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Question. In the context of QM, I hear the phrases 'complete set of states' and 'complete basis' (among other similar expressions) thrown around rather a lot. What exactly is meant by 'complete'?

Further remarks. I understand the term 'complete set', vaguely speaking, to mean a 'set from which all elements of our space can be constructed by linear combination'. However, to me this seems completely indistinct from the term basis. I thought at first that perhaps the word basis wasn't applicable to the infinite dimensional vector spaces that we often meet in QM, but having come across the existence of Schauder bases I no longer believe this to be the case. Is it then so that 'complete set of states' is a poorly-defined, somewhat redundant expression, or does it have a precise meaning distinct from 'constitutes a basis'?

Two definitions that I have seen before (in the context of function spaces) are as follows: the functions $\{\phi_n\}$ are a 'complete set' or 'complete basis' if for all functions $f(x)$ there exists a set $\{a_n\}$ such that

$$ \int_a^b \left| f(x) - \sum_n a_n \phi_n(x) \right|^2 w(x)\, dx = 0 \,, $$

where $w(x)$ is the weight function used in defining the norm on the space. The second definition is: $$ \sum_n \phi_n(x) \phi_n^*(x') = \frac{1}{w(x)}\delta(x-x') \,.$$

So now I ask: are these definitions correct, and are they equivalent? Further, why are these definitions useful? When we talk about complete sets of states in QM, the relevance (in so far as I understand it) is that such sets can be used to construct all other states. If this is so, is not the term 'basis' more appropriate, since it directly expresses such a property of a set? Do the definitions above coincide with the definition of a Schauder basis for an infinite dimensional function space? Or are they different in some subtle way?

I've asked several mathematicians this question; none have known the precise meaning in the sense I've described, but rather only in the sense of Cauchy sequence convergence in metric spaces. Hence my asking this on physics.SE. Thank you for reading.

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What is $w$ in your first identity? –  Valter Moretti Feb 9 at 18:34
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Sorry, I wasn't explicit about that. It's the weight function, as appears in the kernel of the integral defining the norm: $$ \langle \phi_n | \phi_m \rangle = \int_a^b \phi_n^*(x) \phi_m(x) w(x) \,dx$$ –  gj255 Feb 9 at 18:38
    
I wrote a quite complete, I hope, answer to your question about the various notions of basis and completeness. Please have a look at it. –  Valter Moretti Feb 10 at 12:41
    
I've seen it, don't worry. Thank you very much for your response. I have a couple of questions, but I'm preoccupied at the present time so will comment in a little while if that is alright. –  gj255 Feb 10 at 15:24
    
I am here, comment when you can... –  Valter Moretti Feb 10 at 15:27

3 Answers 3

up vote 8 down vote accepted

There are at least three notions of basis depending on the mathematical structure you are considering. I will quickly discuss three cases relevant in physics (topological vector spaces are relevant too, but I will not consider them for the shake of brevity).

(1) Pure algebraic structure (i.e. vector space structure over the field $\mathbb K=$ $\mathbb R$ or $\mathbb C$, actually the definition applies also to modules).

Basis in the sense of Hamel.

Given a vector space $V$ over the field $\mathbb K$, a set $B \subset V$ is called algebraic basis or Hamel basis, if its elements are linearly independent and every $v \in V$ can be decomposed as: $$v = \sum_{b \in B} c_b b$$ for a finite set of non-vanishing numbers $c_b$ in $\mathbb K$ depending on $v$.

Completeness of $B$ means here that the set of finite linear combinations of elements in $B$ includes (in fact coincide to) the whole space $V$.

Remarks.

  • This definition applies to infinite dimensional vector spaces, too. Existence of algebraic bases arises from Zorn's lemma.

  • It is possible to prove that all algebraic bases have same cardinality.

  • Decomposition of $v$ over the basis $B$ turns out to be unique.

(2) Banach space structure (i.e. the vector space over $\mathbb K$ admits a norm $||\:\:|| : V \to \mathbb R$ and it is complete with respect to the metric topology induced by that norm).

