Linear Spectroscopy

The goal of this section is to show that the linear absorption spectrum $\sigma(\omega)$ of a system is determined by the regression of spontaneous polarization fluctuations at equilibrium as follows:

$$\sigma(\omega) = \beta 2 \omega^2 f_0^2 \bar{\varepsilon}^2 \int_{0}^{\infty} <\delta A(0) \delta A(t')> (\omega t') dt',$$ (450)

where $A(t)$ is the time-dependent polarization

$$A(t) = \sum_j p_j \langle \phi_j(t) \mid \hat{A} \mid \phi_j(t) \rangle,$$ (451)

where $\phi_j(t)$ evolves according to the unperturbed system Hamiltonian $H_0$ as follows:

$$i \hbar \frac{\partial}{\partial t} \phi_j(t) = \hat{H}_0 \phi_j(t).$$ (452)

In order to derive Eq. (450), consider that the system is perturbed by a monochromatic electric field,

$$f(t)=f_0 \bar{\varepsilon} (e^{-i \omega t} + e^{i \omega t}),$$ (453)

where $f_0$ is the amplitude of the field and $\bar{\varepsilon}$ is the unit vector along the spatially uniform electric field.

In the linear regime, the interaction between the system and the field is

$$\Delta \hat{H}(t) = - f(t) \hat{A}.$$ (454)

The total energy of the system is

$$E(t) = \sum_j p_j \langle \phi_j(t) \mid \hat{H}_0 + \Delta \hat{H}(t) \mid \phi_j(t) \rangle,$$ (455)

and the differential change of energy per unit time is

$$\dot{E}(t) = \sum_j p_j \langle \phi_j(t) \mid \frac{\partial \Delta \hat{H}(t)}{\partial t} \mid \phi_j(t) \rangle = - \dot{f(t)} A(t),$$ (456)

since according to Eq. (452),

$$\langle \frac{\partial \phi_j(t)}{\partial t} \mid \hat{H}_0 + \D......at{H}_0 + \Delta \hat{H}(t) \mid \frac{\partial \phi_j(t)}{\partial t} \rangle.$$ (457)

Therefore, the total energy change $\sigma$ due to the interaction with the external field for time T is

$$\sigma = \frac{1}{T} \int_0^T (-\dot{f}(t)) A(t),$$ (458)

where, according to Eq. (453),

$$\dot{f}(t) = -i f_0 \bar{\varepsilon} \omega [e^{-i \omega t} - e^{+i \omega t}].$$ (459)

Substituting Eq. (459) into Eq. (458) we obtain,

$$\sigma = - \frac{i \omega}{T} \int_0^T f_0 \bar{\varepsilon} [ e^...... t} - e^{-i \omega t} ] (<A> + \int_{-\infty}^{\infty} \chi (t-t') f(t') dt' ).$$ (460)

Note that

$$\int_{-\infty}^{\infty} \chi (t-t') f(t') dt' = \int_{-\infty}^{\infty} \chi(-t') f(t'-t) dt' = \int_{-\infty}^{\infty} \chi(t') f(-t'-t) dt',$$ (461)

thus Eq. (460) becomes

$$\sigma = - \frac{i \omega}{T} \int_0^T dt f_0 \bar{\varepsilon} [......a t} - e^{-i \omega t} ] (<A> + \int_{-\infty}^{\infty} \chi(t') f(-t'-t) dt').$$ (462)

In order to simplify Eq. (462), we note that

$$\frac{1}{T} \int_0^T e^{int'} dt' = \begin{cases}1, \hspace{.2cm}......hen} \hspace{.2cm} \text{n=0} \\ 0, \hspace{.2cm} \text{otherwise}, \end{cases}$$ (463)

therefore, Eq. (462) becomes

$$\sigma = - \frac{i \omega}{T} \int_0^T dt f_0 \bar{\varepsilon} [......i(t') f_0 \bar{\varepsilon} [e^{+i \omega (t+t')} + e^{-i \omega (t+t')} ] dt',$$ (464)

or

$$\sigma = - i \omega f_0^2 \bar{\varepsilon}^2 \int_{-\infty}^{\infty} \chi(t') [e^{+i \omega t'} - e^{-i \omega t'} ] dt'.$$ (465)

Therefore,

$$\sigma = 2 \omega f_0^2 \bar{\varepsilon}^2 \int_{-\infty}^{\infty} \chi(t') (\omega t') dt'.$$ (466)

Substituting Eq. (449) into Eq. (466) we obtain

$$\sigma = -\beta 2 \omega f_0^2 \bar{\varepsilon}^2 \int_{0}^{\infty} \frac{d}{dt} <\delta A(0) \delta A(t')> (\omega t') dt'.$$ (467)

Finally, integrating Eq. (467) by parts we obtain Eq. (450), since Eq. (467) can be written as $\int dt' u(t') dv/dt' = u(t') v(t') - \int dt' v(t') du/dt'$, with $u(t') =$   sin$(\omega t')$ and $v(t')= <\delta A(0) \delta A(t')>$.