Английская Википедия:Common integrals in quantum field theory
Common integrals in quantum field theory are all variations and generalizations of Gaussian integrals to the complex plane and to multiple dimensions.[1]Шаблон:Rp 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. <math display="block"> G \equiv \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx</math>
In physics the factor of 1/2 in the argument of the exponential is common.
Note: <math display="block"> 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} r e^{-{1 \over 2} r^2}\,dr = 2\pi \int_{0}^{\infty} e^{- w}\,dw = 2 \pi.</math>
Thus we obtain <math display="block"> \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx = \sqrt{2\pi}. </math>
Slight generalization of the Gaussian integral
<math display="block"> \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = \sqrt{2\pi \over a} </math> where we have scaled <math display="block"> x \to {x \over \sqrt{a}}. </math>
Integrals of exponents and even powers of x
<math display="block"> \int_{-\infty}^{\infty} x^2 e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^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}</math> and <math display="block"> \int_{-\infty}^{\infty} x^4 e^{-{1 \over 2} a x^2}\,dx = \left ( -2{d\over da} \right) \left ( -2{d\over da} \right) \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^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}</math>
In general <math display="block"> \int_{-\infty}^{\infty} x^{2n} e^{-{1 \over 2} a x^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 )!! </math>
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
<math display="block"> \int_{-\infty}^{\infty} \exp\left( - \frac 1 2 a x^2 + Jx\right ) dx </math>
This integral can be performed by completing the square: <math display="block"> \left( -{1 \over 2} a x^2 + Jx\right ) = -{1 \over 2} a \left ( x^2 - { 2 Jx \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 } </math>
Therefore: <math display="block">\begin{align} \int_{-\infty}^\infty \exp\left( -{1 \over 2} a x^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} a w^2 \right) \, dw \\[8pt] &= \left ( {2\pi \over a } \right ) ^{1\over 2} \exp\left( { J^2 \over 2a }\right ) \end{align}</math>
Integrals with an imaginary linear term in the argument of the exponent
The integral
<math display="block"> \int_{-\infty}^{\infty} \exp\left( -{1 \over 2} a x^2 +iJx\right ) dx = \left ( {2\pi \over a } \right ) ^{1\over 2} \exp\left( -{ J^2 \over 2a }\right ) </math>
is proportional to the Fourier transform of the Gaussian where Шаблон:Mvar is the conjugate variable of Шаблон:Mvar.
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 Шаблон:Mvar is, the narrower the Gaussian in Шаблон:Mvar and the wider the Gaussian in Шаблон:Mvar. 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)
<math display="block"> \int_{-\infty}^{\infty} \exp\left( {1 \over 2} i a x^2 + iJx\right ) dx. </math>
We now assume that Шаблон:Mvar and Шаблон:Mvar may be complex.
Completing the square
<math display="block"> \left( {1 \over 2} i a x^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}. </math>
By analogy with the previous integrals
<math display="block"> \int_{-\infty}^{\infty} \exp\left( {1 \over 2} i a x^2 + iJx\right ) dx = \left ( {2\pi i \over a } \right ) ^{1\over 2} \exp\left( { -iJ^2 \over 2a }\right ). </math>
This result is valid as an integration in the complex plane as long as Шаблон:Mvar 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]
<math display="block">\int \exp\left( - \frac 1 2 x \cdot A \cdot x +J \cdot x \right) d^nx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( {1\over 2} J \cdot A^{-1} \cdot J \right)</math>
Here Шаблон:Mvar is a real positive definite symmetric matrix.
This integral is performed by diagonalization of Шаблон:Mvar with an orthogonal transformation
<math display="block">D= O^{-1} A O = O^T A O</math>
where Шаблон:Mvar is a diagonal matrix and Шаблон:Mvar is an orthogonal matrix. This decouples the variables and allows the integration to be performed as Шаблон:Mvar 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
<math display="block">\int \exp\left( - \frac 1 2 A_{ij} x^i x^j \right) d^2x = \sqrt{\frac{(2\pi)^2}{\det A}}</math>
where Шаблон:Mvar is a two-dimensional symmetric matrix with components specified as
<math display="block"> A = \begin{bmatrix} a&c\\ c&b\end{bmatrix}</math>
and we have used the Einstein summation convention.
