I am interested in calculating some response properties, namely, susceptibility and polarizability.

In principle, susceptibility should be the functional derivative of the electron density to a perturbing potential

$$ \chi(r,r') = \frac{\partial n(r)}{\partial \phi(r')} $$

Polarizability then should be the functional derivative of the polarization density to a perturbing field

$$ \alpha_{ab}(r,r') = \frac{\partial P_a(r)}{\partial E_b(r')} $$

Now assuming I can calculate the charge and the polarization density of an arbitrary system, then I could just place a Dirac delta potential at some point $r'$, calculate the charge density at $r$, and approximate $\chi(r,r')$ from finite differences.

I was thinking of doing the same thing for $\alpha_{ab}(r,r')$, but then I ran into the question of the infinitesimal field. Is there such a concept? I assume it must have a direction (of course it is a vector field...), but also be conservative. Doesn't conservativeness imply that it's defined at every point in space? Or can I take a very local potential, express the field as its gradient (making it conservative), but accept the fact that the field is defined everywhere?

Or am I thinking in a completely wrong direction, and there's a different way to evaluate functional derivatives numerically on a grid?

  • $\begingroup$ "infinitesimal field." are not all differential calculations based on infinitessimal fields? $\endgroup$
    – anna v
    May 25 at 6:17
  • $\begingroup$ Hi @LaBelleCroissant. Welcome to Phys.SE. Where did you see your 2 equations? Are you following a reference? Which page? $\endgroup$
    – Qmechanic
    May 25 at 11:28
  • $\begingroup$ @Qmechanic no, there's no reference I'm following. In fact, I don't know too much about functional analysis at all, so it might very well be that my question doesn't make any sense. Do you maybe know a reference that I could check on similar things? $\endgroup$
    – user336292
    May 25 at 14:48

1 Answer 1


The word infinitesimal may be misleading due to its long history as a mathematical concept. However, it is just a synonym for a quantity that eventually goes to zero in modern mathematics. In particular, when translating differentials or functional differentials into numerical calculations, the real meaning of infinitesimal is *a quantity small enough to justify the first-order approximation.

Going to a more formal discussion, let's assume that $F[\phi]$ is a functional of a function $\phi({\bf r})$ ($F$ may also depend on the position in the space). The functional derivative of $F$, usually indicated in Physics as $\frac{\delta F}{\delta \phi({\bf r})}$ can be defined as the function of the point $ {\bf r}$ that approximates $F[\phi +\delta \phi]-F[\phi]$, with the exception of second order terms: $$ F[\phi +\delta \phi]-F[\phi] = \int {\rm d}{\bf r} \frac{\delta F}{\delta \phi({\bf r})} \delta \phi({\bf r}) + O\left({(\delta \phi({\bf r}))^2}\right). \tag{1} $$

In a numerical treatment of functional derivatives, one has to approximate the space integrals in terms of finite sums over a finite grid of points: $$ \int {\rm d}{\bf r} \frac{\delta F}{\delta \phi({\bf r})} \delta \phi({\bf r}) \simeq \sum_i \frac{\delta F}{\delta \phi({\bf r}_i)} \delta \phi({\bf r}_i) \Delta V_i, $$ where $\Delta V_i$ is a volume around the point ${\bf r}_i$, small enough to ensure that inside the volume, the integrand does not vary too much.

Therefore, the discretized form of the equation $(1)$, with a function $\delta \phi$ different from zero only at the grid point ${\bf r}_i$, can be used to obtain a numerical (controlled) approximation of the functional derivative at the same point (with such a $\delta \phi$, the sum reduces to a single term).

In practice, if we want to evaluate numerically the functional derivative at the point ${\bf r}_i$ of $F[\phi]$, we start with a space discretization (then the function $\phi$ is represented by the values $\phi_j=\phi({\bf r}_j)$) and, depending on the geometry of the grid, there is a small volume $\Delta V_j$ associated to each grid point. The functional derivative at ${\bf r}_i$ can be numerically obtained with good precision by using a symmetric finite difference formula as $$ \frac{\delta F}{\delta \phi({\bf r}_i)} = \frac{1}{\Delta V_i}\frac{F[\phi+\delta \phi]-F[\phi-\delta \phi]}{2\delta \phi_i} $$

Notice that the two ingredients are necessary:

  1. a spatial grid fine enough (then a $\delta V_i$ small enough) to allow a safe substitution of the integral with a finite sum;
  2. a variation of the function $\phi$ small enough to ensure that the contributions of higher-order terms in $\delta \phi$ are negligible.

All the above discussion is valid independently on additional structures of the functionals. In particular, it holds in the same way if the functional is a component of a vector field at a point ${\bf r}$ and if the variable the functional depends on is, in turn, a component of a vector field. There is no constraint on the vector field. In particular, it may or may not be conservative.

  • $\begingroup$ So, in practice, can I take the perturbation to be $\delta(\pmb{r_i}) \pmb{\hat{x}}$ in the $x$ direction, for example? The fact that there's no potential corresponding to this field does not hurt the numerics in any way? $\endgroup$
    – user336292
    May 25 at 16:01
  • $\begingroup$ @LaBelleCroissant I would say that the important thing is that the perturbed field ($\phi$) should satisfy the constraints. The perturbation required for the evaluation of the functional derivative is localized at one point (actually a small volume) and it does not require special constraints $\endgroup$
    – GiorgioP
    May 25 at 21:08
  • $\begingroup$ Which constraints to you refer to when you say that the perturbing field should satisfy them? Is it only that it should be small in the small volume? $\endgroup$
    – user336292
    May 25 at 21:23
  • $\begingroup$ @LaBelleCroissant I wrote that the perturbation doesn't have constraints. It is the field around which the perturbation is performed that may have some constraint. For example, if it is an electrostatic field, the components of the field should correspond to an irrotational electric field. $\endgroup$
    – GiorgioP
    May 25 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy