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**
that 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 have the following (again familiar) properties:

+ **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**.

$$\mathbb{V}$$ has **dimension** $$N$$
if only up to $$N$$ of its vectors can be linearly independent.
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 a (generally not unique)
valid set of basis vectors of $$\mathbb{V}$$,
then any vector $$V$$ in that space can be **expanded**
in that basis according to unique weights $$v_n$$,
called the **components** of $$V$$ in that basis:

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

Using these components,
the operations of vector addition and scalar multiplication
can then be implemented as follows:

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

It is straightforward to see that this implementation satisfies the properties above.



## 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).

The implementation of the inner product in terms of components and basis vectors
is as follows, which can easily be shown to satisfy the properties above:

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

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

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

This suggests a way to calculate the components $$v_n$$:
taking the inner product of $$V$$ with a basis vector $$\vu{e}_n$$
"picks out" the corresponding component $$v_n$$.
Let $$\delta_{nm}$$ be 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 on the interval $$x \in [a, b]$$
satisfying the boundary condition $$f(a) = f(b) = 0$$, form such a vector space.
In this case, every value of $$f(x)$$ is the component of
an abstract vector $$\Ket{f}$$ with respect to a basis vector $$\Ket{x}$$:

$$\begin{aligned}
    f(x) = \Inprod{x}{f}
\end{aligned}$$

The inner product $$\Inprod{f}{g}$$ must be redefined as follows,
effectively turning the sum over a discrete basis
into an integral over a continuous basis:

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

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.
So how to proceed?

The rationale in this case is that the 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 $$x = \xi$$ then 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.