# Relationship between electric dipole moment and polarization?

$$\vec{p}_\text{ind} = \alpha \vec{E}$$

The induced dipole moment is the polarizability times the electric field vector.

$$\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}(\omega)$$

The polarization is proportional to the susceptibility times the electric field vector.

In spectroscopy we used the dipole approximation. Unfortunately, I often mix up the different formulas. Is there a relationship between both formulas, or do both equations tell us exactly the same thing?

In fact, the first equation is applicable on a microscopic scale (for an atom or a molecule): $\space$ $\vec{p}_\text{ind}=\alpha \vec{E}_\text{local}$
While the second one is applicable on a macroscopic scale (for the bulk): $\space$ $\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}_\text{ext}(\omega)$ So, the polarisability is a microscopic quantity, while the susceptibility is a macroscopic quantity.
• Indeed, the microscopic / macroscopic versions carry forward also for $\beta$ and $\chi^{(2)}$, $\gamma$, etc. – Geoff Hutchison Jun 30 '15 at 23:33
• @YomenAtassi I like your answer and as well the additional page. However, I've still confused, because in Spectroscopy both have been used and we only took a look into small molecules, not bulk. I also have no idea to which of the equations (eq. 1 or eq. 2) the formula for the dipole approximation $\hat{V}=-\hat{\mu}\overrightarrow{E}$ is best assigned. (related to my two weak old, but still unsolved question: chemistry.stackexchange.com/questions/33170/…) – laminin Jul 1 '15 at 18:38
• In the scope of our lecture of Spectroscopy (atoms, small molecules, aggregates) we have restricted to linear spectroscopy which means there is no dependence of quadratic terms in $\overrightarrow{E}$ in eq(2). I'm not sure but may be it's possible, you get also quadratic dependency in $\overrightarrow{E}$ for eq(1) if we deal with non-linear spectroscopy? – laminin Jul 1 '15 at 19:49