New answers tagged quantum-information
0
Quantum teleportation and super dense coding use entanglement. You can teleport a qubit, storing A LOT of information in the coefficients $\alpha \vert0> + \beta \vert 1>$ by communication only 2 classical bits of information.
-2
"Purity" of "mixedness" (if you permit these words) is a property of the system and not the observer. A system is said to be in a pure state if it is in one of the allowed states $|\psi_i\rangle$, $i = 1 \ldots n$ or in a linear superposition $|\phi\rangle = \sum_{i=1}^{i=n}\alpha_i|\psi_i\rangle$ of such states. If it is in a mixed state, then such a ...
1
I found this question by chance yesterday while looking for articles on Werhl entropy. I may have found a possible answer after reviewing properties of the Wigner quasiprobability distribution on http://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution#The_Wigner.E2.80.93Weyl_transformation.
Consider property 7 under the section "Mathematical ...
-1
Well, this should be feasible.
You generate entanglate EPR pairs of qubit. For each pair you keep the qubit A, you send the other qubit B on a spaceship one light-year away.
You have to prepare that in advance because it will take more than one year to reach this place.
Today you mesure your qubits and XOR your message with the result to encode it.
The ...
0
Dear Dan with your suggestions I trying a one answer. Can you verify this please? . (I build to $R_D = H^{\otimes n}(2|0\rangle\langle0|-I)H^{\otimes n}$.). The first and the last column represented $H^{\otimes n}$. For the expression $(2|0\rangle\langle0|-I)$ while the input has form $|000\cdots00\rangle$ tha phase is the same, otherwise the phase change. I ...
0
The only related idea I'm aware of is position based quantum key distribution, which means that you can only distribute keys to a partner located in a specific place. But they only work as long as eavesdroppers have limited resources. You can find more information in "Position-Based Quantum Cryptography: Impossibility and Constructions" ...
1
You are halfway there, because you already wrote down the Hadamard gates. The remaining part, $I-2\left|0\right>\left<0\right|$ (I negated it from what you have, since this gives a simpler solution), is diagonal in the computational basis. Write down these diagonal entries and say out loud to yourself what this operator "does".
2
Quantum cryptography is pretty technical, but here is an elementary argument.
The Heisenberg uncertainty principle implies the no-cloning theorem for quantum-mechanical systems. If you could make an exact copy of a quantum-mechanical system, you could use one copy to measure its position, and the other to measure its momentum. This would violate the ...
3
Because the very act of viewing the code changes it. Maybe not uncrackable but any sniffing will be evident because it will be changed. From RSA Laboratories:
Quantum cryptography has a special defense against eavesdropping: If an enemy measures the photons during transmission, he will use the wrong basis half the time, and thus will change some of the ...
0
This is exactly the question to which I was seeking with a colleague of mine to give some sort of answer. We considered a game played by a team of two - say Alice and Bob - in which the value of a random variable $x$ is revealed to Alice only, who cannot freely communicate with Bob. Instead, she is given a quantum $n$-level system, respectively a classical ...
0
Actually the story of the tensor product, which is typically told as an axiom in the world of quantum information, comes from the relativistic electron theory of Dirac.
In relativistic quantum mechanics one wavefunction is replaced by 4 wavefunctions, and a suitable contraction leads to a two-element array (1,0) or (0,1) which we call the spin.
In the case ...
0
This is only true if A and B are independent. In that case,
$\mathbb{E}[AB] = \mathbb{E}[A]\mathbb{E}[B]$ and your calculations are correct.
Without knowing the distribution for B, this is the strongest supposition needed for your calculations to be correct.
6
This was too long for a comment.
There are two papers that have a nice characterization of positivity in terms of the Bloch vector: quant-ph/0301152 and quant-ph/0302024. You might be able to use dynamic programming to speed up a check of positivity by using the conditions in these papers. Basically, they reduce the positivity constraint to a constraint on ...
4
You are right that there's not much you can do with a single qubit. However it doesn't take many before you can do things that can't be done with classical bits. Here are a few illustrative examples.
Communication
Imagine I make this offer to you and a friend of yours: I will give you each a classical bit, and then you will each give me a classical bit. If ...
1
You can obtain the coordinates in 3-space corresponding to the Bloch sphere, by taking expectation values of the Pauli spin operators:
$$\begin{align*} x &= \langle\psi | \sigma_x |\psi\rangle \\ y &= \langle\psi | \sigma_y |\psi\rangle \\ z &= \langle\psi | \sigma_z |\psi\rangle \end {align*}$$
and in the case of the state you describe, we have ...
