Editing Directional derivative (section)

== In group theory ==

===Translations===
In the [[Poincaré algebra]], we can define an infinitesimal translation operator '''P''' as
<math display="block">\mathbf{P}=i\nabla.</math>
(the ''i'' ensures that '''P''' is a [[self-adjoint operator]]) For a finite displacement '''λ''', the [[Unitary operator|unitary]] [[Hilbert space]] [[Group representation|representation]] for translations is<ref>{{cite book| last1=Weinberg|first1=Steven|title=The quantum theory of fields|date=1999|publisher=Cambridge Univ. Press| location=Cambridge [u.a.]| isbn=9780521550017|edition=Reprinted (with corr.).|url-access=registration| url=https://archive.org/details/quantumtheoryoff00stev}}</ref> 
<math display="block">U(\boldsymbol{\lambda})=\exp\left(-i\boldsymbol{\lambda}\cdot\mathbf{P}\right).</math>
By using the above definition of the infinitesimal translation operator, we see that the finite translation operator is an exponentiated directional derivative:
<math display="block">U(\boldsymbol{\lambda})=\exp\left(\boldsymbol{\lambda}\cdot\nabla\right).</math>
This is a translation operator in the sense that it acts on multivariable functions ''f''('''x''') as
<math display="block">U(\boldsymbol{\lambda}) f(\mathbf{x})=\exp\left(\boldsymbol{\lambda}\cdot\nabla\right) f(\mathbf{x}) = f(\mathbf{x}+\boldsymbol{\lambda}).</math>

{{math proof|title=Proof of the last equation
|proof=
In standard single-variable calculus, the derivative of a smooth function ''f''(''x'') is defined by (for small ''ε'')
<math display="block">\frac{df}{dx} = \frac{f(x+\varepsilon) - f(x)}{\varepsilon}.</math>
This can be rearranged to find ''f''(''x''+''ε''):
<math display="block">f(x+\varepsilon)=f(x)+\varepsilon \,\frac{df}{dx}=\left(1+\varepsilon\,\frac{d}{dx}\right)f(x).</math>
It follows that <math>[1+\varepsilon\,(d/dx)] </math> is a translation operator. This is instantly generalized<ref>{{cite book |last1=Zee|first1=A.|title=Einstein gravity in a nutshell|date=2013|publisher=Princeton University Press| location=Princeton| isbn=9780691145587}}</ref> to multivariable functions ''f''('''x''')
<math display="block">f(\mathbf{x}+\boldsymbol{\varepsilon}) = \left(1+\boldsymbol{\varepsilon}\cdot\nabla\right) f(\mathbf{x}).</math>
Here <math> \boldsymbol{\varepsilon}\cdot\nabla</math> is the directional derivative along the infinitesimal displacement '''''ε'''''. We have found the infinitesimal version of the translation operator:
<math display="block">U(\boldsymbol{\varepsilon}) = 1 + \boldsymbol{\varepsilon}\cdot\nabla.</math>
It is evident that the group multiplication law<ref>{{cite book | last1=Cahill |first1=Kevin Cahill | title=Physical mathematics | date=2013 | publisher=Cambridge University Press | location=Cambridge | isbn=978-1107005211 | edition=Repr.}}</ref> ''U''(''g'')''U''(''f'')=''U''(''gf'') takes the form
<math display="block">U(\mathbf{a})U(\mathbf{b})=U(\mathbf{a+b}).</math>
So suppose that we take the finite displacement '''''λ''''' and divide it into ''N'' parts (''N''→∞ is implied everywhere), so that '''''λ'''''/''N''='''''ε'''''. In other words,
<math display="block">\boldsymbol{\lambda} = N \boldsymbol{\varepsilon}.</math>
Then by applying ''U''('''''ε''''') ''N'' times, we can construct ''U''('''''λ'''''):
<math display="block">[U(\boldsymbol{\varepsilon})]^N = U(N\boldsymbol{\varepsilon}) = U(\boldsymbol{\lambda}).</math>
We can now plug in our above expression for U('''ε'''):
<math display="block">[U(\boldsymbol{\varepsilon})]^N = \left[1+\boldsymbol{\varepsilon}\cdot\nabla\right]^N = \left[1+\frac{\boldsymbol{\lambda}\cdot\nabla}{N}\right]^N.</math>
Using the identity<ref>{{cite book | first1 = Ron | last1 = Larson | last2 = Edwards | first2 = Bruce H. | title = Calculus of a single variable | date = 2010 | publisher = Brooks/Cole | location =Belmont | isbn = 9780547209982 | edition = 9th}}</ref>
<math display="block">\exp(x)=\left[1+\frac{x}{N}\right]^N,</math>
we have
<math display="block">U(\boldsymbol{\lambda})=\exp\left(\boldsymbol{\lambda}\cdot\nabla\right).</math>
And since {{math|1=''U''('''''ε''''')''f''('''x''') = ''f''('''x'''+'''''ε''''')}} we have
<math display="block">[U(\boldsymbol{\varepsilon})]^N f(\mathbf{x}) = f(\mathbf{x}+N\boldsymbol{\varepsilon}) = f(\mathbf{x}+\boldsymbol{\lambda}) = U(\boldsymbol{\lambda})f(\mathbf{x}) = \exp\left(\boldsymbol{\lambda}\cdot\nabla\right)f(\mathbf{x}),</math>
Q.E.D.

