In The universe in a Helium droplet, Grigory Volovik relates the stability of a fermi surface to topology of a Green function. There he gives the example of a Fermi gas and says that the Green function for Fermi gas has form

$$G=\frac{1}{iw-v_{f}(P-P_{f})}$$ which has a singularity at $iw=0$ and $P=P_f$. The Volovik proposes that a topologial invariant for a Fermi surface can be found by calculating the winding number about this singularity using the invariant formula

$$N_1=\frac{1}{2\pi i}\oint dl G^{-1}\partial_lG $$

where $l$ corresponds to contour winding the singularity.

My questions is that singularity of Green function corresponds to the pole of a Green function, so if this is the case then $N_1$ also will not be zero for insulators. I cannot understand this.


The Green function you gave above is not the Green function for an insulator, which should generically has no pole if there is no Fermi surface.

The Volovik argument is a bit circular, since you know from the beginning that $p=p_{F}$ is the Fermi momentum, defining the Fermi surface. When $p\approx p_{F}$ you can infer the form of the Green function as you wrote in your question, and then from complex analysis you can write $N_{1}$. Then you realise $N_{1}$ is topologically non trivial since a perturbation $G=G_{0}+\delta G$ in $N_{1}$ will let it invariant (say differently $\delta N_{1}=N_{1}\left(G_{0}+\delta G\right)-N_{1}\left(G_{0}\right)=\mathcal{O}\left(\delta G^{2}\right)$ has no term linear in $\delta G$). So you promote $N_{1}$ to characterise the Fermi surface. In the case you gave $G^{-1}\left(z\right)=z+v_{F}\left(p-p_{F}\right)$ most certainly $N_{1}=\pm 1$ am I correct ?

I'm now wondering about the generic form of the Green function for an insulator. I feel it's something like $G^{-1}\left(z\right)=1$, but I do not understand why right now. I'd say it should be something sufficiently trivial analytically to have no density of state. The simplest way is to have no pole at all on the real axis. Clearly $N_{1}=0$ in this case.

Edit: Well, I wanted to see this in more details in fact. So let us try a simplistic model when a gap separates electron and hole bands.

A simple model

Suppose the Hamiltonian


with $p$ the momentum (instead of $p/\sqrt{2m}$), $\Delta$ a gap and $\mu$ a parameter (when going to many-body, it becomes the chemical potential). One sees that there is two bands (I adopt the condensed-matter terminology), say for electron and hole, both having the same effective mass and separated by the gap $\Delta$, which I suppose always positive. When $\mu>\Delta$, the Fermi surface is of electronic nature, whereas for $\mu<-\Delta$ only holes are present at the Fermi surface.

The associated Green functions are


with $\sigma=\pm$ representing the electron/hole ambivalence. Usually in condensed-matter, we prefer to discuss energy with the chemical potential taken as a reference, so we introduce $z=\omega-\mu$ (I drop $\hbar$). Then the Fermi surface is defined as the locus of the $p$'s lying at the chemical potential $\omega=\mu$. By construction, these loci are the pôles of $G_{\sigma}\left(z=0\right)$ above. One finds easily


for the pôles. The $\pm$ distinctions comes from the quadratic dispersion, so for each right-moving particle we have a left-moving one. This doubling is obviously essential in order to preserve the Galilean invariance, but it has nothing to do with the gap and the difference between metallic and insulating behaviour, so I drop the $\pm$ distinction from now.

Now we see the essential: for $\sigma = +1$ and $\mu>0$, there is no pôle along the real axis when $\mu<\Delta$. The same is true for $\mu<0$ and the hole sector $\sigma = -1$ if $\left|\mu\right|>\Delta$. In conclusion there is no pôle along the real axis in the gap. There are pôles along the real axis in the usual situation when the chemical potential lies above the gap $\mu > \Delta > 0$ in the electron sector and $-\mu>\Delta>0$ in the hole sector.

This is a generic argument : the Green function associated to an insulator has no pôle along the real axis, since the chemical potential lies inside a gap of forbidden momentum by construction. The Green function might have imaginary pôle inside the gap (as above in fact) which requires the momentum to be ill-defined (i.e. the momentum becomes imaginary). This can only happen by imposing boundary conditions since the "wave-function" is then evanescent. Such states are called for this reason edge states, or surface states.

