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I was reading on Wikipedia in this article about the n-dimensional and functional generalization of the Gaussian integral. In particular, I would like to understand how the following equations are derived:

$$

\begin{eqnarray}

& {} \quad \int x^{k_1}\cdots x^{k_{2N}} \, \exp\left( – \frac 1 2 \sum_{i,j=1}^{n}A_{ij} x_i x_j \right) \, d^nx \\

& =

\sqrt{\frac{(2\pi)^n}{\det A}} \, \frac{1}{2^N N!} \, \sum_{\sigma \in S_{2N}}(A^{-1})^{k_{\sigma(1)}k_{\sigma(2)}} \cdots (A^{-1})^{k_{\sigma(2N-1)}k_{\sigma(2N)}}

\end{eqnarray}

$$

and

$$

\begin{eqnarray}

\int f(\vec x) \, \exp\left( – \frac 1 2 \sum_{i,j=1}^{n}A_{ij} x_i x_j \right) d^nx

=

\sqrt{(2\pi)^n\over \det A} \, \left. \exp\left({1\over 2}\sum_{i,j=1}^{n}(A^{-1})_{ij}{\partial \over \partial x_i}{\partial \over \partial x_j}\right)f(\vec{x})\right|_{\vec{x}=0}

\end{eqnarray}

$$

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The problem is that I can’t find a book or internet source that shows a complete derivation of this .. any reference tip would be highly appreciated!

Another thing that I am not so sure about is what the complete symmetrization means in the above equation, if someone could give a short explanation (or a more detailed equation) of what the following expression means that would be great:

$$

\begin{eqnarray}

\sum_{\sigma \in S_{2N}}(A^{-1})^{k_{\sigma(1)}k_{\sigma(2)}} \cdots (A^{-1})^{k_{\sigma(2N-1)}k_{\sigma(2N)}}

\end{eqnarray}

$$

Many thanks !!

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The presentation here is typical of those used to model and motivate the infinite dimensional Gaussian integrals encountered in quantum field theory.

I will use subscripts instead of superscripts to indicate components.

**I. Wick’s theorem**

First consider the integral

$$Z_0 = \int d^n x \exp\left(-\frac{1}{2} x^\mathrm{T} A x\right),$$

where $d^n x = \prod_i d x_i$ and $A$ is symmetric and positive definite.

Diagonalize $A$ with an orthogonal transformation.

Letting $D = M^\mathrm{T} A M = \mathrm{diag}(\lambda_1,\ldots,\lambda_n)$ and $z = M^\mathrm{T} x$, and noting that the Jacobian is the identity, we find

$$\begin{eqnarray}

Z_0 &=& \int d^n z \exp\left(-\frac{1}{2} z^\mathrm{T} D z\right) \\

&=& \prod_i \int d z_i \exp\left(-\frac{1}{2} \lambda_i z_i^2\right) \\

&=& \prod_i \sqrt{\frac{2\pi}{\lambda_i}} \\

&=& \sqrt{\frac{(2\pi)^n}{\det A}}.

\end{eqnarray}$$

Add a source term

$$Z_J = \int d^n x \exp\left(-\frac{1}{2} x^\mathrm{T} A x + J^\mathrm{T} x\right).$$

Complete the square to eliminate the cross term.

Letting $x = y + A^{-1}J$, we find

$$\begin{eqnarray}

Z_J &=& \int d^n y \exp\left(-\frac{1}{2} {y}^\mathrm{T} A y

+ \frac{1}{2} J^\mathrm{T}A^{-1}J\right) \\

&=& \sqrt{\frac{(2\pi)^n}{\det A}}

\exp\left(\frac{1}{2} J^\mathrm{T}A^{-1}J\right).

\end{eqnarray}$$

There is always a factor of $\sqrt{\frac{(2\pi)^n}{\det A}}$.

For convenience, let’s define

$$\langle x_{k_1} \cdots x_{k_{2N}}\rangle = \frac{1}{Z_0}

\int d^n x \ x_{k_1} \cdots x_{k_{2N}} \exp\left(-\frac{1}{2} x^\mathrm{T} A x\right).$$

(Notice that $\langle x_{k_1} \cdots x_{k_{2N+1}}\rangle = 0$ since the integral is odd.)

Roughly speaking, these are free field scattering amplitudes.