As a technical note, this procedure is only possible because the translation group forms an [[abelian group|Abelian]] [[subgroup]] ([[Cartan subalgebra]]) in the Poincaré algebra. In particular, the group multiplication law ''U''('''a''')''U''('''b''') = ''U''('''a'''+'''b''') should not be taken for granted. We also note that Poincaré is a connected [[Lie group]]. It is a group of transformations ''T''(''ξ'') that are described by a continuous set of real parameters <math>\xi^a</math>. The group multiplication law takes the form
<math display="block">T(\bar{\xi})T(\xi) = T(f(\bar{\xi},\xi)).</math>
Taking <math>\xi^a = 0</math> as the coordinates of the identity, we must have
<math display="block">f^a(\xi,0)=f^a(0,\xi)=\xi^a.</math>
The actual operators on the Hilbert space are represented by unitary operators ''U''(''T''(''ξ'')). In the above notation we suppressed the ''T''; we now write ''U''('''λ''') as ''U''('''P'''('''λ''')). For a small neighborhood around the identity, the [[power series]] representation 
<math display="block">U(T(\xi))=1+i\sum_a\xi^a t_a+\frac{1}{2}\sum_{b,c}\xi^b\xi^c t_{bc}+\cdots</math>
is quite good. Suppose that U(T(ξ)) form a non-projective representation, i.e.,
<math display="block">U(T(\bar{\xi}))U(T(\xi))=U(T(f(\bar{\xi},\xi))).</math>
The expansion of f to second power is
<math display="block">f^a(\bar{\xi},\xi)=\xi^a+\bar{\xi}^a+\sum_{b,c}f^{abc}\bar{\xi}^b\xi^c.</math>
After expanding the representation multiplication equation and equating coefficients, we have the nontrivial condition
<math display="block">t_{bc}=-t_b t_c-i\sum_a f^{abc}t_a.</math>
Since <math> t_{ab}</math> is by definition symmetric in its indices, we have the standard [[Lie algebra]] commutator:
<math display="block">[t_b, t_c]=i\sum_a(-f^{abc}+f^{acb})t_a=i\sum_a C^{abc}t_a,</math>
with ''C'' the [[structure constant]]. The generators for translations are partial derivative operators, which commute:
<math display="block">\left[\frac{\partial}{\partial x^b},\frac{\partial }{\partial x^c}\right]=0.</math>
This implies that the structure constants vanish and thus the quadratic coefficients in the f expansion vanish as well. This means that ''f'' is simply additive:
<math display="block">f^a_\text{abelian}(\bar{\xi},\xi)=\xi^a+\bar{\xi}^a,</math>
and thus for abelian groups,
<math display="block">U(T(\bar{\xi}))U(T(\xi))=U(T(\bar{\xi}+\xi)).</math>
Q.E.D.
}}

===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>