# Continuous value of crystal momentum in Bloch theorem

The reciprocal lattice of any Bravais lattice can be interpreted as a set of possible $$k$$ values representing the wave vector of a standing wave with the periodicity of the Bravais lattice. Therefore, $$k$$ is discrete and the smallest non-zero $$k$$ value is the primitive reciprocal lattice vector ($$2\pi/a$$ for a cubic, for example).

However, the Bloch Theorem says that the energy of an electron in a periodic potential is continuously changed as $$k$$ continuously varies even in the first Brillouin zone, which is "energy band" (generally written as $$E_{nk}(k)$$) as I understand.

My question is

How can continuous $$k$$ values exist even within the first Brillouin zone?

As explained above, my understanding is that $$k$$ is discrete and the smallest non-zero value determines the edge (?) of the first Brillouin zone, so there will be no possible $$k$$ values within the first Brillouin zone.

The reciprocal lattice of any Bravais lattice can be interpreted as a set of possible $$k$$ values representing the wave vector of a standing wave with the periodicity of the Bravais lattice. Therefore, $$k$$ is discrete and the smallest non-zero $$k$$ value is the primitive reciprocal lattice vector ($$2\pi/a$$ for a cubic, for example).

This is correct $$-$$ that is indeed the definition of the Bravais lattice. The reason that Bloch wavefunctions can exist for smaller crystal momentum $$k$$ than the smallest Bravais-lattice element is that the Bloch wavefunctions' crystal momenta is not restricted to the Bravais lattice.

More specifically, the Bravais lattice consists of those crystal momenta $$\mathbf k$$ such that $$\mathbf k \cdot \mathbf a = 2\pi n$$ (i.e. $$e^{i\mathbf k \cdot \mathbf a}=1$$), with $$n$$ an integer, for every displacement $$\mathbf a$$ in the real-space lattice $$\mathcal L$$.

Bloch wavefunctions are independent of this condition. A wavefunction is termed a Bloch wavefunction if it can be written in the form $$\psi_\mathbf{k}(\mathbf r) = e^{i\mathbf k \cdot \mathbf r}u(\mathbf r)$$ where $$u(\mathbf r)$$ is periodic over the lattice, i.e. $$u(\mathbf r+\mathbf a) = u(\mathbf r)$$ for all displacements $$\mathbf a \in \mathcal L$$. This means that if you displace $$\psi_\mathbf k(\mathbf r)$$ itself, you get $$\psi_\mathbf{k}(\mathbf r+\mathbf a) = e^{i\mathbf k \cdot (\mathbf r+\mathbf a)}u(\mathbf r+\mathbf a) = e^{i\mathbf k \cdot \mathbf a}e^{i\mathbf k \cdot \mathbf r}u(\mathbf r) = e^{i\mathbf k \cdot \mathbf a}\psi_\mathbf{k}(\mathbf r), \tag{*}$$ i.e., $$\psi_\mathbf{k}(\mathbf r)$$ multiplied by a phase. This is a slightly different state (i.e. it does have a different phase), but it's also very similar (i.e. experimentally indistinguishable, since experiments don't care about global phases), so this property makes $$\psi_\mathbf k(\mathbf r)$$ special.

The implication of this property is that the Bravais lattice consists of those $$\mathbf k$$ such that $$\psi_\mathbf{k}(\mathbf r+\mathbf a)$$ returns to $$\psi_\mathbf{k}(\mathbf r)$$ itself without any phase at all. That's even nicer, of course! but it wasn't required to begin with, and the quasi-periodicity of $$(*)$$ is also perfectly acceptable.

I think perhaps you're confusing the crystal momentum $$k$$ with the reciprocal lattice vectors $$K$$.

Bloch's theorem: states that solutions to the Schrodinger equation in a periodic potential $$U(x+a)=U(x)$$ can be written as Bloch functions $$\psi$$ of the form $$\psi_k(x) = e^{ikx}u(x)$$, and $$u(x)$$ is periodic $$u(x+a)=u(x)$$.

According to this statement of the theorem, no periodicity condition is placed on $$\psi(x)$$, and at first glance one may choose the crystal momentum $$k$$ to be any continuous value whatsoever. However, consider the following.

Let $$K = n \frac{2\pi}{a}$$ be the reciprocal lattice vectors, then an alternative way to state Bloch's theorem (Ashcroft & Mermin Chapter 8) is:

Bloch's Theorem: The eigenstates can be chosen so that associated with each $$\psi$$ is a $$k$$ such that $$\psi_k(r+R) = e^{ika}\psi_k(r)$$.

So Bloch's theorem states that $$\psi_k$$ is periodic up to a phase $$e^{ika}$$. Now note that if we consider crystal vectors $$k' = k+K$$ for any reciprocal lattice vector $$K$$, then

$$\psi_{k'}(r+R) = e^{iKa} e^{ika}\psi_{k'}(r)$$

But $$e^{iKa} = 1$$ since $$R$$ is a lattice vector and $$K$$ is a reciprocal lattice vector, so what we find is that $$\psi_{k'}$$ also satisfies Bloch's theorem. Therefore, the Bloch functions are not unique. Functions $$\psi_{k+n2\pi/a}$$ satisfy Bloch's theorem just as well as $$\psi_k$$. Choosing $$n=0$$ means restricting ourselves to the first Brillouin zone.