quantum mechanics - Linear purity and von Neumann entropy

Given a density matrix $\rho$, its linear purity is $\mathrm{Tr} \rho^2$, its von Neumann entropy is $-\mathrm{Tr} \rho \log \rho$. Knowing one how do I calculate the other?

Edit:

My thought is that the linear purity and the von Neumann entropy are a parallel set of benchmarks for mixedness. Their inter-conversion should be like unit conversion.

See it this way, as a state goes from a pure state to a maximally mixed state, $\mathrm{Tr} \rho^2$ goes from $1$ to $0$, and $-\mathrm{Tr} \rho \log \rho$ goes from $0$ to $\log d$, like two rulers. A point on one should correspond to a point on the other.

Answer

$$\textbf{Updated answer :}$$

$\textbf{Summary :}$ It is not always that knowing $\mathrm{Tr}\rho^2$ is enough to calculate $-\mathrm{Tr}\rho\log\rho$. i.e., $-\mathrm{Tr}\rho\log\rho$ cannot be determined solely from the knowledge of $\mathrm{Tr}\rho^2$ alone. For a qubit case (2-level system with $2 \times 2$ density matrix) it is always possible. For general $\textbf{n}$-level systems ($n>2$), more information is required and hence $-\mathrm{Tr}\rho\log\rho$ is not always an unique function of $\mathrm{Tr}\rho^2$. This will be shown below. Hence the claim made (on intuitive grounds) by OP is only true for a qubit system (nevertheless an useful case), in general it is not feasible. The general reason will be provided towards the end after treating the problem in a hard way for qubit case.

$$\textbf{Qubit case (demonstrating possibility) :}$$

$\textbf{Hard way :}$ Consider an arbitary $2 \times 2$ density matrix (Positive semi-definite Hermitian matrix with unit trace) :

$$\begin{pmatrix}\rho_{11}^{} & \rho_{12}^{}e^{i\phi}_{} \\ \rho_{12}^{}e^{-i\phi}_{} & (1-\rho_{11}^{}) \end{pmatrix}$$ here unknown real variables $0 \leq \rho_{11}^{} \leq 1$, $\rho_{12}^{}$ and $\phi$.

Consider the explicit expression for $\mathrm{Tr}\rho^2$ :

