User:Espen180/Quantum Mechanics/Preliminary Mathematics

The erader is expected to be familiar with the contents of the Linear Algebra wikibook.

Hilbert Space

A Hilbert space is a generalized complex vector space $V$, which may have an (uncountably) infinite number of dimensions, and on which the following inner product is defined: For any $u,v,w\in V$ and $\alpha ,\beta \in \mathbb {C}$, we define their inner product $\langle u,v\rangle$ such that

i) $\langle u,v\rangle =\langle v,u\rangle ^{*}$,

ii) $\langle u,\alpha v+\beta w\rangle =\alpha \langle u,v\rangle +\beta \langle u,w\rangle$, and

iii) $\langle u,u\rangle \geq 0$, and $\langle u,u\rangle =0$ if and only if $u=0$.

Furthermore, we require that $V$ is complete with respect to the norm $||u||^{2}=\langle u,u\rangle$. Let $\{u_{i}\}_{i=1}^{\infty }$ be a sequence such that for every real number $\epsilon >0$, there exists an integer $N>0$ such that for all integers $n,m>N$,

$||u_{n}-u_{m}||<\epsilon$.

Then the sequence is called Cauchy, and the completeness axiom states that every Cauchy sequence of vectors $\{u_{i}\}_{i=1}^{\infty }$ in $V$ converges to a vector $u_{\infty }$ in $V$.

Two vectors $u,v\in V$ are called orthogonal if $\langle u,v\rangle =0$. A set of vectors orthogonal to one another is neccesarily linearly independent. The proof is left to the reader as an excercise.

A linearly independent subset $B$ of $V$ is called a basis set if all vectors in $V$ have a unique linear expansion in terms of the basis vectors.

Example 1: Let $L_{2}[0,\pi )$ be the set of all square-integrable functions on the real line segment $[0,\pi )$, and let $f(x)$ be any such function. Then, since $f(x)$ has a unique Fourier expansion, the set $B=\{sin(cx),cos(cx)|c\in \mathbb {N} \}$ is a basis set for $L_{2}[0,\pi )$.

Hilbert spaces are frequently taken to be function spaces, that is, spaces whose elements are functions of some kind. The kind of Hilbert space we will be using is called a rigged Hilbert space, in which we generalize to spaces of distributions. In effect, this allows the use of the Dirac delta function $\delta (x)$, defined by

$\int _{-\epsilon }^{\epsilon }\delta (x)\mathrm {d} x=1$ for any real $\epsilon >0$.

Since a function is given uniquely by specifying its value at all elements of its domain, the set $B=\{\delta (x-c)|c\in \mathrm {R} \}$ is a basis for any function space on the real line.

The Dual Space

Associated with every Hilbert space $V$ is the corresponding dual space $V^{*}$, consisting of linear functionals on $V$. A linear functional on $V$ is a linear function $f\,:\,V\rightarrow \mathbb {C}$ such that $f(\alpha u+\beta v)=\alpha f(u)+\beta f(v)$.

Let $B$ be an orthogonal basis set in $V$. We then construct the set $B^{*}$ in $V^{*}$ by sending $b\in B$ to $b^{*}\in B^{*}$, where $b^{*}$ is the functional given by $b^{*}(v)=\langle b,v\rangle$ for all $v\in V$.

Operators

An operator on a Hilber space $V$ is a linear transformation $A\,:\,V\rightarrow V$.

Given two operators $A$ and $B$ on $V$, we can define their composition $AB$ by $(AB)v=A(Bv)$ for all $v\in V$.

The identity function $\mathrm {id} \,:\,V\rightarrow V$ defined by $\mathrm {id} (v)=v$ for all $v\in V$ is an operator on $V$.

Given an operator $A$ on $V$, the inverse operator, if it exists, is the operator $A^{-1}$ on $V$ such that $AA^{-1}=A^{-1}A=\mathrm {id}$.

Given an operator $A$ on $V$, we define its Hermitian adjoint, or simply adjoint, as the unique operator $A^{\dagger }$ such that for any $u,v\in V$, we have

$\langle u,Av\rangle =\langle A^{\dagger }u,v\rangle$

An operator is called Hermitian if $A^{\dagger }=A$. It is called unitary if $A^{\dagger }=A^{-1}$.

Given two operators $A$ and $B$ on $V$, define their commutator $[A,B]=AB-BA$.

An operator A is called normal if $[A,A^{\dagger }]=0$.

It is trivi al to show that if an operator is either Hermitian of unitary, then it must neccesarily be normal.

Eigenvalues and Eigenvectors

Let $A$ be an operator on a Hilbert space $V$ and concider the equation

$Av=\lambda v$.

This is called an eigenvalue equation. $\lambda$ is called an eigenvalue of $A$, and $v$ an eigenvector. We assume the reader to be familar with the eigenvalue problem in the finite-dimensional case. We will now prove a very useful theorem. If $A$ is Hermitian, then the eigenvectors of $A$ constitute a basis for $V$.