Explanation of the 2.60 equation page 76 in the Nielsen and Chuang

Question

In the Nielsen and Chuang book page 76, equation 2.60 says that we can rewrite the trace $$Tr(A \left|\psi\right>\left<\psi\right|)$$ as follow :

$$Tr(A \left|\psi\right>\left<\psi\right|) = \sum_{i}^{}\langle i| A | \psi \rangle \langle \psi | i \rangle = \langle \psi | A | \psi \rangle$$

I understand the idea behind, but I don't manage to do the calculus with the gram-schmidt procedure can lead to such a result.

On addition, I don't understant how $$\sum_{i}^{}\langle i| A | \psi \rangle \langle \psi | i \rangle = \langle \psi | A | \psi \rangle$$

Answer 1

In the Nielsen and Chuang book page 76, equation 2.60 says that we can rewrite the trace $$Tr(A \left|\psi\right>\left<\psi\right|)$$ as follow :

$$Tr(A \left|\psi\right>\left<\psi\right|) = \sum_{i}^{}\langle i| A | \psi \rangle \langle \psi | i \rangle = \langle \psi | A | \psi \rangle$$

On addition, I don't understant how $$\sum_{i}^{}\langle i| A | \psi \rangle \langle \psi | i \rangle = \langle \psi | A | \psi \rangle$$

The $|i\rangle$ are assumed to be a complete set. Use the completeness relation: $$ 1 = \sum_{i}|i\rangle\langle i| $$

Thus: $$\sum_{i}^{}\langle i| A | \psi \rangle \langle \psi | i \rangle =\sum_{i}^{}\langle \psi| i \rangle\langle i| A | \psi \rangle $$ $$ =\langle \psi| \sum_{i}^{}| i \rangle\langle i|A | \psi \rangle =\langle \psi| 1\times A | \psi \rangle = \langle \psi | A | \psi \rangle$$

Answer 2

To add to the answer by @hft, see the discussion at Can we combine the square roots inside the definition of the fidelity?

In particular the invariance of trace when you do cyclic rearrangement of the product terms.