Basis in the sense of Schauder.

Given an infinite dimensional Banach space $V$ over the field $\mathbb K = \mathbb C$ or $\mathbb R$, a countable ordered set $B := \{b_n\}_{n\in \mathbb N} \subset V$ is called Schauder basis, if every $v \in V$ can be uniquely decomposed as: $$v = \sum_{n \in \mathbb N} c_n b_n\quad (2)$$ for a set, generally infinite, of numbers $c_n \in \mathbb K$ depending on $v$ where the convergence of the sum is referred both to the Banach space topology and to the order used in labelling $B$. Identity (2) means: $$\lim_{N \to +\infty} \left|\left|v - \sum_{n=1}^N c_{n} b_n\right|\right| =0$$

Completeness of $B$ means here that the set of countably infinite linear combinations of elements in $B$ includes (in fact coincide to) the whole space $V$.

Remarks.

  • The elements of a Schauder basis are linearly independent (both for finite and infinite linear combinations).

  • An infinite dimensional Banach space also admits Hamel bases since it is a vector space too. However it is possible to prove that Hamel bases are always uncountable differently form Schauder ones.

  • Not all infinite dimensional Banach space admit Schauder bases. A necessary, but not sufficient, condition is that the space must be separable (namely it contains a dense countable subset).

(3) Hilbert space structure (i.e. the vector space over $\mathbb K$ admits a scalar product $\langle \:\:| \:\:\rangle : V \to \mathbb K$ and it is complete with respect to the metric topology induced by the norm $||\:\:||:= \sqrt{\langle \:\:| \:\:\rangle }$).

Basis in the sense of Hilbert (Riesz- von Neumann).

Given an infinite dimensional Hilbert space $V$ over the field $\mathbb K = \mathbb C$ or $\mathbb R$, a set $B \subset V$ is called Hilbert basis, if the following conditions are true:

(1) $\langle z | z \rangle =1$ and $\langle z | z' \rangle =0$ if $z,z' \in B$ and $z\neq z'$, i.e. $B$ is an orthonormal system;

(2) if $\langle x | z \rangle =0$ for all $z\in B$ then $x=0$ (i.e. $B$ is maximal with respect to the orthogonality requirment).

Hilbert bases are also called complete orthonormal systems (of vectors).

The relevant properties of Hilbert bases are fully encompassed within the following pair of propositions.

Proposition. If $H$ is a (complex or real) Hilbert space and $B\subset H$ is an orthonormal system (not necessarily complete) then, for every $x \in H$ the set of non-vanishing elements $\langle x| z \rangle$ with $z\in B$ is at most countable.

Theorem. If $H$ is a (complex or real) Hilbert space and $B\subset H$ is a Hilbert basis, then the following identities hold, where the order employed in computing the infinite sums (in fact countable sums due to the previous proposition) does not matter: $$||x||^2 = \sum_{z\in B} |\langle x| z\rangle|^2\:, \qquad \forall x \in H\:,\qquad (3)$$

$$\langle x| y \rangle = \sum_{z\in B} \langle x|z \rangle \langle z| y\rangle\:, \qquad \forall x,y \in H\:,\qquad (4)$$

$$\lim_{n \to +\infty} \left|\left| x - \sum_{n=0}^N z_n \langle z_n|x \rangle \right|\right| =0\:, \qquad \forall x \in H \:,\qquad (5)$$

where the $z_n$ are the elements in $B$ with $\langle z|x\rangle \neq 0$.

If an orthonormal system verifies one of the three requirements above then it is a Hibertian basis.

Completeness of $B$ means here that the set of infinite linear combinations of elements in $B$ includes (in fact coincide to) the whole space $H$.

Remarks.

  • The elements of a Hilbert basis are linearly independent (both for finite and infinite linear combinations).

  • All Hilbert spaces admit corresponding Hilbert bases. In a fixed Hilbert space all Hilbert bases have the same cardinality.

  • An infinite dimensional Hilbert space is separable (i.e. it contains a dense countable subset) if and only if it admits a countable Hilbert basis.

