# Quantum Field Theory – How to Generalize Wick Theorem for Gaussian Integral of a Function with Non-Zero Mean

integrationpath-integralwick-theorem

From the wikipedia article, for a Gaussian integral of an analytic function we have that

This is equivalent to the Wick theorem when f(x) is a polynomial.

Now I'm trying to obtain a similar formula when there is a linear term in the Gaussian (ie the Gaussian has a nonzero mean).

My guess is that
$$\int f(x) \exp \left( – \frac{1}{2} x^T A x + B^T x \right) d^n x = \\ = \sqrt{\frac{(2 \pi)^n}{\det A}} \exp \left[ \frac{1}{2} \left( B^T + \frac{d}{dx_i} \right) A^{-1} \left( B + \frac{d}{dx_j} \right) \right] f(x) \bigg|_{x=0}$$

but I can't prove it. Is this equation correct? How can I prove it?

The case with the linear term is obtained from the original one by a simple shift, i.e. the substitution $$x = X + A^{-1} B$$ Substitute it to the exponent in your more general integral: $$-\frac 12 x^T A x + B^T x = -\frac 12 (X^T+B^T A^{-1}) A(X+A^{-1}B)+B^T (X+A^{-1}B)=\dots$$ I used $A=A^T$. Now, all the terms that are schematically $BX$ i.e. linear in $X$ cancel: the coefficient is $-1/2-1/2+1=0$. The remaining terms give $$- \frac 12 X^T A X + \frac 12 B^T A^{-1} B$$ The coefficient $+1/2$ in the second term came from $-1/2+1$. The second term only gives a simple factor (the exponential of that), and it's a part of your result – except that the last $B^T$ in your result should be simply $B$.
The hard, quadratic/Gaussian part of the expression may be rewritten in the Wick way from your first identity. It could be enough if you were satisfied with the evaluation of the $x$-derivatives not at $X=0$ but at the right original value $x=0$ which means, thanks to my substitution $$X = -A^{-1}B.$$ However, if you want to use the values of the derivatives at $X=0$, you have to Taylor-expand the shift operator from $X=0$ to $X=-A^{-1}B$. The shift is the operator $$\exp(B^T A^{-1} \frac d{dx})$$ which is exactly what you get from the mixed terms in your guessed exponent, up to an overall sign perhaps that you will surely be able to catch correctly.