Common integrals in quantum field theory are all variations and generalizations of
Gaussian integrals to the
complex plane and to multiple dimensions.
[1] : 13–15 Other integrals can be approximated by versions of the Gaussian integral. Fourier integrals are also considered.
Variations on a simple Gaussian integral
Gaussian integral
The first integral, with broad application outside of quantum field theory, is the Gaussian integral.
G
≡
∫
−
∞
∞
e
−
1
2
x
2
d
x
{\displaystyle G\equiv \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx}
In physics the factor of 1/2 in the argument of the exponential is common.
Note:
G
2
=
(
∫
−
∞
∞
e
−
1
2
x
2
d
x
)
⋅
(
∫
−
∞
∞
e
−
1
2
y
2
d
y
)
=
2
π
∫
0
∞
r
e
−
1
2
r
2
d
r
=
2
π
∫
0
∞
e
−
w
d
w
=
2
π
.
{\displaystyle G^{2}=\left(\int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx\right)\cdot \left(\int _{-\infty }^{\infty }e^{-{1 \over 2}y^{2}}\,dy\right)=2\pi \int _{0}^{\infty }re^{-{1 \over 2}r^{2}}\,dr=2\pi \int _{0}^{\infty }e^{-w}\,dw=2\pi .}
Thus we obtain
∫
−
∞
∞
e
−
1
2
x
2
d
x
=
2
π
.
{\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx={\sqrt {2\pi }}.}
Slight generalization of the Gaussian integral
∫
−
∞
∞
e
−
1
2
a
x
2
d
x
=
2
π
a
{\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx={\sqrt {2\pi \over a}}}
where we have scaled
x
→
x
a
.
{\displaystyle x\to {x \over {\sqrt {a}}}.}
Integrals of exponents and even powers of x
∫
−
∞
∞
x
2
e
−
1
2
a
x
2
d
x
=
−
2
d
d
a
∫
−
∞
∞
e
−
1
2
a
x
2
d
x
=
−
2
d
d
a
(
2
π
a
)
1
2
=
(
2
π
a
)
1
2
1
a
{\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-{1 \over 2}ax^{2}}\,dx=-2{d \over da}\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx=-2{d \over da}\left({2\pi \over a}\right)^{1 \over 2}=\left({2\pi \over a}\right)^{1 \over 2}{1 \over a}}
and
∫
−
∞
∞
x
4
e
−
1
2
a
x
2
d
x
=
(
−
2
d
d
a
)
(
−
2
d
d
a
)
∫
−
∞
∞
e
−
1
2
a
x
2
d
x
=
(
−
2
d
d
a
)
(
−
2
d
d
a
)
(
2
π
a
)
1
2
=
(
2
π
a
)
1
2
3
a
2
{\displaystyle \int _{-\infty }^{\infty }x^{4}e^{-{1 \over 2}ax^{2}}\,dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\left({2\pi \over a}\right)^{1 \over 2}=\left({2\pi \over a}\right)^{1 \over 2}{3 \over a^{2}}}
In general
∫
−
∞
∞
x
2
n
e
−
1
2
a
x
2
d
x
=
(
2
π
a
)
1
2
1
a
n
(
2
n
−
1
)
(
2
n
−
3
)
⋯
5
⋅
3
⋅
1
=
(
2
π
a
)
1
2
1
a
n
(
2
n
−
1
)
!
!
{\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-{1 \over 2}ax^{2}}\,dx=\left({2\pi \over a}\right)^{1 \over {2}}{1 \over a^{n}}\left(2n-1\right)\left(2n-3\right)\cdots 5\cdot 3\cdot 1=\left({2\pi \over a}\right)^{1 \over {2}}{1 \over a^{n}}\left(2n-1\right)!!}
Note that the integrals of exponents and odd powers of x are 0, due to
odd symmetry.
Integrals with a linear term in the argument of the exponent
∫
−
∞
∞
exp
(
−
1
2
a
x
2
+
J
x
)
d
x
{\displaystyle \int _{-\infty }^{\infty }\exp \left(-{\frac {1}{2}}ax^{2}+Jx\right)dx}
This integral can be performed by completing the square:
(
−
1
2
a
x
2
+
J
x
)
=
−
1
2
a
(
x
2
−
2
J
x
a
+
J
2
a
2
−
J
2
a
2
)
=
−
1
2
a
(
x
−
J
a
)
2
+
J
2
2
a
{\displaystyle \left(-{1 \over 2}ax^{2}+Jx\right)=-{1 \over 2}a\left(x^{2}-{2Jx \over a}+{J^{2} \over a^{2}}-{J^{2} \over a^{2}}\right)=-{1 \over 2}a\left(x-{J \over a}\right)^{2}+{J^{2} \over 2a}}
Therefore:
∫
−
∞
∞
exp
(
−
1
2
a
x
2
+
J
x
)
d
x
=
exp
(
J
2
2
a
)
∫
−
∞
∞
exp
−
1
2
a
(
x
−
J
a
)
2
d
x
=
exp
(
J
2
2
a
)
∫
−
∞
∞
exp
(
−
1
2
a
w
2
)
d
w
=
(
2
π
a
)
1
2
exp
(
J
2
2
a
)
{\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+Jx\right)\,dx&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left[-{1 \over 2}a\left(x-{J \over a}\right)^{2}\right]\,dx\\[8pt]&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}aw^{2}\right)\,dw\\[8pt]&=\left({2\pi \over a}\right)^{1 \over 2}\exp \left({J^{2} \over 2a}\right)\end{aligned}}}
Integrals with an imaginary linear term in the argument of the exponent
The integral
∫
−
∞
∞
exp
(
−
1
2
a
x
2
+
i
J
x
)
d
x
=
(
2
π
a
)
1
2
exp
(
−
J
2
2
a
)
{\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+iJx\right)dx=\left({2\pi \over a}\right)^{1 \over 2}\exp \left(-{J^{2} \over 2a}\right)}
is proportional to the
Fourier transform of the Gaussian where
J is the
conjugate variable of
x .
