How to understand Hilbert spaces:
I have read this thread and I would like to add the following explanations on the Hilbert spaces used in QM theory.
I think it is important to understand why Hilbert spaces, which are the generalisation to infinite dimension of the finite dimension vector space, have been introduced (e.g. generalisation of the basis concept to the infinite dimension spaces).
The definition of general hilbert spaces Hilbert is based on the following assumption:
-- A vector space (on K= R or C) called H (dimension finite or infinite)
-- A hermitian product (= an inner product if vector space on R): <,>
-- The vector space is complete
No other stuff is required. No L2, l2 or other spaces are required (they are only examples of what can be abstract Hilbert spaces, and more precisely they are some examples of <,>).
K= R or C
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First, the selection of C instead of R is only a matter of convenience: for example, a real anti symmetric matrix can be made diagonal only if the field number K=C of the vector space.
Completeness:
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The main difficulty on Hilbert spaces rises from the definition of complete that is associated to the hermitian product: The hermitian product, defines a norm |psi|^2=<psi|psi> on the vector space. This norm induces a topology on the vector space that allows defining continuity/limit and thus completeness properties:
-- if H is included in another set with the same topology (i.e. induced by <,>), the completeness property is very simple: it says that the closure of H is H (all the limit points of a sequence belongs to H: H is a closed set).
Example: how to get a hilbert space:
If preHilbert= Vect (xi, i e I) = {sum_ ieI ci.xi, sum is FINITE} we thus have
Hilbert= closure(prehilbert)= {sum_ ieI ci.xi, sum Cauchy convergent}
** e: belongs to
-- When there is no “outside set” containing H, we use the “inside H” definition for the completeness property: The limit (a point=vector) of any Cauchy sequence (of points= vectors) belongs to H, i.e. any Cauchy sequence converge on H.
The important point of the completeness of an Hilbert space is the introduction of vectors that are the limit of a sequence of vectors. The completeness property allows one to write any vector as a sum of other vectors of H: It is the introduction to the existence of the generalisation of a basis (of infinite dimension) in the Hilbert space.
We thus have a trivial result: finite dimensional vector spaces are Hilbert space: we do not need to introduce the concept of Hilbert spaces in finite dimension vector spaces as they have already the property we want (an existence of a basis for the set of vectors).
Now, we can consider the additional properties of a Hilbert space, the existence of a basis and the existence of a countable basis.
In fact, the existence of a countable basis (the existence of basis with the countable property) is directly linked to topology of H (thus the property of the hermitian product): whether H is separable or not.
H Separable – existence of a countable basis:
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H is separable if H is the closure of a countable set (e.g. xn, n e|N) therefore included in H. We thus have:
H= Closure(xn, n e|N)= (sum_ ie|N ci.xi, sum Cauchy convergent) thanks to the linearity of the limit of a sequence and the linearity of a vector space).
The fact, H is separable, is equivalent to the one where each vector of H may be written as a countable sequence of vectors (x eH, x= sum_ i ci.xi, i e|N).
Now if we use the Schmidt orthogonal procedure (construction of an isomorphism), we can transform the family of vectors (e.g. xn, n e|N) into an orthogonal basis (e.g. an, n e|N, <ai,aj>=delta_ij):
We see that we have found a countable Hilbert basis for a SEPARABLE Hilbert space:
H SEPARABLE <=> H has a countable Basis !
How H is separable:
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Now, the important fact is the separable property of H that allows the existence of countable basis. So where does come from the separable property?
From the hermitan product! : H is separable if H is the closure of a countable set (e.g. xn, n e|N), therefore included in H.
Therefore we know that, because the closure property depend on the topology that is defined by the hermitian product, the separable property is also defined by the hermitian product!
The separable property with the existence of countable basis shows the isomorphism between Hilbert spaces that depends only on the properties of the hermitian product.
Example:
L2 is isomorphic to l2 because L2, as a Hilbert space, is separable.
We see that the separable property of L2 space comes from the hermitian product that is the Lebesgues integral. It is the property of the Lebesgues’ integral (the measure on the sigma-algebra– general theory of integration) that induces the separable property.
Conclusion:
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We have seen that the definition of the hermitian product defines all the main properties of the Hilbert space (completeness, separable, and others such as projector theorems).
The advantage of abstract Hilbert space is that they do not require the definition of an “integral”, just the abstract Hermitian product. In that sense, it is the hermitian product that defines the integral (as a particular linear application). It is at the heart of the QM theory tools.
As an example, take two operators a,a+ satisfying the relation [a,a+]=1 (harmonic oscillator) on an unknown abstract SEPARABLE Hilbert space.
The operator relation [a,a+]=1 thus define a countable basis (|n>,n e|N). |n> are the abstract eigenvalues of the N=a+a operator. This countable basis may be used (restriction) to define the hermitian product and the QM separable Hilbert space itself (i.e. the set of vectors induced by |n> with finite norm).
Up to now, we do not need any integral to compute values on the QM Hilbert spaces. However, this Hilbert space is separable, we thus have an isomorphism between this quantum Hilbert spaces and the L2 or l2 spaces: we have the connection between the integral and the abstract hermitian product.
To end, recall that the choice of QM separable Hilbert spaces is equivalent to the restriction of Hilbert spaces generated by the operators N=a+a (we do not take vectors outside).
Therefore, if we do not require the Hilbert space to be separable (for example, one wants to extend the domain of validity of QM theory), we question in fact the existence of an Hilbert basis. The existence of a basis is thus more difficult to demonstrate and more fundamental also. : )
In fact, the existence of a general abstract Hilbert basis is directly connected to the zermelo-frankel set theory: in brief, we may postulate the existence of basis if we accept the zermelo-frankel choice axiom.
And for the one’s who knows a little the zermelo-frankel set theory, he knows that we can choose a consistent set theory without this choice axiom (changing also other important math properties that we currently use)!
Seratend.
The basis functions are real, you can ignore the hermitean conjugate in this case, I was illustrating the general power of the formalism.
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