@SanchayanDutta by the way, I still don't think it makes sense to use complex dimensions (referring to this comment chain)
sure, DMs don't form a vector space, but the question here is what is the smallest embedding space in which they exist. And this embedding space is that of Hermitian matrices, which are a real vector space, not a complex one
Anonymous
@glS But why would you choose the smallest embedding space in which they exist as the "natural choice" of a vector space in which they exist? The vector space $\mathcal{L}(\mathcal{H})$ seems far more natural to me, given that we normally write density matrices as $d\times d$ matrices
Either way, it mostly seems like a semantics issue. As long as we know what we're doing, I doubt it matters...
Anonymous
5:46 PM
@glS Now that I think about it, there's some merit in calling it $d^2-1$ dimensional after all. For instance, we represent states (density matrices) of qubits in a Bloch sphere, which has three degrees of freedom.
Anonymous
6:02 PM
Here, they denote the manifold of density matrices in terms of the group action of $\mathrm{SU}(N)$ on $\Bbb CP^{N^2-1}$. Interestingly, the stratification has dimensions ranging from $N^2-1$ to $2N-2$. In case of $N=2$ (qubits), the maximum is $3$ (for pure states) and the minimum is $2$ (for mixed states). Perhaps I can write a better answer when I study Lie group action and orbits properly. :)
Anonymous
(* Sorry, I meant it the other way round. It's dimension 2 for pure states and dimension 3 for mixed states. Can't edit the message anymore.)
@SanchayanDutta uh... yes of course, $2N-2$ is the dimension for pure states and $N^2-1$ for general states. The Bloch sphere thing doesn't just work for qubits, the Bloch representation works for states of arbitrary dimensions, there are multiple posts on the site about this already
@SanchayanDutta it's the natural choice in the sense that it doesn't make sense to ask for a larger embedding space, there are trivially infinite of those. $\mathcal L(\mathcal H)$ is fine but if you want to find the true dimension you have to renounce the complex structure for the vector space (at least that's how it's always done, I don't know if there is some weird way to still give it a complex vector space structure)
@MoreAnonymous can't you upload it somewhere?
Anonymous
@glS It's a bit more subtle than that. The dimensions of the density matrices range from $N^2-1$ to $2N-2$ depending on the Schmidt rank
Anonymous
That is, there are several strata for qudits. For qubits there are only two though
@SanchayanDutta but then you are changing the question again. The set of all density matrices has dimension/is embedded in a real vector space of dimension $N^2-1$. If you just consider subsets sure, the dimension can be lower
Anonymous
@glS True, yup. I guess we need to be a bit careful while stating these things. The term dimension isn't well defined in this context
@SanchayanDutta the strata thing is a fancy way to talk about purity I think. If you restrict to pure or near-pure (i.e. low rank) states, you can assume lower dimensionality. This is also used to perform efficient tomography under the assumption of near purity, see e.g. arxiv.org/abs/0909.3304
@SanchayanDutta just to be clear, it's not mine, this is standard notation
I think the standard reference here is Geometry of Quantum States: An Introduction to Quantum Entanglement, Book by Ingemar Bengtsson and Karol Życzkowski
though it's not exactly a light read..
Anonymous
@glS Ah, I'll check...thanks !
Anonymous
@glS Makes sense, but do all mixed states of the same rank have equivalent purity in some sense? (Well purity is a function of radial distance, which is true)
@SanchayanDutta mh.. same rank surely does not imply same purity: the rank has integer values so changing a state a bit will change the purity but not the rank. If on the other hand two states have the same purity, that means that $\sum_i p_i^2=\sum_j q_j^2$ if $p_i$ and $q_j$ are the eigenvalues of the two states... does this imply same rank? I'm not so sure
I would say no, it doesn't, because it's like asking whether for two vectors $\|x\|=\|y\|$ implies that $x,y$ have the same number of non-zero elements. I don't know if the fact that $\sum_i p_i=\sum_i q_i=1$ changes things though
Anonymous
Yeah, that's my confusion as well because the strata thing is directly correlated with rank but not purity. However, from a geometrical perspective I'd expect states having nearby values of purity to be grouped in the same strata...
Anonymous
Anyway, I actually have to read through the math...
I think it's actually a great question to ask whether same purity implies same rank. On the one hand it makes sense because they both sort of measure the same thing, and it would seem that purity is just a "finer" version of rank. It's not so obvious to see though. I can see that it is true for two-dimensional systems because $p^2+(1-p)^2=q^2+(1-q)^2$ is not possible if $q=1$ and $p<1$ or the other way around
Anonymous
Right, exactly. I'll try to think about this tomorrow (it's bedtime here now, cya!). Meanwhile, one of us could ask it on the main site...which has been swamped by Qiskit questions lately ;)