This integral can be found by taking derivatives of $Z_J$,

$$\langle x_{k_1} \cdots x_{k_{2N}}\rangle =

\left.\frac{1}{Z_0} \frac{\partial}{\partial J_{k_1}} \cdots \frac{\partial}{\partial J_{k_{2N}}} Z_J\right|_{J=0}.$$

(Notice that every factor of $\partial/\partial_{J_{k_i}}$ brings down one factor of $x_{k_i}$ in the integral $Z_J$.)

Using this formula it is a straightforward exercise to work out, for example, that

$$\begin{eqnarray}

\langle x_{k_1} x_{k_{2}}\rangle

&=& \left.\frac{\partial}{\partial J_{k_1}} \frac{\partial}{\partial J_{k_{2}}}

\exp\left(\frac{1}{2} J^\mathrm{T}A^{-1}J\right)\right|_{J=0} \\

&=& \frac{\partial}{\partial J_{k_1}} \frac{\partial}{\partial J_{k_{2}}}

\frac{1}{2} J^\mathrm{T}A^{-1}J \\

&=& \frac{1}{2} ( A^{-1}_{k_1 k_2} + A^{-1}_{k_2 k_1}) \\

&=& A^{-1}_{k_1 k_2},

\end{eqnarray}$$

which agrees with the formula for $N=1$.

This is the “free field propagator.”

(In the last step we have used the fact that $A^{-1}$ is symmetric.)

It is possible to see by inspection that the formula for general $N$ is

$$\langle x_{k_1}\cdots x_{k_{2N}}\rangle = \frac{1}{2^N N!}

\sum_{\sigma \in S_{2N}}(A^{-1})_{k_{\sigma(1)}k_{\sigma(2)}} \cdots (A^{-1})_{k_{\sigma(2N-1)}k_{\sigma(2N)}}.$$

In fact

$$\begin{eqnarray}

\langle x_{k_1}\cdots x_{k_{2N}}\rangle &=&

\left.\frac{\partial}{\partial J_{k_1}} \cdots \frac{\partial}{\partial J_{k_{2N}}}

\exp \left(\frac{1}{2} J^\mathrm{T}A^{-1}J \right) \right|_{J=0} \\

&=& \frac{\partial}{\partial J_{k_1}} \cdots \frac{\partial}{\partial J_{k_{2N}}}

\frac{1}{N!} \left(\frac{1}{2} J^\mathrm{T}A^{-1}J\right)^{N} \\

&=& \frac{1}{2^N N!} \frac{\partial}{\partial J_{k_1}} \cdots \frac{\partial}{\partial J_{k_{2N}}}

\left(J^\mathrm{T}A^{-1}J\right)^{N}.

\end{eqnarray}$$

Note that the derivative $\partial/\partial J_{k_1}$ will operate on all possible $J$s.

Likewise for the other derivatives.

Thus, we will get a sum over all $(2N)!$ possible permutations of the $k_i$.

These permutations are denoted by $\sigma$ and they live in the symmetric group $S_{2N}$.

Thus, we arrive at the result, known as Wick’s theorem.

(If this seems too vague, it is straightforward to find the result by induction on $N$.

We have the formula for $N=1$ above.)

Let’s unwind the scattering amplitude for $N=2$.

We have

$$\langle x_{k_1}x_{k_2}x_{k_3}x_{k_4}\rangle = \frac{1}{2^2 2!}

\sum_{\sigma \in S_{4}}(A^{-1})_{k_{\sigma(1)}k_{\sigma(2)}} (A^{-1})_{k_{\sigma(3)}k_{\sigma(4)}}.$$

There are $4! = 24$ permutations in $S_4$ and thus $24$ terms in the sum.

For example $(12)$ takes $1$ to $2$ and $2$ to $1$.

Not all permutations give independent terms.

In fact, the degeneracy is $2^N N!$, since $A^{-1}$ is symmetric and the order of the $A^{-1}$s doesn’t matter.

Thus, there will only be

$$\frac{(2N)!}{2^N N!} = (2N-1)!!$$

independent terms.

