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The set of quantum states $\rho$ in $d$ dimensions is the set of positive semidefinite operators living in a Hilbert space of dimension $d$. Let us denote this set by $\text{Pos}(X)$ and note that this is a convex set. Convexity is also seen in other interesting sets, for instance, the set of separable states $S$, the set of PPT states $P$ and so on.

I have this intuition that by some symmetry arguments, the maximally mixed state i.e. the identity operator (which is in $S$, $P$ and $\text{Pos}(X)$) should be at the geometric center of these sets, say if we use reasonable things like the 2-norm to define distance. After all, the maximally mixed state is (roughly speaking) a mixture of all the states in the Hilbert space.

For a single qubit, looking at the Bloch sphere, this is indeed true! The maximally mixed state is at the center of the sphere.

Unfortunately, this picture seems to be misleading in more general cases. According to this, you will get a very nice geometric picture of concentric manifolds for separable states, PPT states and all quantum states and this (as far as I can tell) isn't the case. See this image where the author deliberately makes the sets non concentric which is at odds with my intuition. enter image description here

Does someone have a better intuition for the shape of convex sets of quantum states? And would there be a good reason to see why the maximally mixed state isn't at the center of the set of states?

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  • $\begingroup$ Why wouldn't it be the case? (And do you assume trace 1?) $\endgroup$ – Norbert Schuch Feb 27 at 17:53
  • $\begingroup$ @NorbertSchuch, yes, I'm assuming trace 1 for all the states I talk about. Could you elaborate on why that matters? I only thought this was so because of the non-concentricity of the image I attached. You're saying that the correct picture is one where all the sets shown have the same center which corresponds to the maximally mixed state? $\endgroup$ – user1936752 Feb 27 at 19:04
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    $\begingroup$ If it is not trace 1, the maximally mixed state (with trace 1) is certainly not in the center. The number 1 is by no means special. Why not 17*the maximally mixed state. But I understand your question better now, the picture helped. Basically, you are asking "How can I reconcile the max mixed state being in the center with those non-concentric pictures?" -- is this correct? $\endgroup$ – Norbert Schuch Feb 27 at 19:59
  • $\begingroup$ Yes, this was the question. Thank you for the answer! $\endgroup$ – user1936752 Feb 27 at 21:51
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Density operators live in a real vector space. Consider operators $\rho$, $\rho\ge0$, $\mathrm{tr}\,\rho=1$ from either of the sets $\mathcal S$ you mention (all density operators, separable, or PPT).

Given a state $\rho\in \mathcal S$, we can write it as $$ \rho = \tfrac1d1\!\!1 + X\ , $$ with a traceless hermitian operator $X$. If $\rho$ were in the center of the set, then also the state $$ \rho' = \tfrac1d1\!\!1 - X $$ should be in the set.

Now consider a (bipartite) pure product state $\rho=|00\rangle\langle00|$, which is contained in all three sets (and, in fact, at the boundary of all three sets, as in the bottom of your picture). In that case, $$ \rho'=\tfrac2d1\!\!1-|00\rangle\langle00|\ , $$ which has a negative eigenvalue $\tfrac2d-1$ (note that $d\ge4$ -- and even if we don't look at bipartite systems, $d\ge3$, since you already observed that everything works for the Bloch sphere). We therefore see that $\rho'$ is not in either of the sets, i.e., the sets are not symmetric around the center!

It should be instructive to work out the PPT criterion, or separability, for this or related examples along such a line.

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