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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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2 Answers
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 $A = E_{ij} = |i\rangle\langle j|$.
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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 $A = E_{ij} = |i\rangle\langle j|$.
8,1311313 silver badges2525 bronze badges
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