Magic Gaussian integral

Partial function


The $\varepsilon$ prescription in the definition is just there so that one can evaluate the integral for certain complex matrices $A$ where it wouldn't exist otherwise. For example if $A$ has imaginary eigenvalues, then the naive integral will not be finite, while if we use $A_\varepsilon:=A-\varepsilon\,\mathrm{1}$, then we get an additional term $\mathrm{e}^{-\varepsilon\,\left\langle\phi\left|\right.\phi\right\rangle}$ which makes the integral converge.

That whole integral is just a showoff generalization of the one dimensional integral

$I_a:=\int_{-\infty}^\infty{\mathrm e}^{-\tfrac{1}{2}a\,\phi^2}\mathrm d\phi =(2\pi)^{1/2}a^{-1/2}$

which is obtained, for example, with the basic trick of switching to polar coordinates.

(Side remark: Having found the normalization of ${\mathrm e}^{-\tfrac{1}{2}a\,\phi^2}$ lets us define the function ${\mathrm{erf}}$ with $\lim_{d\to\infty}{\mathrm{erf}}(d)=1$. Then we can show that $\frac{1}{2}\pi^\frac{1}{2}{\mathrm{erf}}(d) = \sum_{k=0}^\infty\frac{1}{k!}(-1)^k\frac{d^{2k+1}}{2k+1}$ and with $\int_0^d(-\phi^2)^k{\mathrm d}\phi = (-1)^k\frac{d^{2k+1}}{2k+1}$ we can evaluate the initial integral on $[-d,d]$, not just all of $\mathbb R$.)

Adding a imaginary linear term in the exponent results in the statement that the Fourier transform of a Gaussian curve is again a Gaussian curve

$\int_{-\infty}^\infty\mathrm e^{-\tfrac{1}{2}a\,\phi^2+i\,\phi\,j}\mathrm d\phi =I_a\cdot\mathrm e^{-\tfrac{1}{2}j^2\,a^{-1}}$

which can be obtained via an integration variable shift.


$ A_\varepsilon $ … $n\times n$ square matrix over $\mathbb C$ $ \lim_{\varepsilon\to 0}A_\varepsilon = A $
$Z_f(J):=\lim_{\varepsilon\to 0}\int_{-\infty}^\infty\,f(\phi)\,\mathrm e^{\frac{1}{2} \left\langle\phi\left|\,A_\varepsilon\,\right|\phi\right\rangle +i\,\left\langle\phi\left|\right.J\right\rangle}\prod_{i=1}^m \mathrm d\phi_i $

Here $\langle u|v\rangle$ denotes the inner product in ${\mathbb C}^n$ as vector space.

Notice that via diagonalization of the matrix and knowledge of basic Gaussian integral above, we get

$Z_1(0):=\int_{-\infty}^\infty \mathrm e^{-\tfrac{1}{2}\left\langle\phi\left|\,A\,\right|\phi\right\rangle} \prod_{i=1}^m \mathrm d\phi_i = (2\pi)^{1/2}(\det A)^{-1/2} $

Taking care of the vector $J$, we can obtain

$Z_1(J):=\int_{-\infty}^\infty \mathrm e^{-\tfrac{1}{2}\left\langle\phi\left|\,A\,\right|\phi\right\rangle +i\,\left\langle\phi\left|\right.J\right\rangle}\prod_{i=1}^m\mathrm d\phi_i = Z_1(0)\cdot\mathrm e^{-\frac{1}{2}\left\langle J\left|\,A^{-1}\,\right|J\right\rangle }$

Now from a physical perspective, it's actually better to write this as $\propto\mathrm e^{-\frac{1}{2}\left\langle A^{-1}J\left|\,A\,\right|A^{-1}J\right\rangle }$. In the path integral treatment of the diffusion equation, the propagator involves quantity $\phi\equiv p,J\equiv q$ and roughly speaking $A\propto \delta t,\ A^{-1}\propto\tfrac{\mathrm d}{\mathrm dt}$. The above integral plays a role in passing from the Hamiltonian perspective to the Lagrangian one: $\left\langle\phi\left|\,A\,\right|\phi\right\rangle\propto p^2\Delta t$ to a sort of conjugate $\left\langle A^{-1}J\left|\,A\,\right|A^{-1}J\right\rangle\propto {\dot q}^2\Delta t$.

Lastly, notice that $-i\frac{\partial}{\partial J_i}e^{i\,\left\langle\phi\left|\right.J\right\rangle}=\phi_i\, \mathrm e^{i\,\left\langle\phi\left|\right.J\right\rangle}$ and therefore

$ Z_f(J) = Z_1(0)\ f\left(-i\frac{\partial}{\partial J}\right)\,\mathrm e^{-\frac{1}{2}\left\langle J\left|\,A^{-1/2}\,\right|J\right\rangle }$

We are interested in that expression as the solution of the integral, because in quantum field theory, the path integral is often an infinite dimensional variant it. There the exponent in the defining integral is the action functional, the operator $A$ involves a hard to invert differential operator (the inverse being strongly related to the response function/green function) and $f$ encodes the type of process and the interaction. The terms from the expansion of $f$ are encoded by Feynman diagrams.





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