By again completing the square we see that the Fourier transform of a Gaussian is also a Gaussian, but in the conjugate variable. The larger a is, the narrower the Gaussian in x and the wider the Gaussian in J . This is a demonstration of the
uncertainty principle .
This integral is also known as the
Hubbard–Stratonovich transformation used in field theory.
Integrals with a complex argument of the exponent
The integral of interest is (for an example of an application see
Relation between Schrödinger's equation and the path integral formulation of quantum mechanics )
∫
−
∞
∞
exp
(
1
2
i
a
x
2
+
i
J
x
)
d
x
.
{\displaystyle \int _{-\infty }^{\infty }\exp \left({1 \over 2}iax^{2}+iJx\right)dx.}
We now assume that a and J may be complex.
Completing the square
(
1
2
i
a
x
2
+
i
J
x
)
=
1
2
i
a
(
x
2
+
2
J
x
a
+
(
J
a
)
2
−
(
J
a
)
2
)
=
−
1
2
a
i
(
x
+
J
a
)
2
−
i
J
2
2
a
.
{\displaystyle \left({1 \over 2}iax^{2}+iJx\right)={1 \over 2}ia\left(x^{2}+{2Jx \over a}+\left({J \over a}\right)^{2}-\left({J \over a}\right)^{2}\right)=-{1 \over 2}{a \over i}\left(x+{J \over a}\right)^{2}-{iJ^{2} \over 2a}.}
By analogy with the previous integrals
∫
−
∞
∞
exp
(
1
2
i
a
x
2
+
i
J
x
)
d
x
=
(
2
π
i
a
)
1
2
exp
(
−
i
J
2
2
a
)
.
{\displaystyle \int _{-\infty }^{\infty }\exp \left({1 \over 2}iax^{2}+iJx\right)dx=\left({2\pi i \over a}\right)^{1 \over 2}\exp \left({-iJ^{2} \over 2a}\right).}
This result is valid as an integration in the complex plane as long as a is non-zero and has a semi-positive imaginary part. See
Fresnel integral .
Gaussian integrals in higher dimensions
The one-dimensional integrals can be generalized to multiple dimensions.
[2]
∫
exp
(
−
1
2
x
⋅
A
⋅
x
+
J
⋅
x
)
d
n
x
=
(
2
π
)
n
det
A
exp
(
1
2
J
⋅
A
−
1
⋅
J
)
{\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+J\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\exp \left({1 \over 2}J\cdot A^{-1}\cdot J\right)}
Here A is a real positive definite
symmetric matrix .
This integral is performed by
diagonalization of A with an
orthogonal transformation
D
=
O
−
1
A
O
=
O
T
A
O
{\displaystyle D=O^{-1}AO=O^{\text{T}}AO}
where
D is a
diagonal matrix and
O is an
orthogonal matrix . This decouples the variables and allows the integration to be performed as
n one-dimensional integrations.
This is best illustrated with a two-dimensional example.
Example: Simple Gaussian integration in two dimensions
The Gaussian integral in two dimensions is
∫
exp
(
−
1
2
A
i
j
x
i
x
j
)
d
2
x
=
(
2
π
)
2
det
A
{\displaystyle \int \exp \left(-{\frac {1}{2}}A_{ij}x^{i}x^{j}\right)d^{2}x={\sqrt {\frac {(2\pi )^{2}}{\det A}}}}
where
A is a two-dimensional symmetric matrix with components specified as
A
=
a
c
c
b
{\displaystyle A={\begin{bmatrix}a&c\\c&b\end{bmatrix}}}
and we have used the
Einstein summation convention .
Diagonalize the matrix
The first step is to
diagonalize the matrix.
[3] Note that
A
i
j
x
i
x
j
≡
x
T
A
x
=
x
T
(
O
O
T
)
A
(
O
O
T
)
x
=
(
x
T
O
)
(
O
T
A
O
)
(
O
T
x
)
{\displaystyle A_{ij}x^{i}x^{j}\equiv x^{\text{T}}Ax=x^{\text{T}}\left(OO^{\text{T}}\right)A\left(OO^{\text{T}}\right)x=\left(x^{\text{T}}O\right)\left(O^{\text{T}}AO\right)\left(O^{\text{T}}x\right)}
where, since
A is a real
symmetric matrix , we can choose
O to be
orthogonal , and hence also a
unitary matrix .
