I got a question concerning the band strucutre of solids. The reference I'm using is the book on solid state physics by Ashcroft/Mermin.

My problem is that I don't completely understand the reason why there exists a band-index n. On the one hand Ashcroft gives a quite plausible explanation saying that the general solution of the given eigenvalue problem H $\psi_k(r)$ = $E_{k}$ $\psi_k(r)$ can be decomposed into $\psi_k(r) = e^{i*k*r} u_{k}(r)$. Plugging this ansatz into the Schroedinger-equation and applying H to the $e^{i*k*r}$ first we obtain a new Hamiltonian $H_{k}$ that depends on the "wavevector" k and a new eigenvalue-problem $H_{k} u_{k}(r) = E_{k} u_{k}(r)$. Now we fix k. The operator $H_{k}$ acts on r and therefor produces a certain number of solution u_{n,k}(r). These solution can be counted using the index n.

So far so good.

Then Ashcroft states that the second proof of Bloch's theorem shows that $u_{k}(r)=\sum_{G} c_{k-G} e^{i G r}$. The sum is taken over all reciprocal lattice vectors G. He says that from this way of writing the function $u_{k}$ it is obvious that multiple solution do exists. I don't really understand why that is the case. First i thought that you may think of the n-th solution to be $u_{k,n}(r)=c_{k-G_{n}} e^{i G_{n} r}$ where $G_{n}$ denotes the n-th reciprocal lattice vector. However this doesn't seem plausible to me, since you need to have the sum to proof that $\psi_{k}$ = $\psi_{k+G}$ and therefor E(k)=E(k+G).

As you can see I'm quite confused about all this. Actually I'm also a bit confused why the reduction to the first brillpuin zone doesn't produce more than one Energy value for a given k.

Anyway, I'd be more than happy if someone could help me. Thanks in advance!!

See you.

  • $\begingroup$ I don't see any question, could you made it explicit? Only thing you have written is that it is possible to express $u_{k,n}$ in two different ways. One of them is in the usual spatial basis and the other is the reciprocal basis. But the correct expression for $u_{k,n}$ in the second case would be $u_{k,n} = \sum_G c_{n,k - G} e^{iGr}$. $c_{n,m}$ are to be thought as Fourier modes and of course depend on $n$. $\endgroup$ – Marek Jan 8 '11 at 14:11

Ashcroft and Mermin are occasionally not too careful about notation, and so the resolution is, as Marek mentioned in a comment, related to the fact that $$ u_k = \sum_G c_{k-G}e^{iGr} $$ admits several solutions -- every set of coefficients $\{c\}$ corresponds to some function $u_k$, but we're only interested in the $n$ solutions that solve the Schroedinger equation.

Every delocalized band arises because of the hybridization of quantum states between unit cells. If there are $n$ states that the electron can occupy, their will be $n$ (possibly degenerate, depending on the symmetry of the $G$-space point) bands. I say this to clarify that there is no sense in which the band index corresponds to the set reciprocal lattice vectors.

However, to answer your last question, you're absolutely right that reduction to the first Brillouin zone does give you multiple values of $E$ for a given $k \in IBZ$. If I remember correctly A&M introduce band theory by starting with the nearly free electron model and should show explicitly that this is the case. In the tight binding picture (sometimes called Linear Combination of Atomic Orbitals) the energies for $k$ outside the IBZ fall back onto their values in the IBZ.


Specifically, I wanted to point out that $E_n(k)=E(k+G_n)$ simply doesn't make any sense. Where did you read that? However the point is that the eigenvalues that make up a particular band are periodic in $G$, so it is true that $E_n(k)=E_n(k+G)$. Now, what this means is that a description with any more than the first Brillouin zone is redundant. The fact that you have multiple bands does not come from the reduction, though it can look that way at first, as A&M present it. I suggest skipping ahead to the chapter on tight binding to understand how multiple bands arise.

