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11

Let's start by finding a complementary channel for any channel given by a Kraus representation $$\Phi(X) = \sum_{k=1}^n A_k X A_k^{\dagger}.$$ To make the necessary equations clear, let us assume that the channel has the form $\Phi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Y})$ for finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{... 7 You probably shouldn't be thinking of the Quantum Fourier Transform as being something where you want to extract the outcoming probability amplitudes. As you say, when you start measuring, you destroy the superposition. The only way to extract the amplitudes is to make the same state many, many times, and keep repeating your measurements until you get enough ... 7 An observable only needs to be Hermitian, and can have any real eigenvalues. They don't even need to be distinct eigenvalues: if there are repeated eigenvalues, we say that the eigenspace for that eigenvalue is degenerate. (In the case of observables on a qubit, having a repeated eigenvalue makes the observable rather uninteresting, because absolutely all ... 5 No. As far as we know, entangled states do not permit faster than light communication. You might be able to use them for things like doubling bandwidth (see superdense coding) or sending quantum states, but that will all happen at the speed of light (or slower). It is true that entangled states do seem to know something about the constituent elements faster ... 5 There is far more structure in entanglement than your simplistic definition seems to be implying. Yes, you could ask a question Is there any entanglement in this pure state$|\psi\rangle$of$n$qubits? Which can be answered by whether or not the entire state can be expressed as $$|\psi\rangle=|\phi_1\rangle|\phi_2\rangle\ldots|\phi_n\rangle$$ but as ... 5 Your mistake is that you assume that$\rho$and$\sigma$are classical-quantum in the same classical basis on$X$. However, there is no need to do so -- all which is necessary is that there exists such a basis, which can however depend on the state. As soon as you choose a different classical basis for the two states, your argument breaks down. 5 I think that I can explain the definition through the following simple example: Suppose that you perform two experiments in the same house in two separate rooms. In the first you measure the observable$A$and instantaneously in the second you measure the observable$B$. The measurements are afflicted with noise, so you do not get a definite answer every ... 5 Alice receives a quantum state$|\psi\rangle$, which is an element of some basis$\mathcal{B}$, though she does not know what$\mathcal B$is. She then teleports this to Bob, who is told by someone else what$\mathcal B$is. Furthermore, it seems that specifically either$\mathcal{B} = \{ \lvert 0 \rangle, \lvert 1 \rangle \}$or$\mathcal{B} = \{ \lvert + \...

4

The only two quasi-particle quanta for which I know there to be active research in quantum computing are phonons and anyons. Phonons: That state-of-the-art is given my answer here: Phononic Quantum Computing Anyons: Synthesizing the first anyonic qubit is still an outstanding goal, but major milestones have been discussed very recently. This paper was the ...

4

Firstly, that sphere that you've pictured is convenient for thinking about what's going on, but remember that it is not what is actually happening. So the fact that you don't visualise light as having a little arrow pointing somewhere doesn't matter. The fact of that matter is that for an electron spin, having the two possible states "up" and "down", we ...

4

In quantum theory, the pure states are associated with the unit vectors of the Hilbert space. A pure state of a quantum bit can be represented as $$| \psi \rangle = \alpha | 0 \rangle + \beta | 1 \rangle$$ where $|\alpha|^2 + |\beta|^2 = 1$. The basis $| 0 \rangle$ and $| 1 \rangle$ can be viewed as two orthogonal polarization directions. Bloch sphere is ...

4

An orthonormal maxinammly entangled basis for two quNits is easily defined: $$|\Psi_{xy}\rangle=\frac{1}{\sqrt{N}}\sum_{i=0}^{N-1}\omega^{iy}|i,i+x\rangle,$$ where $\omega$ is an $N$-th root of unity, and $x,y=0,1,\ldots,N-1$. I don't believe that there is a rotationally invariant maximally entangled state, except in the case $N=2$. You may want to look ...

4

Collective measurements are normal measurements. You just need to be clear on the setting under which they are operating. I haven't delved deeply into the specific paper you mention (so it's always possible they make marginally different assumptions), but I expect it goes like this: You are looking at using many copies of the same channel. Encoding will, ...

4

It seems that using more rounds will not be such helpful for us to get something more powerful from complexity perspectives. There are a few comments about the number of rounds and the number of players for $\mathsf{MIP}^*$ in Thomas Vidick's lecture note regarding quantum mutli-prover interactive proofs. Note that the non-local games are $\mathsf{MIP^*}$ ...

4

I assume the question refers to how LOCC is used to distinguish the two states $$(|00\rangle+|11\rangle)/\sqrt{2}\qquad(|01\rangle+|10\rangle)/\sqrt{2}$$ when Alice and Bob each hold one qubit, and are separated by a great distance. There are many different measurement operators that could achieve the same task. If Alice just held both qubits herself, she ...

4

The first question that we have to deal with is what is meant by "maximally entangled" in this context. There's no single straightforward notion. In particular, for three qubits, there are two inequivalent classes of entangled state that cannot be interconverted by SLOCC (stochastic local operations and classical communication). Each has a maximally ...

