path: root/source/know/concept/hilbert-space/index.md

---
title: "Hilbert space"
sort_title: "Hilbert space"
date: 2021-02-22
categories:
- Mathematics
- Quantum mechanics
layout: "concept"
---

A **Hilbert space**, also called an **inner product space**,
is an abstract **vector space** with a notion of length and angle.



## Vector space

An abstract **vector space** $$\mathbb{V}$$ is a generalization
of the traditional concept of vectors as "arrows".
It consists of a set of objects called **vectors**
which support the following (familiar) operations:

+ **Vector addition**: the sum of two vectors $$V$$ and $$W$$, denoted by $$V + W$$.
+ **Scalar multiplication**: product of a vector $$V$$ with a scalar $$a$$, denoted by $$a V$$.

In addition, for a given $$\mathbb{V}$$ to qualify as a proper vector
space, these operations must obey the following axioms:

+ **Addition is associative**: $$U + (V + W) = (U + V) + W$$
+ **Addition is commutative**: $$U + V = V + U$$
+ **Addition has an identity**: there exists a $$\mathbf{0}$$ such that $$V + 0 = V$$
+ **Addition has an inverse**: for every $$V$$ there exists $$-V$$ so that $$V + (-V) = 0$$
+ **Multiplication is associative**: $$a (b V) = (a b) V$$
+ **Multiplication has an identity**: There exists a $$1$$ such that $$1 V = V$$
+ **Multiplication is distributive over scalars**: $$(a + b)V = aV + bV$$
+ **Multiplication is distributive over vectors**: $$a (U + V) = a U + a V$$

A set of $$N$$ vectors $$V_1, V_2, ..., V_N$$ is **linearly independent**
if the only way to satisfy the following relation
is to set all the scalar coefficients $$a_n = 0$$:

$$\begin{aligned}
    \mathbf{0} = \sum_{n = 1}^N a_n V_n
\end{aligned}$$

In other words, these vectors cannot be expressed in terms of each other.
Otherwise, they would be **linearly dependent**.

A vector space $$\mathbb{V}$$ has **dimension** $$N$$
if only up to $$N$$ of its vectors can be linearly indepedent.
All other vectors in $$\mathbb{V}$$ can then be written
as a **linear combination** of these $$N$$ **basis vectors**.

Let $$\vu{e}_1, ..., \vu{e}_N$$ be the basis vectors,
then any vector $$V$$ in the same space can be **expanded**
in the basis according to the unique weights $$v_n$$,
known as the **components** of $$V$$ in that basis:

$$\begin{aligned}
    V = \sum_{n = 1}^N v_n \vu{e}_n
\end{aligned}$$

Using these, the vector space operations can then be implemented as follows:

$$\begin{gathered}
    V = \sum_{n = 1} v_n \vu{e}_n
    \quad
    W = \sum_{n = 1} w_n \vu{e}_n
    \\
    \quad \implies \quad
    V + W = \sum_{n = 1}^N (v_n + w_n) \vu{e}_n
    \qquad
    a V = \sum_{n = 1}^N a v_n \vu{e}_n
\end{gathered}$$



## Inner product

A given vector space $$\mathbb{V}$$ can be promoted to a **Hilbert space** or **inner product space**
if it supports an operation $$\Inprod{U}{V}$$ called the **inner product**,
which takes two vectors and returns a scalar,
and has the following properties:

+ **Skew symmetry**: $$\Inprod{U}{V} = (\Inprod{V}{U})^*$$, where $${}^*$$ is the complex conjugate.
+ **Positive semidefiniteness**: $$\Inprod{V}{V} \ge 0$$, and $$\Inprod{V}{V} = 0$$ if $$V = \mathbf{0}$$.
+ **Linearity in second operand**: $$\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$$.

The inner product describes the lengths and angles of vectors,
and in Euclidean space it is implemented by the dot product.

The **magnitude** or **norm** $$|V|$$ of a vector $$V$$ is given by
$$|V| = \sqrt{\Inprod{V}{V}}$$ and represents the real positive length of $$V$$.
A **unit vector** has a norm of 1.

