unitaryUnitaryTransformation2001-11-27 22:10:442007-11-15 18:33:08Definitionunitary spaceunitary matrixunitary transformationunitary operatorunitary group\usepackage{amsmath}
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\newcommand{\pln}[2]{\PMlinkname{{#1}}{#2}}\subsection{Definitions}
\begin{itemize}
\item A {\bf unitary space} $V$ is a complex vector space with a
distinguished positive definite Hermitian form,
$$
\langle -,-\rangle: V\times V \rightarrow \cnums,$$
which serves as
the inner product on $V$.
\end{itemize}
\begin{itemize}
\item A {\bf unitary transformation} is a surjective linear transformation
$T:V\rightarrow V$ satisfying
\begin{equation}
\label{eq:def}
\langle u,v \rangle = \langle Tu,Tv\rangle,\quad u, v \in V.
\end{equation}
These are isometries of $V$.
\end{itemize}
\begin{itemize}
\item More generally, a {\bf unitary transformation} is a surjective linear transformation $T:U \longrightarrow V$ between two unitary spaces $U,V$ satisfying
\begin{displaymath}
\langle Tv , Tu \rangle_{V} = \langle v , u \rangle_{U},\quad\; u,v \in U
\end{displaymath}
In this entry will restrict to the case of the first \PMlinkescapetext{definition}, i.e. $U = V$.
\end{itemize}
\begin{itemize}
\item A {\bf unitary matrix} is a square complex-valued matrix, $A$, whose inverse
is equal to its conjugate transpose:
$$A^{-1}=\bar{A}^t.$$
\end{itemize}
\begin{itemize}
\item When $V$ is a Hilbert space, a bounded linear operator $T:V \longrightarrow V$ is said to be a {\bf unitary operator} if its inverse is equal to its adjoint:
\begin{displaymath}
T^{-1} = T^*
\end{displaymath}
In Hilbert spaces unitary transformations correspond precisely to unitary operators.
\end{itemize}
\subsection{Remarks}
\begin{enumerate}
\item
A standard example of a unitary space is
$\cnums^n$ with inner product
\begin{equation}
\label{eq:cprod}
\langle u,v \rangle = \sum_{i=1}^n u_i\, \overline{v_i},\quad
u,v \in\cnums^n.
\end{equation}
\item Unitary transformations and unitary matrices
are closely related. On the one hand, a unitary matrix defines a
unitary transformation of $\cnums^n$ relative to the inner product
\eqref{eq:cprod}. On the other hand, the representing matrix of a
unitary transformation relative to an orthonormal basis is, in fact, a
unitary matrix.
\item A unitary transformation is an automorphism. This follows from
the fact that a unitary transformation $T$ preserves the
inner-product norm:
\begin{equation}
\label{eq:def1}
\Vert T u \Vert= \Vert u\Vert,\quad u\in V.
\end{equation}
Hence, if $$Tu=0,$$
then by the definition \eqref{eq:def}
it follows that
$$\Vert u \Vert = 0,$$
and hence by the inner-product axioms that
$$u=0.$$
Thus, the kernel of $T$ is trivial, and therefore it is an
automorphism.
\item Moreover, relation \eqref{eq:def1} can be taken as the definition
of a unitary transformation. Indeed, using the polarization
identity it is possible to show that if
$T$ preserves the norm, then \eqref{eq:def} must hold as well.
\item A simple example of a unitary matrix is the change of
coordinates matrix between two orthonormal bases. Indeed, let
$u_1,\ldots, u_n$ and $v_1,\ldots,v_n$ be two orthonormal bases, and
let $A=(A^i_j)$ be the corresponding change of basis matrix
defined by
$$v_j = \sum_i A^i_j\, u_i,\quad j=1,\ldots, n.$$
Substituting the above relation into the defining relations for an
orthonormal basis,
\begin{eqnarray*}
\langle u_i,u_j\rangle &=& \delta_{ij},\\
\langle v_k,v_l\rangle &=& \delta_{kl},
\end{eqnarray*}
we obtain
$$\sum_{ij} \delta_{ij} A^i_k \overline{A^j_l} = \sum_i A^i_k \overline{A^i_l} =
\delta_{kl}.$$
In matrix notation, the above is simply
$$A \bar{A}^t = I,$$
as desired.
\item Unitary transformations form a group under composition. Indeed, if $S, T$ are unitary transformations then $ST$ is also surjective and
\begin{displaymath}
\langle STu, STv \rangle = \langle Tu, Tv\rangle = \langle u, v\rangle
\end{displaymath}
for every $u,v \in V$. Hence $ST$ is also a unitary transformation.
\item Unitary spaces, transformations, matrices and operators are of fundamental
importance in quantum mechanics.
\end{enumerate}