Continuous value of crystal momentum in Bloch theorem

Question

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.

Please comment about what I am misunderstanding.

Answer 1

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.

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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.

Answer 2

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.