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What I have understood from the concept of reducing a Bloch wave $\psi_{n,k}(x)=e^{ikx}u_{n,k}(x)$ to the first Brillouin zone is the following: $$\psi_{n,k+\frac{2 \pi}{a}m}(x)=\psi_{n-m,k}(x),\tag1$$ with $m$ and $n$ integers and lattice period $a$. I can see this easily mathematically by repeating the argument in the answer here, and I also come to the same intuitive conclusion: the answer there states: "I.e. if you increase $k$ so far as to go out of first Brillouin zone, you just end up in another band. I.e. you don't get new states this way."

Now comes my question. It's not clear to me if and/or why in addition to $(1)$, $$\psi_{n,k+\frac{2 \pi}{a}m}(x)=\psi_{n-m,k}(x)=\psi_{n,k}(x).\tag2$$ The latter seems to be suggested by e.g. the answer to this question. I have two problems.

  1. It seems contradictory to the statement above. Instead, it suggests that if you increase $k$ so far as to go out of first Brillouin zone, you just end up in the same band.
  2. When I look at a band diagram, the statement $\psi_{n-m,k}(x)=\psi_{n,k}(x)$ makes no sense to me. The whole point of the diagram is I'd say, that for one $k$ value, there exist different Bloch waves $\psi_{n-m,k}(x) \neq \psi_{n,k}(x)$ with different energies, no?
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2 Answers 2

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We can write $$\psi_{k + \frac{2\pi}{a}}(x) = e^{i(k + \frac{2\pi}{a})x}u(x) = e^{ikx}e^{2\pi i\frac{x}{a}}u(x)$$ But $e^{2\pi i\frac{x}{a}}$ has period $a$, so the function $$u'(x) \equiv e^{2\pi i\frac{x}{a}}u(x)$$ is also periodic with period $a$. Our original wave is thus equivalent to $$ \psi'_k(x) = e^{ikx}u'(x)$$ which is a Bloch wave at wavevector $k$.

The energies of $ \psi'_k $ and $\psi_{k + \frac{2\pi}{a}} $ are the same, since they're the same wave. The set of possible energies at $k + \frac{2\pi}{a}$ and $k$ are also the same. $n$ just describes where the energy of $\psi$ lies in the ordered list of possible energies at $k$, so $n$ must be the same for $ \psi'_k $ and $\psi_{k + \frac{2\pi}{a}} $.

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  • $\begingroup$ "wavevector $k$" Do you mean wavenumber? A vector would be noted as $\vec{k}$ or $\mathbf{k}$. Sorry to sound pedantic. $\endgroup$
    – Gert
    Jul 2, 2020 at 5:04
  • $\begingroup$ Well, in this case it's a vector of dimension 1. :) $\endgroup$
    – Daniel
    Jul 2, 2020 at 5:05
  • $\begingroup$ Looks like something of dimension $[k]=L^{-1}$. Does it relate to the wave number as used in wave theory and the wave equation in QM? (TISE) $\endgroup$
    – Gert
    Jul 2, 2020 at 5:14
  • $\begingroup$ That doesn’t solve it for me, sorry. This to me looks like the derivation leading to (1), not (2). Also, I am always extremely confused when the words ‘is equivalent to’ are used. (And they are, in this context, very often.) Does it mean = ? Or does it mean they have a certain parameter which is equal? If so, which parameter do these waves share that is equal? $\endgroup$
    – Laura
    Jul 2, 2020 at 5:22
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    $\begingroup$ @Gert Daniel means a one-dimensional (1D) vector $\endgroup$
    – Laura
    Jul 2, 2020 at 5:31
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First, a brief review of Bloch's theorem, in my preferred style.


Consider a particle in a periodic 1D potential, and let $T$ be the discrete lattice translation operator, $$\big(T\psi\big)(x) = \psi(x+a)$$ where $a$ is the lattice constant. Because $[H,T]=0$ for a periodic Hamiltonian, then we can find a basis of eigenfunctions of $H$ which are also eigenfunctions of $T$. Furthermore, since $T$ is unitary, we know that its eigenvalues must lie on the unit circle in the complex plane. Therefore, we can write the eigenvalues as $\lambda_k = e^{ika}$ for some $k$. Without loss of generality, we can take $k\in (-\pi/a, \pi/a]$, because the true eigenvalue is the phase factor $e^{ika}$, not $k$ itself.

Next, observe that $e^{-ikx}\psi_k$ is periodic in $x$ with period $a$, because $$e^{-ik(x+a)}\psi(x+a) = e^{-ikx}e^{-ika}\big(T\psi\big)(x) = e^{-ikx}e^{-ika} e^{ika}\psi(x) = e^{-ikx}\psi(x)$$ It follows that we can find a basis of eigenfunctions of the Hamiltonian of the form $\psi_k(x) = e^{ikx} u_k(x)$. If we plug this into the Hamiltonian and cancel the phase factor, we obtain $$ H_k u_k(x) \equiv \left[\frac{\hbar^2}{2m}\big(-i\nabla +k)^2 + V(x)\right]u_k(x) = E_k u_k(x)$$ subject to the condition that $u_k$ is periodic with period $a$. For each (fixed) choice of $k\in (-\pi/a,\pi/a]$, the operator $H_k$ can be shown to be self-adjoint with a discrete spectrum. Therefore, we are guaranteed to be able to find a basis of functions $u_{nk}$ which are eigenfunctions of each $H_k$ with eigenvalues $E_{nk}$. In other words, the eigenspace of $T$ corresponding to the eigenvalue $e^{ika}$ is spanned by the eigenfunctions $u_{nk}$ of $H_k$.


I can see this easily mathematically by repeating the argument in the answer here, and I also come to the same intuitive conclusion: the answer there states: "I.e. if you increase k so far as to go out of first Brillouin zone, you just end up in another band.

From a certain perspective, it doesn't really make sense to increase $k$ to beyond the first Brillouin zone. Given an eigenspace of $T$ with eigenvalue $\lambda_k$, we factor the vectors in that eigenspace into the product of a phase factor and a periodic function. The way that we do this is non-unique; the standard choice is $e^{ikx} u_k(x)$ with $k\in(-\pi/a, \pi/a]$, but $e^{i(k+G)x} \big(e^{-iGx}u_k(x)\big)$ is obviously equal to $e^{ikx}u_k(x)$ and has the same desired properties (phase factor times a periodic function).

Consider the free particle.

enter image description here

If you start at the (0,0) point at the bottom of the bold parabola, and I ask you to increase $k$ from $0$ to $2\pi/a$, what does that mean? Do you simply hop across to the vertex of the next parabola over, or do you run your finger along the bold parabola until you get to the point $\left(\frac{2\pi}{a}, \frac{4\pi^2\hbar^2}{2ma^2}\right)$? In the former case, you stay in the same band; in the latter, you move to the next band up.

enter image description here

When the Hamiltonian has a non-zero periodic potential, the band gaps open. When you say you want to increase $k$ by $G$, what do you mean? Do you mean tracing your finger along a continuous curve, or do you mean following along the dashed parabola (as closely as possible, with jumps at the zone edges)?

The point is that increasing $k$ to a value outside of the Brillouin zone doesn't really make sense without explaining what you mean by that. If you mean that you start with a free particle, follow the parabola to the desired value of $k$, and then turn on the periodic potential (thereby opening the band gaps and distorting the bands), then increasing $k$ by $2\pi/a$ takes you to the next energy band. But that's not the only interpretation - and if you mean solve for a Bloch wave $\psi_{nk}(x)$ and then simply replace $k$ everywhere you see it with $k+G$, then that will bring you back to the same band.

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