5
$\begingroup$

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.

Edit:

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?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.