log det on density matrix plus identity A very naive question: given a pure quantum state $|\phi\rangle$, and the associated density matrix $\rho=|\phi\rangle\langle\phi|$, does there exist an efficient quantum operator/procedure that gives me $$\log\operatorname{det}(I+\rho)\quad?$$ Would I need any oracles?
Best,
Whoopy

Assuming that the state space is finite dimesional there is a more pedestrian proof than the very nice once by Cosmas Zachos:
In a basis where $|\phi\rangle$ is the first basis vector the matrix $\rho=|\phi\rangle\langle\phi|$ is the projection matrix
$$
\rho=\begin{pmatrix}1&0&\dots&0\\0&0&\dots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\dots&0 \end{pmatrix}.
$$
Then clearly $\operatorname{det}(I+\rho)=2$ and $\log\operatorname{det}(I+\rho)=\log 2$.