O can be obtained from the
eigenvectors of
A . We choose
O such that:
D ≡ OT AO is diagonal.
Eigenvalues of A
To find the eigenvectors of A one first finds the
eigenvalues λ of A given by
a
c
c
b
u
v
=
λ
u
v
.
{\displaystyle {\begin{bmatrix}a&c\\c&b\end{bmatrix}}{\begin{bmatrix}u\\v\end{bmatrix}}=\lambda {\begin{bmatrix}u\\v\end{bmatrix}}.}
The eigenvalues are solutions of the
characteristic polynomial
(
a
−
λ
)
(
b
−
λ
)
−
c
2
=
0
{\displaystyle (a-\lambda )(b-\lambda )-c^{2}=0}
λ
2
−
λ
(
a
+
b
)
+
a
b
−
c
2
=
0
,
{\displaystyle \lambda ^{2}-\lambda (a+b)+ab-c^{2}=0,}
which are found using the
quadratic equation :
λ
±
=
1
2
(
a
+
b
)
±
1
2
(
a
+
b
)
2
−
4
(
a
b
−
c
2
)
.
=
1
2
(
a
+
b
)
±
1
2
a
2
+
2
a
b
+
b
2
−
4
a
b
+
4
c
2
.
=
1
2
(
a
+
b
)
±
1
2
(
a
−
b
)
2
+
4
c
2
.
{\displaystyle {\begin{aligned}\lambda _{\pm }&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {(a+b)^{2}-4(ab-c^{2})}}.\\&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {a^{2}+2ab+b^{2}-4ab+4c^{2}}}.\\&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {(a-b)^{2}+4c^{2}}}.\end{aligned}}}
Eigenvectors of A
Substitution of the eigenvalues back into the eigenvector equation yields
v
=
−
(
a
−
λ
±
)
u
c
,
v
=
−
c
u
(
b
−
λ
±
)
.
{\displaystyle v=-{\left(a-\lambda _{\pm }\right)u \over c},\qquad v=-{cu \over \left(b-\lambda _{\pm }\right)}.}
From the characteristic equation we know
a
−
λ
±
c
=
c
b
−
λ
±
.
{\displaystyle {a-\lambda _{\pm } \over c}={c \over b-\lambda _{\pm }}.}
Also note
a
−
λ
±
c
=
−
b
−
λ
∓
c
.
{\displaystyle {a-\lambda _{\pm } \over c}=-{b-\lambda _{\mp } \over c}.}
The eigenvectors can be written as:
1
η
−
a
−
λ
−
c
η
,
−
b
−
λ
+
c
η
1
η
{\displaystyle {\begin{bmatrix}{\frac {1}{\eta }}\\[1ex]-{\frac {a-\lambda _{-}}{c\eta }}\end{bmatrix}},\qquad {\begin{bmatrix}-{\frac {b-\lambda _{+}}{c\eta }}\\[1ex]{\frac {1}{\eta }}\end{bmatrix}}}
for the two eigenvectors. Here
η is a normalizing factor given by,
η
=
1
+
(
a
−
λ
−
c
)
2
=
1
+
(
b
−
λ
+
c
)
2
.
{\displaystyle \eta ={\sqrt {1+\left({\frac {a-\lambda _{-}}{c}}\right)^{2}}}={\sqrt {1+\left({\frac {b-\lambda _{+}}{c}}\right)^{2}}}.}
It is easily verified that the two eigenvectors are orthogonal to each other.
Construction of the orthogonal matrix
The orthogonal matrix is constructed by assigning the normalized eigenvectors as columns in the orthogonal matrix
O
=
1
η
−
b
−
λ
+
c
η
−
a
−
λ
−
c
η
1
η
.
{\displaystyle O={\begin{bmatrix}{\frac {1}{\eta }}&-{\frac {b-\lambda _{+}}{c\eta }}\\-{\frac {a-\lambda _{-}}{c\eta }}&{\frac {1}{\eta }}\end{bmatrix}}.}
Note that det(O ) = 1 .
If we define
sin
(
θ
)
=
−
a
−
λ
−
c
η
{\displaystyle \sin(\theta )=-{\frac {a-\lambda _{-}}{c\eta }}}
then the orthogonal matrix can be written
O
=
cos
(
θ
)
−
sin
(
θ
)
sin
(
θ
)
cos
(
θ
)
{\displaystyle O={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}}
which is simply a rotation of the eigenvectors with the inverse:
O
−
1
=
O
T
=
cos
(
θ
)
sin
(
θ
)
−
sin
(
θ
)
cos
(
θ
)
.