  • An infinite dimensional Hilbert space also admits Hamel bases, since it is a vector space as well.

  • In a separable infinite dimensional Hilbert space a Hilbert basis is also a Schauder basis (the converse is generally false).

FINAL COMMENTS.

  • Identities like this: $$ \sum_n \phi_n(x) \phi_n^*(x') = \frac{1}{w(x)}\delta(x-x') \,\qquad (6)$$ stay for the completeness property of a Hilbert basis in $L^2(X, w(x)dx)$: Identity (6) is nothing but a formal version of equation (4) above. However such an identity is completely formal and, in general it does not hold if $\{\phi_n\}$ is a Hilbert basis of $L^2(X, w(x)dx)$ (also because the value of $\phi_n$ at $x$ does not make any sense in $L^2$ spaces, as its elements are defined up to zero measure sets and $\{x\}$ has zero measure). That identity sometime holds rigorously if (1) the functions $\phi_n$ are sufficiently regular and (2) the identity is understood in distributional sense, working with suitably smooth test functions like ${\cal S}(\mathbb R)$ in $\mathbb R$.

  • In $L^2(\mathbb R, d^nx)$ spaces all Hilbert bases are countable. Think of the basis of eigenvectors of the Hamiltonian operator of an Harmonic oscillator in $L^2(\mathbb R)$ (in $\mathbb R^n$ one may use a $n$ dimensional harmonic oscillator). However, essentially for for practical computations it is convenient also speaking of formal eigenvectors of, for example, the position operator: $|x\rangle$. In this case, $x \in \mathbb R$ so it could seem that $L^2(\mathbb R)$ admits also uncountable bases. It is false! $\{|x\rangle\}_{x\in \mathbb R}$ is not an orthonormal basis. It is just a formal object, (very) useful in computations. If you want to make rigorous these objects, you should picture the space of the states as a direct integral over $\mathbb R$ of finite dimensional spaces $\mathbb C$, or as a rigged Hilbert space. In both cases however $\{|x\rangle\}_{x\in \mathbb R}$ is not an orthonormal Hilbertian basis. And $|x\rangle$ does not belong to $L^2(\mathbb R)$.

  • Hilbert bases are not enough to state and prove the spectral decomposition theorem for normal operators in a complex Hilbert space. Normal operators $A$ are those verifying $AA^\dagger= A^\dagger A$, unitary and self-adjoint ones are particular cases. The notion of Hilbert basis is however enough for stating the said theorem for normal compact operators or normal operators whose resolvent is compact. In that case, the spectrum is a pure point spectrum (with only a possible point in the continuous part of the spectrum). It happens, for example, for the Hamiltonian operator of the harmonic oscillator. In general one has to introduce the notion of spectral measure or PVM (projector valued measure) to treat the general case.

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That's all folks. It's my longest and most synthetic answer I have ever written here. I hope you'll appreciate it. I apologize for disasters with English, but I don't have spare time any more to correct my answer, sorry. –  Valter Moretti Feb 10 at 11:20
    
Why is this a community wiki? I wanted to upvote, but it doesn't make sense to upvote a community wiki since no one gets reputation points for this. –  Ruslan Mar 18 at 13:49
    
@Ruslan "community wiki", what is this? Id does not depend on me. –  Valter Moretti Mar 18 at 13:54
    
Ah. Seems editing a post too many times leads to this ugly effect. In your revision history there's added 51 characters in body; [made Community Wiki by V. Moretti editing at least 10 times] at the pre-last edit. The implications of this result can be read if you hover mouse cursor in the "community wiki" next to "edited Feb..." under the post. –  Ruslan Mar 18 at 13:56
    
Is there anything I can do now? I see. I tried to be very precise and I edited my post too many times.... –  Valter Moretti Mar 18 at 13:58

Complete set is a well defined expression.

The reason why people sometimes differentiate between complete orthonormal set (COS) and a basis, is that any vector can be written as a finite linear combination of elements of the basis (if you use basis in the linear algebra sense). While for the COS you need an infinite linear combination.

