4
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ Do you have a reason to believe you actually can calculate one from the other without knowing $\rho$? If so, please include it into your question. $\endgroup$
    – ACuriousMind
    Commented Oct 18, 2017 at 11:48
  • $\begingroup$ @ACuriousMind, I cleared things up a bit, maybe it sounds feasible now? $\endgroup$
    – 2ub
    Commented Oct 18, 2017 at 14:16

3 Answers 3

11
$\begingroup$

$$\textbf{Updated answer :}$$

$\textbf{Summary :}$ 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).

Graphical demonstration of non-uniqueness for 3-level system: Parametric plot of $-\mathrm{Tr}[\rho\log\rho]$ vs $\mathrm{Tr}[\rho^{2}]$ as a function of $p_{1}$ for $p_{2}=0.1$ (red) and $p_{2}^{}=0.2$ (blue)


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

$\endgroup$
22
  • $\begingroup$ I've added some explanation, see if you have more ideas? $\endgroup$
    – 2ub
    Commented Oct 18, 2017 at 14:15
  • $\begingroup$ @L.Quen Seems like your intuition is correct. I haven't analyzed the above expressions correctly. $\endgroup$
    – Sunyam
    Commented Oct 18, 2017 at 14:41
  • $\begingroup$ You imply that you need to deduce the density matrix from the purity to calculate the entropy, I believe this to be not necessary. Like for example, if one know the entanglement of a state in one measure, shouldn't it be possible to convert it to the value of another measure? $\endgroup$
    – 2ub
    Commented Oct 18, 2017 at 14:43
  • 2
    $\begingroup$ These traces are mere functions of their eigenvalues. So work in the diagonal space around a normalized identity; in the 2x2 case there is only one free parameter, but, already at the 3x3 case there are 2 parameters. Do you imagine there is "magic" there? $\endgroup$ Commented Oct 18, 2017 at 15:20
  • 1
    $\begingroup$ @L.Quen To conclude, your proposed claim, although interesting, holds feasible only for the simple (ofcouse relevant) case of a qubit. In geneneral von Neumann entropy cannot be uniquely determined by linear purity. $\endgroup$
    – Sunyam
    Commented Oct 18, 2017 at 16:11
5
$\begingroup$

There is no one-to-one relation already for three-dimensional systems.

The core reason is that the level sets of the entropy and those of the purity are different.

More precisely, focusing on the probability distribution generated by a given set (i.e. the sets $(\rho_{kk})_k$), the set of probabilities compatible with a given purity $\newcommand{\tr}{\operatorname{Tr}}p$ is given by the intersection of the probability simplex with the $(n-1)$-sphere of radius $\alpha$: $$\left\{(p_1,...,p_n) : \sum_k p_k=1\text{ and } \sum_k p_k^2 =\alpha \right\}.$$ Projecting on the simplex, these are therefore $n$-spheres (or subsets of $n$-spheres).

On the other hand, the level sets of the entropy are not spheres.

We can visualise this in the $n=3$ case as follows:

Here, I'm using a parametrisation of the $2$-simplex to project it into two dimensions (more details on how this projection works can be found in this question). The contour lines give the level sets of the entropy, with warm colours corresponding to higher entropy and colder colours to lower entropy. The purple thick circle is the set of states/probability vectors corresponding to a fixed purity (here $0.4$).

As clear from the figure, the set of probabilities corresponding to a fixed purity correspond to different values of the entropy.


Mathematica code to generate the figure:

par[t_, s_] = {1, 0, 0} + t /Sqrt@2 {-1, 1, 0} + s/Sqrt[3/2] {-1/2, -1/2, 1} // Simplify;
ContourPlot[
    ShannonEntropy@par[t, s], {t, 0, Sqrt@2}, {s, 0, Sqrt[3/2]}, 
    PlotRange -> All,
    ColorFunction -> "TemperatureMap", PlotRangePadding -> None,
    Contours -> 10,
    FrameStyle -> Directive[Large, Black, FontFamily -> "Latin Modern Math"],
    FrameLabel -> (MaTeX[#, Magnification -> 2] & /@ {"t", "s"}),
    AspectRatio -> Sqrt[3/2] / Sqrt[2]
] ~ Show ~ Graphics[{
    [email protected], Purple,
    Circle[{1/Sqrt@2, 1/Sqrt@6}, 0.4]
}]

A related question about the relationship between purity and rank can be found here.

$\endgroup$
3
  • $\begingroup$ Nice. It could also be nice to show a plot of, say, entropy vs. purity - but instead of a unique curve (as implied by the original question) there would be a shaded area, indicating the allowed values of entropy given a certain purity. I'd be curious about how this plot then changes for larger and larger Hilbert spaces. $\endgroup$ Commented Aug 15, 2020 at 15:21
  • $\begingroup$ @FlorianMarquardt I agree, so I ended up doing it. In the $n=3$ case this is what I find. Blue is minimum entropy corresponding to the purity and orange the maximum one. Note that the kink at $\text{purity}=1/2$ corresponds to the intersection between sphere and simplex not being completely inside the simplex. The small noise in the curves is because I found the values by partitioning the intersection between sphere and simplex (it's a circle in this case) and choosing min and max. Using a finer partition gives smoother curves $\endgroup$
    – glS
    Commented Aug 15, 2020 at 17:31
  • 1
    $\begingroup$ @gIS I forgot to thank you, I only saw your answer now. This is great! Your plot should be in textbooks when discussing purity and entropy of quantum states! And then there would be the exercise for the reader to see how it changes for higher dimensions. $\endgroup$ Commented Nov 29, 2020 at 9:44
3
$\begingroup$

To give more closure I decided to give a more visual presentation to the more algebraic approach of @confused.

As has been said, for $d>2$ there is no longer a bijection between $\mathrm{Tr} \rho^2$ and $-\mathrm{Tr} \rho \log \rho$ because they no longer have one-to-one correspondences with density matrices. Instead, each given value of the two measures relates to an equivalence class of density matrices $[\rho]_L$ and $[\rho]_{vN}$. Each class possess the same value for a measure but are generally different states admitting different spectrum. Even if a state $\rho$ from $[\rho]_L$ is also a state in $[\rho]_{vN}$, the rest of the states in $[\rho]_L$ are most likely not in $[\rho]_{vN}$.

However, for pure states and the maximally mixed state, there do indeed exist one-to-one correspondences due to the uniqueness in their eigenvalue. This can be seen in the figure below.

enter image description here

The two lines represent the two measures, the gray dots each represents a value for the corresponding measure and the respective equivalence class. As can be seen, even though one state translates from one gray dot to the other, the rest usually do not, unless that is they are mere permutations of the said state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.