4

I think you're doing things a little bit backwards. You probably shouldn't be calculating $P(a|x)$ or $P(b|y)$ in advance, because you're simply trying to ask: Given a set of $\{P(ab|xy)\}$, do there exist assignments to $P(a|x)$ and $P(b|y)$ that satisfy $P(ab|xy)=P(a|x)P(b|y)$ for all $a,b,x,y$? So, how do you evaluate the probability of getting ...

4

Assume this works. Then, nothing prevents Alice from applying the same protocol to a quantum state that is known to her, such as $|0\rangle$ or $|1\rangle$. This way, she could send information to Bob instantaneously. Thus, it violates faster-than-light communication and thus is impossible.

4

Let $S_x$ be a permutation between qubits 2,4 and 6, and acts as identity on the other qubits. Use $x$ to index all possible permutations of that form, of which there are 6. Then you do indeed have a mixed state overall, and it's of the form $$\rho=\frac{1}{6}\sum_{x=1}^6S|\Psi\rangle\langle\Psi|S^T$$ (Note that the $S$ matrix will be real so I can use the ...

3

To be clear: You have a Hilbert space $\mathcal{H}_A\otimes\mathcal{H}_B\otimes\mathcal{H}_C$. The initial state is $\rho_\text{tot}(0)=\rho_A\otimes\rho_B\otimes\rho_C$. There is a Hamiltonian acting on the system of the form $H_{\text{tot}}=H_{AB}\otimes\mathbb{I}_C+\mathbb{I}_A\otimes H_{BC}$ Instead of directly calculating the effect of the Hamiltonian ...

3

One definition for whether a state $\psi \in V_1 \otimes \cdots V_n$ for $n$ qudits with various values of $d_i = \text{dim} V_i$: $\psi$ is not entangled if it can be written as a single summand $\psi = v_1 \otimes \cdots v_n$. Conversely it is entangled if it requires more than 1 summand. Let $\sigma \in S_n$ be a permutation and $R_\sigma$ be the ...

3

First of all, you inverted $a,b$ with $x,y$ when trying to draw the analogy. In Bell's original paper, $\vec a,\vec b$ are used to denote the measurement directions, so the underlying probability distribution should be written as $$p(x,y|\vec a,\vec b,\lambda)=p(x|\vec a,\lambda)p(y|\vec b,\lambda).$$ The expectation values $A(\vec a,\lambda)$ and $B(\vec b,... 3 I don't have a complete answer, but perhaps others can improve on this starting point. There are probably 3 things to ask about the code: How degenerate is it? How hard is it to perform the classical post-processing of the error syndrome in order to determine which corrections to make? What are its error correcting/fault-tolerant thresholds? I suppose a ... 3 This reminds me of another question we had here: What's the difference between a set of qubits and a capacitor with a subdivided plate? Let me try to answer your question separately though: One of the benefits I'm reading about qubits is that they can be in an infinite number of states. Yes, qubits don't have to be in state 0 or state 1, but can ... 3 The Bell states$|\beta_k\rangle$all satisfy$Y\otimes Y|\beta_k\rangle=\pm|\beta_k\rangle$. Hence,$\tilde\rho=\rho$. Thus, the matrix$\sqrt{\sqrt{\rho}\tilde\rho\sqrt{\rho}}=\rho$, given that all the$\lambda_k$are real and non-negative. Hence, the set of eigenvalues$\lambda_i$is the same as the set of the$\tilde \lambda_k$, the Bell-diagonal ... 3 As an initial matter, let's ask "what is the classical computational complexity of solving 'mate-in-$n$' type games?" For example, is it even in$\mathcal{NP}$to know, given a certain chess position, that white can mate in$10$or fewer moves? It's been known for a while that we can consider such questions as a "quantified boolean formula" (QBF) question. ... 3 Speculatively expanding on previous answer Quantum computers tend to outperform classical computers in determining global properties of functions. Further, the properties tend to be some measure of global symmetry. Based on the global symmetry, the probability amplitudes can constructively (and destructively) interfere, in ways that a classical computer ... 2 Use of the Inverse Symbolic Calculator (https://isc.carma.newcastle.edu.au) suggested the possible exact two-qubit separability probability of$\frac{629}{8580} =\frac{17 \cdot 37}{2^2 \cdot 3 \cdot 5 \cdot 11 \cdot 13} \approx 0.07331002\$, quite close to the current estimate of ours, based on the largest sample, of 0.0733116 (employing 4,945,000,000 ...

2

the measurement is carried out on each individual qubit and the measurement on one qubit will not change the state of another qubit This is an incorrect statement. If the state that you are measuring is entangled (which it very much is for the 2D cluster state), measuring the state of one qubit absolutely changes the state of another qubit. The trivial ...

2

I do understand that the sum of these three probabilities is greater than one because there are some constraints already involved; like if we uncover all three coins at least two have to be the same. So naturally, there's some redundancy leading to a sum of probabilities that is greater than one! I would say the explanation is simpler than that. ...

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