# Don't understand proof of Bloch theorem

• B
Gold Member
2019 Award
The potential inside the crystal is periodic ##U(\vec{r} + \vec{R}) = U(\vec{r})## for lattice vectors ##\vec{R} = n_i \vec{a}_i##, ##n_i \in \mathbb{Z}## (where the ##\vec{a}_i## are the crystal basis), and Hamiltonian for an electron in the crystal is ##\hat{H} = \left( -\frac{\hbar^2}{2m} \nabla^2 + U(\vec{r}) \right)##. The book defined a translation operator ##T_{\vec{R}}##, and proved that ##T_{\vec{R}}## and ##\hat{H}## are commuting operators, so have simultaneous eigenstates,$$\hat{H} \psi = E\psi, \quad T_{\vec{R}} \psi = c(\vec{R}) \psi$$They also prove that ##c(\vec{R} + \vec{R}') = c(\vec{R})c(\vec{R}')## for two lattice vectors ##\vec{R}## and ##\vec{R}'##. But I don't understand the next bit, which says ##c(\vec{a}_i)## can always be written in the form$$c(\vec{a}_i) = e^{2\pi i x_i}$$for some ##x_i##. I feel like I'm missing something obvious, why can we do this? Thanks!

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JD_PM

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Twigg
Gold Member
##T_{\vec{R}}\psi(\vec{x}) = \psi(\vec{x}-\vec{R})## by definition, so the statement ##T_{\vec{R}}\psi(\vec{x}) = c(\vec{R})\psi(\vec{x})## implies that ##|c(\vec{R})|^2 = 1## to preserve normalization of ##\psi(\vec{x}-\vec{R})##. That means it has to lie on the complex unit circle.

etotheipi
Gold Member
2019 Award
Ah, yeah that makes sense, thanks! My book defines the translation operator as ##T_{\vec{R}}(\vec{x}) = \vec{x} + \vec{R}##, but that doesn't make any difference to the logic. So you'd say$$\int_{\mathbb{R}^3} d^3 \vec{x} \lVert c(\vec{R}) \psi(\vec{x})\rVert^2= \int_{\mathbb{R}^3} d^3 \vec{x} \lVert \psi(\vec{x} + \vec{R}) \rVert^2 = 1 \implies \lVert c(\vec{R}) \rVert = 1 \implies c(\vec{R}) = e^{2\pi i x_i}$$And we can set ##n_i = 1, n_j = 0 \, (\forall j \neq i)##, i.e. ##\vec{R} = \vec{a}_i##, so also holds for ##\vec{a}_i## (for some ##x_i##). Thanks!

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Twigg
Gold Member
2019 Award
Also, when I was reading the Wikipedia page here, I notice they write$$\vec{n} \cdot \vec{a} = n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3 \vec{a}_3\quad (=\vec{R})$$What is the "##\cdot##" operation here, and what are ##\vec{a}## and ##\vec{n}##? It can't be the scalar product, since the result is a vector! `Perhaps this is a mistake?

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Twigg
Gold Member
I'm no crystallographer, but my guess is that they made a knock-off "dot product" to shorten the notation and look like a cool kid. There's a "dot product" like this for the Pauli matrices too. In a crystal with low symmetry, a1,a2,a3 need not be orthogonal (like in a triclinic? I'm not sure on the nomenclature don't quote me), so really this "dot product" is pure notation.

Like with the Pauli matrix "dot product", ##\vec{a}## here is just a "vector" whose "elements" are the vectors ##\vec{a_1}##,##\vec{a_2}##,##\vec{a_3}##. ##\vec{n}## is just a vector of integers here.

etotheipi
Gold Member
2019 Award
Yeah, I think you're exactly right. I think they've done something along the lines of defining a tuple ##\vec{n} = (n_1, n_2, n_3)## and another tuple ##\vec{a} = (\vec{a}_1, \vec{a}_2, \vec{a}_3)##, and also defining the "##\cdot##" operation to be the sum of the pairwise products, ##\vec{n} \cdot \vec{a} = n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3 \vec{a}_3##. So it's likely there for notational convenience, like you say, and not at all related to an actual inner product or anything.