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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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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 '20 at 5:04
  • $\begingroup$ Well, in this case it's a vector of dimension 1. :) $\endgroup$ – Daniel Jul 2 '20 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 '20 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 '20 at 5:22
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    $\begingroup$ @Gert Daniel means a one-dimensional (1D) vector $\endgroup$ – Laura Jul 2 '20 at 5:31

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