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May 28, 2017 at 12:06 comment added Alexey Sokolik Thank you! But I don't understand what do you mean by "resolving second-order pole by first-order poles". For example, if $A_\Pi(\omega)=(a/\pi)/[(\omega-\omega_0)^2+a^2]$ (Lorentzian spectral function), then $\Pi(\omega)=(1/2\pi)(\omega-\omega_0+ia)^{-1}$, i.e. $\Pi(\omega)$ has the first-order pole at $\omega=\omega_0-ia$, which means existence of a damped excitation. At the same time, $\Pi(\omega)=(1/2\pi)^2(\omega-\omega_0+ia)^{-2}$ still has the second-order pole at $\omega=\omega_0-ia$, as can be obtained either by direct squaring of $\Pi(\omega)$ or by using the function $A_\Pi^{(2)}$.
May 26, 2017 at 20:00 history edited Everett You CC BY-SA 3.0
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May 26, 2017 at 19:50 history answered Everett You CC BY-SA 3.0