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diff --git a/source/know/concept/hilbert-space/index.md b/source/know/concept/hilbert-space/index.md new file mode 100644 index 0000000..d2b9770 --- /dev/null +++ b/source/know/concept/hilbert-space/index.md @@ -0,0 +1,196 @@ +--- +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. |