Question:
I am studying alone and I found p.76 of this book that: $Tr(M |\psi\rangle\langle\psi)=\langle\psi| M |\psi\rangle$.
I want to be sure that my understanding of the formula is correct by proving a more general expression: $Tr(M |\psi\rangle\langle\phi|)=\langle\phi| M |\psi\rangle$, with $M$ an operator is correct.
Answer:
1- We know that $Tr(A) = \sum_i \lambda_i$, with $\lambda_i$ all the eigenvalues of $A$. Hence with $|i\rangle$ the eigenvector associated to the eigenvalue $\lambda_i$ we can write that $Tr(A) = \sum_i \langle i|A|i\rangle$ as $A|i\rangle=\lambda_i |i\rangle$.
Rem: To simplify the scripture I suppose that to each eigenvalue match only one eigenvector (I do not think that it will change a lot my demonstration in the case this is not the case).
2- By identifying $A=M |\psi\rangle \langle \phi|$ we can then write (with $|i\rangle$ the eigenvectors of $M |\psi\rangle \langle\phi|$), $Tr(M |\psi\rangle\langle\phi|)= \sum_i \langle i|M |\psi\rangle \langle \phi|i\rangle = \sum_i \langle \phi|i\rangle \langle i|M |\psi\rangle = \langle \phi| (\sum_i |i\rangle \langle i| ) M |\psi\rangle $. The term $\sum_i |i\rangle \langle i| = I$ is the identity operator since it represents a sum over a complete set of projectors onto an orthonormal basis.
3- Hence we get $Tr(M |\psi\rangle \langle \phi|)=\langle \phi| M |\psi\rangle $
Is this correct? Did I forgot to write some conditions in order to make my prove more precise?
Answer by https://quantumcomputing.stackexchange.com/users/15795/pierre-paul-t
Your proof is not general, it assumes implicitly that the operator $M|\psi\rangle\langle \phi|$ is diagonalizable (there is a basis of eigenvectors).
You can instead just use basic fact of Trace, namely $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$.
So $\mathrm{Tr}(M|\psi\rangle\langle \phi|)=\mathrm{Tr}(\langle \phi|M|\psi\rangle)$
But $\langle \phi|M|\psi\rangle \in \mathbb{C}$ is just a scalar so
$\mathrm{Tr}(\langle \phi|M|\psi\rangle)=\langle \phi|M|\psi\rangle$