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Revision difference : unit vector
Version current Version 11
A \emph{unit vector} is a unit-length element of Euclidean space. A \emph{unit vector} is a unit-length element of Euclidean space.
Equivalently, one may say that the norm of a unit vector is equal Equivelently, one may say that the norm of a unit vector is equal
to $1$, and write $\Vert \bu\Vert=1$, where $\bu$ is the vector in to $1$, and write $\Vert \bu\Vert=1$, where $\bu$ is the vector in
question. question.
Let $\bv$ be a non-zero vector. To \emph{normalize} $\bv$ is to find Let $\bv$ be a non-zero vector. To \emph{normalize} $\bv$ is to find
the unique unit vector with the same direction as $\bv$. This is done the unique unit vector with the same direction as $\bv$. This is done
by multiplying $\bv$ by the reciprocal of its length; the by multiplying $\bv$ by the reciprocal of its length; the
corresponding unit vector is given by $\bu=\frac{\bv}{\Vert corresponding unit vector is given by $\bu=\frac{\bv}{\Vert
\bv\Vert}$. \bv\Vert}$.
\paragraph{Note:} The concept of a unit vector and normalization makes \paragraph{Note:} The concept of a unit vector and normalization makes
sense in any vector space equipped with a real or complex norm. sense in any vector space equipped with a real or complex norm.
Thus, in quantum mechanics one represents states as unit vectors Thus, in quantum mechanics one represents states as unit vectors
belonging to a (possibly) infinite-dimensional Hilbert space. To belonging to a (possibly) infinite-dimensional Hilbert space. To
obtain an expression for such states one normalizes obtain an expression for such states one normalizes
the results of a calculation. the results of a calculation.
\paragraph{Example:} Consider $\mathbb{R}^3$ and the vector \paragraph{Example:} Consider $\mathbb{R}^3$ and the vector
$\bv=(1,2,3)$. The norm (length) is $\sqrt{14}$. Normalizing, we obtain $\bv=(1,2,3)$. The norm (length) is $\sqrt{14}$. Normalizing, we obtain
the unit vector $\bu$ pointing in the same direction, namely the unit vector $\bu$ pointing in the same direction, namely
$\bu=\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right)$. $\bu=\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right)$.