Continued again:

Okay, I didn't address this as clearly as I'd hoped. The reason a new quantum number appears at all is that we know from experience that $E(k)$ is not single-valued. The reason for this is that each atom has a set of discrete orbitals, and mathematically we can describe the motion of an electron in a lattice as tunnelings between these orbitals... However, these orbitals have very different symmetries, and we can not, for example, hop from an s-orbital on site-1 to a p-orbital on site-2. Thus each flavor of orbital symmetry will basically correspond to a new band (with a lot of technical complications that will only muddy the point).

Lastly, you're right about $E_n$ being bounded everywhere, and your reasoning is right as well. What was the question?

  • $\begingroup$ Okay. Thanks. This answers the second and third point of my question. However, it doesn't answer the question in the first paragraph of my posting, which is in short: "Why at all does the quantum number n appear in the coefficients? (This is not clear when one is making the ansatz)". For a more detailed formulation of this question, please see my last posting. Thanks in advance again! $\endgroup$ – Solidz Jan 8 '11 at 19:34
  • $\begingroup$ Well, of course it should appear in the ansatz in order to get the result of the book. But when you would first adress this problem without knowing about n, I think it wouldn't be clear why to choose a discrete index n for solutions. Just one other question: I've heard that it follows from $E_{n}(k) = E_{n}(k+G)$ that the function $E_{n}$ is bounded. Here I would argue like this: $E_{n}$ is a continuous functions (of the PDE's coefficients). The first Brillouin zone is a closed subset. Hence $E_{n}$ is bounded on the first BZ. Hence (because of periodicity) it's pounded everywhere. $\endgroup$ – Solidz Jan 8 '11 at 19:55
  • $\begingroup$ Sorry I might have accidentaly deleted your comment. $\endgroup$ – Solidz Jan 8 '11 at 19:59
  • $\begingroup$ I just realized what you were asking -- I'm sorry, I'll be adding it to my answer! $\endgroup$ – wsc Jan 8 '11 at 20:02
  • $\begingroup$ Okay! I think this answers my question. Thanks! You really gave me much more insight! $\endgroup$ – Solidz Jan 8 '11 at 20:45

Thanks for your respose wsc. Let me maybe go a bit more into detail now. When we start solving the given eigenvalue problem of an single electron in a periodic potential we assume a solution of the form $\psi_{k}(r)=\sum_{k} C_{k} e^{ikr}$. Plugging this in leads to a set of algebraic equation. However one notices that for a fixed k only the coefficients $C_{k}$, $C_{k+G}$, ... are coupled. Hence we obtain a solution of the form $\psi_{k}(r)=\sum_{G} C_{k-G} e^{i(k-G)r}$. So if we wanted the coefficient to depend on a quantum number n we should've known that beforehand and should've added this fact to our first ansatz. However I would say that at the beginning this is not so obvious.

Secondly is there a specific formula for the eigenvalues $E_{n}(k)$. I've read in a book that $E_{n}(k)=E(k+G_{n})$ where $G_{n}$ is the reciprocal lattice vector corresponding to the energy $E_{n}$. However this formula seems wrong to me and I also don't understand the reasoning here. Don't missunderstand me, I'm fine with 'existence'. I don't need an explicit formula. Still it would be kind of interesting to know.

So thirdly concerning the reduction to the first Brillouin zone. When I'm looking at a usual E-k-diagramm with $\pi / a < k < \pi / a$ as usual and I'm fixing a certain k. Then I see that there is in general more than one energy corresponding to this k-vector. Now is this the result of the different energybands or is it the result of the reduction process?

To me it seems as though that these additional energies corresponding to one k vector are the same anyway. For example: Fix k. Then the energies corresponding to this vector are $E_{k}$, $E_{k+G}$, ... where G is a reciprocal lattice vector. However Ashcroft derives the formule that $E_{k} = E_{k+G}$. So the energies are the same anyway.

Thanks for your response in advance!


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