# Tag Info

9

The expression \begin{align} \alpha|1\rangle+\beta|0\rangle \end{align} is the state of a single qubit written as a linear combination of the state $|1\rangle$ and the state $|0\rangle$. If you were to make a measurement on this qubit, then you would either return $1$ or $0$ with probabilities $|\alpha|^2$ and $|\beta|^2$ respectively. The expression ...

4

Your error seems to be the misconception that entanglement will magically make the results of any experiment of photon B exactly mimic those of a similar experiment done on its entangled partner. Entanglement is more subtle than that and must be treated carefully. In particular, there are many different types of entanglement. For example, photons may be ...

4

What are those qubit in essence? Are they some kind of ultimate thing that build up our world? Yes. In the string-net picture of elementary particles, the qubits are the ultimate things that build up our world. We live inside a quantum qubit world (ie a quantum information world) (see http://blog.sciencenet.cn/blog-1116346-736093.html ) Such an emergence ...

3

The two particles $m_s$ and $m_I$ live in different vector spaces, so you are actually not picking the same basis vectors (because the basis vectors of the different particles belong to two separate vector spaces). Secondly, the tensor product between the basis vectors of the two different vector spaces will form the basis vectors of a new $3 \times 3 = 9$ ...

2

Okay, I don't quite get the details of what you are doing, but since this is linear algebra, I'd advise you to use linear algebra. You can then easily transfer between bra-ket notation and matrices. First, let's fix what we are talking about: You have one system $A$ containing one spin, so the system is a space $\mathbb{C}^2$ with basis states ...

2

The group $U(n)$ has a unique Haar measure, both right and left invariant, since it is unimodular as it is compact. Now consider the complex projective space $$\cal{P}(\mathbb C^n)= \left({\mathbb C^n} / \sim\right) - [0]\quad \mbox{when}\quad v \sim v' \quad \mbox{iff}\quad v = cv'\:, \quad c\in \mathbb C -\{0\} \:.$$ The group $U(n)$ acts smoothly (with ...

2

Performing explicit but trivial computations, it turns out that (assuming $\hbar=1$): $$\tau/2 = (\langle S_x\rangle_\rho, \langle S_y\rangle_\rho, \langle S_z\rangle_\rho)$$ So $\tau/2$ describes the expectation values of the three components of the spin when the system is in the, generally, mixed, state $\rho$. Indeed: $$\langle S_k \rangle_\rho := ... 2 Couder and Fort's experiment is based on a mathematical analogy between the Hilbert space of a particle moving in two dimensions, and the two surface of a vibrating oil bath, which interacts with an oil droplet bouncing on top of it. Naïvely, one might try to extend this analogy a two-particle system by having two oil droplets bouncing on a single ... 2 The flaw in your argument is that the claim "entanglement will instantly replicate the photons' paths and such the pattern onto Bob's screen [sic]" is incorrect. The statistics for the measurement outcomes of any experiment performed on one subsystem of a maximally entangled pair is independent of what goes on with the other subsystem. In your case, from ... 1 The answer to your first question is no. Consider the three-dimensional Hilbert space \mathbb C^4, and let B_1 be the canonical basis and$$ B_2=\left\{ \frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ i\\-1\\-i\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1 \\ -1\\1\\-1\end{pmatrix}, \frac{1}{2}\begin{pmatrix}1\\ ...

1

This can be shown directly. The definition you posted can be seen as the single-qubit version of your target, $$H=\frac{1}{\sqrt{2}}\sum_{x_1,y_1}(-1)^{x_1 y_1}| x_1 \rangle\langle y_1 |.$$ The $n$-qubit case is then \begin{align} H^{\otimes n} &=\frac{1}{\sqrt{2^n}} \left(\sum_{x_1,y_1}(-1)^{x_1 y_1}| x_1 \rangle\langle y_1 |\right) ... 1 Well here is an attempt at the answer based on your comments and question. Consider your state  \left | \psi \right>  be \left | \psi \right> = \frac{1}{\sqrt2}(\left | 0 \right> +\left | -1 \right>) $$Now int the \mathbb{C}^3 representation, this becomes$$ [\left | \psi \right> ] = \frac{1}{\sqrt2} \left[ {\begin{array}{c} 0 ...

1

The rotation operator normalization you have chosen, $R_{\boldsymbol n}(\theta)=e^{-i\theta \boldsymbol \sigma\cdot \boldsymbol n/2}$, means that for a rotation by $2\pi$ about the $z$ axis, $$R_{\hat z}(2\pi)=e^{-i\pi \sigma_z} =(-1)^{\sigma_z}=-1,$$ because $(-1)^1=(-1)^{-1}=-1$. Thus, your rotation operator has indeed rotated you back to the same point ...

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