# Local phase gauge in momentum space of Bloch state

We know Bloch state has a phase undetermined, so $\Psi_k \to \Psi_k' = e^{i\theta(k)}\Psi_k$ is still the same eigenstate.

My question: Are there some restriction on $\theta(k)$ except to be a real function? Should $\theta(k+G) = \theta(k)$? where $G$ is reciprocal lattice vector.

My guess is it should. Otherwise $\Psi_{k+G}' \neq \Psi_k'$. But I saw on a famous paper which doesn't impose this restriction and still call it a gauge transform.

• Why would any non-constant $\theta$ be a symmetry at all? True, all $\lvert k \rangle$ are invariant under this transformation, but $\lvert k_1 \rangle + \lvert k_2 \rangle$ clearly is not if $\theta(k_1) \neq \theta(k_2)$! Commented Jan 20, 2015 at 15:52
• @ACuriousMind $\left|k_1\right> + \left|k_2\right>$ is not invariant, but why we want it invariant? We only want observables invariant.
– Tim
Commented Jan 20, 2015 at 21:37
• Don't know what I was thinking in that comment. See my answer for a different take. Commented Jan 20, 2015 at 21:58
• In 1105.4867, this is a gauge transform of the Hamiltonian. The Hamiltonian (bottom of the lhs of pg. 2) is a quadratic form in $c_k$ and $c_k^\dagger$, so the phases cancel. Anyway, you can safely assume that $k$ is in the BZ unless it is explicitly mentioned otherwise. Commented Jan 28, 2015 at 19:57

For the question itself, since in the Brillouin zone $k$ and $k+G$ are simply the same point, $\theta(k)$ has to be the same as $\theta(k+G)$. However, in the paper arxiv.org/pdf/1105.4867v2.pdf, the reason why they performed a gauge transformation $c_k\rightarrow c_k e^{i(k_x-k_y)/2}$ is that the Hamiltonian $h(k)$ they took as the starting point (see the bottom of the left column on the same page) has all these $k_x/2, k_y/2$ in it, and I suppose they wanted to get rid of these multi-valued functions. Actually, it is not clear to me why they had to start from there (which was taken from arxiv.org/pdf/1012.5864.pdf). $h(k)$ defines a tight-binding model on a checkerboard lattice, or a square lattice with two orbitals per unit cell, and if one just directly Fourier transforms the real-space hopping terms, one should naturally end up with something like Eq. (1).

• Because they want $h(k+G)=h(G)$, which they call in Bloch form. Directly Fourier transform real-space hopping term cannot get Eq.(1). By the way, the reason of doing this transform is explained a bit in this book page 72 books.google.com/books?id=_7r_UqFN0IEC using graphene as an example.
– Tim
Commented Jan 21, 2015 at 3:30
• I don't think it is impossible to get this form by direct Fourier transform, since I've checked it for both graphene and the checkerboard lattice models, unless I seriously missed something. Commented Jan 21, 2015 at 3:45
• It is interesting now. I checked also. Do you mean you can get the same $h_{12}(k)$ in Eq.(1) by Fourier transform? There is $k/2$ in Eq. (2) in arxiv.org/pdf/1012.5864.pdf How did you get rid of 2 in the denominator?
– Tim
Commented Jan 21, 2015 at 4:19

Naively, we would say: Since the Bloch waves are an energy eigenbasis, and the unitary operator $\psi_k \to \mathrm{e}^{\mathrm{i}\theta(k)}\psi_k$ defined on that basis is obviously diagonal, hence commutes with the Hamiltonian, and therefore is a symmetry, without any restriction on $\theta$.

Now, the "the Bloch waves" in the first sentence needs a qualification. The basis is not given $\mathcal{B} = \{\psi_k \vert k \in \mathbb{R}^3\}$, since $\psi_{k+G} = \psi_k$ trivially shows that these are not all independent. Hence, to get a basis, we need to divide out the relation $\psi_k \sim \psi_k' \Leftrightarrow k - k' = G$ for any reciprocal lattice vector $G$. Therefore, every function $\theta$ that produces a symmetry is not defined on the whole of $\mathbb{R}^3$, but only on these equivalence classes of vectors, which shows that you are correct: Indeed, $\theta$ must obey $\theta(k + G) = \theta(k)$ for any reciprocal lattice vector $G$.

• yes, when I said basis, $k \in BZ$ is always implied, because it is Bloch wave.
– Tim
Commented Jan 20, 2015 at 22:10
• @Tim: Then I evidently have again failed to understand the question. Were you not asking whether or not $\theta$ has to be restricted, and does this not answer it? Commented Jan 20, 2015 at 22:12
• It seems that the restriction is not necessary. At least from this paper arxiv.org/pdf/1105.4867v2.pdf equation (1)
– Tim
Commented Jan 20, 2015 at 22:19
• @Tim: I think that's most probably an "oversight" and they forgot to mention that you are to take this definition of $\theta$ only on the Brillouin zone and periodically extend it to all $k$. Or do they actually use the value of this $\theta$ for $k$ outside the zone for anything? Commented Jan 20, 2015 at 22:25