# Spectral properties in Solid state physics

So assume we have a periodic 1d Schrödinger operator $$- f'' + V(x) f(x)= \lambda f(x)$$ and we want $V$ to be periodic.

Now if we assume that we are on a finite interval and that we have periodic boundary conditions where $R$ denotes the period of the potential, then we have eigenvalues $E_0 < E_1 \le E_2 < E_3 \le E_4...$ and so on.

Okay, this is clear to me. Then, there is the case that such an operator is defined on the full interval. First question: Do we then need any boundary conditions? In my physics lecture we used so-called Born von Karmann boundary conditions (saying that $f(x+R) =f(x)$) in order to "prove" the Floquet or Bloch theorem which says that we can decompose $f(x) = e^{ikx} u_k(x)$. This theorem says that we can decompose the wavefunction in a periodic part$u_k(x+R) = u_k$ and a complex exponential $e^{ikx}$.

I somehow feel as if these Born von Karmann boundary conditions are not necessary in the sense that any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true?- In that case: Why do we want Born von Karmann boundary conditions?- My problem with the Born von Karmann conditions is that I find that they are not really boundary conditions, as they don't act on some boundary. So what about the domain of such an operator?

2.) In my physics lecture we noticed that due to these Born von Karmann conditions the possible $k'$s for the problem (appearing in the exponentials) are discrete. Not sure if this is automatically satisfied, even if we don't assume Born von Karmann boundary conditions? Then we said that for every $k$, the Schrödinger operator equation that you get by pluggin in the ansatz from the Bloch or Floquet thoerem has a discrete spectrum. Is this true? If so, what does this all have to do with bands, if everything is so nicely discrete? - Or do we just cal these things bands, since the $k$'s get so close, that we cannot really resolve the discrete structure?

3.) Is there any relationship between the finite-interval problem and the infinite interval problem or are these two completely different things?

• The use of the Schroedinger equation in solid state physics is an ad-hoc modeling approach that bypasses the technical difficulty of "correct" effective field theories for the same systems. The effective potentials used in these models are toy potentials. Luckily, many of these simplified systems agree rather well with the observed phenomenology. The additional effects that stem from the discretization of the spectra due to finite (or periodic) boundary conditions can be seen as an additional numerical trick to avoid the mathematical complexity of continuous spectra of linear operators. – CuriousOne Dec 11 '14 at 23:58

Then, there is the case that such an operator is defined on the full interval

I assume that by "full interval" you mean the whole real line.

First question: Do we then need any boundary conditions?

Yes, as noted by Sam Bader, boundary conditions are part of the Hamiltonian.

In my physics lecture we used so-called Born von Karmann boundary conditions ... in order to "prove" the Floquet or Bloch theorem

Indeed, Bloch's theorem requires translational symmetry. If you limit your domain to some finite interval, then such a symmetry is only achieved with periodic boundary conditions.

But, when your domain is the whole real line, there's no need to impose special boundary conditions to achieve translational symmetry: just the usual condition of boundedness of the solution is enough. See e.g. a special case — free particle Hamiltonian: It requires that the solution is bounded at infinity, and that's enough to get the solution as plane waves. And it does have translational symmetry, thus has conserved quasimomentum (which in this case coincides with momentum because period of this symmetry is arbitrarily small).

I somehow feel as if these Born von Karmann boundary conditions are not necessary in the sense that any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true?

This is wrong, and explained properly by Steve B.

My problem with the Born von Karmann conditions is that I find that they are not really boundary conditions, as they don't act on some boundary. So what about the domain of such an operator?

They do act on the boundaries of the domain: if your finite domain is $[a,b]$, then you have

$$f(a)=f(b),\\ f'(a)=f'(b).$$

This is precisely what Born — von Karmann conditions are about. If you try to implement these conditions in a matrix representation of your kinetic energy operator, you'll get a circulant matrix, which explicitly shows the translational symmetry. (Try to play with finite differences approximation of kinetic energy operator, you'll see this directly).

In my physics lecture we noticed that due to these Born von Karmann conditions the possible k′s for the problem (appearing in the exponentials) are discrete. Not sure if this is automatically satisfied, even if we don't assume Born von Karmann boundary conditions?

Indeed, even if you don't take Born — von Karmann boundary conditions, and instead take e.g. homogeneous Dirichlet or Neumann boundary conditions, you'll end up with discrete spectrum. That's the result of finiteness of the domain and is a general feature of Sturm—Liouville problem.

