Let us consider Bloch wave function solutions for a particle confined to a 2D square lattice with a potential of the form $V=V(x) + V(y)$ (that is, one that can be factorized).

In this case we can factorize the Hamiltonian to $\hat{H} = \hat{H}_x + \hat{H}_y$ and the band spectrum structure (the energy dispersion for states) can be presented as $E( \vec{q}) = E_x(q_x) + E_y(q_y)$.

One-dimensional dispersion spectrum has gaps between the bands for quasimomentum values of $q_{x} = n \frac{\pi}{a}$, here $n$ is integer. It means that the energy gaps between the bands are situated at the quasimomentum border values of $q_{x,y} = n \frac{\pi}{a}$.

Am I correct? Is it a specific case of some general principle regarding the positions of band gaps I need to know about? What can be said about the band gap positions generally?

  • $\begingroup$ What do you mean when you say "the band gaps are situated at some quasimomentum"? I cannot make sense of this. Could you clarify your question? $\endgroup$ Jul 4 '11 at 5:08
  • $\begingroup$ @BebopButUnsteady: what he is doing is solving the Schrodinger equation for a periodic potential using the Ansatz $\psi(x)=e^{ikx}\eta(x)$. This is periodic on the lattice translations, but the lattice periodicity is unchanged when you add multiples of $2\pi/a$ to $k_x$ and $k_y$. Nevertheless, by continuity with the zero potential problem, you can label the states by the k value, so long as you are careful to understand that there is a discontinuity in the energy at the boundary of the fundamental momentum region. He is asking if this discontinuity can be shown to be the only one. $\endgroup$
    – Ron Maimon
    Sep 3 '11 at 20:24

Yes, you are correct, and you don't need to assume factorizability--- the situation for a general square-lattice periodic potential is just the same, the bangaps occur at the edge of the k-space domain defined by lattice periodicity.

The reason is all symmetry: the translation group is broken to lattice translations. This means momentum is not conserved, but violations can only come by adding multiples of the k-space reciprocal lattice. At high electronic energies (assuming a smooth V), you don't notice the potential and the bandgaps go to zero, but where V is important, there are bandgaps only when either the x and y momentum are an integer multiple of ($2\pi$ times) the reciprocal lattice size.

Momentum conservation

On the lattice, a momentum state $|k\rangle$ can mix off the potential with momentum states $|k'\rangle$ only when $|k'\rangle$ has the same eigenvalue for the operators $T_x$ and $T_y$, where

$$T_x= e^{iaP_x}$$ $$T_y = e^{iaP_y}$$

Since $T_x$ and $T_y$ commute with the full Hamiltonian. This means that a momentum state only mixes with other states which acquire the same phase for x and y translations by a, and this justifies Bloch's form for the eigenstates--- they are periodic in x and y up to a phase.

The mixing in general will split the energy of the states, which gives rise to the bands. The bandgaps occur at the exact point where the phase acquired when you translate one of x,y by a becomes equal to (an integer multiple of) $2\pi$.

To see why, it is easiest to start with a tight-binding model. Consider a potential which is a deep square well with very high walls, but periodic in x and y. The bound states are only mixed by tunneling, so that the bands are very narrow, their energy is nearly the same as the bound state energy, and these states are mixed with a phase which varies in a linear way from well to well.

If you give a periodic boundary condition on x which says that $\psi(x+a,y)=e^{i\phi}\psi(x)$ where $\phi$ is close to $2\pi$, but slightly less, then $\psi$ in each well is very close to an energy eigenstate, say the ground state, but with a little twist in the tunneling region. The moment this phase $\phi$ equals $2\pi$, however, there is the requirement that the result must be orthogonal to $0$ phase shift, and this condition was absent just before.

It is impossible to satisfy this condition when $\phi=2\pi$ by using a state which is close to the ground state. You must use the first excited state to continue past this value of k. This gives rise to the gap discontinuity.

The discontinuity appears immediately when you add a periodic potential. It always occurs at the first point where you move from the solution being the lowest energy state with the given lattice phase condition to the next lowest energy state with this phase condition. This happens exactly at the boundary of the k-space of the lattice.

Separable potentials

In this case, there is a zero-counting theorem for arranging the eigenstates. The n-th eigenstate has n zeros. In the Bloch case, the parameter "n" is the number of $2\pi$ twists of the phase as you scan from left to right. The first band has no twists, the second band has 1 twist, etc.

The additivity of the energy means that the 2-d bands are sums of the bands of each motion separately. I don't know what more there is to say.

General theorem

it should be possible to prove that the energy as a function of k is continuous except at the boundary of the lattice k-space region It is obvious physically, but I didn't give a proof.

  • $\begingroup$ I'm confused but I think I don't agree with either your answer or the question. Not knowing anything about QM, I know that a 2D square eigenvalue problem will have solutions for all positive values of $n$ and $m$ where those are the numbers for the nodes in both directions, meaning that the possible momentum values are $q_{x,y}=n m \pi/a$, and I don't understand how anything here has shown otherwise yet. $\endgroup$ Sep 2 '11 at 17:52
  • 1
    $\begingroup$ @Zassounotsukushi They are speaking of a square lattice, not a 2D square well! :-) $\endgroup$
    – mmc
    Sep 2 '11 at 22:19
  • $\begingroup$ @Zassounotsukushi: I think you are assuming that the 2d wavefunctions are in a square box with periodic conditions. The boundary conditions are for Bloch waves, with an extra phase, and then you get the phenomenon of bands. $\endgroup$
    – Ron Maimon
    Sep 3 '11 at 0:16

$q_{x} = n \frac{\pi}{a}$ is only correct solution here for a free electron model afaik, but depending of the actual form of your potential and the direction you view the crystal, you can and will get different positions of the gap or not at all one. Also bands can overlap if they get energetically wide enough.

But I'm no expert in this, as calculation of electronic band structures is a highly complex topic with lots of different electron models and approximation methods.

From my phenomenological knowledge you distinguish into halfmetals, semiconductors, isolators, metals. Metals like Cu or Al dont have at all energy gaps when you view their band structure scheme. So i highly doubt there are general properties of positions one could derive. The term "bandgap" mostly relates to materials showing a continuous gap like in Germanium independent of directional view.

This isnt really deeply coverd in most exp. orientated standard solid states books IMO, but search for some theoretical physics solid state scripts on the net if you are deeply intersted. Enough available IMO. Maybe there are some general properties derived by lattice symmetries and potential forms.

  • $\begingroup$ q is not restricted to be an integer in a crystal, that's just wrong. This doesn't answer anything at all, but makes it sound like a complicated problem, when it's not. $\endgroup$
    – Ron Maimon
    Sep 3 '11 at 19:13

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