0
$\begingroup$

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 $$\sum_{i}^{}\langle i| A | \psi \rangle \langle \psi | i \rangle = \langle \psi | A | \psi \rangle$$

$\endgroup$
0

2 Answers 2

0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Thanks a lot for the explanation of this part, but now how can we proove $$Tr(A|ψ⟩⟨ψ|)=∑i⟨i|A|ψ⟩⟨ψ|i⟩$$ ? $\endgroup$
    – Matodo
    Oct 4, 2023 at 17:47
  • $\begingroup$ This follows from the definition of the trace as the sum of diagonal elements: $Tr(M) = \sum_i M_{ii}$. $\endgroup$
    – hft
    Oct 4, 2023 at 18:36
  • $\begingroup$ It is basic linear algebra that the trace is the sum of diagonal elements in any orthonormal complete basis and the trace is independent of the basis chosen. $\endgroup$
    – hft
    Oct 4, 2023 at 18:37
0
$\begingroup$

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.

$\endgroup$

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