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## Hot answers tagged state-space-geometry

22

For pure states, there is a reasonably simple way to make a "2 qubit bloch sphere". You basically use the Schmidt decomposition to divide your state into two cases: not entangled and fully entangled. For the not-entangled part, you just use two bloch spheres. And then the entangled part is isomorphic to the set of possible rotations in 3d space (the rotation ...

10

The conditional von Neumann entropy is a concave function: if $\rho$ and $\sigma$ are states of a pair of registers $(\mathsf{X},\mathsf{Y})$ and $\lambda\in[0,1]$ is a real number, then $$\mathrm{H}(\mathsf{X}|\mathsf{Y})_{\lambda\rho + (1-\lambda)\sigma} \geq \lambda\, \mathrm{H}(\mathsf{X}|\mathsf{Y})_{\rho} + (1-\lambda)\,\mathrm{H}(\mathsf{X}|\mathsf{Y}... 9 Geometric characterization (as any other characterization) of subsets of the quantum state space in relation with their locality and entanglement properties becomes very complicated as the number of qubits rises. The geometry of the space of negative conditional entropy two qubit states, which are also locally maximally mixed (Weyl states) is known; it is ... 9 Since a spin j irreducible representation of SU(2) has a dimension 2j+1 (j is half integer), any finite dimensional Hilbert space can be obtained as a representation space of SU(2). Moreover, since all irreducible representations of SU(2) are symmetric tensor products of the fundamental spinor representation, therefore every finite dimensional ... 8 Recall the law of cosines for two unit vectors \mathbf{u} and \mathbf{v} in \mathbb R^2:$$ \|\mathbf{u}-\mathbf{v}\|^2 = 2-2\cos\theta, $$where \theta is the angle between the vectors. Similarly, you'll recall the definition of the inner product,$$ \langle \mathbf u|\mathbf {v}\rangle = \cos\theta. $$So,$$ \|\mathbf{u}-\mathbf{v}\| = \sqrt{2}\...

7

A density matrix $\rho$ has the properties of being Hermitian, non-negative and has trace 1. Any $2\times 2$ matrix can be written in the form $$\rho=\frac{n_0\mathbb{I}+\vec{n}\cdot\vec{\sigma}}{2}.$$ The trace being 1 fixes that $n_0=1$, while the Hermitian property imposes that $\vec{n}\in\mathbb{R}^3$, where $\vec{\sigma}$ is the vector of the 3 Pauli ...