Now to come back to the $N_{1}$ construction, we should define a generalised $N_{\sigma}$ which select whether we calculate the $N_1$ in the question using $G_{+}$ or $G_{-}$, and we will have

$$\begin{cases} \left|N_{1}\right|=+1\;;\;N_{-1}=0 & \mu>\Delta\\ N_{1}=N_{-1}=0 & \left|\mu\right|<\Delta\\ N_{1}=0\;;\;\left|N_{-1}\right|=+1 & \left|\mu\right|>\Delta \end{cases}$$

The sign in the metallic sector is of no real importance, it depends how one turns around the pôle. The important point is that $N_{\sigma}=0$ in the insulator sector, as is trivial the hole invariant when the chemical potential lies in the electron sector, and vice-versa $N_{1}=0$ when $\left|\mu\right|>\Delta$.

On the pôles of the Green functions

It seems the following remark is welcome (see the comment below from Meng-Cheng).

Usually, the spectrum properties of the system are obtained from the pôles of the Green function, taken as a function of $z=\omega-\mu$. For instance the Green function above $G_{\sigma}\left(z\right)=\left(z-\sigma\left(p^{2}+\Delta\right)+\mu\right)^{-1}$ has only one pôle $\hat{\omega}_{\sigma}=\sigma\left(p^{2}+\Delta\right)$ along the $\omega$-axis which corresponds to the eigen-energies of the system (the dispersion relation), and appears in the so-called spectral properties of the Green function [Economu]. As also remarked by Meng-Cheng, the difference between insulator and metal is then given by the possibility to have arbitrary low energy excitations. In the example, when $\Delta>0$ the lowest energy is $\Delta$ (i.e. $\hat{\omega}_{\sigma=1}\left(p=0\right)=\Delta$, which goes to zero for ungapped system (so for a normal metal).

In contrary, all the machinery developed in the previous section are concerned with the pôles of the function $G_{\sigma}\left(z=0\right)$ with respect to $p$. Only these later pôles of $G_{\sigma}\left(z=0\right)$ with respect to $p$ are associated to the Fermi surface [Horava].


  • $\begingroup$ I think Green function will also have poles for an insulator because the poles of the green function gives you only knowledge about the spectrum of the system it doesn't tell you about the system is insulating or not?@ FraSchelle am i right? $\endgroup$
    – 079
    Apr 2 '15 at 2:03
  • $\begingroup$ @079 The Green function clearly has no pole along the real axis in the insulating region, see edit about a simple model to see how this happens. In fact you're right the Green function gives you knowledge about the spectrum of the Schrödinger equation. More precisely the Green function has a pole along the real axis for each discrete energy, eventually it has branch cuts for bands. So how could a Green function has a pole when there is no associated state, as for a trivial insulator ? $\endgroup$
    – FraSchelle
    Apr 2 '15 at 5:19
  • $\begingroup$ The Green function, as a function of frequency, have poles corresponding to single-particle excitations. In your example, the pole is $z=\sigma(p^2+\Delta)-\mu$. The difference between metal and insulator is whether that the location of the pole can be arbitrarily small frequency (just gapless v.s. gapped). $\endgroup$
    – Meng Cheng
    Apr 2 '15 at 5:53
  • $\begingroup$ @MengCheng You're perfectly right, but for the Volovik construction, the integral is defined in momentum, and so the poles are the poles with respect to p, not z. The spectral properties correspond to the poles in z indeed, as you say. I developed the above (boring) machinery to be sure to say no stupidity in the case of Volovik $N_{1}$ invariant (it has been done by Horava in fact, see arxiv.org/abs/hep-th/0503006 ). But any way to prove the stability of the Fermi surface in a better way is indeed welcome :-) Thanks for your comment $\endgroup$
    – FraSchelle
    Apr 2 '15 at 9:45
  • $\begingroup$ @MengCheng Ok, I understand now your previous comment. Indeed, the comment I made to user:079 was perfectly unclear... sorry for that. So let me state it correctly in the answer. Please check the edit, and tell me if you agree. $\endgroup$
    – FraSchelle
    Apr 2 '15 at 11:46

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