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In Negele's book "quantum many-particle systems", the completeness formula for coherent state is $$\int {\frac{{d{\phi ^*}d\phi }}{{2\pi i}}{e^{ - {\phi ^*}\phi }}} \left| \phi \right\rangle \left\langle \phi \right| = I\tag{1}$$ and it says the integration measure is $$\frac{{d{\phi ^*}d\phi }}{{2\pi i}} = \frac{{d\left( {{\mathop{\rm Re}\nolimits} \phi } \right)d\left( {{\mathop{\rm Im}\nolimits} \phi } \right)}}{\pi }.\tag{2}$$ How to understand this integration measure equality?

As I understand, If I view $\phi=x+iy$, then $${d\left( {{\mathop{\rm Re}\nolimits} \phi } \right)d\left( {{\mathop{\rm Im}\nolimits} \phi } \right)}=dxdy.\tag{3}$$ But

$$d{\phi ^*}d\phi = \left( {dx - idy} \right)\left( {dx + idy} \right) = d{x^2} + d{y^2}\tag{4}$$

I am pretty sure this understanding is wrong. What is the right way?

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Concerning OP's last formula (4): Recall that there is an implicitly written antisymmetric wedge $\wedge$ in the integral measure. Therefore OP's last equality sign should read $$( \mathrm{d}x - i\mathrm{d}y )\wedge ( \mathrm{d}x + i\mathrm{d}y ) ~=~ 2i\mathrm{d}x\wedge\mathrm{d}y. \tag{4'}$$ See also e.g. this Phys.SE post.

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This alternative derivation (without Grassman algebra) uses the transformation rule for integrals. We have an integral over $\mathbb {R}^2$ with the measure $dx\ dy $ (I dont care about $\pi$). We consider a change of variables $(x,y)\mapsto (z, z*):=(x+iy, x-iy) $ This is a diffeomprphism onto its image. (A subset of $\mathbb {C}^2$ seen as a 4d real vectorspace) . The Jacobian is $$ J=\begin{pmatrix} 1 & i \\ 1 & -i \end {pmatrix}$$. The "absolute value" (for real and complex part individually) of the determinant is $2i$, the factor we need.

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    $\begingroup$ Why should one take the absolute value of the Jacobian for real and imaginary part individually? Seems arbitrary. $\endgroup$ – Nephente Feb 24 '17 at 12:29
  • $\begingroup$ It is arbitrary. The normal change of variables formula is not really suited for dealing with complex domains. I assume to make this mathematically precise one would have to consider real and imaginary parts of the measure separatly and the domains as actually real. $\endgroup$ – Adomas Baliuka Feb 24 '17 at 13:20

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