If you use the basis in Schauder's sense, both definitions are equivalent.

See my related question.

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I've never heard someone use the word basis in that Hamel sense (in physics). It seems unnatural. Take some Hamel basis $\{ | \psi_n\rangle \}$ and assume that the $| \psi_n\rangle$ have unit norm. Now what about the state $\sum e^{-n} | \psi_n \rangle$? –  Vibert Feb 9 at 15:54
    
@Vibert At least my ODE's teacher insisted in the difference. In Wikipedia Note that by definition, a linear combination involves only finitely many vectors. For example, by Zorn's lemma every Hilbert space has a Hamel basis, but not Schauder (or something like that). –  jinawee Feb 9 at 16:01

I believe this is just a matter of vocabulary. Here's how it goes in mathematics:

A basis is a linearly independent spanning set of the vector space, ie, a set of vectors such that any vector in the space can be expressed uniquely as a finite linear combination.

In an infinite dimensional Hilbert space, such bases aren't so convenient: due to the Baire category theorem, a basis must be uncountable.

So we use orthonormal bases (aka Hilbert bases), which are countable sets of orthonormal vectors such that any vector can be expressed as a countable linear combination of elements of the orthonormal basis.

For convenience, we drop the "Hilbert"/"orthonormal" and call it just a basis; we have to add "algebraic" or "Hamel" to a true basis to distinguish the two.

--

What physicists refer to as "a complete set of states" is just what a mathematician calls "an orthonormal basis."

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You stated in the third paragraph that any basis for an infinite dimensional vector space must be uncountable, but then go on to talk about countable basis for just such vector spaces. Furthermore, the term 'Hilbert' basis is not one I've ever come across, and seems rather a strange term given that our infinite dimensional vector space may well not be a Hilbert space. You also use the term 'orthonormal basis' to refer to countable sets of vectors, which is just not the definition of an orthonormal basis (which may well be finite in size). –  gj255 Feb 10 at 0:16
    
I said any basis for an infinite dimensional Hilbert space is uncountable, which is true. I proceeded to talk about countable orthonormal bases for them, not countable bases (as I just said, those do not exist). As I explained, Hilbert basis is a mathematical term -- I don't know what kind of math background you have, but I'm not shocked to hear you've never heard the term before. AFAIK, in QM, your state space is always a Hilbert space; I could be wrong, but, regardless, I mentioned I was only speaking about Hilbert spaces, and not infinite-dimensional vector spaces in general. –  daunpunk Feb 10 at 0:54
    
I didn't say an orthonormal basis was any countable set of vectors, and I'm not sure how that's what you took away from what I wrote, but I've very slightly edited my answer to clarify. "Orthonormal basis" means one thing in finite dimensions: a (algebraic) basis whose elements are all unit length and orthogonal to each other. It means another thing for infinite-dimensional Hilbert spaces: namely, exactly what I said in my answer. Lastly, an infinite-dimensional Hilbert space can't have a finite orthonormal basis. –  daunpunk Feb 10 at 1:03
    
OK, thanks for the comments. Could you please provide some references defining a) Hilbert basis, b) orthonormal basis - and in particular expressing how these two terms are used interchangeably (I have only come across the term 'orthonormal basis' in finite dimensions, where it seems quite unrelated from your definition). Further, you distinguish 'countable orthonormal basis' from 'countable basis' --- what is the difference? You say that for the latter (in paragraph 4 of your answer), we can express any vector as a linear combination of its members. So it is a countable basis, no? –  gj255 Feb 10 at 9:29
    
The third paragraph here. A "countable basis" is just a Schauder basis: a basis such that every vector can be expressed as a countable linear combination of elements of it. These are for normed vector spaces, which aren't necessarily equipped with an inner product. Hilbert/orthonormal bases require an inner product -- there'd be no notion of orthogonality without one. So a Hilbert basis is a Schauder basis whose elements are orthonormal. –  daunpunk Feb 11 at 7:48

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