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This question is not so much related to the physics as it is the integral itself. In terms of the canonical coherent states, $|z\rangle$, one encounters a resolution of the identity of the form:

$$\hat{I}=\displaystyle\int \dfrac{d^2z}{\pi} |z\rangle\langle{z}|$$

Where the integral is taken over the entire complex plane. My question is, often this is re-expressed in the forms:

$$\displaystyle\int \dfrac{dzd\bar{z}}{2\pi i}|z\rangle\langle{z}|$$

$$\displaystyle\int \dfrac{dxdy}{\pi} |z\rangle\langle{z}| \qquad \text{where } x=\Re z, \ y=\Im z$$

How does one show that the $z-\bar{z}$ integral is equivalent to $x-y$ integral? How does one show that either of these are equivalent to the original integral? I presume this has something to do with the properties of differential forms inducing the area element but I'm unsure how to formulate these in terms of differential forms (and back again) in the first place.

For instance, it's clear to me that IF I assume that I can assume a correspondence of the form:

$$\int dzd\bar{z} = \int dz \wedge d\bar{z}$$

Then I could use the standard properties of the exterior product to come to the relation $dz \wedge d\bar{z} = -2idx \wedge dy$. Whilst close, this is still not quite right even if one were to use the above unjustified correspondence to return to the element $dxdy$. I presume this issue is related to orientation, but I have yet to see a satisfactory explanation for what's going on here.

How does one relate the area element $dzd\bar{z}$ to the differential 2-form $dz\wedge d\bar{z}$ in the context of the integral? Why is this a valid step to take?

Thanks in advance!

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  • $\begingroup$ It's fairly common to omit wedge products (as a matter of notation) when integrating coordinate forms, but you're right that there's a sign discrepancy between your two integrals: $i\, dz\, d\bar{z} = 2\, dx\, dy$. Offhand I'd guess the complex integral should read $\int \frac{i\, dz\, d\bar{z}}{2\pi} [...]$, but if the notation is common in the physics literature, then presumably something else is going on. $\endgroup$ – Andrew D. Hwang Mar 21 '17 at 11:25
  • $\begingroup$ @AndrewD.Hwang Unfortunately, the complex integral reads as written in the OP. The other sticking point for me is that I'm happy to read $\int_{\mathbb{R}}\int_{\mathbb{R}} \frac{dxdy}{\pi} [...] = \int_{\mathbb{R}^2} \frac{dx \wedge dy}{\pi}[...]$ but significantly less happy with doing this in the complex version, as $z$ and $\bar{z}$ are not independent here. $\endgroup$ – FH93 Mar 21 '17 at 11:37
  • $\begingroup$ Ah, well (about the sign). Regarding the (in)dependence of $z$ and $\bar{z}$, the linearly independent sets $\{x, y\}$ and $\{z, \bar{z}\}$ of complex-valued functions generate the same $2$-dimensional subspace; see also (the second comment on my answer to) How exactly does $\frac{\partial f}{\partial\bar{z}}$ work?. $\endgroup$ – Andrew D. Hwang Mar 21 '17 at 11:48
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    $\begingroup$ This is actually a (not too uncommon) error. The correct formula should be $$\dfrac i2 dz\wedge d\bar z = dx\wedge dy.$$ Sometimes people get too carried away with the $\dfrac 1{2\pi i}$. $\endgroup$ – Ted Shifrin Mar 22 '17 at 21:04

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