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I'm just wondering something which seems to be true: is the Fermi surface given by $A(k,\omega=0)$?

This spectral function is the imaginary part of the diagonal of the retarded Green function at zero temperature, so the above expression tells you the number of gapless states connected to the ground state with wave vector $\mathbf k$. I'd think this is related to the Fermi surface.

I've read a few papers which plots these guys on different graphs, so now I am unsure. Thanks for any help!

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Answer: Right, for Fermi liquid. Maybe No, for strong-correlated system.

According to Fermi liquid theory, Green's function has the general expression: $$G(k, i\omega_n)=\frac{Z_{k}}{i\omega_n-\epsilon_{k}-\Sigma(k, i\omega_n)}$$ (retarded Green's function will replace $i\omega_n$ as $\omega+i\delta$ here) where $\Sigma(k, i\omega_n)$ is the self-energy. One of important property for Fermi liquid is that: the quasi-particles have infinite life-time at Fermi surface, i.e. they are free from scattering when $\omega=0$. From the view of Green's function, "life time" of quasiparticles is corresponding to the imaginary part of self-energy, which means: $$\operatorname{Im} \Sigma(k, \omega\rightarrow 0) \rightarrow 0$$ Also, it's important to note that the relation between "life time" and weight $Z_k$:

$$Z_{k}=\frac{1}{1+\frac{\partial Im \Sigma \left(k_{F}, \omega_n\right)}{\partial \omega_n}|_{\omega_n=0}}$$ Therefore, the decay to zero at a faster rate for the imaginary part of self-energy is equivalent to the non-zero $Z_k$ at Fermi surface. From view of physical picture, non-zero $Z_k$ at $\omega=0$ actually means that there exists well-define quasiparticles at at Fermi surface. And we know that spectral function $A(k,\omega) \propto Z_k$ represents the amplitude of finding quasiparticle at energy $\omega$ and momentum $k$ If so, your argument is right: you can find the Fermi surface by calculate the $A(k,\omega=0)$ since all the quasiparticles are well-define at Fermi surface in this case.

However, for strong-correlated system, the above property for Fermi liquid, i.e. all the quasi-particles have infinite life-time at Fermi surface, is not necessarily correct, i.e. $\operatorname{Im} \Sigma(k, \omega\rightarrow 0) \neq 0$ at some momentum. Thus, $Z_k$ is possible to vanish at some momentum near Fermi surface. As the result, if you still calculate $A_(k,\omega=0)$, you may find Fermi surface is strange, i.e. strength is different for different $k_F$. One of extreme examples is the "Fermi arc" in cuprates, i.e. phenomenological YRZ Green's function[1] can only gives truncated arc at Fermi surface.

If you really care about Fermi surface, maybe you can first calculate the singularity of Green's function and obtain the dispersion $\xi(k)$ for quasi-particle. Then, calculate the spectral function of $G=\frac{1}{\omega+i\delta-\xi(k)}$.

Reference:

1.Yang, K.-Y., Rice, T. M. & Zhang, F.-C. Phenomenological theory of the pseudogap state. Phys Rev B 73, 174501 (2006).

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  • $\begingroup$ Thanks for the info. My question came from a discrepancy in plots for a paper on the Pair Density wave state, so I'd say this probably is true for the cuprates (as you pointed out). $\endgroup$
    – scruby
    Mar 12 '21 at 19:28

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