# Localized states and density of states

• A
Hello,

Let's suppose we have a two dimensional lattice which is periodic along certain direction, say x-direction, allowing us to define a quasi momentum k_x. The lattice is not periodic along the y-direction (perpendicular to x-direction). Therefore, we are able to obtain the band structure of the system as a function of k_x. If the system is finite along y-direction, for some values, we can observe edge states in such band structure. The question is the following, can we get information about the localization of the wave functions of the system (in particular, localization properties of edge states) by looking at the density of states along the y-direction? If so, how can that be done? I thought this after reading some articles at where the localization of wave functions is indicated by colors that correspond to the density of states along y (see, for example, https://arxiv.org/abs/1407.7747, Fig. 3).

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In the density of states along the y direction, quasi-momentum would no longer be a good quantum number, and you would have to switch to something like chemical potential. So you could calculate the DOS numerically (I would just use a simple if loop to manually count the states). I think you would first plot the dispersion relation, maybe chemical potential vs. E, and then could extract a DOS from that (the Hamiltonian could be partially Fourier transformed, so you have kx in the x direction and j in the y direction, or something like that). I'm not sure about your wavefunction question. Are you looking for zero energy states? If so you could use the single particle Hamiltonian and see what comes out at zero energy. I don't think they will be localized unless you have finite boundaries in both directions, but they should be chiral.

In the density of states along the y direction, quasi-momentum would no longer be a good quantum number, and you would have to switch to something like chemical potential. So you could calculate the DOS numerically (I would just use a simple if loop to manually count the states). I think you would first plot the dispersion relation, maybe chemical potential vs. E, and then could extract a DOS from that (the Hamiltonian could be partially Fourier transformed, so you have kx in the x direction and j in the y direction, or something like that). I'm not sure about your wavefunction question. Are you looking for zero energy states? If so you could use the single particle Hamiltonian and see what comes out at zero energy. I don't think they will be localized unless you have finite boundaries in both directions, but they should be chiral.
Thanks for your answer DethbyGreen. You're right, along the y-direction the system is not periodic, hence I have to calculate the DOS numerically, what I already did. On the other hand, I am not looking for zero edge modes, instead I am looking for edge states, which are not necessarily zero energy states. Actually, I already found such edge states. What I want is to prove if they are chiral edge modes or not. To do that, I was thinking in plotting the band structure using a color code. The color code must be related with the localization properties of the system's wave functions. I used the inverse participation ratio and found that the edge states are indeed localized. However, the inverse participation ratio did not shed light about where the edge states are localized. For solving that I did a research on internet and I found that it is possible to obtain the localization of wave functions using the DOS along the y-direction (which is not periodic), unfortunately, no paper I read say how to do that. So the question, in a nutshell, is: once that the DOS has been obtained, can I get some information about the wave functions' localization from it, I mean, is it possible to know where the wave function is localized using only the DOS?

DrDu
When you calculate the DOS for fixed k_x, any discrete eigenvalue must correspond to a localized state.
Maybe you can do a simple model: let ##e_0## be the energy of an unperturbed surface state, approximately independent of ##k_x##.
Let ##e_{k_x,k_y}\ge e_{k_x} ## be a continuum of unperturbed bulk states and ##v## be a coupling between the surface and bulk states.
Now try to find the discrete eigenenergy and eigenstate of this simple hamiltonian in dependence of ##e_0##, ##e_{k_x}## and ##v##.

When you calculate the DOS for fixed k_x, any discrete eigenvalue must correspond to a localized state.
Maybe you can do a simple model: let ##e_0## be the energy of an unperturbed surface state, approximately independent of ##k_x##.
Let ##e_{k_x,k_y}\ge e_{k_x} ## be a continuum of unperturbed bulk states and ##v## be a coupling between the surface and bulk states.
Now try to find the discrete eigenenergy and eigenstate of this simple hamiltonian in dependence of ##e_0##, ##e_{k_x}## and ##v##.
Thanks for your answer DrDu. Since the system is finite, as you said, any discrete eigenvalue must correspond to a localized state. But I consider that a localized state near the center of the sample is a bulk state. Whereas localized states near the edges are edge states. I hope this make sense. At least is what I have found in some papers (see, for exmple, Fig 3 in https://arxiv.org/abs/1407.7747), maybe I am missing something. As before, my question is as follows: is there a way to know at where is a state localized just by looking at the DOS along y-direction? Thanks again for your answers.

