The word "complete" seems to be used in several distinct ways. Perhaps my confusion is as much linguistic as mathematical?

  1. A basis, by definition, spans the space; some books call this "complete" -- though then the phrase "complete basis" is redundant.

  2. In physics/engineering, "complete" seems to be reserved for orthogonal/orthonormal bases -- which necessarily means not merely a vector space, but specifically an inner product space. A complete basis in this QM sense does more than merely span the space: the concept of orthogonality allows for Parseval's relation, non-overlapping projections, Gram-Schmidt, etc. Is it even possible to have a complete basis (in this QM sense) that is NOT orthogonal?

  3. Though complete in the sense of Hilbert space and Cauchy sequences seems to be a different use of the term, the convergence of sequences within the space seems not so far afield, conceptually, from Parseval. So is it really so different?

  • 2
    $\begingroup$ Yes, there are non-equivalent uses of the word "complete". We have to live with that. $\endgroup$
    – ACuriousMind
    Commented Jun 3, 2015 at 18:53
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    $\begingroup$ 2.) is not true, in my opinion. We are using "complete" in exactly the same sense as mathematicians. One of the first things they taught us in my first math classes for physicists was indeed Gram-Schmidt diagonalization. It can't be any other way because many physics problems do not come with orthogonal/orthonormal bases built in and in many systems (crystal lattices) they wouldn't even be a good natural choice. Moreover, as a physicist I am constantly aware that the existence of inner products and in case of Hilbert spaces completeness for infinite series of elements are a special property. $\endgroup$
    – CuriousOne
    Commented Jun 3, 2015 at 19:03
  • $\begingroup$ Yes, inner products abound in physics. But this is additional structure not required of vector spaces or their bases. $\endgroup$
    – gilonik
    Commented Jun 3, 2015 at 19:14
  • $\begingroup$ Let me put it this way: what's the completeness relation for a non-orthogonal basis? $\endgroup$
    – gilonik
    Commented Jun 3, 2015 at 19:16
  • $\begingroup$ Related question physics.stackexchange.com/questions/98462/… $\endgroup$
    – Timaeus
    Commented Jun 3, 2015 at 21:21

3 Answers 3


As noted, many people use "complete" where perhaps they ought to say "complete and orthogonal and orthonormal" or the like. I'm not sure what I can tell you besides confirming that usage is not always ideal. I'll answer one question you brought up, but I'm worried I may have gotten confused myself by what kind of "complete" you meant:

Is it even possible to have a complete basis (in this QM sense) that is NOT orthogonal?

Yep! Consider, for instance, the coherent states. They're not orthogonal, since $\langle \alpha | \beta \rangle$ isn't zero for $\alpha \neq \beta$. But they are complete--indeed, "overcomplete."

  • $\begingroup$ By complete in the "QM sense" I mean the sum/integral over ket-bra being unity. $\endgroup$
    – gilonik
    Commented Jun 3, 2015 at 19:11
  • $\begingroup$ Ah. The integral over all the coherent states is $\pi I$. I'll think about your actual question, then, maybe I'll come up with a weird degenerate case (or a counterproof). $\endgroup$
    – zeldredge
    Commented Jun 3, 2015 at 19:15
  • $\begingroup$ Could you not just add a factor of 1/sqrt(pi) to one of the states? the states would then be complete, and still not orthogonal? $\endgroup$ Commented Jun 3, 2015 at 19:17
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    $\begingroup$ They would be complete -- but then they'd be neither orthogonal nor normalized! $\endgroup$
    – gilonik
    Commented Jun 3, 2015 at 19:24
  • $\begingroup$ An overcomplete set is a spanning set. But it still seems to me that (over)complete in the QM sense requires an inner product space $\endgroup$
    – gilonik
    Commented Jun 3, 2015 at 20:15

You need to be careful with the word span. A mathematician will say that the span of a set of vectors is the set of finite linear combinations, so you can only add linear combinations of finitely many at a time to get something in the span. So there are sets that are mutually orthogonal and all normalized but not enough to span the space with finite linear combinations. But we call them complete if the span is large enough so that its completion (by filling in any holes) is the whole space. So in a sense we call an orthonormal set of vectors complete if the infinite linear combinations make up the whole space.

But for infinite linear combinations we need a metric, like that from an inner product. So the whole notion doesn't make sense in an arbitrary vector space but can make sense in a Hilbert Space. In a Hilbert Space not only is there an inner product (and hence a metric) so that we cab talk about the limit, but Cauchy sequences have things to converge to, so there is something to be the limit of your sum.

So when you say complete for an orthonormal basis you are talking about infinite sums. And saying that the sun if the projections is the identity is usually how you express it, but that requires limits of operators not just vectors, so technically you then need to put a metric on your space of operators if you want to characterize it that way, so then completeness of an orthonormal basis now depends on how you define distances and take limits of operators. But you do need to define that if you want to talk about the exponential of an operator.

And as long as we've brought up completeness and operators. I should warn you that when a mathematician says state, as in quantum state, they might mean an operator such as a density operator.

Now this is a matter of terminology, but the word complete is usually reserved as a matter of definition to orthonormal seta of vectors. And as a definition there is nothing deep about it.