For $N=2$ there are three independent terms,

$$\langle x_{k_1}x_{k_2}x_{k_3}x_{k_4}\rangle

= A^{-1}_{k_1 k_2}A^{-1}_{k_3 k_4}

+ A^{-1}_{k_1 k_3}A^{-1}_{k_2 k_4}

+ A^{-1}_{k_1 k_4}A^{-1}_{k_2 k_3}.$$

**II. Central identity**

Consider

$$I_J = \frac{1}{Z_0} \int d^n x \exp\left(-\frac{1}{2}x^\mathrm{T}A x + J^\mathrm{T}x \right) f(x).$$

We are interested in $I_0$.

The presence of the source allows us to take $f$ out of the integral if we replace its argument with $\partial/\partial J$,

$$\begin{eqnarray}

I_0 &=& \left.f(\partial_J) \frac{1}{Z_0} \int d^n x \exp\left(-\frac{1}{2}x^\mathrm{T}A x + J^\mathrm{T}x \right)\right|_{J=0} \\

&=& \left.f(\partial_J) \exp\left(\frac{1}{2} J^\mathrm{T}A^{-1}J\right)\right|_{J=0}.

\end{eqnarray}$$

This is a typical trick.

In fact, it is equivalent to what Anthony Zee calls the “central identity of quantum field theory.”

Usually $f(x) = \exp[-V(x)]$, where $V(x)$ is the potential.

There is a nice graphical interpretation of the formula in the form of Feynman diagrams.

The process of calculating

$$\left.e^{-V(\partial_J)} \exp\left(\frac{1}{2} J^\mathrm{T}A^{-1}J\right)\right|_{J=0}$$

is equivalent to “tying together” the propagators (the $A^{-1}$) with vertices represented by the operator $-V(\partial_J)$.

Of course, there is a lot more to it than that!

Now we wish to show

$$I_0 = \left.\exp\left(\frac{1}{2}\partial_x^\mathrm{T} A^{-1}\partial_x\right) f(x)\right|_{x=0}.$$

If we consider a Taylor expansion for $f(\partial_J)$ and $f(x)$ we can see this is equivalent to

$$\left.\partial_{J_{k_1}}\cdots \partial_{J_{k_{2N}}} \exp\left(\frac{1}{2} J^\mathrm{T}A^{-1}J\right)\right|_{J=0}

= \left.\exp\left(\frac{1}{2}\partial_x^\mathrm{T} A^{-1}\partial_x\right) x_{k_1}\cdots x_{k_{2N}}\right|_{x=0}.$$

The left hand side is $\langle x_{k_1}\cdots x_{k_{2N}}\rangle$, by previous arguments.

But

$$\begin{eqnarray}

\left.\exp\left(\frac{1}{2}\partial_x^\mathrm{T} A^{-1}\partial_x\right) x_{k_1}\cdots x_{k_{2N}}\right|_{x=0} &=&

\frac{1}{2^N N!} \left(\partial_x^\mathrm{T} A^{-1}\partial_x\right)^N x_{k_1}\cdots x_{k_{2N}} \\

&=& \frac{1}{2^N N!} \sum_{\sigma \in S_{2N}}(A^{-1})_{k_{\sigma(1)}k_{\sigma(2)}} \cdots (A^{-1})_{k_{\sigma(2N-1)}k_{\sigma(2N)}} \\

&=& \langle x_{k_1}\cdots x_{k_{2N}}\rangle.

\end{eqnarray}$$

For example, for $N=1$,

$$\begin{eqnarray}

\left.\exp\left(\frac{1}{2}\partial_x^\mathrm{T} A^{-1}\partial_x\right) x_{k_1} x_{k_{2}}\right|_{x=0}

&=& \left(\frac{1}{2} \partial_x^\mathrm{T} A^{-1}\partial_x\right) x_{k_1} x_{k_2} \\

&=& \frac{1}{2} A^{-1}_{ij}(\delta_{i k_1}\delta_{j k_2} + \delta_{i k_2}\delta_{j k_1}) \\

&=& \frac{1}{2} (A^{-1}_{k_1 k_2} + A^{-1}_{k_2 k_1}) \\

&=& A^{-1}_{k_1 k_2}.

\end{eqnarray}$$

**References**

Many texts on quantum field theory deal with such finite dimensional integrals on the way to treating the infinite dimensional case. See, for example,

A. Zee. *Quantum field theory in a nutshell*

J. Zinn-Justin. *Quantum Field Theory and Critical Phenomena*

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