I am trying to come to grip with some solid state theory. Bloch waves, energy eigenstates for hamiltonians with lattice periodic potential in $\mathbb R^d$, are frequently written as $$\phi_{n,k}(r)=e^{2\pi ik\cdot r}u_{n,k}(r)$$ with $u_{n,k}$ periodic in lattice vectors $R\in\mathbb R^d$. This is 1 version of Bloch's theorem. It implies that $k$ can be restricted to the 1st Brillouin zone. Thus multiplying $\phi_{n,k}(r)$ by a plane wave with frequency $K$ in the reciprocal lattice yields another Bloch wave with larger momentum -i.e. $p=\partial_r$ expectation value- corresponding to some other index $n'$, that is $$\phi_{n,k+K}(r)=e^{2\pi iK\cdot r}\phi_{n,k}(r)=\phi_{n',k}(r).$$ My question is then why do people (e.g. Ashcroft-Mermin) often set $\phi_{n,k+K}(r)=\phi_{n,k}(r)$ against the notation that seems much more natural to me that I used above $\phi_{n,k+K}(r)=e^{2\pi iK\cdot r}\phi_{n,k}(r)$ which attributes essentially momentum $k+K$ to $\phi_{n,k+K}(r)$?

According to their notation all $\phi_{n,K}(r)$ get equal momenta near 0 (more precisely $\langle u_{n,0},pu_{n,0}\rangle$) but if we turn momentum on from 0 to $K$, and beyond to multiples of $K$, continuously in momentum space our Bloch wave varying adiabatically will get larger and larger momentum and they will correspond to my notation for $\phi_{n,k+K}(r)$, not to nearly 0 momentum states. That implies also that to increase momentum while keeping the wavevector $k$ in the 1st Brillouin zone one must increase the index $n$.

A related question is: It seems that Fermi energy is ground state energy if we assume that arbitrarily high momenta get the same energy. We simply pick $\phi_{0,K}(r)$ as wavefunctions for our electrons with arbitarily high momentum $K$ in the reciprocal lattice with $\phi_{0,0}=u_0$ the ground state. Those all have the same energy, increasingly negative potential energy compensates increasing kinetic energy. This is related to having standing waves as energy eigenstates in perfect crystals -i.e. superconductivity when many electrons in are such states. The way out seems to be to consider that real crystals have boundaries. There is also scattering of electrons on phonons.

In real crystals the increasing kinetic energy with increasing momenta $K$ would not be compensated by an infinite periodic potiential. After the boundary the kinetic energy would not be compensated by further potential barriers, thus large momenta Bloch waves would actually be more energetic and electrons would tend to keep low momenta, essentially in the first Brillouin zone.

I find that many lectures on Bloch waves/band structure do not treat momentum crystal(-)clearly. Thanks for any hint.

Update: I am really curious so please share if anybody has any comment. This is related to umklapp processes and other things which are often explained handwavingly.

Update 2: Actually I have thought about my questions. The problem is that momentum cannot be increased continuously. A way to see this is in a simple 1D Kronig-Penney model. Approaching a reciprocal wavevector $K$ there is no Bloch wave with that wavevector so starting from $k=0$ if 1 passes above $K$, $u_{n',k}$ will change drastically/discontinuously to account for the increased momentum while $k$ is taken in the 1st Brillouin zone but is then farther from momentum. To have an accurate picture would require solving Schrödinger equations dynamically. The adiabatic varying eigenstate method I tried to picture myself fails. I may write more with more time.

Update 3: Actually I think I understand what puzzled me. It is that as wavevector $k$ is increased $u_k$ changes so as to compensate oscillations increase due to the $e^{ikr}$ factor. I was visualizing multiplying an almost fixed $u_k$ by $e^{ikr}$ with large k. But continuing from 0 to K in the same band yields exactly $\phi_{n,0}$ physically because $u_k$ unwinds/compensates the more oscillating factor.