Then we said that for every k, the Schrödinger operator equation that you get by pluggin in the ansatz from the Bloch or Floquet thoerem has a discrete spectrum. Is this true?

Again, it is true as long as you have a finite domain. And Bloch theorem makes direct sense only in presence of translational symmetry, i.e. with Born—von Karmann boundary conditions in this case.

If so, what does this all have to do with bands, if everything is so nicely discrete? - Or do we just cal these things bands, since the k's get so close, that we cannot really resolve the discrete structure?

The continuous bands appear when we take the limit of crystal size $L\to\infty$ (i.e. take number of lattice cells to infinity). In this limit Born—von Karmann conditions just automatically transform into the conditions of boundedness of the solution at infinity, and as you increase number of lattice cells, the spectrum becomes more and more dense (i.e. the discrete levels corresponding to some $k$ come closer and closer to each other while their number increases), and in the limit of $L\to\infty$ it becomes continuous.

Note that on infinite domain you can't apply Born—von Karmann conditions — it doesn't make sense to say $f(-\infty)=f(+\infty)$, for instance, so you use the natural conditions of boundedness.

This is an approximation to the real crystal, in which there's a huge number of atoms in all (or some, for e.g. graphene) directions, so that we can really just suppose that it's infinite in the first approximation.

In real crystals the spectral lines are so dense, that they indeed can't be resolved — but this is not only because of instruments: it's because of natural widening of spectral lines: the uncertainty of the level energy due to finite lifetime because of spontaneous emission.

Is there any relationship between the finite-interval problem and the infinite interval problem or are these two completely different things?

Yes, the relationship is as I noted above: via a limit. There appear some subtle differences though, like the fact that eigenfunctions in the infinite domain become unnormalizable, and the spectrum becomes continuous, but physically it's still quite similar. Again, you can play with empty lattice approximation (i.e. Hamiltonian with constant potential) to better understand the properties of these problems.

• I like the mathematical approach that you take here: There are just a few things that I want to ask you now in order to be sure that I got this: 1.) In principle we could show that Bloch's theorem even holds if we take as the domain the whole real line and just demand that our solutions are bounded everywhere? 2.) Why aren't we looking for proper $L^2$ eigenfunctions instead, if we are looking at the full real line, as bounded functions do not have to be an element of any proper Hilbert space? 3.) Do you know how these periodic potentials look like in practice? – Xin Wang Dec 16 '14 at 20:07
• 4.) Assume that you solved the finite-interval problem with periodic boundary conditions( so $H \psi = E\psi$ on some finie domain $[0,L]$ with periodic Boundary conditions). How do you construct from this the solution for mutiple cells? 5.) So if we are on the full real lines, do we get continuity only in the $k-$values or is also the energy spectrum for each individual $k$ itself continuous? ( I guess in the finite-interval case are both discrete) – Xin Wang Dec 16 '14 at 20:08
• 1) Right 2) Because we only require physically preparable states to be in Hilbert space. Eigenstates of continuous spectrum are usually treated using rigged Hilbert space formalism. Without these states you'd find that there're no eigenstates at all in these problems, and even if there were, you'd not be able to expand most of the wavefunctions in that basis, so it's not complete without continuous spectrum states. 3) Take $A \cos(x)$ for example. With it Schrödinger equation is solvable in terms of Mathieu functions. – Ruslan Dec 17 '14 at 4:46
• See also Kronig-Penney model for introduction in simple models in solid state physics. 4) Your initial domain can already have multiple cells (say, $N$). You just apply the Bloch theorem and reduce the problem to the domain of length $L/N$. After that you may consider the problem for any $N$ solved — just use the corresponding values of $k$. 5) The energy spectrum is always in the form $E_n(k)$ where $k$ is quasiwavevector, and $n$ is number of band. If $k$ varies continuously, then of course $E$ is also continuous. But – Ruslan Dec 17 '14 at 4:51
• for given $k$, $E$ as a function of $n$ is discrete — the bands are still discrete entities. This is because of the finite length of lattice cell. You can see this if you apply Bloch's theorem and solve the problem for the case of $k=0$. It'd be the same as original problem, but constrained to one lattice cell. – Ruslan Dec 17 '14 at 4:53

any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true?

