10
$\begingroup$

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?
$\endgroup$
8
$\begingroup$

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}

$\endgroup$
2
  • $\begingroup$ Beautiful! In eq. 3, the sum should go over $k$, right? $\endgroup$
    – A-V Labs
    Jul 7 at 18:42
  • $\begingroup$ Yes! Fixed, thanks $\endgroup$
    – wcw
    Jul 7 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.