Why is the inner product not an element of the Hilbert space?

Question

What I know about Hilbert space is that, elements in that space can be complex numbers.

But I was confused to read this statement from a book:

The inner product, being a complex number, is not an element of the Hilbert space.

Can someone elaborate this?

Answer 1

(i) A Hilbert space $H$ is a vector space with some additional structure.

(ii) The elements $x$ of a vector space are "vectors", not numbers.

(iii) Vectors can be added and can be multiplied ("scaled") by real or complex numbers. The result in both cases is a vector.

(iv) In a Hilbert space $H$ the scalar product $x\cdot y$ of any two vectors is defined. This scalar product is a real or complex number. The number $|x|:=\sqrt{x\cdot x}$ is called the norm of the vector $x$, and $d(x,y):=|x-y|$ is a metric on $H$.

(v) In addition it is postulated that Cauchy sequences in $H$ are convergent.

Answer 2

Hilbert space is a vector space. As such its elements are vectors. The complex numbers are scalars, not vectors and so they live in a whole other place. In the underlying field, rather than in the vector space.

Inner product takes in two vectors and returns a scalar, and so the result of an inner product is not an element of the vector space, but rather of the underlying field.

Answer 3

Hilbert spaces are vector spaces in some field, possibly $\mathbb{C}$, with some properties. They must have and inner product and be complete with the metric induced by this inner product. The set $\mathbb{C}$ satisfies this properties with the usual inner product $(a, b)=a^*b$. But in general, the vectors in the Hilbert space are not complex numbers as for example the set of (equivalence classes) of square integrable complex valued functions, $L^2(\mathbb{R},dx)$. The inner product from two vectors is a complex number, that is not in this case a element of the Hilbert space.

Answer 4

The inner product of any vector space $V$ over a field $\mathbb F$ is doomed to be a map from $V\times V\to\mathbb F$ and as such is not an element of $V$, but of $\mathbb F^{V\times V}$. The question is not related to “Hilbert” and the book's statement could be replaced by: “A whale, as a living being of the sea, is not bound to fly.” Well, only in rare cases.

EDIT: The book's statement is sort of sick. Consider $\mathbb C$ as a vector space over itself and define the multiplication as inner product. Then this inner product itself belongs to $\mathbb C^{\mathbb C\times\mathbb C}$, but it's results to $\mathbb C$.