Functional integration: Difference between revisions

Revision as of 00:38, 18 April 2023

Revision as of 21:32, 24 April 2023

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Whereas standard [[Riemann integral|Riemann integration]] sums a function ''f''(''x'') over a continuous range of values of ''x'', functional integration sums a [[functional (mathematics)|functional]] ''G''[''f''], which can be thought of as a "function of a function" over a continuous range (or space) of functions ''f''. Most functional integrals cannot be evaluated exactly but must be evaluated using [[perturbation methods]]. The formal definition of a functional integral is

<math display="block">

\int G[f]\; \mathcal{D}[Dff] \equiv \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty G[f] \prod_x df(x).

</math>

However, in most cases the functions ''f''(''x'') can be written in terms of an infinite series of [[orthogonal functions]] such as <math>f(x) = f_n H_n(x)</math>, and then the definition becomes

<math display="block">

\int G[f] \; \mathcal{D}[Dff] \equiv \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty G(f_1, f_2, \ldots) \prod_n df_n,

</math>

Line 40 ⟶ 41:

:<math>

\frac{\displaystyle\int e^{\exp\left\lbrace-\frac{i1}{2} \int f(x) \cdot K(x,y) \cdot f(y) \,dx\,dy + \int J(x) \cdot f(x) \,dx}\right\rbrace \mathcal{D}[Dff]}

{\displaystyle\int e^{\exp\left\lbrace-\frac{i1}{2} \int f(x) \cdot K(x,y) \cdot f(y) \,dx\,dy}\right\rbrace \mathcal{D}[Dff]} =

e^{-\exp\left\lbrace\frac{i1}{2}\int J(x) \cdot K^{-1}(x,y) \cdot J(y) \,dx\,dy}\right\rbrace.

</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:

:<math>

\fracdfrac{\displaystyle\int f(a) f(b) e^{i \int f(x) \Box f(x) \,dxd^44x} \mathcal{D}[Dff]}

{\displaystyle\int e^{i \int f(x) \Box f(x) \,dxd^44x} \mathcal{D}[Dff]} =

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

</math>

Line 56 ⟶ 57:

:<math>

\int e^{i\exp\left\lbrace \int f(x) g(x)dx\right\rbrace \,dxmathcal{D} [Dff] = \delta[g] = \prod_x\delta\big(g(x)\big),

</math>