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Trace ${\rm Tr}(\left |a\rangle \langle a|X|b\rangle \langle b \right|) =$? Where $a$ and $b$ are two orthogonal state and $X$ is any operator.

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All you need is ${\rm tr}\{X\}=\sum_a \langle a|X|a\rangle$, where $|a\rangle$ is a compete set of orthonormal states.

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It turns out that the cyclic property of the trace, i.e., $$ \operatorname{Tr}(ABC) = \operatorname{Tr}(CAB) = \operatorname{Tr}(BCA), $$ holds when the matrices in question are not square. In particular, interpreting the kets and bras as row and column vectors, respectively, one can write $$ \operatorname{Tr}(|a\rangle\langle a|X|b\rangle\langle b|) = \operatorname{Tr}(\langle b|a\rangle\langle a|X|b\rangle) = \operatorname{Tr}(\delta_{ab}\langle a|X|b\rangle) =\delta_{ab}\langle a|X|b\rangle, $$ since the trace of a "number" is just the number (in the sense that it's a one-by-one matrix). So, if the states are orthogonal, then that trace is zero.

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A less subtle alternative to using the cyclic property is "brute forcing" it, by plugging in the definition of the trace. It goes like this, $$\begin{aligned} {\rm Tr}(\left |a\rangle \langle a|X|b\rangle \langle b \right|) &= \sum_n\langle n | \big(|a\rangle \langle a|X|b\rangle \langle b| \big) |n\rangle \\ &= \sum_n\langle n |a\rangle \langle a|X|b\rangle \langle b |n\rangle\\ &= \langle a|X|b\rangle \sum_n\langle n |a\rangle \langle b |n\rangle\\ &= \langle a|X|b\rangle \sum_n \langle b |n\rangle \langle n |a\rangle \\ &= \langle a|X|b\rangle \langle b| \bigg( \sum_n |n\rangle \langle n| \bigg )|a\rangle \\ &= \langle a|X|b\rangle \langle b|\hat 1|a\rangle \\ &= \langle a|X|b\rangle \langle b|a\rangle \\ {\rm Tr}(\left |a\rangle \langle a|X|b\rangle \langle b \right|) &= \langle a|X|b\rangle \delta_{ba} \end{aligned}$$ where the $\{|n\rangle\}$ form an orthonormal set. The "trick" is to recognize the resolution of identity $\sum_n|n\rangle \langle n|=\hat 1 $ after rearanging the scalar products and the sum.

Another simplification that has not been mentioned yet, is also realizing that $\langle a|X|b\rangle=X_{ab}$ is just a complex number, and since the trace is a linear operation, we can pull this constant number out before evaluating the trace $$\begin{aligned} {\rm Tr}(\left |a\rangle \langle a|X|b\rangle \langle b \right|) &={\rm Tr}(\left |a\rangle X_{ab} \langle b \right|) \\ &=X_{ab}{\rm Tr}(\left |a\rangle \langle b \right|) \\ &=X_{ab}\langle a|b\rangle \\ &=X_{ab}\delta_{ab} \end{aligned}$$

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