$$ (206)

which is consistent with the FDT.