{\displaystyle O^{-1}=O^{\text{T}}={\begin{bmatrix}\cos(\theta )&\sin(\theta )\\-\sin(\theta )&\cos(\theta )\end{bmatrix}}.}
Diagonal matrix
The diagonal matrix becomes
D
=
O
T
A
O
=
λ
−
0
0
λ
+
{\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\[1ex]0&\lambda _{+}\end{bmatrix}}}
with eigenvectors
1
0
,
0
1
{\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}}
Numerical example
A
=
2
1
1
1
{\displaystyle A={\begin{bmatrix}2&1\\1&1\end{bmatrix}}}
The eigenvalues are
λ
±
=
3
2
±
5
2
.
{\displaystyle \lambda _{\pm }={3 \over 2}\pm {{\sqrt {5}} \over 2}.}
The eigenvectors are
1
η
1
−
1
2
−
5
2
,
1
η
1
2
+
5
2
1
{\displaystyle {1 \over \eta }{\begin{bmatrix}1\\[1ex]-{1 \over 2}-{{\sqrt {5}} \over 2}\end{bmatrix}},\qquad {1 \over \eta }{\begin{bmatrix}{1 \over 2}+{{\sqrt {5}} \over 2}\\[1ex]1\end{bmatrix}}}
where
η
=
5
2
+
5
2
.
{\displaystyle \eta ={\sqrt {{5 \over 2}+{{\sqrt {5}} \over 2}}}.}
Then
O
=
1
η
1
η
(
1
2
+
5
2
)
1
η
(
−
1
2
−
5
2
)
1
η
O
−
1
=
1
η
1
η
(
−
1
2
−
5
2
)
1
η
(
1
2
+
5
2
)
1
η
{\displaystyle {\begin{aligned}O&={\begin{bmatrix}{\frac {1}{\eta }}&{\frac {1}{\eta }}\left({1 \over 2}+{{\sqrt {5}} \over 2}\right)\\{\frac {1}{\eta }}\left(-{1 \over 2}-{{\sqrt {5}} \over 2}\right)&{1 \over \eta }\end{bmatrix}}\\O^{-1}&={\begin{bmatrix}{\frac {1}{\eta }}&{\frac {1}{\eta }}\left(-{1 \over 2}-{{\sqrt {5}} \over 2}\right)\\{\frac {1}{\eta }}\left({1 \over 2}+{{\sqrt {5}} \over 2}\right)&{\frac {1}{\eta }}\end{bmatrix}}\end{aligned}}}
The diagonal matrix becomes
D
=
O
T
A
O
=
λ
−
0
0
λ
+
=
3
2
−
5
2
0
0
3
2
+
5
2
{\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\0&\lambda _{+}\end{bmatrix}}={\begin{bmatrix}{3 \over 2}-{{\sqrt {5}} \over 2}&0\\0&{3 \over 2}+{{\sqrt {5}} \over 2}\end{bmatrix}}}
with eigenvectors
1
0
,
0
1
{\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}}
Rescale the variables and integrate
With the diagonalization the integral can be written
∫
exp
(
−
1
2
x
T
A
x
)
d
2
x
=
∫
exp
(
−
1
2
∑
j
=
1
2
λ
j
y
j
2
)
d
2
y
{\displaystyle \int \exp \left(-{\frac {1}{2}}x^{\text{T}}Ax\right)d^{2}x=\int \exp \left(-{\frac {1}{2}}\sum _{j=1}^{2}\lambda _{j}y_{j}^{2}\right)\,d^{2}y}
where
y
=
O
T
x
.
{\displaystyle y=O^{\text{T}}x.}
Since the coordinate transformation is simply a rotation of coordinates the
Jacobian determinant of the transformation is one yielding
d
2
y
=
d
2
x
{\displaystyle d^{2}y=d^{2}x}
The integrations can now be performed:
∫
exp
(
−
1
2
x
T
A
x
)
d
2
x
=
∫
exp
(
−
1
2
∑
j
=
1
2
λ
j
y
j
2
)
d
2
y
=
∏
j
=
1
2
(
2
π
λ
j
)
1
/
2
=
(
(
2
π
)
2
∏
j
=
1
2
λ
j
)
1
/
2
=
(
(
2
π
)
2
det
(
O
−
1
A
O
)
)
1
/
2
=
(
(
2
π
)
2
det
(
A
)
)
1
/
2
{\displaystyle {\begin{aligned}\int \exp \left(-{\frac {1}{2}}x^{\mathsf {T}}Ax\right)d^{2}x={}&\int \exp \left(-{\frac {1}{2}}\sum _{j=1}^{2}\lambda _{j}y_{j}^{2}\right)d^{2}y\\[1ex]={}&\prod _{j=1}^{2}\left({2\pi \over \lambda _{j}}\right)^{1/2}\\={}&\left({(2\pi )^{2} \over \prod _{j=1}^{2}\lambda _{j}}\right)^{1/2}\\[1ex]={}&\left({(2\pi )^{2} \over \det {\left(O^{-1}AO\right)}}\right)^{1/2}\\[1ex]={}&\left({(2\pi )^{2} \over \det {\left(A\right)}}\right)^{1/2}\end{aligned}}}
which is the advertised solution.
