Examples Functional derivative
1 examples
1.1 thomas–fermi kinetic energy functional
1.2 coulomb potential energy functional
1.3 weizsäcker kinetic energy functional
1.4 entropy
1.5 exponential
1.6 functional derivative of function
1.7 functional derivative of iterated function
examples
thomas–fermi kinetic energy functional
the thomas–fermi model of 1927 used kinetic energy functional noninteracting uniform electron gas in first attempt of density-functional theory of electronic structure:
t
t
f
[
ρ
]
=
c
f
∫
ρ
5
/
3
(
r
)
d
r
.
{\displaystyle t_{\mathrm {tf} }[\rho ]=c_{\mathrm {f} }\int \rho ^{5/3}(\mathbf {r} )\,d\mathbf {r} \,.}
since integrand of ttf[ρ] not involve derivatives of ρ(r), functional derivative of ttf[ρ] is,
δ
t
t
f
δ
ρ
(
r
)
=
c
f
∂
ρ
5
/
3
(
r
)
∂
ρ
(
r
)
=
5
3
c
f
ρ
2
/
3
(
r
)
.
{\displaystyle {\begin{aligned}{\frac {\delta t_{\mathrm {tf} }}{\delta \rho ({\boldsymbol {r}})}}&=c_{\mathrm {f} }{\frac {\partial \rho ^{5/3}(\mathbf {r} )}{\partial \rho (\mathbf {r} )}}\\&={\frac {5}{3}}c_{\mathrm {f} }\rho ^{2/3}(\mathbf {r} )\,.\end{aligned}}}
coulomb potential energy functional
for electron-nucleus potential, thomas , fermi employed coulomb potential energy functional
v
[
ρ
]
=
∫
ρ
(
r
)
|
r
|
d
r
.
{\displaystyle v[\rho ]=\int {\frac {\rho ({\boldsymbol {r}})}{|{\boldsymbol {r}}|}}\ d{\boldsymbol {r}}.}
applying definition of functional derivative,
∫
δ
v
δ
ρ
(
r
)
ϕ
(
r
)
d
r
=
[
d
d
ε
∫
ρ
(
r
)
+
ε
ϕ
(
r
)
|
r
|
d
r
]
ε
=
0
=
∫
1
|
r
|
ϕ
(
r
)
d
r
.
{\displaystyle {\begin{aligned}\int {\frac {\delta v}{\delta \rho ({\boldsymbol {r}})}}\ \phi ({\boldsymbol {r}})\ d{\boldsymbol {r}}&{}=\left[{\frac {d}{d\varepsilon }}\int {\frac {\rho ({\boldsymbol {r}})+\varepsilon \phi ({\boldsymbol {r}})}{|{\boldsymbol {r}}|}}\ d{\boldsymbol {r}}\right]_{\varepsilon =0}\\&{}=\int {\frac {1}{|{\boldsymbol {r}}|}}\,\phi ({\boldsymbol {r}})\ d{\boldsymbol {r}}\,.\end{aligned}}}
so,
δ
v
δ
ρ
(
r
)
=
1
|
r
|
.
{\displaystyle {\frac {\delta v}{\delta \rho ({\boldsymbol {r}})}}={\frac {1}{|{\boldsymbol {r}}|}}\ .}
for classical part of electron-electron interaction, thomas , fermi employed coulomb potential energy functional
j
[
ρ
]
=
1
2
∬
ρ
(
r
)
ρ
(
r
′
)
|
r
−
r
′
|
d
r
d
r
′
.
