# Tag Info

21

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 ...

12

Three real parameters are sufficient due to the constraint that $$|\alpha|^2 + |\beta|^2 = 1\tag1$$ where $\alpha$ and $\beta$ are the two components of a 2D complex vector describing the qubit state. This constraint ultimately derives from the fact that $|\alpha|^2$ and $|\beta|^2$ are probabilities of the two possible outcomes of the computational basis ...

12

Mathematically a qubit's coefficients $c_1$, $c_2$ must have the following properties: \begin{align} |c_1|^2 + |c_2|^2 =1 \tag{1}\\ |c_1|, |c_2| \in [0,1], \tag{2} \end{align} because Born's rule tells us that the modulus squared is a (classical) probability, and classical probabilities must add up to exactly 1 and must be 0, 1 or in between 0 and 1. However ...

11

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 $\alpha\in\mathbb R$,...

10

$|1\rangle$ 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 ...

10

Let $(x,y,z)$ be a point in the unit ball 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 $2\... 9 The Bloch sphere only represents the state of a single qubit. What you’re talking about is taking a multi-qubit state, and representing the state of just one of those qubits on the Bloch sphere. If the multi-qubit state is a product state (pure and separable), then the state of the single qubit is a pure state, and is represented as a point on the surface ... 9 Since a spin$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 ... 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 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$0\leq \phi\leq 2\pi$. ... 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

For vector representation of any qubit it is true that: it has to be a unit vector global phase does not matter and can be fixed at any value As a result two degress of freedom are eliminated and as a result you are left with only two free parameters. Note that although you represent a qubit on Bloch sphere, the sphere has unit radius. So, actually only 2D ...

6

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$$| \psi \rangle = \cos(\theta/2) | 0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle$$ then North is $|0\rangle$ and South ...

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-$Z$ gate on this and you get: $$|\psi'\rangle = \cos(\theta/... 6 A density matrix \rho has the properties of being Hermitian, non-negative and has trace 1. Any 2\times 2 matrix can be written in the form$$ \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 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 ... 6 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 = ... 6 There are two types of information in physics: Classical information Quantum information Physics doesn't answer the question "What is (classical or quantum) information?". This is philosophic question, and physics never answers questions of this kind. Instead, physics answers another question "How (classical or quantum) information is measured?". The ... 6 That is because we have a condition on the two complex amplitudes. The normalization condition. So it eliminates one real number and hence we can use the Bloch sphere picture. And its not exactly a 3D real space. Its similar but not exactly same. 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

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).

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

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 $\theta$ 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

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 ...

5

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}...

5

Clearly this pair of basis is not an orthonormal basis of $C^2$... $\{\lvert0\rangle,\lvert1\rangle\}$ is an orthonormal basis of $C^2$. $C^2$ is a 2-dimensional complex vector space, which means that every element of the space is essentially a vector of 2 complex numbers. $\lvert0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, and \$\lvert1\rangle = \begin{...

Only top voted, non community-wiki answers of a minimum length are eligible