Integrals with complex and linear terms in multiple dimensions
With the two-dimensional example it is now easy to see the generalization to the complex plane and to multiple dimensions.
Integrals with a linear term in the argument
∫
exp
(
−
1
2
x
⋅
A
⋅
x
+
J
⋅
x
)
d
n
x
=
(
2
π
)
n
det
A
exp
(
1
2
J
⋅
A
−
1
⋅
J
)
{\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+J\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\exp \left({1 \over 2}J\cdot A^{-1}\cdot J\right)}
Integrals with an imaginary linear term
∫
exp
(
−
1
2
x
⋅
A
⋅
x
+
i
J
⋅
x
)
d
n
x
=
(
2
π
)
n
det
A
exp
(
−
1
2
J
⋅
A
−
1
⋅
J
)
{\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+iJ\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\exp \left(-{1 \over 2}J\cdot A^{-1}\cdot J\right)}
Integrals with a complex quadratic term
∫
exp
(
i
2
x
⋅
A
⋅
x
+
i
J
⋅
x
)
d
n
x
=
(
2
π
i
)
n
det
A
exp
(
−
i
2
J
⋅
A
−
1
⋅
J
)
{\displaystyle \int \exp \left({\frac {i}{2}}x\cdot A\cdot x+iJ\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi i)^{n}}{\det A}}}\exp \left(-{i \over 2}J\cdot A^{-1}\cdot J\right)}
Integrals with differential operators in the argument
As an example consider the integral
[1] : 21‒22
∫
exp
∫
d
4
x
(
−
1
2
φ
A
^
φ
+
J
φ
)
D
φ
{\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi }
where
A
^
{\displaystyle {\hat {A}}}
is a differential operator with
φ
{\displaystyle \varphi }
and
J functions of
spacetime , and
D
φ
{\displaystyle D\varphi }
indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is
∫
exp
∫
d
4
x
(
−
1
2
φ
A
^
φ
+
J
φ
)
D
φ
∝
exp
(
1
2
∫
d
4
x
d
4
y
J
(
x
)
D
(
x
−
y
)
J
(
y
)
)
{\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi \;\propto \;\exp \left({1 \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right)}
where
A
^
D
(
x
−
y
)
=
δ
4
(
x
−
y
)
{\displaystyle {\hat {A}}D(x-y)=\delta ^{4}(x-y)}
and
D (x − y ), called the
propagator , is the inverse of
A
^
{\displaystyle {\hat {A}}}
, and
δ
4
(
x
−
y
)
{\displaystyle \delta ^{4}(x-y)}
is the
Dirac delta function .
Similar arguments yield
∫
exp
∫
d
4
x
(
−
1
2
φ
A
^
φ
+
i
J
φ
)
D
φ
∝
exp
(
−
1
2
∫
d
4
x
d
4
y
J
(
x
)
D
(
x
−
y
)
J
(
y
)
)
,
{\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +iJ\varphi \right)\right]D\varphi \;\propto \;\exp \left(-{1 \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right),}
and
∫
exp
i
∫
d
4
x
(
1
2
φ
A
^
φ
+
J
φ
)
D
φ
∝
exp
(
−
i
2
∫
d
4
x
d
4
y
J
(
x
)
D
(
x
−
y
)
J
(
y
)
)
.
{\displaystyle \int \exp \left[i\int d^{4}x\left({\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi \;\propto \;\exp \left(-{i \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right).}
See
Path-integral formulation of virtual-particle exchange for an application of this integral.
Integrals that can be approximated by the method of steepest descent
In quantum field theory n-dimensional integrals of the form
∫
−
∞
∞
exp
(
−
1
ℏ
f
(
q
)
)
d
n
q
{\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over \hbar }f(q)\right)d^{n}q}
appear often. Here
ℏ
{\displaystyle \hbar }
is the
reduced Planck constant and
f is a function with a positive minimum at
q
=
q
0
{\displaystyle q=q_{0}}
. These integrals can be approximated by the
method of steepest descent .
For small values of the Planck constant, f can be expanded about its minimum
∫
−
∞
∞
exp
−
1
ℏ
(
f
(
q
0
)
+
1
2
(
q
−
q
0
)
2
f
′
′
(
q
−
q
0
)
+
⋯
)
d
n
q
.
{\displaystyle \int _{-\infty }^{\infty }\exp \left[-{1 \over \hbar }\left(f\left(q_{0}\right)+{1 \over 2}\left(q-q_{0}\right)^{2}f^{\prime \prime }\left(q-q_{0}\right)+\cdots \right)\right]d^{n}q.}
Here
f
′
′
{\displaystyle f^{\prime \prime }}
is the n by n matrix of second derivatives evaluated at the minimum of the function.
If we neglect higher order terms this integral can be integrated explicitly.
∫
−
∞
∞
exp
−
1
ℏ
(
f
(
q
)
)
d
n
q
≈
exp
−
1
ℏ
(
f
(
q
0
)
)
(
2
π
ℏ
)
n
det
f
′
′
.
