After writing out this question, I actually figured out the answer, but I am surprised I haven't seen this detail mentioned in almost any paper on the topic, so I will answer my own question here in case anyone gets confused by this like I did.

In many polarizable force fields, the polarization is represented by induced dipoles located at each atom in the system. The dipoles are induced by the field produced by permanent multipoles in the system. For simplicity, let's assume the field is only produced by charges.

In many models, the atomic polarizabilities are isotropic, meaning that the polarization is equal in the x, y, and z directions.

In general, one might wish to define a polarizability that is anisotropic. In this case, for some chosen orthogonal reference frame of the molecule, the polarizability tensor need not be diagonal or have equal diagonal elements.

The system of equations that needs to be solved in induced dipole models is,

$$ \begin{bmatrix} \alpha_1^{-1} & -T_{12}^{dd} & \dots & -T_{1N}^{dd} \\ -T_{21}^{dd} & \ddots & \dots & \vdots \\ \vdots & & & \\ -T_{N1}^{dd} & \dots & & \alpha_N^{-1} \\ \end{bmatrix} \begin{bmatrix} \mu_1 \\ \mu_2 \\ \vdots \\ \mu_N \end{bmatrix} = \begin{bmatrix} E_1^q \\ E_2^q \\ \vdots \\ E_N^q \end{bmatrix} $$

Where $T_{ij}^{dd}$ is the dipole-dipole interaction tensor between atoms $i$ and $j$. $\alpha_i^{-1}$ is the inverse polarzability tensor. $\mu_i$ is the dipole induced by the field of permanent charges at that atom, $E_i^q$.

The polarization energy of the induced dipoles is then given by,

$$ E_{pol}=-\frac12\sum_i\mu_i\cdot E_i^q $$

If the polarizability tensor is symmetric, then this will trivially result in a rotationally invariant energy.

However, if the polarizability is anisotropic, then the polarizability elements are defined relative to a local axis frame and some rotation must be made to result in a rotationally invariant polarization energy.

I am somewhat surprised that I can't find a good discussion of this. I have an implementation of this model which I've been playing around with and can't seem to get the energy to be rotationally invariant when using anisotropic polarizabilities.

If anyone can point me towards some references that discuss this specific detail or if anyone knows, then that would be greatly appreciated.


1 Answer 1


The answer is actually extremely simple, but wasn't obvious to me at first. It is clear that the elements of the anisotropic polarizability tensor are easiest to define relative to a local axis system. For instance, if one were parameterizing the polarizability of an oxygen atom in water, then one would choose a specific x, y, and z relative to the atoms in water.

The system of linear equations described is extremely convenient to solve in the global axis system, so we must ask how to transform the polarizability from some arbitrary orientation to the reference axis system.

We can find this rotation matrix, $R$, by computing the direction cosine matrix to compute the transformation from the global axis system to the reference axis system (and back again with the transpose).

With this matrix in hand, we can make the polarization energy rotationally invariant with an appropriate modification of the inverse polarizability, $\alpha_i^{-1}$. Namely, we must compute the similarity transformation of $\alpha_i^{-1}$ as $R^{-1}\alpha_i^{-1}R$.

If you think about this operation being applied to the induced dipoles, the product $R^{-1}\alpha_i^{-1}R\mu_i$, has the effect of rotating the dipoles into the reference axis system, appropriately applying the scaling by the inverse polarizability and then rotating back to the global axis system.

Hopefully this is helpful to someone. It's a bit obvious in retrospect, but I was very confused :)


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