DEFINITION: A Hilbert space is an inner product space which, as a metric space, is complete.
We will not present an exhaustive "mathematical" discussion of this subject. Rather, by using examples and analogies, hopefully you will feel more "at ease" with "Hilbert space" at the end of this short discussion.
REMARKS:
(1) Consider the space of functions, 
, where f is a
"square-integrable" complex-valued function on the real interval [a,b], \ie, 
This space will be denoted as 
. One can directly verify that is a complex vector
space, \ie, 
, etc. We will return to clarify in what sense one can
"visualize a function as a vector".
(2) By introducing a "dot-product", 
this complex vector space
becomes an inner product space. This inner product provides us with a positive definite "norm" for each vector,
(3) Define a "distance" between two functions by 
. This turns into a metric space. We shall explain the fact that is a Hilbert space next.
(1) In dealing with real numbers, it is natural to first start working with integers. One then finds that one needs
rationals, and finally irrationals. To formally introduce "irrationals", one can first introduce a notion of distance,
d(x,y), between two numbers x and y, which of course can be chosen to be the absolute value 
. (With a notion of distance, the space of numbers becomes a "metric
space".) Now one can talk about an infinite sequence of numbers, e.g., {3, 3.1, 3.14, 3.141, 3.1415, . . .}.
Note that (i) the differences between any pairs of numbers further down the sequence become smaller and smaller. (This
is an example of a "Cauchy sequence".) (ii) As finite decimals, every term in the sequence is a rational number. Now,
as you can guess, the limit of this sequence is 
, and it is not
a rational number.
(2) A metric space is not complete if it contains Cauchy sequences whose limit points are not contained in the
space. As the example above shows, the space of rational numbers, with the usual notion of distance, is not a
complete metric space. By including all irrationals, the space of all real numbers is complete. Mathematically,
one states that the real numbers, 
, is the completion of the
set of rational numbers.
(3) We all agree "irrational numbers exist" and they must be included in any sensible usage of numbers. However, for
physics, it is also clear that, for a given problem, it is sufficient to work with rationals only, e.g. by
agreeing to always work to the twentieth decimals. Another way of stating this fact is that "one can always approximate
an irrational number by a rational number to any degree of accuracy one desires." Mathematically, one states that the
set of rational numbers is dense in .
(4) It is a mathematical fact that the space , defined by Eq. I1, as
a metric space with a "distance between functions f and g" defined by , is complete. That is, the "limiting function" of any Cauchy sequence of functions in is also in . Therefore,
is a Hilbert space, which will also be denoted by 
. [Note: The notation 
will be used to denote the Hilbert space of functions
defined over 
].
(5) Let us clarify the situation by drawing an analogy with the case of real numbers. The role of rationals is now
played by continuous functions on the interval [a,b]. However, the condition of "square-integrability", I1, is much
less stringent than continuity. Many functions which are not continuous nevertheless satisfy I1. It can be shown that
these "additional" functions can always be thought of as limits of Cauchy sequences of continuous functions. We
"complete" by adding to the set of continuous functions these
"limiting points". That is: is the completion of the space of
continuous functions on the interval [a,b], with respect to a distance defined by .
(6) Just like the situation with real numbers, any function 
can
always be approximated by continuous functions, to any desired degree of accuracy. Mathematically, one states that the
set of continuous functions on [a,b] is dense in .
(7) Let us end with an example using Fourier series. let us consider defined over the interval 
. Consider a periodic
function f(x). Assume that f(x) is continuous and differentiable except at a point 
, where f(x) has a finite discontinuity. Let us consider the following
sequence of functions, 
, defined by
where
for
all integers n. Note that each 
is continuous at 
where f(x) has a discontinuity.
For any small number 
, we can always choose N large enough so
that
for N' >= N. That is, to the accuracy , f can be approximated by a continuous function, , although f(x) is not.
The set of 
is a Cauchy sequence. Therefore, the space of continuous
functions is not complete. However, the space is. Indeed, it
is precisely in the sense of I.3 that we understand the statement that every periodic function of period 2 can be represented by a Fourier series: 
with 
given by (I.2).
