Basis set in the 𝘎𝘞 calculation

Representation of one-particle wave functions

The wave functions are expands by two types of basis set as follows:

$$|{\Psi }_{𝒌n}⟩=\sum _{𝑹u}{\alpha }_{𝑹u}^{𝒌n}|{\phi }_{𝑹u}^{𝒌}⟩+\sum _{𝑮}{\beta }_{𝑮}^{𝒌n}|P_{𝑮}^{𝒌}⟩,$$

where $|{\phi }_{𝑹u}⟩$ is atomic LMTO basis and $|P_{𝑮}^{𝒌}⟩$ is interstitial plane wave to comprised interstitial area.

Detail of $|{\phi }_{𝑹u}^{𝒌}⟩$

The $|{\phi }_{𝑹u}^{𝒌}⟩$ is generated from bloch sum of atomic orbital $|{\phi }_{𝑹u}⟩$ to satisfy the bloch symmetry on basis functions, i.e.:

$${\phi }_{𝑹u}^{𝒌}(𝒓)=\sum _{𝑻}{\phi }_{𝑹u}(𝒓-𝑹-𝑻)\exp (i𝒌\cdot 𝑻)$$

The radial part of atomic orbital are constructed from ${\phi }_{l\nu }(r)$, ${\dot{\phi }}_{l\nu }(r)$, and ${\phi }_l^z(r)$ based on argumentation of Smooth Henkel functions.

The index of atom is omitted for simplify. Each $l$ has 2 or 3 reference energy of ${ϵ}_{l\nu }$. ${\phi }_{l\nu }(r)$ is the $l$ channel orbital of solution on radial Scrhödinger equation at energy ${ϵ}_{l\nu }$ ,${\dot{\phi }}_{l\nu }(r)$ is energy derivative of ${\phi }_{l\nu }(r)$, ${\phi }_l^z(r)$ is local orbital which is the also solution on radial Scrhödinger equation but the given energy is far from ${ϵ}_{l\nu }$. The energy of ${ϵ}_{l\nu }$ is set as the center of gravity for occupied states. This means that this basis set is not fixed in the QSGW cycle.

?? About radial equation??

• Is LDA used?

$$[-\frac{1}{2}\frac{d^2}{d{r}^2}+V_{\mathrm{eff}}[n(r)]-\frac{Z}{r}+\frac{l(l+1)}{2r^2}-{ϵ}_{nl\nu }]{\phi }_{nl\nu }(r)=0$$$$n(r)=2\sum _{nl}(2l+1)|{\phi }_{nl}(r){|}^2/r^2$$

• In the $n(r)$ in radial equation obtained from self-consistent calculation?
• What is the boundary conditions of radial equation.
• ${ϵ}_{l\nu }$ is not the solution of radial equation, the ${\phi }_{l\nu }(r)$ does not satisfy the some boundary conditions, for example ${\phi }_{l\nu }(r)\to 0(r\to 0)$
• how to set the reference energy in the case of ${\phi }_l^z(r)$?

Detail of $|P_{𝑮}^{𝒌}⟩$

The interstitial plane wave is expressed as follows:

$$\begin{array}{r}{P}_{𝑮}^{𝒌}(𝒓)=\{\begin{array}{ll}0& \text{if}𝒓\in \text{any MT}\\ \exp (i(𝒌+𝑮)\cdot 𝒓)& \text{otherwise}\end{array}\end{array}$$

Since the hollow out the MT region, these basis have overlap matrix between them, i.e.:

$$⟨P_{𝑮}^{𝒌}|P_{{𝑮}^{\prime }}^{𝒌}⟩={\delta }_{𝑮,{𝑮}^{\prime }}-\sum _{{𝑹}_I}{\int }_{M{T}_I}{e}^{i(𝒓+{𝑹}_I)\cdot ({𝑮}^{\prime }-𝑮)}d𝒓$$

where,

$${\int }_{M{T}_I}{e}^{i(𝒓+{𝑹}_I)\cdot ({𝑮}^{\prime }-𝑮)}d𝒓=\frac{4\pi e^{i{𝑹}_I\cdot ({𝑮}^{\prime }-𝑮)}}{|{𝑮}^{\prime }-𝑮|}{\int }_0^{R_I}drr\sin (|{𝑮}^{\prime }-𝑮|r)$$

Formula of inner product

This is for checking the orthonormality of wave function, but it is good for better understanding.

$$⟨{\Psi }_{𝒌n}|{\Psi }_{𝒌m}⟩=\sum _{𝑹u{u}^{\prime }}{\alpha }_{𝑹u}^{𝒌n\ast }{\alpha }_{𝑹u^{\prime }}^{𝒌m}⟨{\phi }_{𝑹u}^{𝒌}|{\phi }_{𝑹u^{\prime }}^{𝒌}⟩+\sum _{𝑮{𝑮}^{\prime }}{\beta }_{𝑮}^{𝒌n\ast }{\beta }_{{𝑮}^{\prime }}^{𝒌m}⟨P_{𝑮}^{𝒌}|P_{{𝑮}^{\prime }}^{𝒌}⟩$$

The cross term are vanished.