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---
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 $V + W$.
+ **Scalar multiplication**: product of a vector $V$ with a scalar $a$, denoted $a V$.
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.
|
