# $k$-dependence of the energy in solid state physics

In a crystal, the electrons are subject to a periodic potential due to the fact that the atoms form a periodic lattice. From this periodicity we can obtain the Bloch theorem, and get a general formula for the electron wave function, known as the Bloch function.

Through the Bloch theorem, it seems that we can show the existence of band structures, i.e. that the electronic energies depend on $$k$$ and on an integer $$n$$ referring to the band index, and do not depend on $$x$$. This means we have a Schrödinger equation of the form (in 1D):

$$\left[ -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right] \Psi_{nk}(x) = E_n(k) \Psi_{nk}(x),$$ with $$V$$ a periodic potential, and $$\Psi_{nk}(x)$$ a Bloch function.

My question is: how can we prove that this non-dependence of the energy on $$x$$ is true for any periodic potential $$V$$? I could find examples for specific $$V$$'s, like for rectangular potential barriers, but not in general.

• This isn't unique to periodic potentials - think about the cases you solved in elementary quantum mechanics - the square well, the harmonic oscillator, the hydrogen atom. – jacob1729 May 21 at 15:43

I think your question is confused, the quantum numbers $$k$$ and $$n$$ denote a wavefunction with a particular spatial dependence $$\psi(x)$$ and energy $$E_{n,k}$$. So in that sense, the energy does sensitively depend on $$x$$. One should note that this is the same result as that of the simple harmonic oscillator, which has states labeled by $$n$$, or the hydrogen atom with states labeled by $$n,l,m$$. In both of these cases, the quantum numbers uniquely denote an eigenstate with a particular dependence on $$x$$.

Perhaps what you are wondering is how to show that the energies of a lattice model are invariant upon translation by a lattice vector. This is shown in Bloch's theorem (see wikipedia link here: https://en.wikipedia.org/wiki/Bloch%27s_theorem). One can see how this works by substituting $$x$$ by $$x+a$$ and recognizing $$V(x+a)=V(x)$$.

After thinking about KF Gauss's answer, I realized the mistake I made when wondering why the energy did not depend on $$x$$, as indeed it should not, by simple definition of the energy in the Schrödinger equation. I am thus putting my own answer here, for those who upvoted the question and might find this helpful.

What got me confused was the fact that everything depends on $$k$$ in solid-state physics, and along the way I forgot that $$k$$ is not equivalent to $$x$$.

Let us take the Schrödinger equation in the Dirac notation: $$H |\Psi\rangle = E |\Psi\rangle.$$ If we project this equation onto the position basis, we get: $$\langle x| H |\Psi\rangle = E \langle x|\Psi\rangle = E \, \Psi(x),$$ i.e. the energy being a scalar, it does not depend on the basis considered, and in particular it does not depend on position.

In contrast, I don't think that the crystal momentum $$k$$ can form such a basis on which $$|\Psi\rangle$$ can be projected, making it possible for the energy to depend on $$k$$. Unfortunately my mathematical demonstration comes short on this point...

My intuition is that, unlike the momentum $$p$$ which corresponds to the momentum-space basis on which we can equivalently project $$|\Psi\rangle$$, $$k$$ is only conserved modulo a lattice vector, and is thus not proportional to $$p$$.