{\displaystyle j[\rho ]={\frac {1}{2}}\iint {\frac {\rho (\mathbf {r} )\rho (\mathbf {r} )}{\vert \mathbf {r} -\mathbf {r} \vert }}\,d\mathbf {r} d\mathbf {r} \,.}
from definition of functional derivative,
∫
δ
j
δ
ρ
(
r
)
ϕ
(
r
)
d
r
=
[
d
d
ϵ
j
[
ρ
+
ϵ
ϕ
]
]
ϵ
=
0
=
[
d
d
ϵ
(
1
2
∬
[
ρ
(
r
)
+
ϵ
ϕ
(
r
)
]
[
ρ
(
r
′
)
+
ϵ
ϕ
(
r
′
)
]
|
r
−
r
′
|
d
r
d
r
′
)
]
ϵ
=
0
=
1
2
∬
ρ
(
r
′
)
ϕ
(
r
)
|
r
−
r
′
|
d
r
d
r
′
+
1
2
∬
ρ
(
r
)
ϕ
(
r
′
)
|
r
−
r
′
|
d
r
d
r
′
{\displaystyle {\begin{aligned}\int {\frac {\delta j}{\delta \rho ({\boldsymbol {r}})}}\phi ({\boldsymbol {r}})d{\boldsymbol {r}}&{}=\left[{\frac {d\ }{d\epsilon }}\,j[\rho +\epsilon \phi ]\right]_{\epsilon =0}\\&{}=\left[{\frac {d\ }{d\epsilon }}\,\left({\frac {1}{2}}\iint {\frac {[\rho ({\boldsymbol {r}})+\epsilon \phi ({\boldsymbol {r}})]\,[\rho ({\boldsymbol {r}} )+\epsilon \phi ({\boldsymbol {r}} )]}{\vert {\boldsymbol {r}}-{\boldsymbol {r}} \vert }}\,d{\boldsymbol {r}}d{\boldsymbol {r}} \right)\right]_{\epsilon =0}\\&{}={\frac {1}{2}}\iint {\frac {\rho ({\boldsymbol {r}} )\phi ({\boldsymbol {r}})}{\vert {\boldsymbol {r}}-{\boldsymbol {r}} \vert }}\,d{\boldsymbol {r}}d{\boldsymbol {r}} +{\frac {1}{2}}\iint {\frac {\rho ({\boldsymbol {r}})\phi ({\boldsymbol {r}} )}{\vert {\boldsymbol {r}}-{\boldsymbol {r}} \vert }}\,d{\boldsymbol {r}}d{\boldsymbol {r}} \\\end{aligned}}}
the first , second terms on right hand side of last equation equal, since r , r′ in second term can interchanged without changing value of integral. therefore,
∫
δ
j
δ
ρ
(
r
)
ϕ
(
r
)
d
r
=
∫
(
∫
ρ
(
r
′
)
|
r
−
r
′
|
d
r
′
)
ϕ
(
r
)
d
r
{\displaystyle \int {\frac {\delta j}{\delta \rho ({\boldsymbol {r}})}}\phi ({\boldsymbol {r}})d{\boldsymbol {r}}=\int \left(\int {\frac {\rho ({\boldsymbol {r}} )}{\vert {\boldsymbol {r}}-{\boldsymbol {r}} \vert }}d{\boldsymbol {r}} \right)\phi ({\boldsymbol {r}})d{\boldsymbol {r}}}
and functional derivative of electron-electron coulomb potential energy functional j[ρ] is,
δ
j
δ
ρ
(
r
)
=
∫
ρ
(
r
′
)
|
r
−
r
′
|
d
r
′
.
{\displaystyle {\frac {\delta j}{\delta \rho ({\boldsymbol {r}})}}=\int {\frac {\rho ({\boldsymbol {r}} )}{\vert {\boldsymbol {r}}-{\boldsymbol {r}} \vert }}d{\boldsymbol {r}} \,.}
the second functional derivative is
δ
2
j
[
ρ
]
δ
ρ
(
r
′
)
δ
ρ
(
r
)
=
∂
∂
ρ
(
r
′
)
(
ρ
(
r
′
)
|
r
−
r
′
|
)
=
1
|
r
−
r
′
|
.
{\displaystyle {\frac {\delta ^{2}j[\rho ]}{\delta \rho (\mathbf {r} )\delta \rho (\mathbf {r} )}}={\frac {\partial }{\partial \rho (\mathbf {r} )}}\left({\frac {\rho (\mathbf {r} )}{\vert \mathbf {r} -\mathbf {r} \vert }}\right)={\frac {1}{\vert \mathbf {r} -\mathbf {r} \vert }}.}
weizsäcker kinetic energy functional
in 1935 von weizsäcker proposed add gradient correction thomas-fermi kinetic energy functional make suit better molecular electron cloud:
t
w
[
ρ
]
=
1
8
∫
∇
ρ
(
r
)
⋅
∇
ρ
(
r
)
ρ
(
r
)
d
r
=
∫
t
w
d
r
,
{\displaystyle t_{\mathrm {w} }[\rho ]={\frac {1}{8}}\int {\frac {\nabla \rho (\mathbf {r} )\cdot \nabla \rho (\mathbf {r} )}{\rho (\mathbf {r} )}}d\mathbf {r} =\int t_{\mathrm {w} }\ d\mathbf {r} \,,}
where
t
w
≡
1
8
∇
ρ
⋅
∇
ρ
ρ
and
ρ
=
ρ
(
r
)
.