Two vectors $$U$$ and $$V$$ are **orthogonal** if their inner product
$$\Inprod{U}{V} = 0$$. If in addition to being orthogonal, $$|U| = 1$$ and
$$|V| = 1$$, then $$U$$ and $$V$$ are known as **orthonormal** vectors.

Orthonormality is desirable for basis vectors, so if they are
not already like that, it is common to manually turn them into a new
orthonormal basis using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method).

As for the implementation of the inner product, it is given by:

$$\begin{gathered}
    V = \sum_{n = 1}^N v_n \vu{e}_n
    \quad
    W = \sum_{n = 1}^N w_n \vu{e}_n
    \\
    \quad \implies \quad
    \Inprod{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \Inprod{\vu{e}_n}{\vu{e}_j}
\end{gathered}$$

If the basis vectors $$\vu{e}_1, ..., \vu{e}_N$$ are already
orthonormal, this reduces to:

$$\begin{aligned}
    \Inprod{V}{W} = \sum_{n = 1}^N v_n^* w_n
\end{aligned}$$

As it turns out, the components $$v_n$$ are given by the inner product
with $$\vu{e}_n$$, where $$\delta_{nm}$$ is the Kronecker delta:

$$\begin{aligned}
    \Inprod{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n
\end{aligned}$$



## Infinite dimensions

As the dimensionality $$N$$ tends to infinity, things may or may not
change significantly, depending on whether $$N$$ is **countably** or
**uncountably** infinite.

In the former case, not much changes: the infinitely many **discrete**
basis vectors $$\vu{e}_n$$ can all still be made orthonormal as usual,
and as before:

$$\begin{aligned}
    V = \sum_{n = 1}^\infty v_n \vu{e}_n
\end{aligned}$$

A good example of such a countably-infinitely-dimensional basis are the
solution eigenfunctions of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).

However, if the dimensionality is uncountably infinite, the basis
vectors are **continuous** and cannot be labeled by $$n$$. For example, all
complex functions $$f(x)$$ defined for $$x \in [a, b]$$ which
satisfy $$f(a) = f(b) = 0$$ form such a vector space.
In this case $$f(x)$$ is expanded as follows, where $$x$$ is a basis vector:

$$\begin{aligned}
    f(x) = \int_a^b \Inprod{x}{f} \dd{x}
\end{aligned}$$

Similarly, the inner product $$\Inprod{f}{g}$$ must also be redefined as
follows:

$$\begin{aligned}
    \Inprod{f}{g} = \int_a^b f^*(x) \: g(x) \dd{x}
\end{aligned}$$

The concept of orthonormality must be also weakened. A finite function
$$f(x)$$ can be normalized as usual, but the basis vectors $$x$$ themselves
cannot, since each represents an infinitesimal section of the real line.

The rationale in this case is that action of the identity operator $$\hat{I}$$ must
be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/):

$$\begin{aligned}
    \hat{I} = \int_a^b \Ket{\xi} \Bra{\xi} \dd{\xi}
\end{aligned}$$

Applying the identity operator to $$f(x)$$ should just give $$f(x)$$ again:

$$\begin{aligned}
    f(x) = \Inprod{x}{f} = \matrixel{x}{\hat{I}}{f}
    = \int_a^b \Inprod{x}{\xi} \Inprod{\xi}{f} \dd{\xi}
    = \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
\end{aligned}$$

Since we want the latter integral to reduce to $$f(x)$$, it is plain to see that
$$\Inprod{x}{\xi}$$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
i.e $$\Inprod{x}{\xi} = \delta(x - \xi)$$:

$$\begin{aligned}
    \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
    = \int_a^b \delta(x - \xi) f(\xi) \dd{\xi}
    = f(x)
\end{aligned}$$

Consequently, $$\Inprod{x}{\xi} = 0$$ if $$x \neq \xi$$ as expected for an
orthogonal set of vectors, but if $$x = \xi$$ the inner product
$$\Inprod{x}{\xi}$$ is infinite, unlike earlier.

Technically, because the basis vectors $$x$$ cannot be normalized, they
are not members of a Hilbert space, but rather of a superset called a
**rigged Hilbert space**. Such vectors have no finite inner product with
themselves, but do have one with all vectors from the actual Hilbert
space.