Consider the brief attached discussion on closure and completeness (used, I think, in the physics sense) of basis functions from Zangwill's Modern Electrodynamics.

Zangwill demonstrates that closure implies completeness, but I can't seem to show how the converse holds. I note that IF I can use orthonormality in the form (7.20) then the development to follow works. Considering any test function $g(v')$, the action of a delta function thereon (assuming everything remains in our domain) should be given by, where I expand the delta function as a function of $v'$ and use that the delta function is even,

$$g(v) = \int_a^bg(v') \delta(v-v')\ dv'$$

while, if we consider the action of $\sum_k \psi_k(v) \psi_k^*(v')$ under the integral sign, we compute that (where we expand $g$ using completeness)

$$ \int_a^b \sum_j G_j \psi_j(v') \sum_k \psi_k(v)\psi_k^*(v')\ dv' = \sum_j \sum_k G_j \left(\int_a^b \psi_j(v')\psi_k^*(v')\ dv' \right) \psi_k(v) = \sum_k G_k \psi_k(v) = g(v)$$

I therefore conclude that indeed $\sum_k \psi_k(v)\psi_k(v') = \delta(v-v')$ insofar as that delta function equality is defined (in the nonrigorous physicist sense with which I am familiar) as meaning that each side of the equation have the same action on an arbitrary test function when under an integral sign.

However, it’s not clear to me that I should be able to use (7.20) since it was derived using closure which is that which I’m trying to prove. How then does one go about establishing completeness implies closure?

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    $\begingroup$ Check-my-work questions are off-topic. $\endgroup$ Jan 13, 2023 at 16:43
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    $\begingroup$ The indexing sets for complete sets of functions on an interval are never finite. $\endgroup$
    – mike stone
    Jan 13, 2023 at 16:56
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    $\begingroup$ If the index set were finite, the space of sums $\{ \sum c_k f_k(x)\}$ would be finite dimensional, but the space of functions on a finite interval is infnite-dimensional. $\endgroup$
    – mike stone
    Jan 13, 2023 at 17:49
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 19, 2023 at 17:57
  • $\begingroup$ @Qmechanic I'm not sure; things certainly aren't very rigorous mathematically here. There's no mention of exactly what's meant by basis (Schauder, Hamel?) or the tacit assumption of unique expansions. If you think so I'm happy to move it though. $\endgroup$
    – EE18
    Feb 20, 2023 at 5:18

2 Answers 2


First, the basis that the author has in mind is an orthonormal basis in a Hilbert space of functions. In particular, this means the topological structure is used in that you are actually talking about infinite sums which must be studied according to convergence properties. Most of the time in Physics one just deals with these issues formally and manipulates all the sums as if they were finite ones. The Hamel bases are not employed in Physics.

Indeed, the author clearly shows that the closure relation implies the basis is orthonormal. As such, if you have a basis that is not orthonormal, the closure relation simply cannot hold. As such, assuming that we are dealing with an orthonormal basis is mandatory to show the equivalence between completeness and closure proposed by the author.

In that case, I'm going to assume the one is working on an interval $[a,b]$ with inner product $$\langle f,g\rangle = \int_a^b dx f^\ast(x)g (x)\tag{1}.$$

The relevant space is then $L^2([a,b],dx)$ the space of square integrable functions on $[a,b]$ with the standard Lebesgue measure $dx$. One could also consider modifying the measure by a weight $\rho(x)$ so that one works with a measure $\rho(x)dx$. In any case, the author is implicitly assuming the standard measure as suggested by (7.20).