No!! The eigenfunctions are Bloch waves $\psi(x) = u(x)e^{ikx}$, where $u$ is periodic (with the period of the lattice). But the product $\psi$ is not periodic (with the period of the lattice) unless $k=0$. I put up an example on Wikipedia recently:

A Bloch wave (bottom) can be broken up into the product of a periodic function (top) and a plane-wave (center). The left side and right side represent the same Bloch wave broken up in two different ways, involving the wave vector k1 (left) or k2 (right). The difference (k1−k2) is a reciprocal lattice vector. In all plots, blue is real part and red is imaginary part.

I made this plot mainly to show why $k$ is ambiguous, which is not relevant to your question. But if you just look at the bottom row, you'll see that Bloch waves are not periodic with the potential's period.

Here is a stupid example. In the vacuum, $V(x)=0$, which is (trivially) periodic with period 1 nanometer (I could have put any other distance here). But its energy eigenstates are plane waves which are not generally periodic with period of 1 nanometer.

do we just call these things bands, since the k's get so close, that we cannot really resolve the discrete structure?

Yes indeed! A macroscopic crystal might have $10^{20}$ allowed values of $k$, equally spaced in the first Brillouin zone. So we can and should think of it as a smooth curve or surface instead of discrete points.

... Is there any relationship between the finite-interval problem and the infinite interval problem or are these two completely different things?

For an infinite interval, there are $\infty$ allowed values of $k$, instead of merely $10^{20}$. So the bands become literally smooth curves or surfaces instead of effectively smooth curves or surfaces. Apart from that, everything is the same.

...if we don't assume Born von Karmann boundary conditions...

...then usually the eigenstates cannot quite be written as $u(x)e^{ikx}$. In 1D, most realistic (non-periodic) boundary conditions will give you a superposition of the state with $+k$ and the state with $-k$, to form a standing wave instead of a traveling wave: $\psi(x) = A u_1(x)e^{ikx} + B u_2(x)e^{-ikx}$ where $A$ and $B$ are complex numbers and it turns out that $u_1=u_2^*$. Again, this $\psi$ does not have a $k$ in the way you're used to.

Does this change the band structure? What does "band structure" even mean if we're not talking about Bloch states? I believe the answer is: The term "band structure" is short for "band structure assuming Born von Karmann boundary conditions", and if your boundary conditions are different, then you must keep in mind that for you, the band structure is not directly a catalog of allowed states and energies, it is a step removed from that. (But it's just a small and straightforward step removed.) :-D

• thank you, you touched some aspects, but not all, so I just want to ask the remaining questions: I noticed that if we have $\psi(x) = e^{ikx} u(x)$, then this function is not square integrable( is an immediate consequence of the periodicity). Isn't this a problem, if we want to talk about actual eigefunctions (closely related to the question about domains of the Hamiltonian). You also indicated that the only difference between the finite-interval and full interval problems is the number of allowed $k$-values. If we assume the potential to be zero, don't we have continuous spectrum then? – Xin Wang Dec 14 '14 at 16:36
• I mean how is there a discrete set of eigenfunctions for $-f'' = \lambda f$ on the whole real line available? – Xin Wang Dec 14 '14 at 16:48
• @XinWang on the whole real line you have continuous bands of allowed energies, so indeed the eigenfunctions are not square-integrable. Special case of this is just free particle (i.e. that in constant potential). – Ruslan Dec 16 '14 at 15:34
• @SteveB in fact we don't use Born-von Karman conditions for the whole real line - our boundary conditions are boundedness at infinities. – Ruslan Dec 16 '14 at 15:37
• @Ruslan you seem to be also quite an expert on this, would you mind talking a little bit about this problem in more detail? (Post an answer by yourself) – Solid State Physicist Dec 16 '14 at 18:07

It's good that you're considering questions like this; I find that this type of questions really forces a student to a deeper understanding of the math involved.

Do we then need any boundary conditions?

Yes, boundary conditions should be considered as part of the definition of the Hamiltonian and its domain. Different boundary conditions can result in different eigenfunctions/eigenvalues.

I somehow feel as if these Born von Karmann boundary conditions are not necessary in the sense that any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true?

No, even with BvK conditions, this is not true (see Steve B's answer for examples). What is true is that (with BvK), the wavefunction is periodic up to an an $e^{ikx}$ phase. Without BvK, neither of those statements is necessarily true.

