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Time correlation function
TCFs describe the time dependence of a fluctuating variable for an ensemble at equilibrium.

$$C_{AA}(t)=\langle A(t) A(0) \rangle $$

$$C(t)=\langle A(t) A(0) \rangle $$

$$=\lim_{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} d\tau \ A(t+\tau)A^*(\tau) $$

$$=\sum_n \frac{e^{-\beta E_n}}{Z}\langle n |A(t) A(0) | n \rangle $$

$$C_{AA}(t) = \int d\operatorname{\textbf{p}} \int d\operatorname{\textbf{q}} \ A(\operatorname{\textbf{p}},\operatorname{\textbf{q}};t) \ A(\operatorname{\textbf{p}},\operatorname{\textbf{q}};0) \ \rho(\operatorname{\textbf{p}},\operatorname{\textbf{q}})$$

$$\tilde{C}_{AA}(\omega) = \int^{\infty}_{-\infty} e^{i \omega t} C_{AA}(t) dt$$

$$\frac{\tilde{C}_{AA}(\omega_{lk})}{\tilde{C}_{AA}(\omega_{kl})} = \frac{p_k}{p_l} = e^{\beta E_{lk}}$$

$$\tilde{C}_{AA}(\omega) = e^{\beta \hbar \omega} \tilde{C}_{AA}(-\omega)$$

Symmetry of correlation function


C(t)=\langle A(t)A(0) \rangle. $$

C(t_1,t_2)=\langle A(t_1)A(t_2) \rangle=\langle A(t_1-t_2)A(0) \rangle=C(t_1-t_2). $$

$\begin{matrix} C(-t) &=& C^{*}(t) \\

C(t) &=& C'(t) + i C''(t) \\

\tilde{C}(\omega) &=& \int^{\infty}_{-\infty} e^{i \omega t} C(t) dt \\ &=&\int^{\infty}_{-\infty} e^{i \omega t} C'(t) dt -i \int^{\infty}_{-\infty} e^{i \omega t} C''(t) dt\\ &=& \tilde{C'}(\omega) + \tilde{C''}(\omega) \\ \tilde{C}(-\omega) &=& e^{-\beta \hbar \omega} \tilde{C}(\omega) \\

\tilde{C''}(\omega) &=& \tanh (\beta \hbar \omega/2) \tilde{C'}(\omega) \\

\tanh(\omega)&=&\omega-\frac{1}{3}\omega^3+\frac{2}{15}\omega^5+ \cdots\\

\frac{d^n f(t)}{dt^n} &\rightleftharpoons &(i\omega)^n \tilde{f}(\omega) \\ \tan(\frac{d}{dt})&=&\frac{d}{dt}+\frac{1}{3}(\frac{d}{dt})^3+\frac{2}{15}(\frac{d}{dt})^5+ \cdots\\ C''(t) &=& \tan \left ( \frac{\beta \hbar}{2} \frac{d}{dt} \right ) C'(t)\\

\end{matrix} $

C'(t) = \sum^{M}_{m=1} B_m e^{z_mt} $$$$C''(t)=\tan(\frac{\beta \hbar}{2} \frac{d}{dt})C'(t) =  \sum^{M}_{m=1} B_m \tan(\frac{\beta \hbar}{2} z_m) e^{z_mt} $$

\tilde{C}(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i \omega t} C(t) dt $$
 * $$\begin{matrix}

\tilde{C}(\omega) & = & \int_{-\infty}^{\infty} \cos( \omega t) C'(t) dt -  \int_{-\infty}^{\infty} \sin( \omega t) C''(t) dt \\ & = & \tilde{C'}(\omega) + \tilde{C''}(\omega). \end{matrix}$$

\begin{matrix} \tilde{C}(-\omega) & = & e^{-\beta \hbar \omega} \tilde{C}(\omega) \\

\end{matrix} $$



\begin{matrix} C'(t) & = & C'(-t) & \mbox{(even function)}, \\ C(t) & = & -C(-t) & \mbox{(odd function)}. \end{matrix} $$



\begin{matrix} \tilde{C}(\omega) & = & \frac{1}{2\pi} \int_{-\infty}^{\infty} \cos( \omega t) C'(t) dt - \frac{1}{2\pi} \int_{-\infty}^{\infty} \sin( \omega t) C''(t) dt \\ & = & \tilde{C}_S(\omega) + \tilde{C}_A(\omega). \end{matrix} $$


 * $$\tilde{C}(-\omega) = e^{-\beta \hbar \omega} \tilde{C}(\omega)$$
 * $$\tilde{G}(-\omega) = e^{-\beta \hbar \omega} \tilde{G}(\omega)$$



\tilde{C}_A(\omega) = \tanh (\frac{\beta \hbar \omega}{2}) \tilde{C}_S(\omega). $$



C''(t) = \tan \left ( \frac{\beta \hbar}{2} \frac{d}{dt} \right ) C'(t). $$

Quantum correction
$$ \tilde{C}(\omega) = Q(\omega) \tilde{C}_{cl}(\omega)$$

standard
$$ \begin{matrix} \tilde{C'}(\omega)&=&(\tilde{C}(\omega)+\tilde{C}(-\omega))/2 \\ &=& \frac{1+e^{-\beta\hbar\omega}}{2} \tilde{C}(\omega) \end{matrix}$$

by assuming $$C_{cl}(t) \cong C'(t)$$

$$\tilde{C}(\omega)=\frac{2}{1+e^{-\beta\hbar\omega}}\tilde{C}_{cl}(\omega)$$

harmonic
$$ H_s=\frac{p^2}{2\mu}+\frac{1}{2} \ \mu \ \omega_0^2 \ q^2$$

$$ H_b = \sum_\alpha \Biggl( \frac{p_{\alpha}^2}{2 m_{\alpha}}+\frac{1}{2} \ m_{\alpha} \ \omega_{\alpha}^2 \ x_{\alpha}^2 \Biggr)$$

