Functional integration: Difference between revisions

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==Examples==

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

:<math>

\intfrac{ \int e^{i \int{ -\frac{1}{2}f(x) \cdot K(x,y) \cdot f(y) dxdy}\,dx\,dy + \int J(x) \cdot f(x) \,dx} [Df] }

\frac{

{\int{ e^{i \int{ -\frac{1}{2}f(x) \cdot K(x,y) \cdot f(y) dxdy} + \int{ J(x) ,dx\cdot f(x) dx,dy} }[Df]} }=

e^{i \frac{1}{2}\int{ J(x) \cdot K^{-1}(x,y) \cdot J(y) dxdy } \,dx\,dy}.

}

{

\int{ e^{i \int{ -\frac{1}{2}f(x) \cdot K(x,y) \cdot f(y) dxdy} } [Df] }

}

=

e^{i \frac{1}{2}\int{ J(x) \cdot K^{-1}(x,y) \cdot J(y) dxdy } }

</math>

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

:<math>

\intfrac{\int f(a) f(b) e^{i \int{ f(x) \Box f(x) \,dx^4}} }[Df] }

\frac{

{\int{ f(a) f(b) e^{i \int{ f(x) \Box f(x) \,dx^4}} }[Df]} =

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

}{

\int{ e^{i \int{ f(x) \Box f(x) dx^4}} }[Df]

}

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

</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{ e^{i \int{ f(x) g(x) \,dx}} }[Df] = \delta[g] = \prod_x\delta\big( g(x) \big),

</math>

which is useful to specify constraints. Functional integrals can also be done over [[Grassmann number|Grassmann-valued]] functions <math>\psi(x)</math>, where <math>\psi(x) \psi(y) = -\psi(y) \psi(x)</math>, which is useful in quantum electrodynamics for calculations involving [[fermions]].

==In symbolic algebra software==