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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 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.

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.

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:

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

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, 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.

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