$$ H_{sb} = -q \sum_{\alpha} g_{\alpha} x_{\alpha} $$

$$ =-qF$$

for

$$ C(t) = \langle q(t) q(0) \rangle$$

Schofield
$$ C(t) = C_{cl}(t+i\beta\hbar/2)$$

Prony

 * $$\hat{f}(t) = \sum_{i=1}^{M} A_i e^{\sigma_i t} \cos(2\pi f_i t + \phi_i)$$



\begin{align} \hat{f}(t) &= \sum_{i=1}^{M} A_i e^{\sigma_i t} \cos(2\pi f_i t + \phi_i) \\ &= \sum_{i=1}^{M} \frac{1}{2} A_i e^{\pm j\phi_i}e^{\lambda_i t} \end{align} $$

where:
 * $$\lambda_i = \sigma_i \pm j \omega_i$$ are the eigenvalues of the system,
 * $$\sigma_i$$ are the damping components,
 * $$\phi_i$$ are the phase components,
 * $$f_i$$ are the frequency components,
 * $$A_i$$ are the amplitude components of the series, and
 * $$j$$ is the imaginary unit ($$j^2 = -1$$).

Prony

 * $$F_n = \hat{f}(\Delta_t n) = \sum_{m=1}^{M} \Beta_m e^{\lambda_m \Delta_t n}, \quad n=0,\dots,N-1.$$


 * $$\begin{align}

\Beta_a &= \frac{1}{2} A_i e^{ \phi_i j}, \\ \Beta_b &= \frac{1}{2} A_i e^{-\phi_i j}, \\ \lambda_a &= \sigma_i + j \omega_i, \\ \lambda_b &= \sigma_i - j \omega_i, \end{align}$$ where
 * $$\begin{align}

\Beta_a e^{\lambda_a t} + \Beta_b e^{\lambda_b t} &= \frac{1}{2} A_i e^{\phi_i j} e^{(\sigma_i + j \omega_i) t} + \frac{1}{2}A_i e^{-\phi_i j} e^{(\sigma_i - j \omega_i) t} \\ &= A_i e^{\sigma_i t} \cos(\omega_i t + \phi_i). \end{align}$$


 * $$\hat{f}(\Delta_t n) = -\sum_{m=1}^{M} \hat{f}[\Delta_t (n - m)] P_m, \quad n=M,\dots,N-1.$$


 * $$ z^M + P_1 z^{M-1} + \dots + P_M = \prod_{m=1}^{M} \left(z - e^{\lambda_m}\right).$$

These facts lead to the following three steps to Prony's Method:

1) Construct and solve the matrix equation for the $$P_m$$ values:

\begin{bmatrix} F_{M} \\ \vdots \\ F_{N-1} \end{bmatrix} = \begin{bmatrix} F_{M-1} & \dots & F_{0} \\ \vdots & \ddots & \vdots \\ F_{N-2} & \dots & F_{N-M-1} \end{bmatrix} \begin{bmatrix} p_1 \\ \vdots \\ p_M \end{bmatrix}. $$

Note that if $$N \ne 2M$$, a generalized matrix inverse may be needed to find the values $$P_m$$.

2) After finding the $$P_m$$ values find the roots (numerically if necessary) of the polynomial
 * $$ z^M + P_1 z^{M-1} + \dots + P_M,$$

The $$m$$-th root of this polynomial will be equal to $$e^{\lambda_m}$$.

3) With the $$e^{\lambda_m}$$ values the $$F_n$$ values are part of a system of linear equations that may be used to solve for the $$\Beta_m$$ values:

\begin{bmatrix} F_{0} \\ \vdots \\ F_{N-1} \end{bmatrix} = \begin{bmatrix} (e^{\lambda_1 \Delta t})^{0} & \dots & (e^{\lambda_M \Delta t})^{0} \\ \vdots & \ddots & \vdots \\ (e^{\lambda_1 \Delta t})^{N-1} & \dots & (e^{\lambda_M \Delta t})^{N-1} \end{bmatrix} \begin{bmatrix} \Beta_1 \\ \vdots \\ \Beta_M \end{bmatrix}, $$ where $$M$$ unique values $$k_i$$ are used. It is possible to use a generalized matrix inverse if more than $$M$$ samples are used.

Note that solving for $$\lambda_m$$ will yield ambiguities, since only $$e^{\lambda_m}$$ was solved for, and $$e^{\lambda_m} = e^{\lambda_m \,+\, q 2 \pi j}$$ for an integer $$q$$. This leads to the same Nyquist sampling criteria that discrete Fourier transforms are subject to:
 * $$ \left|Im(\lambda_m)\right| = \left|\omega_m\right| < \frac{1}{2 \Delta_t}.$$

Model
test on position-position time correlation function of phonon bath

with spectral density

$$C(t)=\langle q(t) q(0) \rangle $$

$$\lim_{\hbar \rightarrow 0}G(t)=G_{cl}(t)$$

$$\tilde{C''}(\omega)=2\lambda \frac{\omega^{\alpha}}{\omega_c^{\alpha+1}} e^{-\omega^2/\omega_c^2}$$

$$\tilde{G_{cl}}(\omega)=\int_{-\infty}^{\infin} \operatorname{d}\!t \ e^{i\omega t} G_{cl}(t)$$

$$\tilde{G}(\omega)=Q(\omega)\tilde{G}_{cl}(\omega)$$