# In the PAW-method how does one come up with the most general form for the projector functions?

The projector augmented-wave (PAW) method as introduced by Blöchl gives an expression for the most general form for the projector functions $$\tag{1} \langle \tilde{p}_i \vert = \sum_j \left( \lbrace \langle f_k \vert \tilde{\phi}_l \rangle \rbrace \right)_{ij}^{-1} \langle f_j \vert,$$ "where $$\vert f_j \rangle$$ form an arbitrary, linearly independent set of functions". This statement is given without any further detail.

My assumption so far is that this satisfies the condition $$\langle \tilde{p}_i \vert \tilde{\phi}_j \rangle = \delta_{ij}$$. Thus I have tried to plug it in but I can't show the relation. I am confused about several things:

• Is the summation index $$j$$ the same one as the one of $$\tilde{\phi}_j$$ in the orthonormality relation?
• I take it that the indices $$ij$$ denote a matrix element and the power of $$-1$$ denotes the inverse. Then where does the condition of linear independence for $$f_j$$ come into play?
• Is this even the right track?

My assumption so far is that this satisfies the condition $$\langle \tilde{p}_i \vert \tilde{\phi}_j \rangle = \delta_{ij}$$.
To make the notation a bit simpler, let's define $$S$$ as the matrix with $$S_{ij} = \langle f_i \vert \tilde{\phi}_j \rangle\tag{1}.$$ Its inverse matrix is $$S^{-1}$$. The expression for the projectors is then $$\langle \tilde{p}_i \vert = \sum_j (S^{-1})_{ij} \langle f_j \vert,\tag{2}$$ such that
$$\begin{eqnarray} \langle \tilde{p}_i \vert \tilde{\phi}_j \rangle &=\sum_k (S^{-1})_{ik} S_{kj} \tag{3}\\ &= (S^{-1} S)_{ij} \tag{4}\\ &= \delta_{ij}.\tag{5} \end{eqnarray}$$
• Beautiful! In eq. 3, the sum should go over $k$, right? Jul 7 '21 at 18:42