Don't understand proof of Bloch theorem

[etotheipi]

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!

 

[Twigg]

$T_{\vec{R}}\psi(\vec{x}) = \psi(\vec{x}-\vec{R})$ 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]

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!

 

[etotheipi]

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?

 

[Twigg]

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]

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.

Thanks for your help!