DrDu
Ok, now I understand your question. So what kind of localised states are possible?
There are surface states, localized states at point defects and localised states at line defects.
As long as there is no reordering of the surface (which can easily be tested experimentally), the surface state will have the same translational symmetry in x direction as the bulk crystal. So the surface states should form a band in x direction with the same Brillouin zone as the bulk.
However for fixed k_x, there should be an isolated energy eigenvalue.
On the other hand, point defects in the bulk or at the surface won't lead to a band structure but to an isolated energy eigenvalue, independent of k_x.
Lastly a line defect. In general, a line defect won't be parallel to the surface, or at least not all line defects will be so. Rather, they are characterised by the Miller indices of the line. Along this dislocation line, the translational invariance will be reduced, which leads to the appearance of sub bands with corresponding singularities at the band gaps in the DOS as a function of k_x.

Furthermore I would expect surface states and line defects to be localised in different regions of the spectrum.

DrDu
Hello,

Let's suppose we have a two dimensional lattice which is periodic along certain direction, say x-direction, allowing us to define a quasi momentum k_x. The lattice is not periodic along the y-direction (perpendicular to x-direction). Therefore, we are able to obtain the band structure of the system as a function of k_x. If the system is finite along y-direction, for some values, we can observe edge states in such band structure. The question is the following, can we get information about the localization of the wave functions of the system (in particular, localization properties of edge states) by looking at the density of states along the y-direction? If so, how can that be done? I thought this after reading some articles at where the localization of wave functions is indicated by colors that correspond to the density of states along y (see, for example, https://arxiv.org/abs/1407.7747, Fig. 3).

Btw, a two dimensional lattice is by definition periodic in two directions. While it can happen that the lattice is periodic along x but not periodic along a perpendicular direction y, I wonder whether this is really what you mean. Or do you simply want to say that once you have a semi infinite lattice, the surface breaks translational symmetry?

DrDu
Thanks for your answer DethbyGreen. You're right, along the y-direction the system is not periodic, hence I have to calculate the DOS numerically, what I already did.
When you calculate the DOS, you also get the wavefunctions, don't you? So it should be possible to check directly whether they are located at the surface?

Ok, now I understand your question. So what kind of localised states are possible?
There are surface states, localized states at point defects and localised states at line defects.
As long as there is no reordering of the surface (which can easily be tested experimentally), the surface state will have the same translational symmetry in x direction as the bulk crystal. So the surface states should form a band in x direction with the same Brillouin zone as the bulk.
However for fixed k_x, there should be an isolated energy eigenvalue.
On the other hand, point defects in the bulk or at the surface won't lead to a band structure but to an isolated energy eigenvalue, independent of k_x.
Lastly a line defect. In general, a line defect won't be parallel to the surface, or at least not all line defects will be so. Rather, they are characterised by the Miller indices of the line. Along this dislocation line, the translational invariance will be reduced, which leads to the appearance of sub bands with corresponding singularities at the band gaps in the DOS as a function of k_x.

Furthermore I would expect surface states and line defects to be localised in different regions of the spectrum.
Thanks for your very detailed answer DrDu, I am looking for surface states and a way of knowing where they are located. I ended up using the wave functions of the system (see my last comment to your answers).

Btw, a two dimensional lattice is by definition periodic in two directions. While it can happen that the lattice is periodic along x but not periodic along a perpendicular direction y, I wonder whether this is really what you mean. Or do you simply want to say that once you have a semi infinite lattice, the surface breaks translational symmetry?
You're right, what I meant is that I have a two dimensional nanoribbon (a graphene nanoribbon, for instance), which is periodic (or infinite) along the x-direction but it is finite along the y-direction. Hence, as you said before, the surface states will form a band as a function of k_x.

When you calculate the DOS, you also get the wavefunctions, don't you? So it should be possible to check directly whether they are located at the surface?
I do, in fact, I have obtained the wave functions for the whole system. Yes, I can check directly if the wave functions are located at the surface or not. I ended up using the inverse participation ratio, which was obtained using the wave functions. I just got confused when I read that it was possible to obtain the localization of a wave functions just by checking the DOS.

Finally, I would like to thank you for your interest and your answers. They were very useful.