I have been studying solid state physics recently. For Bloch theorem which states that for crystal with periodic symmetry, we have $$\psi_{nk}(r)=e^{ikr}u_{nk}(r).$$

According to the textbook, this theorem is deduced under the independent electrons approximation, as the deduction starts from the one electron Schrödinger Equation. But I wonder if this theorem stills holds for correlated electrons. Because for correlated electrons, we should still have the potential $U(r)$ to obey $$U(r)=U(r+R),$$ although in the case of correlated electrons $U(r)$ contains the interaction between different electrons. Therefore, in my opinion, Bloch theorem should stills hold for correlated electrons. Can anyone give some comments on my opinion?


2 Answers 2


Consider a model of two 1D particles in a background periodic potential. When the particles are independent, the potential energy of the system will look like


Since we know that the background potential $f$ is periodic, $U$ is then periodic in both $x_1$ and $x_2$. Moreover, it's periodic in a countable infinity of directions, not only in $x_1$ and $x_2$: e.g., it is periodic in $x_1-x_2$ and $x_1+x_2$ directions. But if you consider interaction between the particles, it'll likely depend on $x_1-x_2$, so the potential energy will now look like


This interaction term $Q$ now spoils most of the periodicity. The only direction in which this potential energy remains periodic is $x_1+x_2$. This is related to the fact that despite the electrons interact, their center of mass (COM) is still "free". So you can use Bloch theorem, but only for the motion of the COM.

This readily generalizes to the case of $N$ particles and $D$ dimensions. After all you'll get that for interacting identical particles you have only $D$ wavenumbers — those related to motion of the COM, and all the other dimensions of configuration space are not periodic, thus not describable by Bloch's theorem.

  • $\begingroup$ Thanks Ruslan for your help! One thing I can't understand is that your model contains two 1D particles. But later on you said that "Moreover, it's periodic in a countable infinity of directions, not only in x1 and x2: e.g., it is periodic in x1−x2 and x1+x2 directions." Here your word "different directions" does it means that different intervals. Because for 1D system, I think there are only two directions one is x+ the other one is x-. Thanks again for your help! $\endgroup$
    – FaDA
    May 22, 2017 at 5:28
  • 1
    $\begingroup$ @FaDA By different directions I mean different angles of the lines drawn in the 2D configuration space (the space spanned by pairs of $x_1$ and $x_2$). $\endgroup$
    – Ruslan
    May 22, 2017 at 5:31
  • $\begingroup$ I still cant fully understand your comments. We are talking about a 2D configuration space spanned by (x1,x2). Just take x1 and x2 as two axes. x1 direction means (x1,0), and x2 direction means (0,x2) in that 2D space. Then what do you mean by saying x1−x2 and x1+x2 directions. Do you mean (x1-x2,x1+x2)? I think you does not mean that. $\endgroup$
    – FaDA
    May 26, 2017 at 8:09
  • $\begingroup$ @FaDA Just change variables: $u=x_1-x_2$, $v=x_1+x_2$. Then the direction of $x_1-x_2$ is the direction in which the $u$ axis will point; similarly for $x_1+x_2$ one. Is this clearer? If not, I'll try to make a drawing. $\endgroup$
    – Ruslan
    May 26, 2017 at 9:21
  • $\begingroup$ Thank Ruslan and @Rococo a lot for the help from both of you! I think I have understood your comments. You mean if I have a many-body wave function $\psi(r_1,r_2...)$, then the translation periodicity with respect to a lattice vector $R$ should be, $\psi(r_1+R,r_2+R...)$ which corresponds to the translation of COM said by Ruslan. $\endgroup$
    – FaDA
    May 27, 2017 at 4:51

I want to answer this by myself. Yesterday, I found some materials about the question. http://www.rug.nl/research/portal/files/2861460/c3.pdf

In general, the answer is yes. Bloch theorem still holds if we consider the interaction between electrons.

  • 3
    $\begingroup$ Note that in this paper they show that the many-electron state obeys the Bloch theorem, not any single-electron state. The translation of this state corresponds to the COM translation that Ruslan discusses. $\endgroup$
    – Rococo
    May 15, 2017 at 15:50

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