In that case: Why do we want Born von Karmann boundary conditions?- My problem with the Born von Karmann conditions is that I find that they are not really boundary conditions, as they don't act on some boundary.

A couple things:

1. The hope in solid-state physics is that, if you have a big enough region, and you are computing bulk properties, then the edges form a tiny part of your system compared to the volume, and so the results should not depend much on which boundary conditions you choose.

2. BvK are boundary conditions: they say that the wavefunction at the left boundary of the space is the same as the wavefunction at the right boundary of the space.

3. BvK are useful, because they keep the discrete translational symmetry of the potential [translate by $a$ and nothing happens]. This is the symmetry which the Bloch waves represent, and if you choose boundary conditions which do not preserve that symmetry, then you will not have solutions which reflect that symmetry.

• If you took some other set of boundary conditions, eg wavefunction goes to zero at both infinities, then your eigenfunctions (if they exist) would not be Bloch waves. They might be able to look a lot like Bloch waves over some finite region, but would have to fall off at some point. By definition, Bloch waves have the same norm in every unit cell.
• BvK provides you with conveniently normalizable eigenfunctions, since your space is finite. You don't have to worry about integrating out to infinity for the unbound states and having the norm explode.

In my physics lecture we noticed that due to these Born von Karmann conditions the possible k′s for the problem (appearing in the exponentials) are discrete. Not sure if this is automatically satisfied, even if we don't assume Born von Karmann boundary conditions? True, this is not a special thing about BvK. If you designed the problem so the periodic lattice actually was large but finite and took the boundary conditions to go to zero at infinity, then you would find a discrete spectrum of bound states in the lattice region that look something like Bloch states deep in the lattice but then fall off at the edges. (I assume the lattice potential is strong enough to have bound states). This is, of course, a more realistic case then BvK, since materials are finite in the real world...but you don't do it because solving for eigenfunctions just became quite ridiculous.

Then we said that for every k, the Schrödinger operator equation that you get by pluggin in the ansatz from the Bloch or Floquet thoerem has a discrete spectrum. Is this true? If so, what does this all have to do with bands, if everything is so nicely discrete? - Or do we just cal these things bands, since the k's get so close, that we cannot really resolve the discrete structure?

Exactly. If we have well-separated nondegenerate bands, then lowest "band" is the set of {for each k, take the lowest eigenvalue}. The second lowest band is then {for each k, take the second lowest eigenvalue}. Etc. Since the allowed k's are so close as to basically be unresolvable, each band looks like a continuum of states.

• I actually have to other questions: Actually the $V=0$ case is a good example, which shows us some problems. In that case, the set of $k's$ should not be discrete, right? So do we have to exclude the period $0$ or can this also happen for non-zero finite periods? Also, you say "BvK provides you with conveniently normalizable eigenfunctions, since your space is finite", are you sure that we only apply BvK boundary conditions, if we consider FINITE intervals? – Solid State Physicist Dec 14 '14 at 19:03
• Regarding your first question. Say you apply BvK to a length $L$ of the real line. Even if $V=0$, BvK will still give you discrete k's: $k_n=2\pi n/L$. Your solutions are just plane waves $e^{ikx}/\sqrt(L)$. – Sam Bader Dec 14 '14 at 19:40
• And I'm not sure what your second question is. Using BvK means that you are basically solving the problem on a finite sized region (ie you can explicitly say it is length $L$) that is "connected" to itself end-to-end. Since the region over which you integrate is finite, a well-behaved wave-function will have a finite norm. (As opposed to those periodic plane wave solutions in an infinite free space which are normalized to $\delta(0)$.) – Sam Bader Dec 14 '14 at 19:41
• Well, actually I did not know so far that BvK is only for finite size. But this is actually a problem, cause if you really want to treat the infinitely extended case, then the Bloch decomposition theorem cannot hold, as your solution is not $L^2$ if it is periodic and infinitel extended, is this true? – Solid State Physicist Dec 14 '14 at 20:20
• Well I don't know if BvK is only for finite size, maybe you can do something similar for an infinite line segment; frankly, I don't know what that would mean. In practice, you generally just treat the infinite-extended case as above, but take $L\rightarrow\infty$, so then your allowed k-values become really close. ie it's just a limit of the finite case. – Sam Bader Dec 14 '14 at 20:37