Editing Directional derivative (section)

===Rotations===
The [[rotation operator (quantum mechanics)|rotation operator]] also contains a directional derivative. The rotation operator for an angle '''''θ''''', i.e. by an amount ''θ'' = |'''''θ'''''| about an axis parallel to <math> \hat{\theta} = \boldsymbol{\theta}/\theta</math> is
<math display="block">U(R(\mathbf{\theta}))=\exp(-i\mathbf{\theta}\cdot\mathbf{L}).</math>
Here '''L''' is the vector operator that generates [[SO(3)]]:
<math display="block">\mathbf{L}=\begin{pmatrix}
 0& 0 & 0\\ 
 0& 0 & 1\\ 
 0& -1 & 0
\end{pmatrix}\mathbf{i}+\begin{pmatrix}
0 &0  & -1\\ 
 0& 0 &0 \\ 
1 & 0 & 0
\end{pmatrix}\mathbf{j}+\begin{pmatrix}
 0&1  &0 \\ 
 -1&0  &0 \\ 
0 & 0 & 0
\end{pmatrix}\mathbf{k}.</math>
It may be shown geometrically that an infinitesimal right-handed rotation changes the position vector '''x''' by
<math display="block">\mathbf{x}\rightarrow \mathbf{x}-\delta\boldsymbol{\theta}\times\mathbf{x}.</math>
So we would expect under infinitesimal rotation:
<math display="block">U(R(\delta\boldsymbol{\theta})) f(\mathbf{x}) = f(\mathbf{x}-\delta\boldsymbol{\theta}\times\mathbf{x})=f(\mathbf{x})-(\delta\boldsymbol{\theta}\times\mathbf{x})\cdot\nabla f.</math>
It follows that 
<math display="block">U(R(\delta\mathbf{\theta}))=1-(\delta\mathbf{\theta}\times\mathbf{x})\cdot\nabla.</math>
Following the same exponentiation procedure as above, we arrive at the rotation operator in the position basis, which is an exponentiated directional derivative:<ref>{{cite book|last1=Shankar|first1=R. | title=Principles of quantum mechanics | date=1994|publisher=Kluwer Academic / Plenum|location=New York|isbn=9780306447907|page=318|edition=2nd}}</ref>
<math display="block">U(R(\mathbf{\theta}))=\exp(-(\mathbf{\theta}\times\mathbf{x})\cdot\nabla).</math>