DEFINITION: A system of functions, 
, defined on [a,b] is said
to be an orthonormal system if 
DEFINITION: A system of functions, , defined on [a,b] and
belonging to , is said to be complete if there exists no
functions different from zero in which is orthogonal to all
functions 
.
Without a loss of generality, let us now shift the interval [a,b] to [-L/2, L/2]. Consider the set of basis functions,
, 
, defined by
We have previously pointed out that this set forms an
orthonormal set:
For each function F in , consider the Fourier series
representation
The fact that any function in can be expanded in this basis can be
used to show that this set is complete. By substituting (II.5) into (II.4), one finds that the completeness can
be expressed mathematically as 
COMMENTS:
(1) This set of complete orthonormal basis functions will be referred to as our standard basis. In terms
of this standard basis, every function F can be expanded as in (II.4). Instead of writing out (II.4) explicitly,
we can also represent the function F by a "column vector", with components: 
. (To save space, let's not write it out as a "column" here.) That is, in
this basis, we can associate a function with a vector with infinite many "components". In this representation, one
"sees" a function as a column vector, just like the situation with a finite dimensional vector space.
(2) To draw a closer analogy with our "standard representation" for a finite dimensional vector space, let's re-write
the inner-product, (II.2), and the norm, (II.3), as follows: Consider two functions, F(x) and F'(x),
described by components 
and 
respectively in this standard basis. It follows from (II.3) that
(3) In this representation, the "vector" nature of the space is
manifest! Instead of representing a vector by its "components", one can also represent a function, F(x), by an
abstract vector in a Dirac notation, 
. If we first introduce a Dirac
notation for our standard basis vectors: 
, , Eq. (II.4) can be written as:
with
The orthonormality condition in a Dirac notation becomes
We will shortly show that the completeness condition in a Dirac notation is
You should now compare these representations with those for a finite dimensional vector space, and convince yourself that these two sets are formally identical.
(4) You should also convince yourself that every linear operator on 
, can be associated with an infinite by infinite matrix as follows:
What is the form of the matrix 
for the momentum operator 
?
"slides/tanIllust/tan
It is convenient to formally define a set of basis vectors, 
,
labelled by a continuous index, 
, satisfying
Dirac-delta orthonormality condition as follows: "slides/tanIllust/tan
In terms of this basis, it is now possible to express each abstract
vector 
in as
Coordinate Representation: 
It follows from (III.1) and (III.2) that f(x) can be recovered by "dotting" a bra vector 
into Eq. (III>2), \ie,
That is, we have achieved a representation where the original function f(x) in now appears as "components" of the abstract vector in a coordinate basis. In particular, the inner product of two vectors,
and 
, through
(III.3) and (III.2), is
in agreement with the original definition, (III.2).
Since every vector in can be expressed by (III.2), they form a
complete basis:
Completeness: 
COMMENTS:
(1) We claim that 
is an eigenstate of our position operator,
, ie,
for . Eq. (III.6) follows if one writes down the components of , \ie,
That is, 
is the coordinate wavefunction for the eigenvector of with an eigenvalue 
.
By applying (III.3) to (III.7), one simply reproduces the orthogonality condition, (III.1).
(2) In this representation, every linear operator can be expressed as
This representation follows by considering 
and use (III.5) twice. For instance,
Whenever 
is proportional to , the operator 
will be called a local operator.
Instead of a "double-sum", Eq. (III.9), one has a single-sum representation:
For local operators, one often simply writes down the matrix for
the operator ,
Better yet, one can drop the factor entirely
if it is understood that one is dealing with a local operator! For instance, instead of writing out
explicitly, it is much simpler to accept the notation
[Convince yourself that the Hamiltonian operator 
is local in a coordinate representation, and make sure you understand
the proper interpretation of the expression 
]
(3) Lastly, let's return to our "standard basis", . It is clear
that the abstract vector, 
, is given by
With this understanding, by sandwiching (II.12) between
and 
, it leads
to (II.6), which proves (II.12), as promised.