Diagonalize the matrix
The first step is to diagonalize the matrix.[3] Note that
<math display="block">A_{ij} x^i x^j \equiv x^TAx = x^T \left(OO^T\right) A \left(OO^T\right) x = \left(x^TO \right) \left(O^TAO \right) \left(O^Tx \right) </math>
where, since Шаблон:Mvar is a real symmetric matrix, we can choose Шаблон:Mvar to be orthogonal, and hence also a unitary matrix. Шаблон:Mvar can be obtained from the eigenvectors of Шаблон:Mvar. We choose Шаблон:Mvar such that: Шаблон:Math is diagonal.
Eigenvalues of A
To find the eigenvectors of Шаблон:Mvar one first finds the eigenvalues Шаблон:Mvar of Шаблон:Mvar given by
<math display="block"> \begin{bmatrix}a&c\\ c&b\end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}=\lambda \begin{bmatrix}u\\ v\end{bmatrix}.</math>
The eigenvalues are solutions of the characteristic polynomial
<math display="block">( a - \lambda)( b-\lambda) -c^2 = 0 </math> <math display="block">\lambda^2 - \lambda(a+b) + ab -c^2 = 0 </math>
which are found using the quadratic equation: <math display="block">\begin{align}
\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{align}</math>
Eigenvectors of A
Substitution of the eigenvalues back into the eigenvector equation yields
<math display="block"> v = -{ \left( a - \lambda_{\pm} \right)u \over c }, \qquad v = -{cu \over \left( b - \lambda_{\pm} \right)}.</math>
From the characteristic equation we know
<math display="block"> {a - \lambda_{\pm} \over c } = {c \over b - \lambda_{\pm}}.</math>
Also note
<math display="block"> { a - \lambda_{\pm} \over c } = -{ b - \lambda_{\mp} \over c}. </math>
The eigenvectors can be written as:
<math display="block">\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} </math>
for the two eigenvectors. Here Шаблон:Mvar is a normalizing factor given by,
<math display="block">\eta = \sqrt{1 + \left(\frac{a -\lambda_{-}}{c} \right)^2 } = \sqrt{1 + \left(\frac{b - \lambda_{+}}{c} \right)^2}.</math>
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
<math display="block">O = \begin{bmatrix} \frac{1}{\eta} & -\frac{b - \lambda_{+}}{c \eta} \\ -\frac{a - \lambda_{-}}{c \eta} &\frac{1}{\eta}\end{bmatrix}.</math>
Note that Шаблон:Math.
If we define
<math display="block"> \sin(\theta) = -\frac{a - \lambda_{-}}{c \eta }</math>
then the orthogonal matrix can be written
<math display="block">O = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix} </math>
which is simply a rotation of the eigenvectors with the inverse:
<math display="block">O^{-1} = O^T = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}.</math>
Diagonal matrix
The diagonal matrix becomes
<math display="block"> D = O^T A O = \begin{bmatrix}\lambda_{-}& 0 \\[1ex] 0 & \lambda_{+}\end{bmatrix}</math>
with eigenvectors
<math display="block">\begin{bmatrix} 1\\ 0\end{bmatrix}, \qquad \begin{bmatrix} 0\\ 1 \end{bmatrix}</math>
Numerical example
<math display="block">A = \begin{bmatrix} 2&1\\ 1 & 1\end{bmatrix}</math>
The eigenvalues are
<math display="block">\lambda_{\pm} = {3\over 2} \pm {\sqrt{ 5} \over 2}.</math>
The eigenvectors are
<math display="block">{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}</math>
where
<math display="block">\eta = \sqrt{{5\over 2} + {\sqrt{5}\over 2}}.</math>
Then
<math display="block">\begin{align} 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{align}</math>
The diagonal matrix becomes
<math display="block">D = O^TAO = \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} </math>
with eigenvectors
<math display="block">\begin{bmatrix} 1\\ 0\end{bmatrix}, \qquad \begin{bmatrix} 0\\ 1 \end{bmatrix}</math>
Rescale the variables and integrate
With the diagonalization the integral can be written
<math display="block">\int \exp\left( - \frac 1 2 x^T A x \right) d^2x = \int \exp\left( - \frac 1 2 \sum_{j=1}^2 \lambda_{j} y_j^2 \right) \, d^2y</math>
where
<math display="block">y = O^T x.</math>
Since the coordinate transformation is simply a rotation of coordinates the Jacobian determinant of the transformation is one yielding
<math display="block"> d^2y = d^2x </math>
The integrations can now be performed.
