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# How to prove that ${\rm tr}(A|\psi\rangle\langle\psi|)=\langle\psi| A|\psi\rangle$?

How can one prove that $$tr(A\mid\psi\rangle\langle\psi\mid)=\langle\psi\mid A\mid\psi\rangle$$? In Nielsen/Chuang they mention this is due to Gram-Schmidt decomposition but I can’t understand how.

You can use the cyclic property of the trace, $${\rm Tr}(XY) = {\rm Tr}(YX)$$.

Another way is to note that both sides are linear over $$A$$. Thus it's enough to prove it for $$A = E_{ij} = |i\rangle\langle j|$$.

You can proceed in the following way: $$\text{Tr}(A|\psi\rangle\langle\psi|) = \sum_i \langle i|A|\psi\rangle\langle\psi|i\rangle = \sum_i \langle\psi|i\rangle\langle i|a|\psi\rangle = \langle \psi|(\sum_i |i\rangle\langle i|)A|\psi\rangle = \langle\psi|A|\psi\rangle$$

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# How to prove that ${\rm tr}(A|\psi\rangle\langle\psi|)=\langle\psi| A|\psi\rangle$?

How can one prove that $$tr(A\mid\psi\rangle\langle\psi\mid)=\langle\psi\mid A\mid\psi\rangle$$? In Nielsen/Chuang they mention this is due to Gram-Schmidt decomposition but I can’t understand how.

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You can use the cyclic property of the trace, $${\rm Tr}(XY) = {\rm Tr}(YX)$$.

Another way is to note that both sides are linear over $$A$$. Thus it's enough to prove it for $$A = E_{ij} = |i\rangle\langle j|$$.

It's just a number. Trace of a number is the number itself. - Danylo Y Feb 23 at 10:36

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