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I am interested in representing an arbitrary charge density (say, of atoms in a molecule) $\rho(r), \; r\in \mathbb{R}^3$ by a finite linear combination of basis functions

$\rho(r) = \sum_{i=1}^N q_i \phi_i (r) $

where $\phi_i (r)$ is normalized to $\int_{\mathbb{R}^3} \phi_i (r) dr = 1$ and has the interpretation of being the shape of some charge distribution (shape) of a unit charge. $\rho$ and the $\phi_i$s are real-valued but may be positive in some regions and negative in others. The basis functions are nonorthogonal and local in space but not strictly compact. Let's say for now that we use spherical Gaussians of the form $\phi_i (r) \propto \exp (-\alpha_i |r-R_i|^2)$, where $R_i$ is where the basis function is centered around. The number of basis functions chosen scales approximately as the number of atoms, as we expect charge to concentrate around atomic nuclei. (We may add additional basis function per atom of different shapes until we achieve a reasonable approximation to the desired shape of the charge distribution around an atom.)

The energy of the system can then be given by

$E = \frac 1 2 \sum_{i,j=1}^N q_i q_j J_{ij}$

where the matrix $J$ has elements

$J_{ij} = \int_{\mathbb{R}^{3\times 2}} \frac{\phi_i(r_1) \phi_j(r_2)}{|r_1 - r_2|} dr_1 dr_2$

and represents the Coulomb interaction between the unit charge distributions $\phi_i$ and $\phi_j$.

One way to look at the matrix $J$ is as a finite dimensional (approximate) representation of the Coulomb operator $\hat J = 1 / {|r_1 -r_2|}$. We know that $\hat J$ has certain nice properties such as positivity, so we expect a "good" representation of $\hat J$ should be a symmetric positive definite matrix.

My question is this: are there conditions on the discrete representation (possibly expressible as conditions on the {$\phi_i$} basis) to detect whether or not a given claimed representation $J$ is "good" in that it preserves such properties? Or asked another way, if I have some matrix $J$ which is claimed to represent $\hat J$, what are necessary and sufficient conditions on its matrix elements for it to be a "good" representation of $\hat J$?

I hope the question makes sense, and that I am not misusing too much terminology.

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  • $\begingroup$ What type of basis functions do you have (e.g. are they orthogonal)? As I recall, how "big" you need your matrix to be so that the truncation is a satisfactory approximation of your operator is dependent on the nature of the basis set you chose. On the other hand, for computational purposes, what one might do is take a sequence of bigger and bigger sections of the matrix ($J_{N}$ for some sequence of $N$), do your computations with each of those truncations ($E_N$ for instance), and then employ an extrapolation algorithm (e.g. Wynn ε) to your new sequence of quantities. $\endgroup$
    – J. M. isn't a mathematician
    Commented Aug 22, 2010 at 22:51
  • $\begingroup$ Oh I completely forgot to specify what basis I'm using! I will update the question. $\endgroup$
    – Jiahao Chen
    Commented Aug 23, 2010 at 0:35
  • $\begingroup$ Here is an example of what I mean by a "bad" representation of J: I add basis functions to improve the shape, but accidentally use a basis function that is linearly dependent with some of the existing functions. Then the matrix representation of J that I have becomes singular and is a "bad" representation because it does not preserve the positivity of the original operator. $\endgroup$
    – Jiahao Chen
    Commented Aug 23, 2010 at 0:45
  • $\begingroup$ Here is another example of a "bad" representation: I have a matrix that claims to be a finite dimensional representation of J, but it is not diagonally dominant and thus may not be positive definite. Although not diagonally dominant =/=> not posdef, I have found from numerical experiments that for some matrices that are not DD, it is possible to come up with an arrangement of basis functions that reproduces the off-diagonal elements correctly, but produces a zero eigenvalue in the J matrix when the diagonal elements are sufficiently small. $\endgroup$
    – Jiahao Chen
    Commented Aug 23, 2010 at 0:50
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    $\begingroup$ Apologies, the last time I did any computational chemistry work was as an undergraduate, so I may be rusty on the names and terms... but is this: dx.doi.org/10.1134/S0030400X06060014 related to what you're trying to do? $\endgroup$
    – J. M. isn't a mathematician
    Commented Aug 23, 2010 at 2:22

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The answer to your question, "are there conditions on the discrete representation ... to detect whether or not a given claimed representation $J$ is 'good'...", is yes. One usually measures "good" in terms of a norm.

I think your operator $\hat{J}$ is actually an integral operator, right?

$$ \hat{J}(u)(r) = \int \frac{u(s)}{|r-s|}~ds $$

To keep things as general as possible, this is a map between two normed linear spaces $\hat{J}: V\to W$. Your finite matrix can be regarded as the representation of $\hat{J}$ on some finite dimensional subspace $V_n\subseteq V$; call it $J_n:V_n \to W$. (Here $n$ is a parameter, maybe the basis size, etc.) Writing $P_n:V \to V_n$ for some appropriate projection operator, your measure of goodness will be

$$ \| \hat{J}u - J_nP_nu\|_W \leq O(n^{-\alpha}), \quad n\to \infty, \quad u\in V,$$

where $\alpha > 0$ and where the asymptotic constant in the big-O notation invovles the quantity $\|u\|_V$ in some way.

This is more an attempt to formulate your question mathematically rather than to answer it. Does it look like something you can work with?

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  • $\begingroup$ Most probably a discretization indeed; that's how it's done in computational chemistry. As an aside, the condition number I was alluding to in the comments is in fact a product of two norms: $\kappa(J)=\|J\|\|J^{-1}\|$ ; as can be seen, this is dependent on which norm you are considering (Frobenius, spectral, etc.), but otherwise, the way it is done in numerical work, only the order of magnitude of the condition number need be considered when assessing how good- or ill-conditioned a discretized operator (matrix) is. $\endgroup$
    – J. M. isn't a mathematician
    Commented Aug 23, 2010 at 7:51
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    $\begingroup$ Having reread the original question, your note about the condition number of $J$ being a measure of how good the representation is is a cogent answer. I would expect that someone working on a problem of this nature has seen something like the formulation in my "answer" before. I'm tying to get ideas on the table for discussion. $\endgroup$
    – Jerry
    Commented Aug 23, 2010 at 8:45

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