Functional integration: Difference between revisions

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Revision as of 23:07, 24 April 2023

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</math>

which is slightly more understandable. The integral is shown to be a functional integral with a capital ''D''. Sometimes it is written in square brackets: [''Df''] or ''D''[''f''], to indicate that ''f'' is a function.<math>

W[J]

</math>

==Examples==

Most functional integrals are actually infinite, but thenoften the limit of the [[quotient]] of two related functional integrals can still be finite. The functional integrals that can be evaluated exactly usually start with the following [[Gaussian integral]]:

:<math>

\frac{\displaystyle\int \exp\left\lbrace-\frac{1}{2} \intint_{\mathbb{R}}\left[\int_{\mathbb{R}} f(x) K(x,;y) f(y) \,dx\,dy + \int J(x) f(x) \,right]dx\right\rbrace \mathcal{D}[f]}

{\displaystyle\int \exp\left\lbrace-\frac{1}{2} \intint_{\mathbb{R}^2} f(x) K(x,;y) f(y) \,dx\,dy\right\rbrace \mathcal{D}[f]} =

\exp\left\lbrace\frac{1}{2}\intint_{\mathbb{R}^2} J(x) \cdot K^{-1}(x,;y) \cdot J(y) \,dx\,dy\right\rbrace.\,,

</math>

in which <math>

By functionally differentiating this with respect to ''J''(''x'') and then setting to 0 this becomes an exponential multiplied by a polynomial in ''f''. For example, setting <math>K(x, y) = \Box\delta^4(x - y)</math>, we find:

K(x;y)=K(y;x)

</math>. By functionally differentiating this with respect to ''J''(''x'') and then setting to 0 this becomes an exponential multiplied by a polynomialmonomial in ''f''. ForTo example,see setting <math>K(xthis, y)let's =use \Box\delta^4(x - y)</math>,the wefollowing findnotation:

<math>

G[f,J]=-\frac{1}{2} \int_{\mathbb{R}}\left[\int_{\mathbb{R}} f(x) K(x;y) f(y)\,dy + J(x) f(x)\right]dx\, \quad,\quad W[J]=\int \exp\lbrace G[f,J]\rbrace\mathcal{D}[f]\;.

</math>

With this notation the first equation can be written as:

<math>

\dfrac{W[J]}{W[0]}=\exp\left\lbrace\frac{1}{2}\int_{\mathbb{R}^2} J(x) \cdot K^{-1}(x;y) \cdot J(y) \,dx\,dy\right\rbrace.

</math>

Now, taking functional derivatives to the definition of <math>

W[J]

</math> and then evaluating in <math>

J=0

</math>, one obtains:

<math>

\dfrac{\delta }{\delta J(a)}W[J]\Bigg|_{J=0}=\int f(a)\exp\lbrace G[f,0]\rbrace\mathcal{D}[f]\;,

</math>

<math>

\dfrac{\delta^2 W[J]}{\delta J(a)\delta J(b)}\Bigg|_{J=0}=\int f(a)f(b)\exp\lbrace G[f,0]\rbrace\mathcal{D}[f]\;,

</math>

<math>

\qquad\qquad\qquad\qquad\vdots

</math>

which is the result anticipated. More over, by using the first equation one arrives to the useful result:

:<math>

\dfrac{\displaystyle\int f(a) f(b) edelta^2}{i \intdelta fJ(xa) \Boxdelta fJ(xb) \,d^4x} \mathcalleft(\dfrac{DW[J]}{W[f0]}\right)\Bigg|_{J=0}=

K^{-1}(a; b)\;;

{\displaystyle\int e^{i \int f(x) \Box f(x) \,d^4x} \mathcal{D}[f]} =

</math>

K^{-1}(a, b) = \frac{1}{|a - b|^2},

Putting these results together and backing to the original notation we have:

<math>

\frac{\displaystyle\int f(a)f(b)\exp\left\lbrace-\frac{1}{2} \int_{\mathbb{R}^2} f(x) K(x,y) f(y)\, dx\,dy\right\rbrace \mathcal{D}[f]}

{\displaystyle\int e^\exp\left\lbrace-\frac{i1}{2} \intint_{\mathbb{R}^2} f(x) \Box fK(x,y) f(y) \,d^4x}dx\,dy\right\rbrace \mathcal{D}[f]} =

K^{-1}(a;b)\,.

</math>

where ''a'', ''b'' and ''x'' are 4-dimensional vectors. This comes from the formula for the propagation of a photon in quantum electrodynamics. Another useful integral is the functional [[delta function]]:

:<math>

\int \exp\left\lbrace \intint_{\mathbb{R}} f(x) g(x)dx\right\rbrace \mathcal{D}[f] = \delta[g] = \prod_x\delta\big(g(x)\big),

</math>