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I know that semiconductors have a band gap, and conductors have bands that cross the Fermi level. However, can a material have a band gap for certain $\vec k$ regions but bands cross the Fermi level only for a certain $\vec k$? Does this mean that a material is a conductor in one direction and a semiconductor in perpendicular direction?

This paper [1] shows that a semiconducting MoS2 can have metallic edges. It also mentions that bulk materials can have two-dimensional metallic surfaces. This paper [2] shows conductive copper channels in a semiconducting black phosphorus. Some bands in its band structure cross the Fermi level (Fig. 5C), so it should be a metal, however the conductivity in the perpendicular direction is lower, as shown in their Fig. 5D.

Most band structures of conductors that I saw cross the Fermi level somewhere and don't cross them in other places, so I am confused.

Can a material be a conductor in one direction and a semiconductor in another direction and how can I see it from the band structure?

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  • $\begingroup$ Regarding the first paper, this sounds like a topological material, this answer by ProfM is a good introduction mattermodeling.stackexchange.com/a/1546/1554. And for the second paper, this looks like a semimetal, you can check mattermodeling.stackexchange.com/a/11913/1554. Other users would have a full answer for sure, but those links might be helpful in this regard. $\endgroup$ Jan 22 at 5:23
  • $\begingroup$ @JaafarMehrez Is it really a semimetal in the second paper? Isn't it like purely metallic lines inside a semiconductor? Which means metal in one direction and semiconductor in another? $\endgroup$ Jan 22 at 6:04
  • $\begingroup$ It is possible that you could observe metallic regions within a semiconductor, they could appear as isolated islands, I didn't go through the whole paper, but by looking at the electronic structure, maybe they came to this conclusion because the density of states is very low at the Fermi level or where the band crossing is happening thus they didn't use the term 'metal'. $\endgroup$ Jan 22 at 6:18

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If I understand your question, the answer is yes. Really, the only requirement is anisotropy. Consider e.g. a half-filled 1d chain. It is metallic along the chain and insulating perpendicular to it. I assume such things exists in e.g. polymer chains. Similarly for 2d and 3d systems.

Basically any text book on solid state physics will cover these things. I like Ziman's book, Ashcroft-Mermin, and have looked quite a bit at Girvin and Yang. Significantly more advanced and specialized, but a good introduction to second quantization, is Altland-Simons. Moreover, David Tong's notes on his website are much more concise and instructional.

One of the simplest Hamiltonians for a crystal is $ H = -\sum t_{ij} c^\dagger_j c_i + \sum_i \epsilon_i c^\dagger_i c_i$. This is "second quantized notation" with $c_i$ an operator that removes an electron from site $i$; $c^\dagger_j$ adds the electron to site $j$. $t_{ij}$ is the so-called hopping amplitude: it is the probability of the electron hopping from site $i$ to $j$. $\epsilon_{i}$ and $t_{ij}$ are parameters that depends on the atom types, crystal lattice, etc. We will assume they are already known in what follows.

$\epsilon_i$ is the on-site energy, $c^\dagger_i c_i \equiv n_i$ is the electron density operator: it counts the number of electrons on site $i$. $\epsilon_i n_i$ is basically the energy for an electron to be placed on site $i$. If all the atoms are identical, $\epsilon_i \equiv \epsilon$ and the on-site terms all add up to a constant that we can ignore (we ignore it later).

$t_{ij}$ is the probability for an electron to tunnel between sites. To make some sense of it, consider the example (I think due to Mott, it is discussed somewhere in Ashcroft-Mermin) where $t_{ij} \ll \epsilon$. Assuming each atom has the same number of electrons on it, $n_i \equiv n$, then $H \approx \epsilon \sum n = N \epsilon n$ where $N$ is just the number of unit cells in the crystal. Notably, this is just an array of isolated atoms with no possibility for an electron to hop between sites. It is a perfect insulator. At some point, with finite $t_{ij}$, the system goes from metal to insulator. The point is that $t_{ij}$, in-large part, determines the nature of the system.

example of 2d lattice model

Now to answer your question. Consider an anisotropic 2d model. See the image above. Assume that it is longer along the $y$-axis so that hopping along that direction is smaller probability. Let the hopping amplitude along $x$ be called $t$ and along $y$ be called $t'$ with $t' \ll t$. I.e. it is difficult for electrons to go along $y$, but easy along $x$. The Hamiltonian is $H = -\sum_i [ t (c^\dagger_i c_{i+x} + c^\dagger_{i+x} c_i) + t' (c^\dagger_i c_{i+y} + c^\dagger_{i+y} c_i)]$ with $c_{i+x}$ being the operator for the next site from $i$ along the $x$ direction and so on. This can be solved by Fourier transforms: $c_i = N^{-1/2} \sum_k c_k \exp(i \bf{k} \cdot \bf{R}_i)$ and so on. The solution is $H = \sum_{\bf{k}} E_{\bf{k}} c^\dagger_k c_k$ with $ E_{\bf{k}} = -2 [t \cos(k_x a_x) + t' \cos(k_y a_y) ] $. This band structure with $t=1$ and $t'=0.1$ is plotted below.

bands

Note that the dispersion crosses the Fermi level $E_f \equiv 0$ along the path from $\Gamma$ to $X$. These electrons have momentum parallel to the $x$ axis and, since there is no gap, can easily be exited into conduction states. Along the $\Gamma$ to $y$ direction, there are no electrons at the Fermi level, i.e. there are no conduction electrons with momentum parallel to $y$.

The conductivity in the Einstein theory is (see chp. 8 in Girvin and Yang) $\sigma=e^2 \rho(E_f) D$ with $D$ the diffusion constant, $e$ the electron charge, and $\rho(E_f)$ the density of states at the Fermi level. $D$ determine how long an electron that is excited into a conduction state remains conducting before being reabsorbed into the lattice. It depends on electron-electron interactions and other things. Assume it is a constant, the same for all electrons. $\rho(E_f)$ tells us how many electrons there are that can be excited. If we try to resolve this into directions, $\sigma_x \sim \rho_x(E_f)$ and $\sigma_y \sim \rho_y(E_f)$ with $\rho_{\bf{k}}(E_f)$ the $k$-resolved density of states (i.e. count of the states with momentum $\bf{k}$ at the Fermi level. It tells us how many electrons there are that can be exited into conduction states with momentum $\bf{k}$. Notably, $\rho_x(E_f) \neq 0$ while $\rho_y(E_f) \equiv 0$. The material is conducting along the $x$ direction, but not along $y$, because there are no available states along $y$. These examples occur in nature, e.g. the doped cuprates are conducting (and even superconducting) in the Cu-O plane, but insulating in the direction perpendicular to the Cu-O planes.

I hope this updated answer helps! Ty

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  • $\begingroup$ This might be the shortest answer we have ever had. Are you able to expand on it so that it's longer than a comment? See: "If I understand your question, the answer is yes. Consider e.g. a half-filled 1d chain. It is metallic along the chain and insulating perpendicular to it. I assume such things exists in e.g. polymer chains. Similarly for 2d systems". "If I understand your question, the answer is yes. Consider e.g. a half-filled 1d chain. It is metallic along the chain and insulating perpendicular to it. I assume such things exists in e.g. polymer chains. Similarly for 2d systems." $\endgroup$ Jan 25 at 19:06
  • $\begingroup$ Where can I read more about them and see how their band structure look? $\endgroup$ Jan 26 at 9:57

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