1
$\begingroup$

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.

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.