Ensemble average of polarization with applied field This question is in regard to a dipole moment fluctuation formula seen in: MOLECULAR PHYSICS, 1983, VOL. 50, NO. 4, 841-858, on page 843.
For a system of polar liquid (water) at equilibrium, the ensemble average of the total dipole moment $M$ is:
$$\left<M\right> = \frac{\int dq M\exp(-\beta U(q))}{\int dq \exp(-\beta U(q))}$$
Where $q$ is a collective variable (for e.g. position and momentum) that dictates the Hamiltonian ($U$) of the system. The denominator is the partition function.
Now consider a small external applied electric field $E_o$. The ensemble average of the total dipole moment is now:
$$\left<M\right>_E = \frac{\int dq M\exp(-\beta (U(q)-ME_o))}{\int dq \exp(-\beta (U(q)-ME_o))}$$
where the Hamiltonian has an extra term that accounts for the field-dipole interaction.
To simplify the above expression, literature often states "apply linearization on $E_o$" and that's where I'm confused. My interpretation of that is:
$$\exp{(\beta ME_o) \approx 1 + \beta ME_o}$$
which makes
$$\left<M\right>_E = \frac{\int dq M\exp(-\beta U(q)) + M^2\beta ME_o \exp(-\beta U(q))}{\int dq \exp(-\beta (U(q)-ME_o))}$$
and I'm stuck here.
In the final expression shown in the paper, they arrived at:
$$\left<M\right>_E = \beta E_o\left<M^2\right>$$
where the bracket around $M^2$ denotes average in equilibrium of no field, i.e. $\left<M^2\right>=\frac{\int dq M^2\exp(-\beta U(q))}{\int dq \exp(-\beta U(q))}$.
Could someone help me explain how they arrived at their final expression?
 A: I've found an answer. My misunderstanding was from the term "linearization". That term was meant for $\left<M\right>_E$ and not $E_o$.
Starting from ensemble average of $M$ with a field $E_o$:
$$\left<M\right>_E = \frac{\int dq M\exp(-\beta (U(q)-ME_o))}{\int dq \exp(-\beta (U(q)-ME_o))}$$
The "linearization" here means to expand $\left<M\right>_E$ to its first order in the Taylor series:
$$\left<M\right>_E \approx \left<M\right> + \frac{\delta\left<M\right>_E}{\delta E_o}$$
Applying product/quotient rule to the expression of $\left<M\right>_E$, we'll find:
$$ \frac{\delta\left<M\right>_E}{\delta E_o}= \frac
{\int dq \beta M^2E_o\exp(-\beta (U(q)-ME_o))\int dq \exp(-\beta (U(q)-ME_o))
- \int dq M\beta E_o\exp(-\beta (U(q)-ME_o))\int dq M\exp(-\beta (U(q)-ME_o))}
{\int dq \exp(-\beta (U(q)-ME_o))^2}$$
$$ \frac{\delta\left<M\right>_E}{\delta E_o}= \frac
{\int dq \beta M^2E_o\exp(-\beta (U(q)-ME_o))}
{\int dq \exp(-\beta (U(q)-ME_o))}-
\frac
{\int dq M\beta E_o\exp(-\beta (U(q)-ME_o))\int dq M\exp(-\beta (U(q)-ME_o))}
{\int dq \exp(-\beta (U(q)-ME_o))^2}$$
Then by following the definition of ensemble average:
$$ \frac{\delta\left<M\right>_E}{\delta E_o}= \beta E_o\left<M^2\right> - 
\beta E_o\left<M\right>\left<M\right>$$
Finally, by also assuming $\left<M\right>=0$ for an isotropic system with no field:
$$\left<M\right>_E = \left<M\right> + \beta E_o (\left<M^2\right> - \left<M\right>\left<M\right>) = \beta E_o\left<M^2\right>$$.
woooooohooo, it only took me 3 days to figure this out :(
