If there exist two initially neutral bodies (say atoms) some distance apart, with no external electric field applied, can they induce dipoles within each other?

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    $\begingroup$ Objects can and do induce temporary dipoles in each other. This is the origin of the London dispersion force. Are you asking if there is a state where permanent dipoles exist that has a lower energy then the unpolarised state? $\endgroup$ – John Rennie May 19 '15 at 9:53

Indeed they can. This can even be seen in a classical set of equations if the interaction between the two polarizable objects is strong enough. Take a simple 1d problem with two polarizable point particles in a line. For low polarizations, we assume a linear relation:

$$ p_1 = \alpha E(r_1) = \alpha (E_2 + E_{ext}) = \alpha (k p_2 + E_{ext})$$

$$ p_2 = \alpha (k p_1 + E_{ext}) $$

where the $p$s are dipole moments, $\alpha$ is the polarizability, $E$s are fields (i.e., $E_1$ is the field at $p_2$'s position due to the polarization at $p_1$), and k relates the polarization of $p_1$ to the field it produces at $p_2$ (scales like the inverse distance cubed). This is solved by

$$ p_1 = p_2 = \alpha \frac{\alpha k +1}{1-\alpha^2 k^2} E_{ext} $$

It's clear that something strange happens as $\alpha k $ approaches 1: polarization diverges. This is because we didn't include the fact that polarization has a finite limit. Let's call the maximum polarizability $p_m$. We can include this in our model as such:

$$ p_2 = \alpha (k p_1 + E_{ext})(1-\frac{p_2^2}{p_m^2}) $$

and vice versa for $p_1$. Solving this is tedious, but doable. What you find is that for $\alpha k < 1$, there is a single real solution. However, as $\alpha k$ increases, this solution becomes unstable and two new solutions emerge that form a hysteresis loop in $E_{ext}$. In other words, the two objects together become "ferro-electric" with a remanent dipole moment $\pm \frac{\alpha k -1}{\alpha k} p_m$ that persists in the absence of an external field.

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