0
I'm not completely sure of the following - if you have any comments/questions about it I'd be very happy to hear them.
Calculating $P(E'|E)$ comes from the standard expression $P(E') = P(E'|E)P(E)$, so this is why we're looking for the ratio of $P(E')/P(E)$.
Start from equation 12:
$P(E) = \prod_{\ell} (1-p) \prod_{\ell} ...
2
If you read the wikipedia article about the Bloch sphere, you will see that any pure state has the form
$|\psi\rangle = \cos\left(\tfrac{\theta}{2}\right) |0 \rangle \, + \, ( \cos \phi + i \sin \phi) \, \sin\left(\tfrac{\theta}{2}\right) |1 \rangle$
with $0 \leq \theta \leq \pi$ and $0 \leq \phi < 2 \pi$. Notice that if you have a complex coefficient ...
0
You need to know that the probability that $\lvert\phi\rangle$ is in state $\lvert a\rangle$ is
$$
P = \lvert\langle a\vert\phi\rangle\rvert^2
$$
And that your states are orthonormal such that $\langle n\vert m\rangle = 0$ if $n\ne m$ and $1$ if $n=m$.
3
Computation of the $S_z$ probability distribution for each of the
manifolds of equal entanglement:
Remark: Notations and references from Kuś and Žyczkowski are used.
Case 1: The separable case: The state vector is parametrized as (equation: 24)
$w = \begin{bmatrix} \cos \alpha \cos \beta e^{i \chi_1},& \cos \alpha \sin\beta e^{i \chi_2}, &\sin ...
0
There is a difference between finding a solution and recognizing a solution. Oracle can recognize the solution or solve a particular instance of the problem but cannot give you the solution for complete problem. Or in other words, oracles gives you a part of solution and you may need to consult oracle a number of times to get the complete solution. Oracle ...
0
The oracle doesn't need to know the desired state in order to verify whether a given state is the desired state.
Grover's algorithm can be applied to NP-complete problems. This is the set of problems for which there is no known way to generate a solution in polynomial time, but a given solution can be in recognized in polynomial time.
1
You made a mistake in calculation. In the last matrix multiplication, when you multiplied the last row with the column vector the result should be 0. I suspect that you accidently took (-i)^2=1 instead of (-i)^2=-1
1
Such statements emerge when trying to capture quantum physics in classical physics terms. In other words, statements like "P or not P is not a valid proposition in a quantum world" are typically made by philosophers who don't really understand quantum physics and stubbornly attempt to map the quantum onto their classical intuition.
A much more insightful ...
3
Edited to add the second part
Edited again, for part 3 and 4
$\newcommand\ket[1]{\left|#1\right>} \newcommand\bra[1]{\left<#1\right|} $
1. Absence of Quantum Loophole
You can easily see that there is no "quantum loophole" in your argument by writing explicitly any pure separable state. With your notations, we have :
$$
...
1
Here's a calculation to get you started.$\def\ket#1{\lvert#1\rangle}$
Define $\ket{h_j} = H\ket{j}$:
$$\begin{align*}
\ket{h_0} &= \tfrac1{\sqrt 2}\Bigl(\ket0 + \ket1\Bigr) \\
\ket{h_1} &= \tfrac1{\sqrt 2}\Bigl(\ket0 - \ket1\Bigr)
\end{align*}$$
Compute (for example) $ \ket{h_0}\ket{h_1}$ by distributing the tensor product over the addition and ...
-2
We have to be careful with the bra-ket formalism and its meaning. Unlike $|x_1>$, I am not sure that the notation $|x_1 x_2>$ where $x_1$ and $x_2$ are positional coordinates makes any sense. In literature [1] the notation $|ab>$ designates the Slater determinant or Hartree-Fock state, i.e.:
$$|ab>=c_a^{\dagger}c_b^{\dagger}|0>=\phi_a(x_1) ...
-2
The question asked is relevant and something that is usually ignored because no-one understands the EPR paradox. (well I have ideas: http://quantummechanics.mchmultimedia.com/
I find that when students come up against this question of non-locality (persistence of entanglement) they are puzzled and, like good scientists, ask their mentors. The mentors ...
Top 50 recent answers are included