{\displaystyle \int _{-\infty }^{\infty }\exp \left[-{1 \over \hbar }(f(q))\right]d^{n}q\approx \exp \left[-{1 \over \hbar }\left(f\left(q_{0}\right)\right)\right]{\sqrt {(2\pi \hbar )^{n} \over \det f^{\prime \prime }}}.}
Integrals that can be approximated by the method of stationary phase
A common integral is a path integral of the form
∫
exp
(
i
ℏ
S
(
q
,
q
˙
)
)
D
q
{\displaystyle \int \exp \left({i \over \hbar }S\left(q,{\dot {q}}\right)\right)Dq}
where
S
(
q
,
q
˙
)
{\displaystyle S\left(q,{\dot {q}}\right)}
is the classical
action and the integral is over all possible paths that a particle may take. In the limit of small
ℏ
{\displaystyle \hbar }
the integral can be evaluated in the
stationary phase approximation . In this approximation the integral is over the path in which the action is a minimum. Therefore, this approximation recovers the
classical limit of
mechanics .
Fourier integrals
Dirac delta distribution
The
Dirac delta distribution in
spacetime can be written as a
Fourier transform
[1] : 23
∫
d
4
k
(
2
π
)
4
exp
(
i
k
(
x
−
y
)
)
=
δ
4
(
x
−
y
)
.
{\displaystyle \int {\frac {d^{4}k}{(2\pi )^{4}}}\exp(ik(x-y))=\delta ^{4}(x-y).}
In general, for any dimension
N
{\displaystyle N}
∫
d
N
k
(
2
π
)
N
exp
(
i
k
(
x
−
y
)
)
=
δ
N
(
x
−
y
)
.
{\displaystyle \int {\frac {d^{N}k}{(2\pi )^{N}}}\exp(ik(x-y))=\delta ^{N}(x-y).}
Fourier integrals of forms of the Coulomb potential
Laplacian of 1/r
While not an integral, the identity in three-dimensional
Euclidean space
−
1
4
π
∇
2
(
1
r
)
=
δ
(
r
)
{\displaystyle -{1 \over 4\pi }\nabla ^{2}\left({1 \over r}\right)=\delta \left(\mathbf {r} \right)}
where
r
2
=
r
⋅
r
{\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} }
is a consequence of
Gauss's theorem and can be used to derive integral identities. For an example see
Longitudinal and transverse vector fields .
This identity implies that the
Fourier integral representation of 1/r is
∫
d
3
k
(
2
π
)
3
exp
(
i
k
⋅
r
)
k
2
=
1
4
π
r
.
{\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}}={1 \over 4\pi r}.}
Yukawa potential: the Coulomb potential with mass
The
Yukawa potential in three dimensions can be represented as an integral over a
Fourier transform
[1] : 26, 29
∫
d
3
k
(
2
π
)
3
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
e
−
m
r
4
π
r
{\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}+m^{2}}={e^{-mr} \over 4\pi r}}
where
r
2
=
r
⋅
r
,
k
2
=
k
⋅
k
.
{\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} ,\qquad k^{2}=\mathbf {k} \cdot \mathbf {k} .}
See
Static forces and virtual-particle exchange for an application of this integral.
In the small m limit the integral reduces to 1 / 4πr .
To derive this result note:
∫
d
3
k
(
2
π
)
3
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
∫
0
∞
k
2
d
k
(
2
π
)
2
∫
−
1
1
d
u
e
i
k
r
u
k
2
+
m
2
=
2
r
∫
0
∞
k
d
k
(
2
π
)
2
sin
(
k
r
)
k
2
+
m
2
=
1
i
r
∫
−
∞
∞
k
d
k
(
2
π
)
2
e
i
k
r
k
2
+
m
2
=
1
i
r
∫
−
∞
∞
k
d
k
(
2
π
)
2
e
i
k
r
(
k
+
i
m
)
(
k
−
i
m
)
=
1
i
r
2
π
i
(
2
π
)
2
i
m
2
i
m
e
−
m
r
=
1
4
π
r
e
−
m
r
{\displaystyle {\begin{aligned}\int {\frac {d^{3}k}{(2\pi )^{3}}}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}={}&\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}du{e^{ikru} \over k^{2}+m^{2}}\\[1ex]={}&{2 \over r}\int _{0}^{\infty }{\frac {kdk}{(2\pi )^{2}}}{\sin(kr) \over k^{2}+m^{2}}\\[1ex]={}&{1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over k^{2}+m^{2}}\\[1ex]={}&{1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over (k+im)(k-im)}\\[1ex]={}&{1 \over ir}{\frac {2\pi i}{(2\pi )^{2}}}{\frac {im}{2im}}e^{-mr}\\[1ex]={}&{\frac {1}{4\pi r}}e^{-mr}\end{aligned}}}
Modified Coulomb potential with mass
∫
d
3
k
(
2
π
)
3
(
k
^
⋅
r
^
)
2
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
e
−
m
r
4
π
r
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
{\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {r}} \right)^{2}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}={\frac {e^{-mr}}{4\pi r}}\left[1+{\frac {2}{mr}}-{\frac {2}{(mr)^{2}}}\left(e^{mr}-1\right)\right]}
where the hat indicates a unit vector in three dimensional space. The derivation of this result is as follows:
∫
d
3
k
(
2
π
)
3
(
k
^
⋅
r
^
)
2
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
∫
0
∞
k
2
d
k
(
2
π
)
2
∫
−
1
1
d
u
u
2
e
