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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

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The oracle is just the Mercator series for the log.

Since you have a normalized pure state, $$ \operatorname {tr} \rho^2= \operatorname {tr} \rho= 1=\operatorname {tr} \rho^k, $$ for any k; so that, recalling this, $$ \operatorname {log ~det }(I+\rho)=\operatorname {tr~log}(I+\rho)\\ = \operatorname {tr} \sum_{n=1}^{\infty} \frac{-(-)^n}{n} \rho^n= \sum_{n=1}^{\infty} \frac{(-)^{1+n}}{n}\operatorname{tr} \rho^n = \sum_{n=1}^{\infty} \frac{(-)^{n+1}}{n}=\log 2~~. $$

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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 $\log\operatorname{det}(I+\rho)=\log 2$.

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