  • $\begingroup$ "...crystal(-)clearly". Pun intended? $\endgroup$
    – march
    Aug 7, 2015 at 20:23
  • $\begingroup$ Yes, double pun -the term crystal momentum is not treated clearly+plus electron momentum is often not treated in a crystal-clear way. Thanks for noticing, someone read my question at least. :) $\endgroup$
    – plm
    Aug 11, 2015 at 19:40

1 Answer 1


With Bloch electrons, the Brillouin zone and all bands comprises a complete space. This means that while it is possible to write down and speak of states outside the Brillouin zone, these states are actually duplicates of the ones inside. In short, the Brillouin zone must be defined with a width in k-space of $2\pi/a$, with $a$ being the lattice constant. Note that technically the only requirement is the width, so you would be free to choose $[-\pi/a, +\pi/a]$, $[0, 2\pi/a]$, $[\pi/a,3\pi/a]$, etc. although the first is the most conventional.

We can see a little more on how this works by considering the Bloch electrons with the Bloch function $u_{nk}$ expanded into a Fourier series. This is taking advantage of the fact the the Bloch function has the periodicity of the lattice:

\begin{equation} \psi_{n,k}(x) = u_{n,k}e^{ikx} = \left[ \sum_{m=-\infty}^{+\infty} u_{n,k}^{(m)} e^{imKx} \right]e^{ikx}, \end{equation}

with $K = 2 \pi/a$ being the reciprocal lattice wavevector and so $u_{n,k}^{(m)}$ are the Bloch function Fourier coefficients. Next let's construct Bloch electrons at a momentum $k+K$, which is outside the original Brillouin zone choice:

\begin{equation} \psi_{n,k+K}(x) = u_{n,k+K}e^{i(k+K)x} = \left[ \sum_{m=-\infty}^{+\infty} u_{n,k+K}^{(m)} e^{imKx} \right]e^{i(k+K)x}. \end{equation}

Next we can bring $e^{iKx}$ inside the sum and do a substitution $m'=m+1$:

\begin{equation} \psi_{n,k+K}(x) = \left[ \sum_{m=-\infty}^{+\infty} u_{n,k+K}^{(m)} e^{i(m+1)Kx} \right]e^{ikx} = \left[ \sum_{m'=-\infty}^{+\infty} u_{n,k+K}^{(m'-1)} e^{im'Kx} \right]e^{ikx}. \end{equation}

Now, we see that it is possible to satisfy $\psi_{n,k+K} = \psi_{n,k}$, as long as:

\begin{equation} u_{n,k+K}^{(m-1)} = u_{n,k}^{(m)}. \end{equation}

This is how the wavefunctions duplicate themselves every $k=2\pi/a$: by shifting the Bloch function Fourier coefficients we can actually write down identical wavefunctions at each one.

  • $\begingroup$ Sorry for the late reply. I have connection troubles. Thanks alot for the reply. You do not answer my questions though. I understood well the wavevector can be restricted to the 1st Brillouin zone. I'll edit my question. $\endgroup$
    – plm
    Aug 15, 2015 at 20:04
  • 1
    $\begingroup$ Actually, I do believe that the meaning of the Brillouin zone is your main confusion. In particular, take the second equation you wrote down: why would you assume that? It seems that would apply only to completely free electrons, in which case the choice of periodicity is arbitrary. $\endgroup$
    – dbq
    Aug 16, 2015 at 17:12
  • $\begingroup$ That equation is wrong, yes. $\cdot e^{2\pi i K}$ will not give a new eigenstate. But the state $\phi_{n,k+K}(r)$ should perhaps not be defined as $\phi_{n,k}(r)$ but as a state $\phi_{n',k}(r)$ with greater energy because we increase kinetic energy. It should perhaps be in the "next band" when we draw the extended diagram for $E(k)$. I have other questions though. I added an updated. But I think I understand things ok. I just find textbook presentations misleading. Increasing wavevector by K "physically" (to account for both the exp and the $u_k$ momentum) increases energy. Thanks alot. $\endgroup$
    – plm
    Aug 17, 2015 at 19:50

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