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I want to know how to calculate the integration(it is from a physics book): $$ \int {\rm d}z{\rm d}\bar z \exp (-z\bar z) = \pi $$ But I even do not know the definition of ${\rm d}z{\rm d}\bar z$. Maybe if you treat it as a 2-form, then, $$ {\rm d}z\wedge{\rm d}\bar z = -2i {\rm d}x\wedge{\rm d}y $$ But the result $\pi$ implies it treat the term as, $$ {\rm d}z{\rm d}\bar z \to {\rm d}x{\rm d}y $$ I have read this question complex integral over $dz\,dz^*$, I get really confused. Can someone give me the definition in math or some references for this.

Really thanks.

nsigma
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  • Why not assume they forgot the $-2i$ constant, does it bring any trouble somewhere else ? – reuns May 14 '19 at 02:44

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Start with $z=x+iy$, compute the (absolute value of the) Jacobian to get a Gaussian integral of the type $$ \int_{-\infty}^\infty dx\,\int_{-\infty}^\infty dy \, e^{-(x^2+y^2)}\, . $$ This kind of notation is common in mathematical physics, especially in the study of coherent states, where it is often rewritten as $d^2\alpha$, with $\alpha=x+ip$, as in this wiki page. The notation using $z$ rather than $\alpha$ is favoured in the early papers of Perelomov and the Russian school.

ZeroTheHero
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  • the Jacobian is $-2\rm i$, so the absolute $|\cdot|$ gives $2$. I think maybe it is just a definition of the notation. – nsigma May 14 '19 at 02:35