$$\mathrm{Tr}\rho^2=(\rho_{11}^{})^2+(1-\rho_{11}^{})^2+2(\rho_{12}^{})^2,$$
and an explicit expression for $-\mathrm{Tr}\rho\log\rho$ : $$\begin{eqnarray} -\mathrm{Tr}\rho\log\rho&=&-\frac{1}{2}\log\left(-\rho_{11}^2+\rho_{11}-(\rho_{12}^{})^2\right)\\ &-&\sqrt{(1-2\rho_{11})^2+4 (\rho_{12}^{})^2} \tanh^{-1}_{}\left(\sqrt{(1-2\rho_{11})^2+4 (\rho_{12}^{})^2}\right) \end{eqnarray}.$$ After a bit of algebra it can be shown that : $$\begin{eqnarray} -\mathrm{Tr}\rho\log\rho&=&-\frac{1}{2}\log\left(\frac{1-\mathrm{Tr}[\rho^2]}{2}\right)-\sqrt{2\mathrm{Tr}[\rho^2]-1} \tanh^{-1}_{}\left(\sqrt{2\mathrm{Tr}[\rho^2]-1}\right) \end{eqnarray}.$$ $$\textbf{Hence for the qubit case : $-\mathrm{Tr}\rho\log\rho$ is an unique function of $\mathrm{Tr}\rho^2$.}$$ $\textbf{Simple way :}$ Since $\rho$ is a (Positive definite) Hermitian matrix (with unit trace) it can always be diagonalised with an unitary matrix to get (the eigevalues of $\rho$ are enough to calculate the observables under consideration as they involve $\mathrm{Tr}$ of functions of $\rho$ alone) : $$\tilde{\rho}=\begin{pmatrix}p & 0 \\ 0 & 1-p\end{pmatrix}$$ where $0 \leq p \leq 1$. Now calculating the quantities under consideration : $$\mathrm{Tr}[\rho^2]=\mathrm{Tr}[\tilde{\rho}^2]=p^{2}_{}+(1-p)^2_{},$$ observe that $\mathrm{Tr}[\rho^2]$ is a function of single variable, which can be inverted (assuming we know$\mathrm{Tr}[\rho^2]$) to get $$p=\frac{1}{2}\pm\frac{\sqrt{2\mathrm{Tr}[\rho^2]-1}}{2}$$ out of which we choose only physical branch (satisfying the condition $0 \leq p \leq 1$ $\Rightarrow$ $0 \leq 2\mathrm{Tr}[\rho^2]-1 \leq 1$). $\textbf{Note : The solution switches between the above two around $p=\frac{1}{2}$.}$ This when used in (irrespective of the branch) we get $$\begin{eqnarray} -\mathrm{Tr}[\rho.\log\rho]&=&-\mathrm{Tr}[\tilde{\rho}.\log\tilde{\rho}]\\ &=&\frac{1}{2} \left(\left(\sqrt{2 \text{Tr}\left[\rho ^2\right]-1}-1\right) \log \left(1-\sqrt{2 \text{Tr}\left[\rho^2\right]-1}\right)\\ -\left(\sqrt{2 \text{Tr}\left[\rho ^2\right]-1}+1\right) \log \left(\sqrt{2 \text{Tr}\left[\rho ^2\right]-1}+1\right)+\log (4)\right) \end{eqnarray}.$$ Again demonstrating the fact that for qubit case OP's proposition is feasible. (Both the results obtained here and above for $\mathrm{Tr}[\rho.\log\rho]$ are equivalent.) $$\textbf{General $n$-level system case (impossibility argument) :}$$ For general $n \times n$ case $\rho$ (again a $n \times n$ positive semi-definite Hermitian matrix with unit trace, can be diagonalized and quantities under consideration can be calculated in eigenbasis), general density matrix in its eigenbasis is given as : $$\tilde{\rho}= \begin{pmatrix} p_{1} & & \\ & \ddots & \\ & & p_{n} \end{pmatrix},$$ here $0 \leq p_{k}^{} \leq 1$ for all $k=1,...,n$ along with the constraint $\sum_{k=1}^{n}p_{k}^{}=1$.

Now calculating the quantities of interest (in basis diagonalizing $\rho_{}^{}$) :

$$\mathrm{Tr}[\rho_{}^{2}]=\mathrm{Tr}[\tilde{\rho}_{}^{2}]= \sum_{k=1}^{n}p_{k}^{2}$$ and $$-\mathrm{Tr}[\rho_{}^{}.\log\rho_{}^{}]=-\mathrm{Tr}[\tilde{\rho}_{}^{}.\log\tilde{\rho}_{}^{}]= -\sum_{k=1}^{n}p_{k}^{}\log p_{k}^{}.$$ Since $\mathrm{Tr}[\rho_{}^{2}]$ (given) imposes only one constraint apart form another constraint $\sum_{k=1}^{n}p_{k}^{}=1$, $\{p_{k}^{}\}$ is an under determined set and cannot be uniquely expressed in terms of $\mathrm{Tr}[\rho_{}^{2}]$ (except for $n=2$ qubit case which is concretely demonstrated above explicitly), hence $-\mathrm{Tr}[\rho_{}^{}.\log\rho_{}^{}]$ cannot be uniquely determined in terms of $\mathrm{Tr}[\rho_{}^{2}]$ alone (more information is needed through imposition of some constraints).

$\textbf{Graphical demonstartion of non uniqueness for $3$-level system :}$

$$\textbf{To conclude, OP's proposed claim, although interesting, holds feasible only for the simple}$$ $$\textbf{(of couse relevant) case of a qubit. In general $n$-level ($n>2$) system case von Neumann}$$ $$\textbf{entropy cannot be uniquely determined by linear purity.}$$