i
k
r
u
k
2
+
m
2
=
2
∫
0
∞
k
2
d
k
(
2
π
)
2
1
k
2
+
m
2
1
k
r
sin
(
k
r
)
+
2
(
k
r
)
2
cos
(
k
r
)
−
2
(
k
r
)
3
sin
(
k
r
)
=
e
−
m
r
4
π
r
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
{\displaystyle {\begin{aligned}&\int {\frac {d^{3}k}{(2\pi )^{3}}}\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {r}} \right)^{2}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}\\[1ex]&=\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}du\ u^{2}{\frac {e^{ikru}}{k^{2}+m^{2}}}\\[1ex]&=2\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}{\frac {1}{k^{2}+m^{2}}}\left[{\frac {1}{kr}}\sin(kr)+{\frac {2}{(kr)^{2}}}\cos(kr)-{\frac {2}{(kr)^{3}}}\sin(kr)\right]\\[1ex]&={\frac {e^{-mr}}{4\pi r}}\left[1+{\frac {2}{mr}}-{\frac {2}{(mr)^{2}}}\left(e^{mr}-1\right)\right]\end{aligned}}}
Note that in the small m limit the integral goes to the result for the Coulomb potential since the term in the brackets goes to 1 .
Longitudinal potential with mass
∫
d
3
k
(
2
π
)
3
k
^
k
^
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
1
2
e
−
m
r
4
π
r
(
1
−
r
^
r
^
+
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
1
+
r
^
r
^
)
{\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\mathbf {\hat {k}} \mathbf {\hat {k}} {\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}={1 \over 2}{\frac {e^{-mr}}{4\pi r}}\left(\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]+\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right)}
where the hat indicates a unit vector in three dimensional space. The derivation for this result is as follows:
∫
d
3
k
(
2
π
)
3
k
^
k
^
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
∫
d
3
k
(
2
π
)
3
(
k
^
⋅
r
^
)
2
r
^
r
^
+
(
k
^
⋅
θ
^
)
2
θ
^
θ
^
+
(
k
^
⋅
ϕ
^
)
2
ϕ
^
ϕ
^
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
e
−
m
r
4
π
r
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
{
1
−
1
2
1
−
r
^
r
^
}
+
∫
0
∞
k
2
d
k
(
2
π
)
2
∫
−
1
1
d
u
e
i
k
r
u
k
2
+
m
2
1
2
1
−
r
^
r
^
=
1
2
e
−
m
r
4
π
r
1
−
r
^
r
^
+
e
−
m
r
4
π
r
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
{
1
2
1
+
r
^
r
^
}
=
1
2
e
−
m
r
4
π
r
(
1
−
r
^
r
^
+
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
1
+
r
^
r
^
)
{\displaystyle {\begin{aligned}&\int {\frac {d^{3}k}{(2\pi )^{3}}}\mathbf {\hat {k}} \mathbf {\hat {k}} {\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}\\[1ex]&=\int {\frac {d^{3}k}{(2\pi )^{3}}}\left[\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {r}} \right)^{2}\mathbf {\hat {r}} \mathbf {\hat {r}} +\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {\theta }} \right)^{2}\mathbf {\hat {\theta }} \mathbf {\hat {\theta }} +\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {\phi }} \right)^{2}\mathbf {\hat {\phi }} \mathbf {\hat {\phi }} \right]{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}\\[1ex]&={\frac {e^{-mr}}{4\pi r}}\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left\{\mathbf {1} -{1 \over 2}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right\}+\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}du{\frac {e^{ikru}}{k^{2}+m^{2}}}{1 \over 2}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\\[1ex]&={1 \over 2}{\frac {e^{-mr}}{4\pi r}}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]+{e^{-mr} \over 4\pi r}\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left\{{1 \over 2}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right\}\\[1ex]&={1 \over 2}{\frac {e^{-mr}}{4\pi r}}\left(\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]+\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right)\end{aligned}}}
Note that in the small m limit the integral reduces to
1
2
1
4
π
r
1
−
r
^
r
^
.
{\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}
Transverse potential with mass
∫
d
3
k
(
2
π
)
3
1
−
k
^
k
^
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
1
2
e
−
m
r
4
π
r
{
2
(
m
r
)
2
(
e
m
r
−
1
)
−
2
m
r
}
1
+
r
^
r
^
{\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\left[\mathbf {1} -\mathbf {\hat {k}} \mathbf {\hat {k}} \right]{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}+m^{2}}={1 \over 2}{e^{-mr} \over 4\pi r}\left\{{2 \over (mr)^{2}}\left(e^{mr}-1\right)-{2 \over mr}\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]}
In the small mr limit the integral goes to
1
2
1
4
π
r
1
+
r
^
r
^
.
