On p22 of "Green's Functions for Solid State Physicists" by Doniach and SondHeimer, there is the following definition:

$$G^0(\omega)=\frac{1}{2M\Omega_0}\left( \frac{1}{\omega-\Omega_0+i\eta} - \frac{1}{\omega+\Omega_0-i\eta} \right)$$

which (according to the author) simplifies to


for $\eta>0$. They then proceed to explain that the poles occur at $\omega=\pm (\Omega_0-i\eta)$. I don't understand this step. Moreover when I calculate the poles of the two different (but apparently equivalent) expressions with Mathematica I get

for the first one and

for the second one. What is going on here? Why are the poles different if these two expressions are equivalent. I guess maybe some approximation has been made because $\eta$ is an infinitesimal constant but ultimately I want to use this same technique with a finite $\eta$ corresponding to damping.

Any insight gratefully received :)

Edit 1: Applying these ideas to a finite (non-infinitesimal) $\eta$

Thanks to some comments and answers I understand the situation. However I still have some confusion which I'll explain here, but perhaps this should be a different question. In the book I reference, the authors explain that instead, if you have some finite value for $\eta$, suppose there is a pole at $\Omega_0-i\Gamma_k$ then the Fourier transform $G^0(t)$ acquires a real exponential factor $\exp(-\Gamma_k t)$ (whereas with an infinitesimal $\eta$ there is no factor left over in the Fourier transform, corresponding to an infinite lifetime). In my work I have derived the Green's function


For this particular $G_q(\omega)$ what will the poles be, and what will the exponential decay factor be? I'm assuming it won't simply be $\exp(-\omega\gamma t)$ because I can't play the same tricks as above because it's a finite, not infinitesimal factor?

  • 4
    $\begingroup$ the actual simplification is $G^0=\frac{1}{M\Omega}\frac{\Omega-i\eta}{\eta^2+\omega^2-\Omega^2+2i\eta\Omega}$. The $-i\eta$ factor in the numerator can be droped (its irrelevant for the poles). The $\eta^2$ can be dropped ("second order infinitesimal"). Finally (and most importantly), the factor $2i\eta\Omega$ is redefined into $i\tilde\eta$, where $\tilde\eta=\Omega\eta$, which has the same sign as $\eta$, and is also infinitesimal. In the end, the author drops the tilde (for notational convenience). $\endgroup$ – AccidentalFourierTransform Jan 1 '16 at 20:03

Your two expressions for $G^0$ are not identical, only when $\eta$ is taken to be a formal infinitesimal constant (so we take $\eta$ and $\text{const.} \eta$ to be the same and $\eta^2 \simeq 0$). This is useful if you only use $\eta$ to locate the zeroes of the denominator on the right place. In reality, the first formula can be rewritten as

$$ G^0(\omega)= \frac{1-i \eta/\Omega_0}{M(\omega^2 - \Omega_0^2 + 2 i \eta \Omega_0 + \eta^2)}.$$

Only if you ignore the $i \eta/\Omega_0$ term in the numerator and the $\eta^2$ term in the denominator to zero, and if you rescale $2 i \eta \Omega_0 \to i \eta$, you recover the second formula.

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$\eta$ does not necessarily represent the same quantity in the two expressions. Since $\eta$ is a positive real infinitesimal, there is no meaningful difference between, say, $\eta$ and $\eta/|\Omega|$. If you keep this in mind and expand the square roots to linear order in $\eta$ (the only order that matters, since $\eta$ is infinitesimal), the results should agree.

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