Independence of $z$ and $z^*$ in coherent states

Question

In the book of Lowell Brown on QFT its mentioned that

$$\int_{\mathbb{R}^2} \frac{dq'dp'}{2\pi} e^{(-z^{*}z + z^{*}_1z + z^{*}z_2)} = e^{z^{*}_1z_2}\tag{1.8.12}$$

where $$z=\frac{q'+ip'}{\sqrt{2}} \qquad \tag{1.8.1}$$ is the eigenvalue of a coherent state.

In the next paragraph, its mentioned that

the transformation function $\langle z^{*}|z\rangle$ and the integration weight $e^{-z^{*}z}$ can both obviously be extended to analytic functions in the separate, distinct variables $z^{*}$ and $z$.

In view of this extended analyticity, one can, in general, make independent translations: $$z \rightarrow z+a \text{ and } z^{*} \rightarrow z^{*}+b^{*}\tag{1.8.13}$$ where $a$ and $b^{*}$ are arbitrary complex numbers.

How is this possible? Won't it always be true that $a = b$ if we use such a transformation?

Answer 1

TL;DR: Eq. (1.8.13) in Ref. 1 is exactly right: Remarkably inside an integral, one may effectively shift the integration variable $z=x+iy$ and its complex conjugate variable $\bar{z}=x-iy$ by two independent complex numbers!

Mathematically, eq. (1.8.13) is essentially the statement that for an arbitrary polynomial $P:\mathbb{C}^2\to\mathbb{C}$ the integral $$\int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}~P(z\!+\!c,\bar{z}\!+\!d)e^{-(z+c)(\bar{z}+d)}\quad\text{does not depend on}\quad c,d~\in~\mathbb{C}. \tag{1}$$ For the definition of the integration (1) in the complex plane, see e.g. this & this related Phys.SE posts. Eq. (1) is equivalent to $$\int_{\mathbb{R}^2} \!\frac{\mathrm{d}x \wedge \mathrm{d}y}{\pi }~P(x\!+\!a,y\!+\!b)e^{-(x+a)^2-(y+b)^2}\quad\text{does not depend on}\quad a,b~\in~\mathbb{C}. \tag{2}$$ Here the relationship between the 4 complex constants are $$ c~=~a+ib, \qquad d~=~a-ib .\tag{3}$$ Due to linearity, it is enough to prove eq. (2) for polynomials that factorizes, i.e. polynomials of the form $P(x\!+\!a)Q(y\!+\!b)$. Then eq. (2) boils down to show that $$\int_{\mathbb{R}} \!\mathrm{d}x ~P(x\!+\!a)e^{-(x+a)^2}~=~\int_{\mathbb{R}+i{\rm Im}(a)} \!\mathrm{d}x ~P(x)e^{-x^2}\quad\text{does not depend on}\quad a~\in~\mathbb{C}. \tag{4}$$ Eq. (4) follows by a simple application of Cauchy's integral theorem. $\Box$

For more on coherent states, complex conjugation and independence of variables, see also e.g. this & this related Phys.SE posts.

References:

1. L.S. Brown, QFT, 1992; eqs. (1.8.12)-(1.8.13).