<math display="block">\begin{align} \int \exp\left( - \frac{1}{2} x^\mathsf{T} A x \right) d^2x ={}& \int \exp\left( - \frac 1 2 \sum_{j=1}^2 \lambda_{j} y_j^2 \right) d^2y \\[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{align}</math>
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
<math display="block">\int \exp\left(-\frac{1}{2} x \cdot A \cdot x +J \cdot x \right) d^nx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( {1\over 2} J \cdot A^{-1} \cdot J \right)</math>
Integrals with an imaginary linear term
<math display="block">\int \exp\left(-\frac{1}{2} x \cdot A \cdot x +iJ \cdot x \right) d^nx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( -{1\over 2} J \cdot A^{-1} \cdot J \right)</math>
Integrals with a complex quadratic term
<math display="block">\int \exp\left(\frac{i}{2} x \cdot A \cdot x +iJ \cdot x \right) d^nx =\sqrt{\frac{(2\pi i)^n}{\det A}} \exp \left( -{i\over 2} J \cdot A^{-1} \cdot J \right)</math>
Integrals with differential operators in the argument
As an example consider the integral[1]Шаблон:Rp
<math display="block">\int \exp\left[ \int d^4x \left (-\frac{1}{2} \varphi \hat A \varphi + J \varphi \right) \right ] D\varphi</math>
where <math> \hat A </math> is a differential operator with <math> \varphi </math> and Шаблон:Mvar functions of spacetime, and <math> D\varphi </math> indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is
<math display="block">\int \exp\left[ \int d^4x \left (-\frac 1 2 \varphi \hat A \varphi +J\varphi \right) \right ] D\varphi \; \propto \; \exp \left( {1\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right)</math>
where
<math display="block">\hat A D( x - y) = \delta^4 ( x - y)</math>
and Шаблон:Math, called the propagator, is the inverse of <math> \hat A</math>, and <math> \delta^4( x - y)</math> is the Dirac delta function.
Similar arguments yield
<math display="block">\int \exp\left[\int d^4x \left (-\frac 1 2 \varphi \hat A \varphi + i J \varphi \right) \right ] D\varphi \; \propto \; \exp \left( - { 1\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right),</math>
and
<math display="block">\int \exp\left[ i \int d^4x \left ( \frac 1 2 \varphi \hat A \varphi + J\varphi \right) \right ] D\varphi \; \propto \; \exp \left( -{ i\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right).</math>
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
<math display="block">\int_{-\infty}^{\infty} \exp\left( -{1 \over \hbar} f(q) \right ) d^nq</math>
appear often. Here <math>\hbar</math> is the reduced Planck's constant and f is a function with a positive minimum at <math> q=q_0</math>. These integrals can be approximated by the method of steepest descent.
For small values of Planck's constant, f can be expanded about its minimum
<math display="block">\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^nq.</math>Here <math> f^{\prime \prime} </math> 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.
<math display="block">\int_{-\infty}^{\infty} \exp\left[ -{1 \over \hbar} (f(q)) \right] d^nq \approx \exp\left[ -{1 \over \hbar} \left( f\left( q_0 \right) \right ) \right] \sqrt{ (2 \pi \hbar )^n \over \det f^{\prime \prime} }.</math>
Integrals that can be approximated by the method of stationary phase
A common integral is a path integral of the form
<math display="block"> \int \exp\left( {i \over \hbar} S\left( q, \dot q \right) \right ) Dq </math>
where <math> S\left( q, \dot q \right) </math> is the classical action and the integral is over all possible paths that a particle may take. In the limit of small <math> \hbar </math> 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]Шаблон:Rp
<math display="block"> \int \frac{d^4 k}{(2\pi)^4} \exp(ik ( x-y)) = \delta^4 ( x-y).</math>
In general, for any dimension <math> N </math>
<math display="block"> \int \frac{d^N k}{(2\pi)^N} \exp(ik ( x-y)) = \delta^N ( x-y).</math>
Fourier integrals of forms of the Coulomb potential
Laplacian of 1/r
While not an integral, the identity in three-dimensional Euclidean space
<math display="block">-{1 \over 4\pi} \nabla^2 \left( {1 \over r} \right) = \delta \left( \mathbf r \right) </math>where<math display="block">r^2 = \mathbf r \cdot \mathbf r</math>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
<math display="block">\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 }.</math>
Yukawa Potential: The Coulomb potential with mass
The Yukawa potential in three dimensions can be represented as an integral over a Fourier transform[1]Шаблон:Rp
<math display="block">\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 } </math>
where
<math display="block">r^2 = \mathbf{r} \cdot \mathbf r, \qquad k^2 = \mathbf k \cdot \mathbf k.</math>
See Static forces and virtual-particle exchange for an application of this integral.