So let $\{\psi_k\}$ be a complete set of functions forming an orthonormal basis in the Hilbert space $L^2([a,b],dx)$. This means that any element of the space can be expanded in the basis and that $$\langle \psi_k,\psi_\ell\rangle = \int_a^b dx \psi^\ast_k(x)\psi_\ell(x)= \delta_{k\ell}.\tag{2}$$

A generic function $F(x)$ then expands as in (7.15)

$$F(x)=\sum_k c_k \psi_k(x).\tag{3}$$

Taking the inner product we observe that $c_k = \langle \psi_k,F\rangle$. Using the definition (1) this means we can express $F(x)$ as $$F(x)=\sum_k \left(\int_a^bdy \psi^\ast_k(y) F(y)\right)\psi_k(x).\tag{4}$$

Rearranging this is $$F(x)= \int_a^bdy \left(\sum_k\psi^\ast_k(y)\psi_k(x) \right)F(y).\tag{5}$$

But recall that the Dirac delta centered at some point in the interval acts on functions by $$ F(x)=\int dy \delta(y-x)F(y)\tag{6}.$$

Equating (5) and (6) then shows that we have the equality in distributional sense

$$\sum_{k} \psi^\ast_k(y)\psi_k(x)=\delta(y-x)\tag{7},$$

which is the closure relation.

  • $\begingroup$ Thank you for your answer. I’m not sure I follow the logic though — if $A \implies B \implies C$ as is the case here, I can’t use C in proving $A \implies B$ right? Unless you’re saying that orthonormality is part of the definition of completeness though it’s not mentioned? $\endgroup$
    – EE18
    Feb 22, 2023 at 13:50
  • $\begingroup$ If $A\Longrightarrow B$ then $\neg B\Longrightarrow \neg A$ where $\neg$ means the negation. This is the so-called contraposition (en.wikipedia.org/wiki/Contraposition). Since closure implies in an orthonormal basis, not having an orthonormal basis implies in the closure condition being violated. So the whole discussion regarding the closure relation is simply meaningless without dealing with orthonormal bases in the first place. $\endgroup$
    – Gold
    Feb 22, 2023 at 13:55
  • $\begingroup$ I see. So the moral of the story is that the original statement of completeness should also say that the basis functions are orthonormal? $\endgroup$
    – EE18
    Feb 23, 2023 at 2:04
  • $\begingroup$ Yes, that's right. Of course, he could also not include it, but then completeness won't imply the closure relation he wrote down. $\endgroup$
    – Gold
    Feb 23, 2023 at 23:55
  • $\begingroup$ "The relevant space is then $L^2([a,b],dx)$ the space of square integrable functions ...". Here you mean functions identified upto measure zero sets, right? $\endgroup$ Mar 5, 2023 at 20:10

I'm not 100% sure I understand the question correctly, but I think the solution may simply be that completeness does not imply closure in the stated sense.

If we suppose completeness as the starting point as in the textbook, i.e. that every function in our space can be written as $$F(\nu) = \sum_k F_k \psi_k(\nu) \,,$$ then (ignoring issues on whether the delta function is also in our space or not for simplicity) $$\delta_{\nu'}(\nu) := \delta(\nu-\nu') = \sum_k \delta_{\nu',k} \psi_k(\nu) \,.$$

This is already rather close to the closure relation. However, it is not clear whether the expansion coefficients $\delta_{\nu',k}$ are equal to the starred expansion basis, i.e. that $$\delta_{\nu',k} \stackrel{?}{=} \psi^*_k(\nu') \,. $$

The reason is simply that this is not true in general for a given complete basis.

However, the OP has already realized in which cases it is true: for bases which are also orthonormal. [I am not sure whether this is an iff or on if relation.]

Besides this argument, I thought it would be fun to give a counterexample. Indeed, there is one from physics: quasinormal modes from resonance theory. They form a complete (even overcomplete) set, but fulfill a rather strange closure relation without the complex conjugate and with a pre-factor [Eq. (7) in the linked review].

So if you allow generalizations of closure relations, that works... but the most general notion of closure in the end is just the statement that you can write the delta-function as some superposition of basis states (see second formula above), which trivially follows from completeness (up to the question whether the delta-function is in your function space or not).


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