Do the source terms multiplying a complex field and its conjugate need to be conjugates for the Gaussian identity to hold? E.g. is

$$\int D({\phi,\psi,b}) e^{-b^\dagger A b +f(\phi, \phi^\dagger,\psi, \psi^\dagger )b +b^\dagger g(\phi, \phi^\dagger,\psi, \psi^\dagger )} = \int D(\phi,\psi) \det(A^{-1}) e^{f(...) A^{-1} g(...)} $$

valid when $f \ne g^* $?

If I change to real and imaginary coordinates in the $b$ it seems fine, but I'm worried that I'm screwing up the measure in $D(...)$ without realizing it.


Let's say $A$ is a $c$-number. To do the integral I can write $b = x +iy$ etc. Then the integral is

$$\int D(...) e^{- Ax^2 - A y^2 +x(f + g) + i y(f-g)} = \frac{\pi}{A}\int D(...) e^{(4A)^{-1}((f+g)^2 - (f-g)^2)}$$ $$=\frac{\pi}{A}\int D(...) e^{A^{-1} fg}.$$

But then this implies that Hubbard Stratonovich transformations don't need to be of squares.. so I can decouple any interaction $$e^{2fg} = \int d \phi d\phi^\dagger e^{-|\phi|^2 +f\phi + \phi^\dagger g}.$$ This can't be right?

  • $\begingroup$ Lets say A is a c-number. To do the integral I can write $b = x +iy$ etc. Then the exponent is $- Ax^2 - A y^2 +x(f + g) + i y(f-g)$, which I can integrate to give me the exponential $(f+g)^2/2A -(f-g)^2/2A = \frac{1}{A}(fg) $. But then this implies that Hubbard Stratonovich transformations don't need to be of squares.. $\endgroup$ – Adam B Jan 2 '19 at 2:14
  • $\begingroup$ okay well you only have to worry about a change in the measure if you are transforming the fields $\endgroup$ – InertialObserver Jan 2 '19 at 2:15
  • $\begingroup$ aren't the $\phi,\psi$ fields being evaluated as constants? I'm free to shift $x$ and $y$ by constants freely? $\endgroup$ – Adam B Jan 2 '19 at 2:33
  • $\begingroup$ I'm not exactly sure about your question.. It seems like a good question though that I could learn from. So I upvoted with hopes that it gets more attention.. I would recommend just further explaining exactly what it is your asking $\endgroup$ – InertialObserver Jan 2 '19 at 7:52

Theorem: Given a normal$^1$ $n\times n$ matrix $A$ where ${\rm Re}(A)>0$ is positive definite, then the complex Gaussian integral is$^2$ $$\begin{align} I&~:=~\int_{\mathbb{R}^{2n}} \! d^nx ~d^ny~ \exp\left\{-z^{\dagger}Az +f^{\dagger}z +z^{\dagger}g\right\}\cr &~=~\exp\left\{f^{\dagger}A^{-1}g\right\}\int_{\mathbb{R}^{2n}} \! d^nx ~d^ny~ \exp\left\{-(z^{\dagger}-f^{\dagger}A^{-1})A(z-A^{-1}g)\right\}\cr &~=~\frac{\pi^n}{\det(A)}\exp\left\{f^{\dagger}A^{-1}g\right\}, \qquad z^k~\equiv~ x^k+iy^k.\end{align}$$

Sketched proof:

  1. The normal matrix $A=U^{\dagger}DU$ can be diagonalized with a unitary transformation $U$. Here $D$ is a diagonal matrix with ${\rm Re}(D)>0$. Next change integration variables$^3$ $w=Uz$. The absolute value of the Jacobian determinant is 1. So it is enough to consider the case $n=1$, which we will do from now on.

  2. There exist two complex numbers $x_0,y_0\in\mathbb{C}$ such that$^4$ $$ x_0-iy_0~=~f^{\dagger}A^{-1}\qquad\text{and}\qquad x_0+iy_0~=~A^{-1}g.$$

  3. We can shift the real integration contour into the complex plane $$\int_{\mathbb{R}} \! dx \int_{\mathbb{R}} \! dy~ \exp\left\{-(z^{\dagger}-f^{\dagger}A^{-1})A(z-A^{-1}g)\right\}$$ $$~=~\int_{\mathbb{R}+x_0} \! dx \int_{\mathbb{R}+y_0} \! dy~ \exp\left\{-z^{\dagger}Az\right\}~=~\frac{\pi}{A},$$ with no new non-zero contributions arising from closing the contour, cf. Cauchy's integral theorem.$\Box$


$^1$ The Gaussian integral is presumably also convergent for a pertinent class of non-normal matrices $A$, but in this answer we consider only normal matrices for simplicity.

$^2$ Recall that the notation $\int_{\mathbb{C}^n}d^nz^{\ast} d^nz$ means $\int_{\mathbb{R}^{2n}} \! d^nx ~d^ny$ up to a conventional factor, cf. my Phys.SE answer here. Here $z^k \equiv x^k+iy^k$ and $z^{k\ast} \equiv x^k-iy^k$.

$^3$ More generally, under a holomorphic change of variables $u^k+iv^k\equiv w^k=f^k(z)$, the absolute value of the Jacobian determinant in the formula for integration by substitution is $$ |\det\left(\frac{\partial (u,v)}{\partial (x,y)} \right)_{2n\times 2n}|~=~ |\det\left(\frac{\partial w}{\partial z} \right)_{n\times n}|^2. $$

$^4$ The underlying philosophy in point 2 is similar to my Phys.SE answer here: One can in a certain sense treat $z$ and $z^{\dagger}$ as independent variables! And therefore it is possible to consider OP's case where $f,g\in\mathbb{C}^n$ are independent complex constants.

  • $\begingroup$ Amazing, H.S. transform is far more general than I realized.. Thanks for the confident answer! $\endgroup$ – Adam B Jan 3 '19 at 20:04

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