Given a lattice in real space with lattice vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$, we can do the following.

We take any function $f(\vec{r})$ and take the Fourier transform to obtain $\tilde{f}(\vec{k})$. The discrete symmetry in $f(\vec{r})$ forces us to have $\vec{k}.\vec{R} = 2\pi n$, where $\vec{R}$ is some integer linear combination of $\vec{a}$, $\vec{b}$ and $\vec{c}$.

This allows us to define the basis vectors of the reciprocal lattice in the "k-space" to be




I see these reciprocal lattice vectors still in real space. Many courses commonly teach that the reciprocal lattice vectors are perpendicular to the crystal planes and this makes sense, when you look at the cross product. However, the "k-space" is also taught to have dimensions of momentum and this is something I'm unable to grasp.

Can anyone show me how and why I should be seeing these vectors (which in my head are still in real space, just perpendicular to the crystal planes) as having dimensions of momentum?


1 Answer 1


You should understand the difference between two related quantities: the wavevector $\vec{k}$ and the momentum $\vec{p}$. The wavevector has dimensions $\frac{1}{length}$ whereas the momentum has dimensions $\frac{mass*length}{time}$. In quantum mechanics, these are linked by the equation $\vec{p} = \hbar \vec{k}$, where Plank's constant $\hbar$ has dimensions $\frac{mass*length^2}{time}$.

In condensed matter literature, the terms "k-space" and "momentum-space" are used interchangeably, because the relationship between $\vec{k}$ and $\vec{p}$ is simple, universal and well-known. Unfortunately, this can be confusing for newcomers to the field.

To address the concrete example in your question, $2\pi\frac{\vec{b}\times\vec{c}}{\vec{a}.(\vec{b}\times\vec{c})}$ is a wavevector. You can check it has dimensions $\frac{length * length}{length * length * length} = \frac{1}{length}$. The associated momentum vector would be $\vec{p}=\hbar \vec{k} = 2\pi\hbar\frac{\vec{b}\times\vec{c}}{\vec{a}.(\vec{b}\times\vec{c})}$.


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