Categories: Mathematics, Quantum mechanics.

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

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:

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

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:

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

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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