{\displaystyle t_{\mathrm {w} }\equiv {\frac {1}{8}}{\frac {\nabla \rho \cdot \nabla \rho }{\rho }}\qquad {\text{and}}\ \ \rho =\rho ({\boldsymbol {r}})\ .}
using derived formula functional derivative,
δ
t
w
δ
ρ
(
r
)
=
∂
t
w
∂
ρ
−
∇
⋅
∂
t
w
∂
∇
ρ
=
−
1
8
∇
ρ
⋅
∇
ρ
ρ
2
−
(
1
4
∇
2
ρ
ρ
−
1
4
∇
ρ
⋅
∇
ρ
ρ
2
)
where
∇
2
=
∇
⋅
∇
,
{\displaystyle {\begin{aligned}{\frac {\delta t_{\mathrm {w} }}{\delta \rho ({\boldsymbol {r}})}}&={\frac {\partial t_{\mathrm {w} }}{\partial \rho }}-\nabla \cdot {\frac {\partial t_{\mathrm {w} }}{\partial \nabla \rho }}\\&=-{\frac {1}{8}}{\frac {\nabla \rho \cdot \nabla \rho }{\rho ^{2}}}-\left({\frac {1}{4}}{\frac {\nabla ^{2}\rho }{\rho }}-{\frac {1}{4}}{\frac {\nabla \rho \cdot \nabla \rho }{\rho ^{2}}}\right)\qquad {\text{where}}\ \ \nabla ^{2}=\nabla \cdot \nabla \ ,\end{aligned}}}
and result is,
δ
t
w
δ
ρ
(
r
)
=
1
8
∇
ρ
⋅
∇
ρ
ρ
2
−
1
4
∇
2
ρ
ρ
.
{\displaystyle {\frac {\delta t_{\mathrm {w} }}{\delta \rho ({\boldsymbol {r}})}}=\ \ \,{\frac {1}{8}}{\frac {\nabla \rho \cdot \nabla \rho }{\rho ^{2}}}-{\frac {1}{4}}{\frac {\nabla ^{2}\rho }{\rho }}\ .}
entropy
the entropy of discrete random variable functional of probability mass function.
h
[
p
(
x
)
]
=
−
∑
x
p
(
x
)
log
p
(
x
)
{\displaystyle {\begin{aligned}h[p(x)]=-\sum _{x}p(x)\log p(x)\end{aligned}}}
thus,
∑
x
δ
h
δ
p
(
x
)
ϕ
(
x
)
=
[
d
d
ϵ
h
[
p
(
x
)
+
ϵ
ϕ
(
x
)
]
]
ϵ
=
0
=
[
−
d
d
ε
∑
x
[
p
(
x
)
+
ε
ϕ
(
x
)
]
log
[
p
(
x
)
+
ε
ϕ
(
x
)
]
]
ε
=
0
=
−
∑
x
[
1
+
log
p
(
x
)
]
ϕ
(
x
)
.
{\displaystyle {\begin{aligned}\sum _{x}{\frac {\delta h}{\delta p(x)}}\,\phi (x)&{}=\left[{\frac {d}{d\epsilon }}h[p(x)+\epsilon \phi (x)]\right]_{\epsilon =0}\\&{}=\left[-\,{\frac {d}{d\varepsilon }}\sum _{x}\,[p(x)+\varepsilon \phi (x)]\ \log[p(x)+\varepsilon \phi (x)]\right]_{\varepsilon =0}\\&{}=\displaystyle -\sum _{x}\,[1+\log p(x)]\ \phi (x)\,.\end{aligned}}}
thus,
δ
h
δ
p
(
x
)
=
−
1
−
log
p
(
x
)
.
{\displaystyle {\frac {\delta h}{\delta p(x)}}=-1-\log p(x).}
exponential
let
f
[
φ
(
x
)
]
=
e
∫
φ
(
x
)
g
(
x
)
d
x
.
{\displaystyle f[\varphi (x)]=e^{\int \varphi (x)g(x)dx}.}
using delta function test function,
δ
f
[
φ
(
x
)
]
δ
φ
(
y
)
=
lim
ε
→
0
f
[
φ
(
x
)
+
ε
δ
(
x
−
y
)
]
−
f
[
φ
(
x
)
]
ε
=
lim
ε
→
0
e
∫
(
φ
(
x
)
+
ε
δ
(
x
−
y
)
)
g
(
x
)
d
x
−
e
∫
φ
(
x
)
g
(
x
)
d
x
ε
=
e
∫
φ
(
x
)
g
(
x
)
d
x
lim
ε
→
0
e
ε
∫
δ
(
x
−
y
)
g
(
x
)
d
x
−
1
ε
=
e
∫
φ
(
x
)
g
(
x
)
d
x
lim
ε
→
0
e
ε
g
(
y
)
−
1
ε
=
e
∫
φ
(
x
)
g
(
x
)
d
x
g
(
y
)
.
