# Tag Info

14

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 ...

8

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 ...

7

It is a convention, chosen so that $\theta$ is the azimuthal angle of the point representing the state in the Bloch sphere. To see where this convention comes from, start from a state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. Remembering the normalisation constraint $|\alpha|^2+|\beta|^2=1$, and assuming without loss of generality $\alpha\in\mathbb R$,...

7

$|1\rangle$ and $-|1\rangle$ are assigned to the same point on the Bloch sphere because they are equal up to global phase. Algebraically: $|1\rangle \equiv -|1\rangle$ where $\equiv$ means "equal up to global phase". Meaning there is some $\theta$ such that $-|1\rangle = e^{i \theta} |1\rangle$. The thing that is confusing you is that, despite the fact that ...

7

The way to think about the Bloch sphere is in terms of the density matrix for the state. $Z$ acting on either $|0\rangle\langle 0|$ or $|1\rangle\langle 1|$ does nothing, as is true for any diagonal density matrix. To see the effect of the rotation, you need to look at how any non-diagonal density matrix is changed by $Z$, such as $|+\rangle\langle +|$.

7

The Bloch sphere only represents the state of a single qubit. What you’re talking about is taking a multi-qubit state, and representing the state of just one of those qubits on the Bloch sphere. If the multi-qubit state is a product state (pure and separable), then the state of the single qubit is a pure state, and is represented as a point on the surface ...

7

Let $(x,y,z)$ be a point in the unit sphere with $x^2+y^2+z^2 \leq 1$. The state associated with this point is \begin{eqnarray*} \rho &=& \frac{1}{2} (I_2 + x \sigma_x + y \sigma_y + z \sigma_z)\\ &=& \frac{1}{2} \begin{pmatrix} 1+z&x-iy\\ x+iy&1-z\\ \end{pmatrix} \end{eqnarray*} This is just a convenient way to parameterize all $2\... 6 A density matrix$\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 ... 6 The Pauli-$Z$gate maps$|0\rangle$to$|0\rangle$and$|1\rangle$to$-|1\rangle$. For Bloch sphere representation, state of a qubit is written like (look at my previous answer for a detailed explanation) $$|\psi\rangle = \cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle$$ Apply the Pauli-$Zgate on this and you get: $$|\psi'\rangle = \cos(\... 6 Yes. The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state |\Psi\rangle of a qubit can be written in the form:$$|\Psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle$$where 0\leq \theta\leq \pi and 0\leq \phi\leq 2\pi. ... 5 If we use the convention$$| \psi \rangle = \cos(\theta) | 0 \rangle + e^{i \phi} \sin(\theta)| 1 \rangle$$then the North (\theta=0) and the South (\theta=\pi) are (physically) the same state |0\rangle; If we use the convention$$| \psi \rangle = \cos(\theta/2) | 0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \ranglethen North is |0\rangle and South ... 5 Let \hat{n}=(\cos\phi\sin\theta,\sin\phi \sin\theta,\cos\theta) i.e. the Cartesian coordinate vector for a point on the unit sphere with polar angle \theta and azimuthal angle \phi. By sending a spin-1/2 particle through a Stern-Gerlach device with orientation \hat{n}, we can measure the observable \begin{align} S_n:=\vec{S}\cdot \hat{n} &=S_x \... 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 The problem you are describing (i.e. finding an approximation of some state given some number of identical copies of it and some set of measurements) is known as quantum state tomography or state tomography for short. In practise, the most efficient schemes for state tomography will depend on a specific experiment's setup and limitations, for which ... 5 Some people really do visualise single qubit operations by thinking about the Bloch sphere, and they work out everything based on that picture. Those who like thinking in that way are often incredibly quick at these sorts of manipulations, and have more intuitive understandings for why things work. But here's one concrete example where I use it: averaging ... 5 The components of the Bloch vector of a state are the expectation values of the X,Y and Z Pauli matrices in that state and it has to be a full three-dimensional vector to capture the interior of the Bloch sphere as well, which represents mixed states. In general a state with density matrix \rho of a single Qbit has Bloch vector \vec{r} when \rho = \... 5 This doesn't really answer the question as it's not an online simulator. It might still be relevant though as it is a way to produce this sort of gifs if one has access to the software. It is relatively easy to do this sort of things using Wolfram Mathematica. As a quick and dirty example, if we just define a couple of relevant helper functions: pauliX = ... 5 The\theta$and the$\phi$angles are not equivalent in the Bloch sphere. First, they have different ranges -- one is$\pi$and the other is$2\pi$. More importantly,$\phi$is a rotation around a fixed axis,$z$, while$\theta$is a rotation around a non-fixed axis that is moving with$\phi$. For$\phi=0$this axis is$y$, for$\phi=\pi/2$it is$x$, and ... 5 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$\...

5

Neither. Phase estimation algorithm does not estimate a property of a qubit state (and the angles $\theta$ and $\varphi$ in your question are exactly that - a property of a given qubit state). Rather, it estimates the eigenvalue of a given unitary $U$ that corresponds to the given eigenvector $|\psi\rangle$ - the angle $\alpha$ in the following equation: ...

5

Prepare a qubit in state $|\psi\rangle=\mathrm{cos}\frac{\theta}{2}|0\rangle+\mathrm{e}^{i\phi}\mathrm{sin}\frac{\theta}{2}|1\rangle$, given the angles $\psi$ and $\theta$. Let's start with a qubit in the $|0\rangle$ state, as is customary for Q#. You can use one of the general library operations to prepare the state, such as PrepareArbitraryState. Or you ...

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 ...

4

Single-qubit unitaries are just 3D rotations, multiplied by a phase. So in order to find the actual angles, you can resort to the theory of rotation matrices, in particular to Euler's rotation theorem, which states that any rotation is a composition of 3 rotations (the theorem proof is constructive, so you get the actual angles).

4

First of all, qutip is not a visualisation library, even though it does provide some visualisation functionalities, mostly leveraging matplotlib. However, because qutip does provide handy functionalities to plot Bloch spheres and points on it, it does make sense to ask how one can tweak such functionalities to for example add tangent vectors to the bloch ...

4

Bloch spheres are a reasonable visualisation of individual qubits, so long as you don't fall in to the trap of thinking that's what they actually, physically look like. It's just a convenient mathematical mapping from a parametrisation of the state into a pretty picture. And it doesn't work so well once entanglement between multiple qubits comes into play. ...

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

You can calculate the probability of a given answer $\pm 1$ to each measurement by evaluating $$\langle B|\frac{I+(-1)^{\eta_1}\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+(-1)^{\eta_2}\vec{n}_2\cdot\vec{\sigma}}{2}|B\rangle$$ Thus, the probability of equal measurement outcomes is  \langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+\vec{...

4

Hadamard gate can be interpreted as a rotation in 3D Euclidean space (on Bloch sphere) by angle $\pi$ around X+Z axis. The qubit rotation by angle $\theta$ around axis pointed by unit vector $\textbf{n}=\{n_x,n_y,n_z\}$ is described by rotation operator ($X$, $Y$ and $Z$ are Pauli matrices) \begin{align} R_{\textbf{n}}(\theta)=&n_xe^{-i\frac{\theta}{2}X}...

3

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 ...

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 $\Bbb R^n$.

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