One common way of calculating a dipole polarizability from electronic structure is by calculating the following derivative by finite difference: $$ \mathbf{\mu}=\left(\frac{\partial U}{\partial \mathbf{E}}\right)_{\mathbf{E}=0} $$ where here $\mathbf{E}$ is an external electric field and $U$ is the electronic energy. Therefore, we can calculate the three components of the dipole moment, $\mathbf{\mu}$, via finite difference where we apply a finite field along the principle axes in both the positive and negative directions. Notice, however, that this calculation is typically done in the case that zero net field is applied.

We can then calculate the dipole polarizability in the same manner as the change in dipole in response to an electric field or the second derivative of the energy with respect to a field, $$ \alpha_{pq}=\left(\frac{\partial^2 U}{\partial \mathbf{E}_p\partial \mathbf{E}_q}\right)_{\mathbf{E}=0} $$

We can then calculate each of the nine polarizability tensor components via second-order finite difference of the energy. $p$ and $q$ are the cartesian axes. Typically, again, this is done with zero net external field.

My question is: Is there any physical reason and/or technical reason why I can't do the same thing in the case of a finite external field?

If one considers a molecule in solution, for example, the dipole response should not be the same as the gas-phase zero-field dipole response because the molecule is in the presence of an electric field created by the surrounding environment. That's the motivation for the calculation.

Additionally, any references that do this or explain why not to do this would be greatly appreciated!


1 Answer 1


There is nothing wrong with taking electric field derivatives at nonzero field strengths. The zero field strength conditions come from taking a Taylor expansion of the internal energy about $\vec{E}=0$:

$$ U=U_0+\mu_0\cdot\vec{E}+\frac{1}{2}\alpha_0:\vec{E}\vec{E}+... $$

where the static dipole/polarizability are defined at zero field strength. A similar expansion could be made for the energy at any field strength, yielding dipoles, polarizabilities, hyperpolarizabilities, etc. for those field strengths.

Note the frequency $\omega$ of the field also affects these properties, but as long as the you are including this frequency in however you represent the field in your calculation, your finite difference scheme should capture this frequency dependence as well.

You should also be careful with taking double numerical derivatives, as it can often require very high precision (and/or a stencil with many points) to achieve reasonable accuracy.


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