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2

All $\mathrm{SU}(2)_k$ with $k>2, k\neq 4$ are universal. For a proof see http://arxiv.org/abs/math/0103200.


2

This is a very complex question you are asking and very much depends on the notion of "operational" you consider. This paper, for instance, shows that any entangled state can help to do some task better than without it. Regarding PPT vs. NPT, e.g. PPT states cannot be distilled, while most (?) NPT states can be distilled. (Whether all NPT states can be ...


2

You have fallen into a very subtle trap, essentially the one I discussed here, but it's probably worth going into the specifics. Lets say that after measurement the combined state was $|00\rangle\langle00|$, so $B$'s actual reduced density operator is $|0\rangle \langle0|$ after measurement The problem with this is that you cannot consider the ...


1

You can use the teleportation protocol to teleport and part of a larger quantum state (which can be arbitrarily entangled), and it will work the way it should: I.e., if initially A+C hold $\vert\psi\rangle_{AC}$, after the protocol B+C hold $\vert\psi\rangle_{BC}$. The same is true if the initially shared state is mixed. This follows from the linearity of ...


0

I was under the impression that the standard teleportation protocol preserves teleportation. That is to say that if Alice shares some entangled state with Bob, and then teleports her part of this state to Carol in the usual way, Carol will now share that entangled state with Bob. Using this teleportation protocol requires Alice and Carol to share a ...


1

Note: There is a short summary at the bottom. This is actually also described in Nielsen&Chuang: You don't learn about general measurements, because they are completely equivalent to projective measurements + unitary time evolution + ancillary systems, all of which is described in your usual QM formalism. The Measurement Postulate Let's start from ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$A state $ \ket \psi \in H_1 \otimes H_2 \otimes H_3$ is said to be entangled if there exist no coefficients $a_i,b_i,c_i$ such that: $$\ket \psi = \sum_{ijk} d_{ijk} \ket{e^1_i} \otimes \ket{e^2_j} \otimes \ket{e^3_j} = \sum_i a_i \ket{e^1_i} \otimes \sum_j b_j \ket{e^2_j} \sum_k c_k \ket{e^3_k} \tag{1}$$ ...


3

You can calculate the density matrix of the state $$\rho = |\varphi\rangle\langle\varphi|$$ And then, the reduced density matrix for one of the particles taking the partial trace $$ \rho_A = \textrm{tr}_{B,C}(\rho)$$ The state is entangled if and only if the reduced matrix is a mixed state. This can be checked, for example, calculating the von Neumann ...


0

I've done a lot more work on this since asking, and now have the answers to my questions. First, I mentioned that it seemed that with single qubit gates you can always change the state given by c-z-z-z... (from the equal superposition) to one with this 'bit-flip' property, but that I hadn't proven it. Turns out this is true for any circuit made up of c-z ...


1

The pentagon and hexagon equations are consistency equations for the fusion and braiding of anyons. They should be satisfied by anyons realized in a gapped, local Hamiltonian. If you do not find solutions, then according to our current understanding of topological phase, yes, such anyons do not exist (in the sense that can not be realized by gapped, local ...


0

Only a partial answer here, since I am not a real expert into the field (especially not adiabatic quantum computation): If such limitations were known, these particular lines of research would be pretty much dead. Since they are not, I conclude that no such limits are known. If we take all possible implementations into account, I do not know a single ...


1

At a fundamental level, if you are asking whether the universe has unknown variables (which Bayesian statistics often teases out in other sciences), the answer is provably not. If you are asking whether any of the math looks the same, the answer is an unequivocal yes, though you must remember that calculating a probability in Bayesian statistics is ...


2

Heisenberg discusses probability in his Physics and Philosophy. He first stresses that quantum mechanics contains objective, frequentist probabilities: Probability in mathematics or in statistical mechanics means a statement about our degree of knowledge of the actual situation. In throwing dice we do not know the fine details of the motion of our ...


6

It's pretty hard to be Bayesian about quantum mechanics without believing in some sort of underlying hidden-variable theory. Such theories are highly unpopular in modern culture (not to mention experimentally falsified in the majority of cases) and so the overwhelming interpretation amongst physicists is an operationalist/frequentist one.


0

The basic approach is to break independent operations to different gates and then calculate the gate operation on all the basis vectors (it's not very pleasant but always work). At your case the representation matrix is - $ G=(\mathbb{I} \otimes \mathbb{H} \otimes \mathbb{I})(C_{1,3}) $ $ C_{1,3}=\begin{pmatrix} 1 & 0 & 0 & 0 & 0 ...


4

If you want to experimentally create a qubit, you need some actualization. One example is the z-component of the spin of a spin-1/2 particle such as an electron. There are two independent states which can be denoted $\left(\matrix{ 1 \\ 0}\right)$ and $\left(\matrix{ 0 \\ 1}\right)$ and each of which can be produced by orienting a stern-gerlach device in ...


