Integrating Gaussians with complex arguments

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  • #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 cant 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 doesnt 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]
 

Answers and Replies

  • #2
mathman
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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
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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 dont 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 cant pull [itex]e^c[/itex] out of the integral.
 
  • #4
mathman
Science Advisor
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I'm getting a similar problem. It looks it will work only if ##J## and ##z## are real.
 
  • #5
mathman
Science Advisor
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461
If the original exponent had an additional term ##+J\bar z##, it might work.
 

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