Categories: Mathematics, Quantum mechanics.

# Hilbert space

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 $$\braket{U}{V}$$ called the inner product, which takes two vectors and returns a scalar, and has the following properties:

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

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 $$|V| = \sqrt{\braket{V}{V}}$$ 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 $$\braket{U}{V} = 0$$. 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.

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

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

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

\begin{aligned} \braket{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} \braket{\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.

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 \braket{x}{f} \dd{x} \end{aligned}

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

\begin{aligned} \braket{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:

\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) = \braket{x}{f} = \matrixel{x}{\hat{I}}{f} = \int_a^b \braket{x}{\xi} \braket{\xi}{f} \dd{\xi} = \int_a^b \braket{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 $$\braket{x}{\xi}$$ can only be a Dirac delta function, i.e $$\braket{x}{\xi} = \delta(x - \xi)$$:

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

Consequently, $$\braket{x}{\xi} = 0$$ if $$x \neq \xi$$ as expected for an orthogonal set of basis vectors, but if $$x = \xi$$ the inner product $$\braket{x}{\xi}$$ 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.

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