{\displaystyle {\begin{aligned}{\frac {\delta f[\varphi (x)]}{\delta \varphi (y)}}&{}=\lim _{\varepsilon \to 0}{\frac {f[\varphi (x)+\varepsilon \delta (x-y)]-f[\varphi (x)]}{\varepsilon }}\\&{}=\lim _{\varepsilon \to 0}{\frac {e^{\int (\varphi (x)+\varepsilon \delta (x-y))g(x)dx}-e^{\int \varphi (x)g(x)dx}}{\varepsilon }}\\&{}=e^{\int \varphi (x)g(x)dx}\lim _{\varepsilon \to 0}{\frac {e^{\varepsilon \int \delta (x-y)g(x)dx}-1}{\varepsilon }}\\&{}=e^{\int \varphi (x)g(x)dx}\lim _{\varepsilon \to 0}{\frac {e^{\varepsilon g(y)}-1}{\varepsilon }}\\&{}=e^{\int \varphi (x)g(x)dx}g(y).\end{aligned}}}
thus,
δ
f
[
φ
(
x
)
]
δ
φ
(
y
)
=
g
(
y
)
f
[
φ
(
x
)
]
.
{\displaystyle {\frac {\delta f[\varphi (x)]}{\delta \varphi (y)}}=g(y)f[\varphi (x)].}
this particularly useful in calculating correlation functions partition function in quantum field theory.
functional derivative of function
a function can written in form of integral functional. example,
ρ
(
r
)
=
f
[
ρ
]
=
∫
ρ
(
r
′
)
δ
(
r
−
r
′
)
d
r
′
.
{\displaystyle \rho ({\boldsymbol {r}})=f[\rho ]=\int \rho ({\boldsymbol {r}} )\delta ({\boldsymbol {r}}-{\boldsymbol {r}} )\,d{\boldsymbol {r}} .}
since integrand not depend on derivatives of ρ, functional derivative of ρ(r) is,
δ
ρ
(
r
)
δ
ρ
(
r
′
)
≡
δ
f
δ
ρ
(
r
′
)
=
∂
∂
ρ
(
r
′
)
[
ρ
(
r
′
)
δ
(
r
−
r
′
)
]
=
δ
(
r
−
r
′
)
.
{\displaystyle {\begin{aligned}{\frac {\delta \rho ({\boldsymbol {r}})}{\delta \rho ({\boldsymbol {r}} )}}\equiv {\frac {\delta f}{\delta \rho ({\boldsymbol {r}} )}}&={\frac {\partial \ \ }{\partial \rho ({\boldsymbol {r}} )}}\,[\rho ({\boldsymbol {r}} )\delta ({\boldsymbol {r}}-{\boldsymbol {r}} )]\\&=\delta ({\boldsymbol {r}}-{\boldsymbol {r}} ).\end{aligned}}}
functional derivative of iterated function
the functional derivative of iterated function
f
(
f
(
x
)
)
{\displaystyle f(f(x))}
given by:
δ
f
(
f
(
x
)
)
δ
f
(
y
)
=
f
′
(
f
(
x
)
)
δ
(
x
−
y
)
+
δ
(
f
(
x
)
−
y
)
{\displaystyle {\frac {\delta f(f(x))}{\delta f(y)}}=f (f(x))\delta (x-y)+\delta (f(x)-y)}
and
δ
f
(
f
(
f
(
x
)
)
)
δ
f
(
y
)
=
f
′
(
f
(
f
(
x
)
)
(
f
′
(
f
(
x
)
)
δ
(
x
−
y
)
+
δ
(
f
(
x
)
−
y
)
)
+
δ
(
f
(
f
(
x
)
)
−
y
)
{\displaystyle {\frac {\delta f(f(f(x)))}{\delta f(y)}}=f (f(f(x))(f (f(x))\delta (x-y)+\delta (f(x)-y))+\delta (f(f(x))-y)}
in general:
δ
f
n
(
x
)
δ
f
(
y
)
=
f
′
(
f
n
−
1
(
x
)
)
δ
f
n
−
1
(
x
)
δ
f
(
y
)
+
δ
(
f
n
−
1
(
x
)
−
y
)
{\displaystyle {\frac {\delta f^{n}(x)}{\delta f(y)}}=f (f^{n-1}(x)){\frac {\delta f^{n-1}(x)}{\delta f(y)}}+\delta (f^{n-1}(x)-y)}
putting in n=0 gives:
δ
f
−
1
(
x
)
δ
f
(
y
)
=
−
δ
(
f
−
1
(
x
)
−
y
)
f
′
(
f
−
1
(
x
)
)
{\displaystyle {\frac {\delta f^{-1}(x)}{\delta f(y)}}=-{\frac {\delta (f^{-1}(x)-y)}{f (f^{-1}(x))}}}
^ (parr & yang 1989, p. 247, eq. a.6).
^ (parr & yang 1989, p. 248, eq. a.11).
^ (parr & yang 1989, p. 247, eq. a.9).
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