Tag Info

New answers tagged

0

For your first questions: Observables correspond to hermitian (or self-adjoint) operators. As such, the eigenvalues are real, and these values are the possible outcomes. Also because of the hermitian/self-adjointness of the operator, when you have eigenvectors with different eigenvalues, the eigenvectors are orthogonal. So you know the values, but what ...


0

I would like to determine the general expansion of $(A+B)^n$, where [A,B]≠0 The expansion of $(A+B)^n$ for non-commuting A and B is the sum of $2^n$ different terms. Each term has the form $$ X_1X_2...X_n\;, $$ where $X_i=A$ or $X_i=B$, for all the different possible cases (there are 2^n possible cases). For example: $$ ...


0

if $[A,B]=0$ then as you know you get the usual $$ (A+B)^n = \sum_{p=0}^n C^n_p A^{n-p}B^p $$ Now if $[A,B]\neq 0$ each term in the sum (for each $p$) splits into a sum of $C^n_p$ terms of all possible permutations of $(n-p)$ $A$s and $p$ $B$s, without regard to the order of $A$s and $B$s. Equivalently to the sum of all possible permutations of $(n-p)$ $A$s ...


1

In quantum mechanics an "observable" like the position of an atom, $\hat X$, is represented by a Hermitian operator. An "observable" is some physical quantity you can actually go out and measure, like the position. The eigenvalues of that operator are the possible measurement results. In quantum mechanics a "system" (some physical entity that has ...


1

Correct, the evolution of a quantum system does not necessarily need to be unitary. However, if the system is closed, it needs to be unitary. The latter is due to the fact that we assume the system evolves due to the Schrödinger equation with a self-adjoint Hamiltonian, which induces a unitary evolution operator via Stone's theorem. For non-closed systems, ...


1

$\renewcommand{ket}[1]{\left| #1 \right\rangle}$ In my question I was trying to define the $\mathrm{NOT}$ operation using set theory. I got confused because I regarded $ \mathrm{NOT}$ as mapping from points in $[a,b]$ onto points in $(-\infty,a,)\cup (b,\infty)$. This is wrong. Let $X = [a,b]$ and let $Y = (-\infty,a,)\cup (b,\infty)$. Consider the set $S ...


4

$\newcommand{\HH}{\mathcal{H}}$As Martin says, entanglement is correlation rather than anything like "determining" the state of the other particle. We don't necessarily even need to talk about correlations, though, although they are one of the primary interesting features of entangled states. More precisely, your quote Quantum entanglement is a physical ...


4

The problem is that entanglement doesn't mean that a measurement on the first particle determines the outcome of the second. Although this is often perpetuated, it's not the gist of entanglement. Entanglement is (mostly?) nonclassical correlations. Bipartite states are correlated, when the outcome of the measurement on one particle tells us something about ...


3

The de Broglie wavelength, $\lambda$, is given by $$\lambda = {h \over p}$$ where $h$ is planck's constant and $p$ is momentum. If we take the mass to be $160$ kg and speed to be $100$ ms$^{-1}$ then we get $\lambda = 4\times10^{-38}$m given a slow light spaceship and only one person (more people and speed would increase the momentum and decrease the value ...


0

See the following examples: $\rho_1 = \frac{1}{2}(|00\rangle + |11\rangle)(\langle 00| + \langle 11|)$ is a maximally entangled state. $\rho_2 = \frac{1}{2} (|0\rangle \langle 0| + |1\rangle \langle 1|)$ is a maximally mixed state. The difference is not related with "maximally". Your question can be changed to : What's the difference ...


3

Suppose we have two Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$. A quantum state on $\mathcal{H}_A$ is a normalized, positive trace-class operator $\rho\in\mathcal{S}_1(\mathcal{H}_A)$. If $\mathcal{H}_A$ is finite dimensinal (i.e. $\mathbb{C}^n$), then a quantum state is just a positive semi-definite matrix with unit trace on this Hilbert space. ...


0

Here's a list of quantum computer simulators, categorized by the programming language in which they were written: http://www.quantiki.org/wiki/List_of_QC_simulators Specifically, http://www.davyw.com/quantum/ allows the full simulation of up to 9 qubits.


0

When the state space for a system can be expressed as a tensor product of the state spaces of individual components of the system, an entangled state is one that can't be expressed as a tensor product of states of those individual components. Thus an entangled state is a particular type of (pure, i.e. non-mixed) state. A mixed state, by contrast, is a ...


2

Property 1 is equivalent to R being a correctable region. Property 2 of course follows from property 1. It is less obvious, but property 2 also implies property 1. (You can prove this by considering different sets of orthogonal codeword states.)


0

Suppose the particle is in a state of $\frac{1}{\sqrt{2}}$up + $\frac{1}{\sqrt{2}}$down. There is a 50% chance of measuring it up and a 50% chance of measuring it down. If you run this through the imperfect cloner multiple times, it won't have half the clones be up and half be down. Suppose the first clone is up. Since it measured the particle spin up, it is ...


1

I have the following two density operators, the paper I am reading says that these two operators have same eigenvalues $$\rho^i = \frac{1}{3} ( |0\rangle \langle 0 | +|1\rangle \langle 1 > |+|2\rangle \langle 2 |+a|0\rangle \langle 1 |+a|1\rangle \langle 0 > |+c|1\rangle \langle 2 |+c|2\rangle \langle 1 |),$$ $$\rho^f = \frac{1}{3} ( ...


3

Hints: What happens to the density operator and its eigenvalues under a change of the orthonormal basisvectors $|0\rangle $, $|1\rangle $, $|2\rangle$ by phase factors? More generally, what happens to the density operator and its eigenvalues under a unitary transformation $\rho\longrightarrow U^{\dagger}\rho U$?


0

It is confusing because they say superposition when they mean entanglement. The yellow beam containing the information about the cutout got it from an entangled red beam which made contact with the cutout. In fact, National Geographic has a much better article: ...


1

The things are simpler than you say. Though, let me work with the polarization singlet of photons, $$|S\rangle = \frac {|+\rangle |+\rangle + |-\rangle |-\rangle}{\sqrt {2}}.$$ Let Bob prepare 4 sets of polarized photons: (1) $|+\rangle$, (2) $|-\rangle$, (3) $|u\rangle = \frac {|+\rangle + |-\rangle}{\sqrt{2}}$, (4) $|v\rangle = \frac {|+\rangle - ...


1

The second density matrix is actually a rank-1 projection (if normalised) hence a dyadic product and therefore a pure state. It is enough then to measure against a state which is perpendicular to this vector (i.e. $(1/\sqrt 2,1/\sqrt 2)$) to say whether the qubit is not coming from the second source.


0

$a$ and $a^\dagger$ are operators. $\alpha$ is a pure imaginary number which we then call $i\alpha$ with $\alpha$ now real, this is why from eq1 to eq2 we use $\alpha→i\alpha$ and $−\alpha^*→i\alpha$. Now from eq2 to eq3 we simply substitute $a+a^\dagger$ for $x$ operator. I don't understand why you want to write $\alpha=\ldots$, its just some real constant. ...


0

Mathematically, Norbert Schuch pointed out "purification of quantum states": Given an ensemble on a Hilbert space $\mathcal{H}$, you can always write it as a pure state on a bigger space $\mathcal{H}\otimes \mathcal{H}^{\prime}$ such that the restriction of the pure state to $\mathcal{H}$ results in the ensemble description. There is maybe a second and ...



Top 50 recent answers are included