In the small m limit the integral reduces to Шаблон:Math.
To derive this result note:
<math display="block">\begin{align} \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{k dk}{(2\pi)^2} {\sin(kr) \over k^2 + m^2} \\[1ex] ={}& {1\over ir} \int_{-\infty}^{\infty} \frac{k dk}{(2\pi)^2} {e^{ikr} \over k^2 + m^2} \\[1ex] ={}& {1\over ir} \int_{-\infty}^{\infty} \frac{k dk}{(2\pi)^2} {e^{ikr} \over (k + i m)(k - i m)} \\[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{align}</math>
Modified Coulomb potential with mass
<math display="block">\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]</math>
where the hat indicates a unit vector in three dimensional space. The derivation of this result is as follows:
<math display="block">\begin{align} &\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{align} </math>
Note that in the small Шаблон:Mvar limit the integral goes to the result for the Coulomb potential since the term in the brackets goes to Шаблон:Math.
Longitudinal potential with mass
<math display="block">\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 ) </math>
where the hat indicates a unit vector in three dimensional space. The derivation for this result is as follows:
<math display="block">\begin{align} & \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{align}</math>
Note that in the small Шаблон:Mvar limit the integral reduces to
<math display="block">{1\over 2} {1 \over 4 \pi r } \left[ \mathbf 1 - \mathbf{\hat r} \mathbf{\hat r} \right]. </math>
Transverse potential with mass
<math display="block">\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]</math>
In the small mr limit the integral goes to
<math display="block">{1\over 2} {1 \over 4 \pi r } \left[\mathbf 1 + \mathbf{\hat r} \mathbf{\hat r}\right].</math>
For large distance, the integral falls off as the inverse cube of r
<math display="block">\frac{1}{4 \pi m^2r^3 }\left[\mathbf 1 + \mathbf{\hat r} \mathbf{\hat r}\right].</math>
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]Шаблон:Rp
<math display="block">\int_0^{2 \pi} {d\varphi \over 2 \pi} \exp\left( i p \cos( \varphi) \right)=J_0 (p)</math>
and
<math display="block">\int_0^{2 \pi} {d\varphi \over 2 \pi} \cos( \varphi) \exp\left( i p \cos( \varphi) \right) = i J_1 (p). </math>
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
<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_0 \left( kr \right)=K_0 (mr). </math>
See Abramowitz and Stegun.[6]Шаблон:Rp
For <math> mr \ll 1 </math>, we have[5]Шаблон:Rp
<math display="block">K_0 (mr) \to -\ln \left( {mr \over 2}\right) + 0.5772.</math>
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]
<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) =I_1 (mr)K_1 (mr).</math>
For small mr the integral becomes
<math display="block">\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].</math>
For large mr the integral becomes
<math display="block">\int_o^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) \to {1\over 2} \left( {1\over mr}\right).</math>
For applications of this integral see Magnetic interaction between current loops in a simple plasma or electron gas.
In general
<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_{\nu}^2 (kr) = I_{\nu} (mr)K_{\nu} (mr) \qquad \Re (\nu) > -1. </math>
Integration over a magnetic wave function
The two-dimensional integral over a magnetic wave function is[6]Шаблон:Rp
<math display="block">{2 a^{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).</math>
Here, M is a confluent hypergeometric function. For an application of this integral see Charge density spread over a wave function.
See also
References
Шаблон:Physics-footer Шаблон:Integrals
