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What is the Physical interpretation of the k-points used in ab initio calculations? Why must the number of k-points be optimized rather than just using some large fixed number of them?

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    $\begingroup$ This previous answer doesn't directly address your question, but it does delve into the practical details of using these quantities. At the moment, you have multiple distinct questions in your post that would require their own answer. It may help if you can edit your post to narrow it down to a single question, with some background of what you already understand about the problem. $\endgroup$
    – Tyberius
    Dec 1 '21 at 14:29
  • $\begingroup$ thanks for your response, can you please explain to me that what is the physical interpretation of the kpoints? why do we need to optimize it instead of taking larger k points in calculation directly? @Tyberius $\endgroup$ Dec 2 '21 at 5:15
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In order to answer this, I will introduce k-points in two different ways.

1. Bloch's theorem

The usual approach to introducing k-points is to start with Bloch's theorem. In a periodic potential $V$, such that,

$V(r+L)=V(r),$

where $L$ is any lattice vector, the Kohn-Sham density $\rho$ should share that periodicity, i.e.

$\rho(r+L)=\rho(r).$

Since,

$\rho(r)=\sum_{bk}f_{bk}|\psi_{bk}(r)|^2$

(where $b$ and $k$ are band and k-point indices, respectively, and $f_{bk}$ is the band occupation), this may be satisfied if the magnitude of the wavefunctions $\psi_{bk}$ also has the requisite periodicity.

Wavefunctions are complex-valued, and although we have determined that the magnitude should be periodic, their phase is undetermined; in fact, there will be solutions to the Kohn-Sham equations for any choice of their phase. We express a particular solution as

$\psi_{bk}(r)=u_{bk}(r)e^{ik\cdot r},$

where $u_{bk}(r)$ is completely periodic in both its phase and magnitude,

$u_{bk}(r+L)=u_{bk}(r),$

and is usually called a Bloch function. $e^{ik\cdot r}$ is called the phase factor, because it has unit magnitude for any choice of $k$, and so only affects the phase of $\psi_{bk}$. $k$ is chosen such that $e^{ik\cdot r}$ is not periodic, except for the special choice $k=(0,0,0)$.

Thus, $\psi_{bk}(r)$ is periodic in its magnitude, but not its phase:

$\psi_{bk}(r+L)=u_{bk}(r+L)e^{ik\cdot(r+L)}=u_{bk}(r)e^{ik\cdot r}e^{ik\cdot L}= \psi_{bk}(r)e^{ik\cdot L}.$

What should $k$ be? There are solutions to the Kohn-Sham equations for any choice of $k$, so the most general solution would be to include a linear combination of all possible choices. Since $k$ is continuous, this means an integral over all the possible choices of $k$, i.e. those for which $k\cdot L \neq 2m\pi$ for any integer $m$; this is an integral over the first Brillouin zone.

We don't know how the Kohn-Sham states vary with $k$, so we can't perform this integral analytically. We choose to evaluate the integral numerically, sampling the Brillouin zone at various points and using it to estimate the integral; these sampling points are the "k-points".

Note that a k-point component of the form $\frac{1}{n}$, for any integer $n$, means that the wavefunction phase will be periodic with a period of $nL$. In other words, each rational k-point may be interpreted as an assumption that the wavefunction is periodic, but with period $nL$ rather than $L$.

2. Finite-sized systems

For simplicity, let's consider an infinite 2D slab, where it is periodic in Cartesian $x$ and $y$, and has a finite thickness in $z$. Let's assume we can separate the variables and write

$\psi_{bk}(x,y,z)=\phi_{bk}(x,y)\chi_{bk}(z).$

What form does $\chi_{bk}(z)$ take? In other words, what do the states look like with respect to $z$?

We could consider our 2D slab to be a potential well, let's say it has thickness $C$ and let's suppose it is a deep enough well that we can treat it as infinitely deep (this is not required, but it simplifies our description). What are the lowest energy states of this infinite potential well in $z$?

The states of an infinitely deep potential well are well-known to anyone who has studied quantum mechanics. The particles are essentially "free" inside the well, since we're approximating the potential as a constant in this region, and they go to zero at the edges (since the potential is infinite; this is why we made that simplifying approximation). If we choose our origin ($z=0$) to be in the middle of the well, then the allowed states have the form,

$\chi^{k_z}_{bk}(z)=A\sin{(k_z z)}$

where $k_z=\frac{m\pi}{C}$ for any integer $m$ (this choice of $k_z$ is required for $\chi_{bk}(z)$ to go to zero at the edges of the slab). Thus, we see that the allowed wavevectors in $z$ are $k_z = \frac{\pi}{C}, \frac{2\pi}{C}, \frac{3\pi}{C}\ldots$. We can solve the Kohn-Sham equations for any particular choice of $k_z$, and we would write a general solution as a sum over all these possible choices,

$\chi_{bk}(r)=\sum_{m=0}^\infty A_m\sin\left(\frac{m\pi z}{C}\right)$

Let's now write this in terms of complex exponentials,

$\chi_{bk}(r) = \sum_{m=-\infty}^\infty A_m e^{i\frac{m\pi z}{C}},$

or, in terms of $k_z$,

$\chi_{bk}(r) = \sum_{k_z}A_{k_z}e^{ik_z z}$

What happens as we make our slab thicker, i.e. increase $C$? A larger value of $C$ means that $\frac{\pi}{C}$ is smaller, so the allowed wavevectors are closer together. As $C\longrightarrow\infty$ the allowed wavevectors form a continuum,

$A_{k_z}\longrightarrow A(k_z)$

and

$\chi_{bk}(r) = \int_{k_z}A(k_z)e^{ik_z z}dk_z$

So our k-points started as the allowed wavevectors for a slab of finite thickness, and as the thickness tended to infinity we recovered the familiar 3D result that they form a continuum.

A particular choice of k-point is, therefore, equivalent to assuming that the wavefunction has the same form as for a slab of a particular thickness. If we use 10 "k-points" in the $z$ direction, it's equivalent to assuming the wavefunction has the same form as a wavefunction for a slab of thickness 5 times the lattice vector (half, because in the finite slab $k_z=\frac{m\pi}{C}=\frac{2m\pi}{2C}$, i.e. periodic with $2C$, rather than just $C$).

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  • $\begingroup$ Thanks a lot for the answer $\endgroup$ Dec 20 '21 at 18:12

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