# 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

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, $$\operatorname {log ~det }(I+\rho)=\operatorname {tr~log}(I+\rho)\\ = \operatorname {tr} \sum_{n=1}^{\infty} \frac{-(-)^n}{n} \rho^n= \sum_{n=1}^{\infty} \frac{(-)^{1+n}}{n}\operatorname{tr} \rho^n = \sum_{n=1}^{\infty} \frac{(-)^{n+1}}{n}=\log 2~~.$$
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 $$\log\operatorname{det}(I+\rho)=\log 2$$.