1

It's not uncommon that popular science articles will refer to paradoxes in physics. However it is extremely important to understand that there are no paradoxes in physics. Our current theories of physics are self consistent and do not contain paradoxes (though there are some conditions not covered by any of our existing theories). Non-physicists tend to use ...


3

There is indeed a way to construct the projection-operators, when you only know the operator itself and its eigenvalues. The derivation can be found in Julian Schwingers "Quantum Mechanics: Symbolism of Atomic Measurement" and leads to $$ P_j = \prod_{i\neq j} \frac{A-a_j}{a_i - a_j}, $$ where the product goes over all distinct eigenvalues $a_i$ of $A$. ...


1

Eigenvectors are undetermined up to a scalar multiple. So for instance if c=1 then the first equation is already 0=0 (no work needed) and the second requires that y=0 which tells us that x can be anything whatsoever. This is fine, and correct, as $x=re^{i\theta}, y=0$ is a fine solution when c=1. If $c\neq 1$ then none of the equations are already 0=0 so ...


0

Bits are just like ordinary qubits, except that they are always entangled with the environment. Each bit has two preferred basis states, |0> and |1>. If somehow you create a superposition between these two states, the bit and the environment will evolve into a state of the form |A>|0> + |B>|1> where |A> and |B> describe the rest of the universe, these ...


0

For the qubits, you have the states that you listed above, but you can also form any superposition of these states. For example the state $\frac{01+e^{i\theta}10}{\sqrt{2}}$ is a valid qubit for any real number $\theta$. The two qubits thus contain more information than their classical counterparts.


5

The website to which you linked doesn't seem to understand the purpose and results of Prof. Schwab's experiment. In fact, it didn't really describe the experiment at all. It just rehashed a lot of quantum mumbo-jumbo to make it look as though some power of "mind" causes quantum effects, rather than the more mundane cause-and-effect of having to use tools ...


1

To take the partial trace you need to build the sum over the matrix elements w.r.t. the same input and output basis, as you probably already used to calculate the partial traces you gave. In Dirac notation this is often written as: $$ tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle ...


1

Let $H_A \otimes H_B$ be your Hilbert space, and $O$ be an operator acting on this composite space. Then $O$ can be written has $$ O = \sum_{i,j} c_{ij} M_i \otimes N_j$$ where the $M_i$'s and $N_j$'s act on $H_A$ and $H_B$ respectively. Then the partial trace over $H_A$ defined as $$tr_{H_A}(O) = \sum_{i,j} c_{ij} tr(M_i) N_j ,$$ and similarly for $H_B$.


-1

It is a direct consequence of the matrix formalization and Born's rule. Check out the quantum Hadamard gate, which is the core example for your question. It takes a superpositioned state (which can be long) to be computed as input, collapses it, and outputs a result. Note that the computation on the superpositioned state is done in one step, no matter how ...


4

The key is a "one time password" - like a "one time pad". In the good old days of manual cryptography, embassies would be sent these one time pads which they could use to encrypt a message. Using modular addition / subtraction, with a key as long as the message, you can make an unbreakable code - as long as the key remains secret. By rotating keys all the ...


2

As suggested by WSA aka RV, I copy my comments into a (partial) answer. The key point is that the theorem says "for any given $\psi$ there exist two bases $\{i_A\}$ and $\{i_B\}$ such that...". This means that the choice of the bases depends on the vector $\psi$ we are considering. So there is not an $n$ dimensional common basis that span the whole ...


2

Denote $|\psi\rangle = \sum\limits_{i = 1}^m \sum\limits_{j = 1}^n h_{ij} |ij\rangle$ as $|\psi\rangle \rightarrow H = (h_{ij})_{m \times n}$. Then we have the following lemma: Lemma: Define matrix $U$ (in the original basis) as a new setting for Alice and $V$ for Bob, then a state $|\psi\rangle$ in the basis of the new settings is $U^* H V^\dagger$. ...


3

There are - even after clarifications in the comments - several ways to interpret the question. So let me start out with the way that fits all parameters here: Suppose we have a Hilbert space $\mathcal{H}$ (for qubits, $\mathcal{H}=\mathbb{C}^2$) and suppose we take a (pure) state in the Hilbert space $\mathcal{H}\otimes \mathcal{H}$. Question 1: If we ...


1

It is a grating used in the system of X-ray differential phase contrast imaging. In front of this grating, there is a phase one, which can modulate the X-ray phase and formed a self-image at the Talbot distance. the absorption grating, generally, composed by the high aspect ratio parallel walls made of Au, used to periodically absorb X-rays. Finally´╝îa moire ...



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