Integrating Gaussians with complex arguments

  • Context: Graduate 
  • Thread starter Thread starter hideelo
  • Start date Start date
  • Tags Tags
    Complex Integrating
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
hideelo
Messages
88
Reaction score
15
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]
 
on Phys.org
mathman said:
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.