According to the Wikipedia page on many-body Green's function, since the spectral density $\rho$ obeys the sum rule

$$ \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega}{2\pi} \rho(\mathbf{k},\omega) = 1, $$

the retarded Green's function given by

$${\displaystyle G^{\mathrm {R} }(\mathbf {k} ,\omega )=\int _{-\infty }^{\infty }{\frac {\mathrm {d} \omega '}{2\pi }}{\frac {\rho (\mathbf {k} ,\omega ')}{-(\omega +\mathrm {i} \eta )+\omega '}}}$$

has the asymptotic behavior

$$ G^{\mathrm{R}}(\omega)\sim\frac{1}{|\omega|} $$

as $\omega\rightarrow\infty$.

Does this result assume certain properties of the spectral density? Pehaps $\rho$ is nonzero only over a finite range of $\omega$?


Yes. It does assume that the spectral function $\rho(\omega)$ is only non-zero over a finite range of $\omega$. This is a reasonable assumption. No physical system can have spectral weight all the way to infinite energy.


Using the definition of the retarded Green's function, one shows that $$\rho({\bf k},t)=\langle [\psi_{\bf k}(t),\psi_{\bf k}^\dagger(0)]\rangle ,$$ possibly up to some signs and/or factor $i$. I will only treat the case of bosonic operators. Here $\rho({\bf k},t)$ is the Fourier transform of $\rho({\bf k},\omega)$.

The large frequency behavior of $G({\bf k},z)=\int_{\omega'} \frac{\rho({\bf k},\omega')}{\omega'-\omega}$ is thus $$G({\bf k},z)=\frac{\int_{\omega'}\rho({\bf k},\omega')}{z}+\frac{\int_{\omega'}\omega'\rho({\bf k},\omega')}{z^2}+\cdots$$ Using $\int_{\omega'}\rho({\bf k},\omega')=\rho({\bf k},t=0)=\langle [\psi_{\bf k}(0),\psi_{\bf k}^\dagger(0)]\rangle =1$, we obtain the asymptotic behavior of the Green's function.

One can use $$\int_{\omega'}\omega'\rho({\bf k},\omega')=\partial_t\rho({\bf k},t)|_{t=0}= \langle [[\psi_{\bf k}(t),H],\psi^\dagger_{\bf k}(t)] \rangle,$$ with $H$ is many-body Hamiltonian, to obtain the next correction (exactly), in terms of averages of fields (and the parameters of the Hamiltonian), which are however hard to compute.


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