{\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}
For large distance, the integral falls off as the inverse cube of r
1
4
π
m
2
r
3
1
+
r
^
r
^
.
{\displaystyle {\frac {1}{4\pi m^{2}r^{3}}}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}
For applications of this integral see
Darwin Lagrangian and
Darwin interaction in a vacuum .
Angular integration in cylindrical coordinates
There are two important integrals. The angular integration of an exponential in cylindrical coordinates can be written in terms of Bessel functions of the first kind
[4]
[5] : 113
∫
0
2
π
d
φ
2
π
exp
(
i
p
cos
(
φ
)
)
=
J
0
(
p
)
{\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\exp \left(ip\cos(\varphi )\right)=J_{0}(p)}
and
∫
0
2
π
d
φ
2
π
cos
(
φ
)
exp
(
i
p
cos
(
φ
)
)
=
i
J
1
(
p
)
.
{\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\cos(\varphi )\exp \left(ip\cos(\varphi )\right)=iJ_{1}(p).}
For applications of these integrals see
Magnetic interaction between current loops in a simple plasma or electron gas .
Bessel functions
Integration of the cylindrical propagator with mass
First power of a Bessel function
∫
0
∞
k
d
k
k
2
+
m
2
J
0
(
k
r
)
=
K
0
(
m
r
)
.
{\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{0}\left(kr\right)=K_{0}(mr).}
See Abramowitz and Stegun.
[6] : §11.4.44
For
m
r
≪
1
{\displaystyle mr\ll 1}
, we have
[5] : 116
K
0
(
m
r
)
→
−
ln
(
m
r
2
)
+
0.5772.
{\displaystyle K_{0}(mr)\to -\ln \left({mr \over 2}\right)+0.5772.}
For an application of this integral see
Two line charges embedded in a plasma or electron gas .
Squares of Bessel functions
The integration of the propagator in cylindrical coordinates is
[4]
∫
0
∞
k
d
k
k
2
+
m
2
J
1
2
(
k
r
)
=
I
1
(
m
r
)
K
1
(
m
r
)
.
{\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)=I_{1}(mr)K_{1}(mr).}
For small mr the integral becomes
∫
o
∞
k
d
k
k
2
+
m
2
J
1
2
(
k
r
)
→
1
2
1
−
1
8
(
m
r
)
2
.
{\displaystyle \int _{o}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)\to {1 \over 2}\left[1-{1 \over 8}(mr)^{2}\right].}
For large mr the integral becomes
∫
o
∞
k
d
k
k
2
+
m
2
J
1
2
(
k
r
)
→
1
2
(
1
m
r
)
.
{\displaystyle \int _{o}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)\to {1 \over 2}\left({1 \over mr}\right).}
For applications of this integral see
Magnetic interaction between current loops in a simple plasma or electron gas .
In general,
∫
0
∞
k
d
k
k
2
+
m
2
J
ν
2
(
k
r
)
=
I
ν
(
m
r
)
K
ν
(
m
r
)
ℜ
(
ν
)
>
−
1.
{\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{\nu }^{2}(kr)=I_{\nu }(mr)K_{\nu }(mr)\qquad \Re (\nu )>-1.}
Integration over a magnetic wave function
The two-dimensional integral over a magnetic wave function is
[6] : §11.4.28
2
a
2
n
+
2
n
!
∫
0
∞
d
r
r
2
n
+
1
exp
(
−
a
2
r
2
)
J
0
(
k
r
)
=
M
(
n
+
1
,
1
,
−
k
2
4
a
2
)
.
{\displaystyle {2a^{2n+2} \over n!}\int _{0}^{\infty }{dr}\;r^{2n+1}\exp \left(-a^{2}r^{2}\right)J_{0}(kr)=M\left(n+1,1,-{k^{2} \over 4a^{2}}\right).}
Here, M is a
confluent hypergeometric function . For an application of this integral see
Charge density spread over a wave function .
See also
References
^
a
b
c
d A. Zee (2003). Quantum Field Theory in a Nutshell . Princeton University.
ISBN
0-691-01019-6 .
^ Frederick W. Byron and Robert W. Fuller (1969). Mathematics of Classical and Quantum Physics . Addison-Wesley.
ISBN
0-201-00746-0 .
^ Herbert S. Wilf (1978).
Mathematics for the Physical Sciences . Dover.
ISBN
0-486-63635-6 .
^
a
b
Gradshteyn, Izrail Solomonovich ;
Ryzhik, Iosif Moiseevich ;
Geronimus, Yuri Veniaminovich ;
Tseytlin, Michail Yulyevich ; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel;
Moll, Victor Hugo (eds.).
Table of Integrals, Series, and Products . Translated by Scripta Technica, Inc. (8 ed.).
Academic Press, Inc.
ISBN
978-0-12-384933-5 .
LCCN
2014010276 .
^
a
b Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley.
ISBN
0-471-30932-X .
^
a
b M. Abramowitz; I. Stegun (1965).
Handbook of Mathematical Functions . Dover.
ISBN
0486-61272-4 .