7

There is a geometric interpretation that you certainly can take seriously, but the geometry that you get is not as clean as you might have hoped. Trace distance between operator states is an example of a Banach norm on a vector space $V$. The rules for such a norm are that $||v|| > 0$ when $0 \ne v \in V$, $||\lambda v|| = |\lambda|\cdot||v||$ for $\... 7 There are many ways to describe a qutrit or a general$N$level system geometrically. There is also a large amount of references either explaining these geometries or applying them to various problems in quantum information. I'll try to explain here one quite general geometrical method, somewhat in detail. This method is a generalization of the Bloch ... 5 For more than 1-qubit visualization, we will need more complex visualizations than a Bloch sphere. The below answer from Physics Stack Exchange explains this concept quite authoritatively: Bloch sphere for 2 and more qubits In another article, the two qubit representation is described as a seven-dimensional sphere, S 7, which also allows for a Hopf ... 5 Upon some reflection, the answer is that no, they most definitely do not. The easiest way to see this is to observe that there are$d^2-1$orthogonal directions in the Bloch representation (i.e. orthogonal Hermitian traceless operators) containing pure states. This means that that the pure states are not contained in any linear subspace of dimension less ... 5 The other answer already gave a counterexample. From a geometrical point of view, the question is about the intersection of hyperplanes with hyperspheres. Indeed, the purity of a state$\rho$with eigenvalues$(p_i)_i$is$\sum_i p_i^2=\|\boldsymbol p\|^2$, whereas these are probabilities and therefore also constrained by$\sum_i p_i=1$. The set of ... 4 Let me supplement the other answer by also showing what happens in the general case of the Bloch representation of generic qudits of dimension$d$. Let$\rho$be an arbitrary state over$d$modes, that is, a$d\times d$positive semi-definite Hermitian matrix with unit trace. As any other Hermitian matrix, we can decompose it in terms of a basis of ... 4 $$\rho = \begin{pmatrix} a & 0 & 0\\ 0 & 1-a & 0\\ 0 & 0 & 0\\ \end{pmatrix}\\ \sigma = \begin{pmatrix} b & 0 & 0\\ 0 & c & 0\\ 0 & 0 & 1-b-c\\ \end{pmatrix}\\ Tr (\rho^2 ) = 2 a^2 - 2a + 1\\ Tr (\sigma^2 ) = b^2 + c^2 + (1-b-c)^2$$ Let's pick$a=1/2$so$2a^2-2a+1=\frac{1}{2}. $$b=\frac{2}{3}\\ c=\frac{1}... 4 "Generalized Bloch" manifolds are synonyms to coherent state manifolds. The points of these manifolds do not correspond, in general, to orthonormal vectors, as there are much more points than the dimension of the system's Hilbert space. Points on the manifold correspond rather to generalized coherent states. These states are actually classical, they ... 4 We have some multiqubit visualizations within Q-CTRL's Black Opal package. These are all fully interactive and are designed to help build intuition about correlations in interacting two-qubit systems. The two Bloch spheres represent the relevant separable states of two qubits. The tetrahedra in the middle visually capture correlations between certain ... 3 I need some useful sources about the geometry of qutrit. The most useful resource I know on the geometries of qutrits is the paper Geometry of the generalized Bloch sphere for qutrits. Specifically related to the Gell-Mann matrix representation. The eight Gell-Mann matrices, which form one of the generalizations of Pauli matrices to 3-level systems, are ... 3 This other answer already gave a nice proof that orthogonal bases are mapped into vectors r_i such that \sum_i r_i=0. Here I'll work out explicitly the coordinates in a few cases, to show what kind of geometrical figure exactly comes out. I'll consider in particular the computational basis in d dimensions: \{\lvert 1\rangle,...,\lvert d\rangle\}. ... 3 One natural generalization of that property is that Bloch vectors for a basis set must sum to 0 vector. Though, this property is not a criterion for basis sets in dimensions higher than 2. If \rho_i, i=1 .. d, are corresponding density matrices for a basis set, then$$ \sum_{i=1}^d \rho_i = I $$Now if \sigma_i, i=1 .. d^2-1, are generalized Pauli ... 3 Almost. You get a manifold with boundary with the Bloch ball. The radius from the origin parameterizing how pure it is. The origin being maximally mixed. This isn't a manifold because a point on the boundary has a neighborhood that is homeomorphic to a half space but not one homeomorphic to \Bbb R^n. 3 You could try solving this numerically using semidefinite programming. We know the trace norm of an operator X can be formulated as$$ \begin{aligned} \|X\|_1 &= \min_{Y,Z}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\quad \mathrm{s.t.} \quad \begin{pmatrix} Y & X \\ X^* & Z \end{pmatrix} \geq 0. \end{aligned} $$Furthermore, we can write your problem ... 3 If perturbations are sufficiently small and \rho_{AB} has sufficiently broad support then a desired global state \rho_{AB}' exists. Define$$ \rho_{AB}' = \rho_{AB} + (\rho_A' - \rho_A) \otimes \rho_B + \rho_A \otimes (\rho_B' - \rho_B) \tag1. $$Note that \rho_{AB}' is Hermitian and trace one, but may not be positive. However, \rho_{AB}' is positive ... 3 Filling out a number of details for the sake of a complete answer — Starting from the linked article, Distance measures to compare real and ideal quantum processes [arXiv:quant-ph/0408063], the definition of fidelity is given in Eqn. (4) as$$ F(\rho,\sigma) = \mathrm{tr}\Bigl( \!\sqrt{\sqrt{\rho} \!\phantom|\sigma \phantom|\!\!\sqrt{\rho}\phantom|}\Bigr)^2$$... 3 I'll try to address the problem from the Riemannian geometry point of view. In this approach, the distances are identified as length of geodesics of Riemannian metrics on spaces of quantum states. The advantage of this approach lies in the fact that the Riemannian distances automatically satisfy the metric axioms of positivity, symmetry and the triangle ... 2 The Bures metric is the limit of the Bures distance for two infinitesimally close density matrices \rho and \rho+d\rho. The Bures distance however, is not unique. It depends on the space on which we integrate the Bures metric. The Bures distance for positive matrices differs from that of density matrices (with a unit trace), however they have the same ... 2 The outer product of two states is a matrix. Here are some often used outer products: |0\rangle \langle 0 | = \begin{pmatrix} 1&0\\0&0 \end{pmatrix} \qquad |0\rangle \langle 1 | = \begin{pmatrix} 0&1\\0&0 \end{pmatrix} \\ |1\rangle \langle 0 | = \begin{pmatrix} 0&0\\1&0 \end{pmatrix} \qquad |1\rangle \langle 1 | = \... 2 Non-singular density matrices \rho have no zero eigenvalues, \lambda=\lambda_{\mathrm{min}}(\rho)>0, with \lambda_{\min} the smallest eigenvalue. Then, the ball$$ \{\rho+M|-\lambda I \le M \le \lambda I\}$will also be in the set of density operators, and thus,$\rho$is an interior point. 2 Your intuition is correct, since rank-2 density matrices will be convex combinations of rank-1 density matrices, but they will be singular and hence will still be on the boundary. We can prove that non-singular$\rho$are in the interior of$\Omega_3$with a fairly direct proof. Let$\epsilon>0$and let$\rho'$be some matrix (the Bloch representation ... 2 A simple way of proving that the purity (or any other property which only depends on the eigenvalues of$\rho$) can only depend on the distance from the center of the Bloch sphere is rotational invariance: Rotating a density matrix$\rho\mapsto U\rho U^\dagger\$ corresponds to a rotation on the Bloch sphere. Thus, any property which only depends on the ...

2

A paper has been published on the subject, called "Bloch sphere model for two-qubit pure states" https://arxiv.org/abs/1403.8069

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