# Alternate derivation of linear response function

I’ve given a derivation before of the density-density response function, but today I want to give an alternate derivation of the more general linear-response function, which will prove to be useful in the derivation of the time-dependent Hartree-Fock equations (TDHF), also known in the nuclear physics community as the Random Phase Approximation (RPA). This derivation is largely taken from McWeeny (1989).

In order to derive the response function — also called frequency-dependent polarizability — we must first partition the Hamiltonian into two parts: the time-independent Hamiltonian, and the time dependent response:

$H = H_0 + H'(t) \ \ \ \ \ (1)$

Furthermore, the time-dependent Schrodinger equation is given as:

$H\Psi = i\hbar\frac{\partial \Psi}{\partial t} \ \ \ \ \ (2)$

In the interaction picture, $${\Psi'_0(t) = \Psi'_0e^{-iE_0t/\hbar}}$$. Thus we can partition the time-dependent wavefunction (expanding over basis of complete eigenstates) as

$\Psi'_0(t) = \sum\limits_n c_n(t)\Psi_n \notag \ \ \ \ \ (3)$ $\Psi'_0e^{-iE_0t/\hbar } =\sum\limits_n c_n(t)e^{-iE_0t/\hbar}\Psi_n \ \ \ \ \ (4)$ $\Psi'_0 = \Psi_0 + \sum\limits_{n\neq0} c_n(t)e^{-i\omega_{0n}t}\Psi_n \ \ \ \ \ (5)$

Where

$\omega_{0n} = (E_n - E_0)/\hbar \ \ \ \ \ (6)$

are real, positive, exact excitation frequencies of unperturbed system. Note that $${c_0(t) = 1}$$. We are assuming that $${H'(t)}$$ is turned on slowly at time $${t \rightarrow -\infty}$$. Substitute 1 and 3 into 2, separate the orders, and impose the boundary conditions that $${c_0 = 1}$$ and $${c_m = 0}$$ $${(n\neq0)}$$ at $${t \rightarrow -\infty}$$. This gives

$i\hbar\dot{c}_n = \langle n \vert H'(t) \vert 0 \rangle e^{i\omega_{0n}t} \ \ \ \ \ (7)$

If we let

$H'(t) = F(t)A \ \ \ \ \ (8)$

Where a ‘fixed’ Hermitian operator $${A}$$ determines the shape’ of the perturbation, while time dependence is confined to the (real) ‘strength’ factor $${F(t)}$$. For a perturbation beginning at time $${t \rightarrow -\infty}$$ up to time $${t}$$,

$c_n(t) = (i\hbar)^{-1}\int\limits_{-\infty}^{t} \langle n \vert A \vert 0 \rangle F(t') e^{i\omega_{0n}t'}dt' \ \ \ \ \ (9)$

Which, to first order, determines the perturbed wavefunction. Now we are interested not in the perturbed wavefunction per se, but rather in the response of an observable $${O}$$ to the perturbation.

$\delta\langle O \rangle = \langle O \rangle - \langle O \rangle_0 = \int\limits_{-\infty}^{t} K(OA\vert t-t')F(t')dt' \ \ \ \ \ (10)$

where

$K(OA\vert t-t') = (i\hbar)^{-1} \sum\limits_{n\neq0} [\langle 0 \vert O \vert n \rangle\langle n \vert A \vert 0 \rangle e^{-i\omega_{0n}(t-t')} - \langle 0 \vert A \vert n \rangle\langle n \vert O \vert 0 \rangle e^{i\omega_{0n}(t-t')}] \ \ \ \ \ (11)$

This is a time correlation function, relating fluctuation of $${\langle O \rangle}$$ at time t to the strength of the perturbation $${A}$$ at some earlier time $${t'}$$. $${K(OA\vert t-t')}$$ is defined only for $${t'<t}$$, in accordance with the principle of causality. Thus, it is a function only of the difference $${\tau = t - t'}$$. Recalling the definitions of the Fourier transform $${f(\omega)}$$:

$f(\omega) = \int\limits_{-\infty}^{\infty} F(t) e^{i\omega t} dt \ \ \ \ \ (12)$

Then instead of 8, we have:

$H'(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} f(\omega) A_{\omega} e^{-i\omega t} d\omega \ \ \ \ \ (13)$

Requiring $${H'(t)}$$ to be Hermitian,

$H'(t) = H'(t)^{\dagger} \ \ \ \ \ (14)$ $\frac{1}{2\pi} \int\limits_{-\infty}^{\infty} f(\omega) A_{\omega} e^{-i\omega t} d\omega = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} f(-\omega) A^{\dagger}_{\omega} e^{i\omega t} d\omega \notag \ \ \ \ \ (15)$

Now,

$A_{-\omega} = A_{\omega}^{\dagger} \ \ \ \ \ (16)$ $f(-\omega) = f(\omega) \ \ \ \ \ (17)$

Which, upon combining the expressions for $${H'(t)}$$ so as to Hermitize’ the expression:

$2H'(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} (f(\omega) A_{\omega} e^{-i\omega t} + f(\omega) A_{-\omega} e^{i\omega t}) d\omega \notag \ \ \ \ \ (18)$ $H'(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} f(\omega) \frac{1}{2} (A_{\omega} e^{-i\omega t} + A_{-\omega} e^{i\omega t}) d\omega \ \ \ \ \ (19)$

Thus

$A = \frac{1}{2}(A_{\omega} + A_{-\omega}) \ \ \ \ \ (20)$

with $${F(t)}$$ real. Instead of working in the time domain, we may also consider the response in terms of a single oscillatory perturbation. This means that

$H'(\omega) = \frac{1}{2} (A_{\omega} e^{-i\omega t} + A_{-\omega} e^{i\omega t}) \ \ \ \ \ (21)$

To ensure $${H'(\omega)}$$ builds smoothly from zero at $${t \rightarrow -\infty}$$, we can introduce a convergence factor $${e^{\eta t}}$$ with the initial condition $${c_0 = 1}$$ and $${c_n = 0}$$, which gives:

$c_n(t) = \lim_{\eta \rightarrow 0} \left(-\frac{1}{2\hbar}\right) \left\{ \frac{\langle n \vert A_{\omega} \vert 0\rangle}{\omega_{0n} - \omega - i\eta} e^{i(\omega_{0n} - \omega - i\eta)t} + \frac{\langle n \vert A_{-\omega} \vert 0\rangle}{\omega_{0n} + \omega - i\eta} e^{i(\omega_{0n} + \omega - i\eta)t} \right\} \ \ \ \ \ (22)$

Then, collecting terms of $${\pm \omega}$$:

$\delta \langle O \rangle = \frac{1}{2} \left[\Pi(OA_{\omega}\vert \omega)e^{-i\omega t} + \Pi(OA_{-\omega}\vert -\omega)e^{i\omega t} \right] \ \ \ \ \ (23)$

Finally:

$\Pi(OA_{\omega} \vert \omega) = \lim_{\eta \rightarrow 0} \left(\frac{1}{\hbar}\right) \sum\limits_{n\neq0} \left\{ \frac{\langle 0 \vert O \vert n\rangle\langle n \vert A_{\omega}\vert 0\rangle}{\omega + i\eta - \omega_{0n}} - \frac{\langle 0 \vert A_{\omega} \vert n\rangle \langle n \vert O \vert 0 \rangle}{\omega + i\eta + \omega_{0n}} \right\} \ \ \ \ \ (24)$

Which is the response function, or frequency-dependent polarizability.