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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?
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1 Answer 1

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You're right that this part is key:

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}

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  • $\begingroup$ Beautiful! In eq. 3, the sum should go over $k$, right? $\endgroup$
    – A-V Labs
    Jul 7, 2021 at 18:42
  • $\begingroup$ Yes! Fixed, thanks $\endgroup$
    – wcw
    Jul 7, 2021 at 19:16

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