Integrating Gaussians with complex arguments

In summary, the author is trying to find the integral of a function that involves conjugation, but is not getting the results he expects. He is suggesting that maybe the problem is with the exponent, and that if the exponent had an additional term, it might work.f
  • #1
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The integral I'm looking at is of the form

[tex]\int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 + \bar{J}z \right)[/tex]

Where [itex]K \in \mathbb{R}[/itex] and [itex]J \in \mathbb{C}[/itex]

The book I am following (Kardar's Statistical Physics of Fields, Chapter 3 Problem 1) asserts that by completing the square this becomes [itex]Z \exp\left( \frac{- |J|^2}{2K} \right) [/itex] where [itex] Z = \int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 \right)[/itex]. I can't seem to reproduce this, and I think the trouble I'm running into arises from [itex]|z|^2[/itex] not being a square, but rather it involves conjugation as well. Therefore, I get the following

[tex]-\frac{1}{2}K|z|^2 + \bar{J}z = -\frac{1}{2}K\left( z\bar{z} -2 \frac{\bar{J}}{K}z \right)= -\frac{1}{2}K\left( z\bar{z} -2 \frac{\bar{J}}{K}z - 2 \frac{J}{K}\bar{z} +2 \frac{J}{K}\bar{z} + 4\frac{ |J|^2}{K^2} - 4 \frac{ |J|^2}{K^2} \right) = [/tex]

[tex] -\frac{1}{2}K\left( z - 2 \frac{J}{K} \right) \left( \bar{z} -2 \frac{\bar{J}}{K} \right) -J\bar{z} +2 \frac{ |J|^2}{K} [/tex]

Which means that I'm getting that

[tex]\int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 + \bar{J}z \right) = \left[ \int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K \left| z-2 \frac{J}{K} \right|^2 - J\bar{z} \right) \right] \exp\left( 2 \frac{ |J|^2}{K} \right) [/tex]

Which doesn't at all seem like

[tex] \left[ \int_\mathbb{C} dz \: \exp \left( -\frac{1}{2}K|z|^2 \right) \right] \exp\left( \frac{- |J|^2}{2K} \right) [/tex]

[itex][/itex]
[tex][/tex]
 
  • #2
I haven't worked through all the details, but it looks like there was a shift in ##z##, i.e. ##z'=z-\frac{2J}{K^2}##.
 
  • #3
I haven't worked through all the details, but it looks like there was a shift in ##z##, i.e. ##z'=z-\frac{2J}{K^2}##.

I don't think shifting [itex]z[/itex] by anything can help. Suppose you sent [itex]z \mapsto z+a[/itex] for any [itex]a[/itex] then I would get the following

[tex]-\frac{1}{2}K |z|^2+\bar{J}z \mapsto -\frac{1}{2}K |z+a|^2+\bar{J}(z + a) = [/tex]
[tex]-\frac{1}{2}K \left( z\bar{z} + a\bar{z} + z\bar{a} +a\bar{a} -\frac{2\bar{J}}{K} z - \frac{2\bar{J}}{K}a \right) = [/tex]
[tex]-\frac{1}{2}K \left( z\bar{z} + a\bar{z} + z \left( \bar{a} - \frac{2\bar{J}}{K} \right) +a\bar{a} - \frac{2\bar{J}}{K}a \right) [/tex]

If I want this to look like [itex]-\frac{1}{2}K|z+b|^2 + c[/itex], then I need to add and subtract terms with [itex]\bar{z}[/itex] which means that I can't pull [itex]e^c[/itex] out of the integral.
 
  • #4
I'm getting a similar problem. It looks it will work only if ##J## and ##z## are real.
 
  • #5
If the original exponent had an additional term ##+J\bar z##, it might work.
 

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