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Can anyone show me why this equality below holds? I understand the matrix form of Kronecker delta is an identity matrix, but why this "coming from nowhere" delta function $\delta_{i,j}$ can have the exact same index $(i, j)$ as the previous terms?

$$\sum_j p_ j|\psi_j\rangle\!\langle\psi_j|\rho^{-1}|\psi_i\rangle= \sum_j p_j|\psi_j\rangle\!\langle\psi_j|\rho^{-1}|\psi_i\rangle\delta_{i,j},$$

where $\rho$ is a diagonal density matrix.

glS
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Guannan Guo
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    Is there any particular reason you used the ibm tags for what seems like a mathematical problem? If there's more context to the problem or some relation to a qiskit use-case, that may be helpful to know to ensure a thorough answer – ryanhill1 Jul 09 '22 at 00:44
  • Since I asked this question in the Slack platform of Qiskit and was directed by one of the qiskit advocates to post this question here with these tags. And I agree with your suggestion :). – Guannan Guo Jul 11 '22 at 18:28

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So I might be missing some detail or context, but if the note about $\rho$ being diagonal means it's diagonal in the $\{ |\psi_i \rangle \}$ orthonormal basis, then $\rho = \sum_k p_k |\psi_k\rangle\langle\psi_k| $ for some set of probabilities $\{p_k\}$ and in the case where those are all nonzero, $\rho^{-1} = \sum_k \frac{1}{p_k} |\psi_k\rangle\langle\psi_k|$.

Then sandwiching $\rho^{-1}$ between $\langle \psi_j |$ and $|\psi_i\rangle$ will result in 0 when $i\neq j$ and we can multiply it by 1 when $i=j$ without changing anything, so we can introduce the $\delta_{ij}$ for free. I'm assuming it's useful for some simplification later on in a derivation, but again I could be missing some context.

Chris E
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