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3:59 PM
A: Relationship between functional derivative and potential value at a position

wzkchem5Yes, the sum of these potentials is zero everywhere with the true ground-state density. But the true ground-state density is generally only expressible when the basis set is complete (if you know the true ground-state density in advance, you can of course construct a finite basis set that reprodu...

+1 for taking care of this one so quickly! I have a comment about "the true ground-state density is generally only expressible when the basis set is complete (if you know the true ground-state density in advance, you can of course construct a finite basis set that reproduces it exactly)". In this case, can't we say that the finite basis set is complete? — Nike Dattani 1 hour ago
@NikeDattani I think the definition of a complete basis set is that you can use the basis set to expand an arbitrary function. A finite basis set in an infinite-dimensional Hilbert space can be exact for a given purpose, but cannot be completewzkchem5 57 mins ago
Okay! I don't think I've seen the definition of "complete basis set" before. I thought a basis set that exactly represents a function could be considered "complete", but I suppose we could also define it to be a basis set that exactly represents all (well-behaved) functions. Do you know any reference that explicitly defines complete basis set that way? — Nike Dattani 53 mins ago
@NikeDattani The Wikipedia page en.wikipedia.org/wiki/Basis_set_(chemistry) mentions the complete basis set a few times. It seems to implicitly assume that completeness implies an infinite basis set, though it does not define completeness directly... — wzkchem5 43 mins ago
The word "complete" seems to be mentioned 18 times on that page, but with no definition of complete basis set. The word "infinite" only appears in "When the finite basis is expanded towards an (infinite) complete set of functions", which is in a paragraph with no references. I'm curious to know now if anyone's ever explicitly defined the term, or if it's just been used vaguely all along! — Nike Dattani 23 mins ago
@NikeDattani I looked up Jensen's famous textbook Introduction to Computational Chemistry (2nd edition), and on page 97 he says "In the limit of a complete basis set (infinite number of basis functions), the results are identical to those obtained by a numerical HF method, and this is known as the Hartree–Fock limit." Again this supports the view that completeness implies infinite basis, but again it is not worded like a formal definition... — wzkchem5 17 mins ago
4:20 PM
The idea of a "basis set" in general is an approximation of the mathematical concept of a basis for a vector space, i.e. a set of vectors that span the entire space. "Complete basis set" should basically just translate to "basis", since we are saying we can actually represent any function in the vector space, rather than just an approximation. To have a complete basis set/basis even for 1-electron functions, we would need an infinite set of basis functions.
This is why expansions like Taylor or Fourier series in general have infinitely many terms.
This Wolfram page gives a similar notion of what it means for a set of functions to be complete: mathworld.wolfram.com/CompleteOrthogonalSystem.html
4 hours later…
8:08 PM
@Tyberius That makes sense in terms of the vector |1> not being enough to span the Hilbert space for 2-level (spin-1/2) systems which would need to be spanned by {|0> , |1>} or {|+> , |->}, or {|i> , |-i>}.
It's a bit fascinating to me, that that MathWorld article on "Complete Orthogonal System" seems to be the closest thing on MathWorld to the definition of "Complete" since he article for Complete set of Functions is empty.
Also, a basis doesn't need to be orthogonal, linearly independent is enough.
8:35 PM
By the way, do you know the history of the term "basis set"? The way you describe it is like it's a "subset of a basis". I always found it awkward when seeing this phrase, because in math class we learned "basis" and chemists talked about "basis sets" and I didn't really see the former group of people using the latter terminology.
8:46 PM
@user1271772 Right, it similar not quite the same. I think the Complete Set of Functions page would be the closest to what we are describing (if it existed)
here's some relevant questions from Math.SE:
Q: When a set of functions becomes complete?

Omar NagibI know that a set of functions are said to form a complete basis on an inteval if any function on that interval can be expressed as a linear combination of the functions in the set. I also know that every function in the set are orthogonal. Now what is what is the condition(s) that a set of fun...

Q: What is the difference between a complete orthonormal set and an orthonormal basis in a Hilbert space

ShinyaSakaiI don't know if it's true but I think in a finite dimensional veccter space, an orthonormal set which is complete becomes a basis. But in a Hilbert space, the books say that the set of finite linear combinations of the elements in a complete orthonormal set is only dense in the Hilbert space. So ...

Q: Complete function spaces and inner products

tisPrimeTimeI am reading the technical report by Leo Breimann entitled "Some Infinity Theory for Predictor Ensembles". On page 4 he has the following: Definition 1 Let $L_2(P)$ be the space of functions on $\mathbb{R}^D$ that are square - integrable wrt $P(dx)$. A set of functions $F$ in $L_2(P)$ will ...

Q: Can I form a complete set of functions using $e^{-nx}$?

Kieran MullenIf I start with the set of functions $e^{-nx}$ for all integers $n>1$, can I use them as basis to create a complete set of orthogonal functions on the interval $(0,+\infty)$? By complete I mean that they can be used in an expansion that will converge in the mean to any square integrable function...

Q: Complete orthonormal families of functions in product spaces

Alex NolteToday I was working on solving a partial differential equation in the unit square, $[0,1] \times [0,1].$ Using the method of separation of variables on the spatial problem gave solutions of the form $f_i(x) g_j(y)$, where $\{f_i\}$ and $\{g_j\}$ each form complete orthonormal families of function...

Q: Complete set of vectors?

user123124The following excerpt is from the book Operators, Functions, and Systems, Vol. 1 by Nikolai K. Nikolski, pages 216-217 What is a "complete set of vectors" in the proof of Lemma 2.4.4? Im putting my money on "A subset of the basis which is linearly independent and whose span is dense is ca...

and probably very many more. I haven't had a chance to go through all of them yet, but I'm leaving here in case someone does want to, and for future reference if I find myself with a burning desire to get to the bottom of this!
Interestingly "completeness of functions" isn't one of the examples here: en.wikipedia.org/wiki/Completeness
@user1271772 I don't, but "subset of a basis" might not be a bad way to describe it. I think basis set is just to avoid referring to a basis, where a mathematician would assume we should be able to represent any function in the space exactly just using these basis functions.
The term "complete basis set" might have been inspired by overompleteness in linear algebra.
which was a term that was introduced in 1952, when basis sets would have started becoming popular in matter modeling too. I mean, the Roothaan equations were developed circa 1951!
@Tyberius I find it fascinating that there's no universal and widely taught term being used (for example in approximation theory) for the "basis functions/vectors" used to represent something, when the set of functions/vectors isn't enough to span the whole space.
For example if we wanted to write a signal in terms of sin(nx) and cos(nx) with only n=1,2,3, we may not have a full Fourier basis, but we can still approximate a lot with those "basis functions".
I guess they are actually elements of the basis, so we can call them basis functions or basis vectors, it's just that the "basis set" we're using doesn't contain a whole basis.

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