Constraints on higher-dimensional Bloch vectors

I'm interested in the constraints on the $$(4^n-1)$$-dimensional generalized Bloch vector (the Bloch vector for $$n$$ qubits). To the best of my knowledge, these are not analytically characterized for the general case, but what if the vector only has support on a fully commuting or fully anticommuting set of generalized Pauli operators (tensor products of $$2\times2$$ Pauli matrices)? Does anyone know if the constraints simplify in either of these cases?

• what do you mean by "constraints" here? The boundary of the set of states can be characterised by saying that (taking $I/d$ as the origin of the state space) the length of each vector should be less than or equal to the inverse of the absolute value of the minimum eigenvalue of the operator corresponding to that direction, see physics.stackexchange.com/a/425101/58382. If the operatorial basis is chosen appropriately, one can even say that the Bloch sphere is a hypersphere with fixed radius.
– glS
Apr 25, 2019 at 21:43
• What do you even mean by "Bloch vector"? Apr 25, 2019 at 23:32
• @NorbertSchuch, by Bloch vector I mean the coefficients of the (generalized) Pauli matrices in the density matrix, so something like $\rho=\frac{1}{2^n}(I+\vec{m}\cdot\vec{P})$, where $n$ is the number of qubits, $I$ is the $2^n\times2^n$ identity, and $\vec{P}$ is the vector composed of the $4^n-1$ Pauli matrices for $n$ qubits. Here $\vec{m}$ is the (generalized) Bloch vector. I say "something like" this, because there are various other conventions that add normalization coefficients to the vector.
– Will
Apr 26, 2019 at 1:26
• @glS, good point, I should be more specific. I am looking for what the surface of the Bloch "sphere" is, i.e., the boundary between vectors $\vec{m}$ (see my previous comment) that give valid density operators and those that do not, for the specific cases in which the vector only has support on fully commuting or fully anticommuting subsets of the Pauli operators. The answer you linked to is very helpful though -- thanks! I guess my question is nearly a duplicate of that one.
– Will
Apr 26, 2019 at 1:35

Given a completely anticommuting set $$\mathcal{S}$$ of Pauli operators on $$n$$ qubits, we know that for any density operator $$\rho$$ we may write $$\rho=\frac{1}{2^n}(I+\vec{m}\cdot\vec{\mathcal{S}}+\cdots),$$ where the ellipsis indicates that in general more Paulis will be necessary to specify the whole density matrix (I've written the set $$\mathcal{S}$$ as a vector $$\vec{\mathcal{S}}$$ so that I can represent the coefficients simply as the vector $$\vec{m}$$.) Then the vector $$\vec{m}$$ satisfies $$|\vec{m}|\le1.$$ This follows fairly straightforwardly from the fact that the observable $$\vec{m}\cdot\vec{\mathcal{S}}$$ has eigenvalues $$\pm|\vec{m}|$$ when $$\mathcal{S}$$ is fully anticommuting.
Furthermore, any (real) vector $$\vec{m}$$ satisfying $$|\vec{m}|\le1$$ corresponds to at least one valid density operator with the form given above, namely $$\rho'\equiv\frac{1}{2^n}(I+\vec{m}\cdot\vec{\mathcal{S}}).$$ Since the density operators form a convex set, it is enough to show that this is true for any $$|\vec{m}|=1$$, which one can do by explicitly constructing the corresponding pure state (which turns out to always exist) and showing that its outer product with itself gives the desired density operator.