1
$\begingroup$

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 $\omega$?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.