If you always think of complete as enough then you are fine. When you have a set of orthonormal vectors so large that you can't add another orthonormal vector then it is complete. When you've filled in so many holes that every Cauchy sequence now has something to converge to then your space is complete.

But that complete set of orthonormal vectors is not as big as it can be if you give up on being orthonormal. As a set if linearly independent vectors there are potentially more vectors that could be added to the set that can't be written as finite linear combinations of the vectors already there. So they are not enough in an algebraic sense it is only in the metric sense that they are enough. And that's morally why you insist on orthogonality, it is only when you insisted on orthogonality that you had a sense where you couldn't add more.

Oh, and a basis is supposed to just barely span, so not have too many. The coherent states from another answer are overcomplete and have too many vectors.

  • $\begingroup$ I'm not sure I'm following. Are you saying, e.g., that a spin j basis spans but is not complete? That may be good mathematical terminology, but the distinction is not sufficiently consequential to be made in QM. $\endgroup$
    – gilonik
    Commented Jun 3, 2015 at 20:52
  • $\begingroup$ @gilonik I'm not sure I'm following you. In a finite dimensional space you don't have to worry about completeness because your inner product space is already complete and your orthonormal basis is already complete. A finite dimensional inner product space is always a Hilbert space but an infinite dimensional ip space might not be and so you need to bring up completeness. I'm saying that span means finite linear combination, and therefore your #1 is technically not true: a complete basis might not technically be a basis because it might not technically span the space. It's confusing terminology $\endgroup$
    – Timaeus
    Commented Jun 3, 2015 at 21:00

Completeness in mathematics is essentially a metric concept (that means that every Cauchy sequence in the metric space converges to an element of the space).

Sometimes (but I think more on a physical standpoint, and I agree is a sort of repetition and not so common) it is used to characterize bases in vector spaces, in the sense that a basis is complete if its linear span is the whole vector space. The axiom of choice implies that every vector space has a basis.

The orthonormal basis on a Hilbert space is actually different: the space has such additional structure that we can afford to do infinite linear combinations of vectors, provided they converge in norm. However, be warned that a (complete) basis of orthogonal vectors in a vector space with inner product $V$ (in the finite linear combination sense), it is not in general a (complete) basis of the metric (Hilbert) space obtained as the completion of $V$ with respect to the metric induced by the inner product. However it may happen that it is a basis in the Hilbert space sense, i.e. if infinite linear combinations are allowed.

Apart from that and mathematical logic, that is probably quite far from what it is intended here, I cannot think of other instances of the word complete in math (but maybe I am forgetting something).

However, since in one case "complete" is associated to a (metric) space, and in the other to bases in vector spaces (possibly with additional care to specify if we allow finite or infinite combinations), I think it reasonably easy to avoid confusion.

  • $\begingroup$ Saying complete means the span is the space is wrong, a basis already spans and when we say complete basis it technically doesn't span. A basis is either a minimal spanning set or a maximal linearly independent set. It always spans the set. However when we call something an orthogonal basis we don't mean a basis that happens to be orthogonal we instead mean a maximal orthogonal linearly independent set, which can then end up not being a spanning set at all and instead spans something dense, spans something whose completion is the space. So we mean the completion of the span is the space. $\endgroup$
    – Timaeus
    Commented Jun 3, 2015 at 21:15
  • $\begingroup$ @Timaeus Every vector space has a basis (it is a proved theorem) that, as per definition, spans the space. This basis is "complete" (I said it was not so good and redundant terminology). And given a vector space $V$ with inner product structure, even if there is an orthogonal basis (there may be or not) that spans the whole space (surely there is at least a basis that spans the whole space, and maybe is of pairwise orthogonal vectors) then it would be not, in general, a basis of the metric space obtained as a completion of the inner product vector space. I stand by the fact that it is correct $\endgroup$
    – yuggib
    Commented Jun 3, 2015 at 22:08
  • $\begingroup$ I agree, however, that the concept of orthogonal bases in Hilbert spaces is different, for it allows for infinite linear combinations, provided they converge in norm. That is, however, another definition of basis that is usual of Hilbert and Banach spaces (Schauder basis). I also agree that complete in that sense is not a so good terminology, however it is completely unrelated to the one that refers to complete metric spaces. $\endgroup$
    – yuggib
    Commented Jun 3, 2015 at 22:11
  • $\begingroup$ @Timaeus As an addition, I remark that a basis on the vector space sense (finite combinations) of a separable Hilbert space indeed exists, but it may be (and probably is) uncountable. On the contrary, the basis in the Schauder sense, i.e. with infinite linear combinations is countable. $\endgroup$
    – yuggib
    Commented Jun 3, 2015 at 22:16
  • $\begingroup$ You say it is a theorem that every vector space has a basis but it is not a theorem in any useful definition of the word. It is a baseless assumption. ZF plus C and certain methods of deduction produce "every vector space has a basis" as a theorem, but on the other hand ZF plus "every vector space has a basis" and the exact same methods of deduction yield C as a theorem. So the so called theorem is equivalent to an axiom that has no basis. I think Hamel bases are unrelated to Physics for that exact reason. $\endgroup$
    – Timaeus
    Commented Jun 3, 2015 at 22:39

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