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I would like to know where the $2\pi$ factors are coming from in the formula for reciprocal vectors in reciprocal lattices.

For example, in a simple cubic lattice the primitive vectors are given by $$ a_1=a_x ,\qquad a_2=a_y ,\qquad a_3=a_z,$$ and the volume of the cube is given by $$ a_1 \cdot (a_2 \times a_3) \equiv v.$$

I could define a reciprocal lattice vector as $$ b_1 \equiv \frac{1}{a_1},$$ which would be expressed in terms of area and volume as $$b_1=\frac{a_2 \times a_3}{a_1 \cdot (a_2 \times a_3)} = \frac{\text{area}}{\text{volume}} = \frac{1}{a_1}.$$

However, the generally used formulas for reciprocal vectors include a $2\pi$, like for example:

$$ b_1= 2\pi \frac{a_2 \times a_3}{a_1 \cdot (a_2 \times a_3)}.$$

Why is it necessary to work with the $2\pi$ factors?

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    $\begingroup$ they are not necessary, but with this choice you can interpret them as a basis for the wave numbers $\mathbf k$ (as in $e^{i\mathbf k\cdot\mathbf r}$). $\endgroup$ – Phoenix87 Jan 29 '15 at 16:32
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They are not necessary, but with this choice you can interpret them as a basis for the wave numbers $\mathbf k$ (as in $e^{i\mathbf k\cdot\mathbf r}$).

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  • $\begingroup$ Why "interpret"? The $2 \pi$'s are there because if you don't include them the reciprocal lattice vectors are not the right basis for the Fourier transform. $\endgroup$ – DanielSank Jan 29 '15 at 17:50
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    $\begingroup$ but they are still a basis of the reciprocal lattice, only difference is that they won't be as nice to treat as the rescaled ones. It's just a matter of mathematical convenience, which also has a physical interpretation. $\endgroup$ – Phoenix87 Jan 29 '15 at 17:52
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Some people like using angular wavevectors, while others like using ordinary wavevectors. For example, angular wavevectors are more common in most area of physics, except for crystallography ... ordinary wavevectors are common in signal processing, etc.

There should be a personality test...

  • If you like writing $\cos(\mathbf{k}\cdot\mathbf{r})$ or $e^{i\mathbf{k}\cdot\mathbf{r}}$ much more than $\cos(2\pi \mathbf{k}\cdot\mathbf{r})$ or $e^{2\pi i\mathbf{k}\cdot\mathbf{r}}$, then you would love angular wavevectors.
  • If you think that "inverse meters" is a simpler and more intuitive unit than "radians per meter", you would love ordinary wavevectors.
  • ...

Anyway, any crystal lattice has special plane waves with the property that they share the crystal's translational symmetry. You can say that the "reciprocal lattice" is the set of wavevectors of those special plane waves ... or the set of angular wavevectors of those special plane waves. In the source you're looking at, it's the latter. If you pick up other textbooks (especially crystallography), you'll find the former. It's not really well standardized.

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You understand this in one dimension and then three dimensions is obvious.

Consider a function $f(x)$. If we define the Fourier transform of this function by $$ \tilde{f}(k) = \int \, dx \, f(x) e^{- i k x}$$ then the original function $f(x)$ is can be written as $$ f(x) = \int \frac{dk}{2\pi} \tilde{f}(k) e^{i k x} \, .$$ The $2 \pi$ is there because the equation simply is not true without it. To prove this you really have to use strict analytical mathematics. Anyway, the point is that if you want to expand a function of time in terms of $\exp[i k x]$ functions (i.e. plane waves) then you need the $2\pi$ in the reverse transform.

If we had defined the Fourier transform like this $$ \tilde{f}(\nu) = \int dx \, f(x) e^{- i 2 \pi \nu x}$$ then the inverse transform would be $$ f(x) = \int d\nu \, \tilde{f}(\nu) e^{i 2 \pi \nu x} \, .$$ Therefore you see that you either need to put the $2\pi$ in both of the exponentials or leave it out of the exponentials but divide the reverse transform by $2\pi$.$^{[a]}$

These two sets of transformations are just related by a change of variables $\nu = k/2\pi$, so you can just remember one of them and easily recover the other. Since the wave number $k$ is defined as $k \equiv 2 \pi / \lambda$ that means that $\nu = 1 / \lambda$.

Physicists tend to prefer the first pair of transformations because they don't like writing the $2 \pi$ in the argument of the trig functions. In other words, they prefer things like $\cos(kx)$ instead of $\cos(2 \pi x / \lambda)$. You see this in solid state physics where wave functions are expanded in terms of functions like $\exp[i \vec{k} \cdot \vec{x}]$, and this is the root of why you're seeing factors of $2\pi$ in the equations that go back and forth between position space and $k$ space.

$[a]$: Actually there are even more options but the two I showed are the most common ones used in physics and engineering.

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