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\question
{Daniel Casimir}                  % Enter your name
{08/03/12}                        % Enter mm/dd/yy
{casimir.daniel@gmail.com}   % Enter e-mail address
{1-10}                          % Enter the number and question.
%question
{Consider the optical electric field of the incident light with
an angular frequency $\omega_0 = 2\pi \nu_0$ and amplitude $E_0$,
$$
E = E_0 \cos \omega_0 t .
$$
Then the dipole moment $P$ of a diatomic molecule is proportional to $E$
such that $P = \alpha E$, in which $\alpha$ is called the polarizability.
When the molecule is vibrating with a frequency $\omega$, then $\alpha$
is also vibrating with the frequency $\omega$,
$$
\alpha = \alpha_0 + \alpha_1 \cos \omega t .
$$ When substituting $\alpha$ into the formula $P
= \alpha E$, show that there are three different
frequencies for the scattered light (or $P$),
$\omega_0$ (elastic, Rayleigh scattering) and
$\omega_0 \pm \omega$ (inelastic, Stokes (-) and
anti-Stokes (+) Raman scattering).}
%answer
{The incoming electric field varies sinusoidally as
\begin{equation}
\mbox{E} = \mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}}
\end{equation}
where $\mbox{E}_{0}$ is the amplitude and $\nu_{0}$ is the frequency of the laser.  For a diatomic molecule the induced dipole moment P is
\begin{equation}
\mbox{P} = \alpha\mbox{E} = \alpha\mbox{E}_0\cos{2\pi\nu_{0}\mbox{t}}
\end{equation}
$\alpha$ is the polarizability.
The vibration of the molecule is also sinusoidal, with frequency $\nu_m$, and described by the nuclear displacement coordinate q,
\begin{equation}
\mbox{q} = \mbox{q}_0\cos{2\pi\nu_m\mbox{t}}
\end{equation}
where $\mbox{q}_0$ is the amplitude.  If the amplitude is small, $\alpha$ can be expanded as a function of the coordinate q as,
\begin{equation}
\alpha = \alpha_{0} + \biggl(\frac{\partial\alpha}{\partial\mbox{q}}\biggr)_{0}\mbox{q}   +  \cdots
\end{equation}
where $\alpha_{0}$ is the polarizability at the equilibirum position, and $\biggl(\frac{\partial\alpha}{\partial\mbox{q}}\biggr)_{0}$ is the rate of change of $\alpha$ with respect to the change in the coordinate q evaluated at the equilibrium position.  Combining the expressions for $\mbox{P}$, $\mbox{q}$, and $\alpha$ yields,
\begin{equation}
\mbox{P} = \alpha\mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}} = \biggl(\alpha_{0} + \biggl(\frac{\partial\alpha}{\partial\mbox{q}}\biggr)_{0}\mbox{q}\biggr)\mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}}
\end{equation}

\begin{equation}
\mbox{P} = \alpha_{0}\mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}} + \biggl(\frac{\partial\alpha}{\partial\mbox{q}}\biggr)_{0}\mbox{q}\mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}}
\end{equation}

\begin{equation}
\mbox{P} = \alpha_{0}\mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}} +  \biggl(\frac{\partial\alpha}{\partial\mbox{q}}\biggr)_{0}\mbox{q}_{0}\mbox{E}_{0}\cos{2\pi\nu_{m}\mbox{t}}\cos{2\pi\nu_{0}\mbox{t}}
\end{equation}

\begin{equation}
\mbox{P} = \alpha_{0}\mbox{E}_{0}\cos{2\pi\nu_{0}\mbox{t}} + \frac{1}{2}\biggl(\frac{\partial\alpha}{\partial\mbox{q}}\biggr)_{0}\mbox{q}_{0}\mbox{E}_{0}\biggl[\cos{2\pi\left(\nu_{0}+\nu_{m}\right)\mbox{t}} + \cos{2\pi\left(\nu_{0}-\nu_{m}\right)\mbox{t}}\biggr]
\label{eq:q1-10-8}
\end{equation}

The first term in Eq.\,(\ref{eq:q1-10-8})
represents the Rayleigh scattered light with the
same frequency as that of the incident laser
light, the second term represents the anti-Stokes
shifted light with frequency
$\left(\nu_{0}+\nu_{m}\right)$, and the last term
represents the Stokes shifted light with a
frequency of $\left(\nu_{0}-\nu_{m}\right)$.


} % Do not forget this.





