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Your $U_f$ must depend on $f$. Let's consider the two trivial examples: $f$ is the zero-function. In this case, $U_f$ is just the identity. $f$ is the one-function ($x\mapsto 1$), then $|x\rangle|y\rangle \mapsto |x\rangle|y\oplus 1\rangle$, then $U_f$ is a NOT-gate on the second qubit. Just a note: The whole idea of the Deutsch-Josza algorithm is that ...


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The above answers correspond correctly to the holographic information storage where surface area limits the storage. Bekenstein bound is given also in other ways- ` $$ I \leq \frac{2 \pi c R m}{\hbar \ln 2} \approx 2.577\times 10^{43} m R $$ (see Bekenstein bound wikipedia), $m$ being the mass in kg and $R$ being the radius So for a given mass of the ...


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Quantum gates are all unitary transformations on a state of qubits. Any unitary transformation can be considered a "gate", although the ones you mention are primitive ones from which others can be constructed. More complex ones are usually referred to as circuits. The two qubit gates, $\mathrm{H}$, $\frac{\pi}{8}$ and $\mathrm{CNOT}$ are considered universal ...


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A really neat and intuitive way to deal with qubit systems like spin is to use the Bloch sphere. The Bloch sphere represents the two dimensional Hilbert space which the spin $\frac{1}{2}$ state vectors live in by a sphere in $\mathbb{R^3}$. Have a look at this wikipedia page for some more info http://en.wikipedia.org/wiki/Bloch_sphere. (Particularly the ...


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Simple answer: Yes it does. The monogamy of entanglement is often stated as "when two particles are maximally entangled, they cannot be entangled by a third particle", while really it means that the more two particles are entangled, the less they can be entangled with a third particle. However, since "more" and "less" are ill-defined, phrasing it in the ...


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DISCLAIMER: I usually use a different definition of the covariance matrix: It should only fulfill $\gamma\geq iJ_{2n}$. The factor 1/2 is irrelevant. I tried to adjust everything, but there might be some problems with factors of 1/2. I don't have a complete answer to this question (still learning myself), but here are some ideas. Let's call the $\kappa_i$ ...


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The two density matrices are expressed in a different basis of the same Hilbert space. If you compute the expectation values on any state you like you will find the same result with both $\rho_1$ and $\rho_2$. Hence by polarization you can conclude they are indeed the same operator, i.e. they're acting in the same way on the Hilbert space in question.


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Since I started writing it, I will complete this answer to add some extra mathematical detail. As stated by lionelbrits, the $d^2-1$ Hermitian generators of $\mathrm{SU}(d)$ are traceless and orthonormal with respect to the Hilbert-Schmidt inner product $$(A,B) = \mathrm{Tr} (A^{\dagger}B).$$ Therefore, along with the identity, these operators form a ...


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The generators you mention form a complete basis for traceless hermitian matrices. Thus, to express a hermitian matrix that has a non zero trace in this basis, you have to augment the basis with one more matrix that is trace orthogonal to the others, and the identity does the trick. You can verify that the number of matrices in your basis agrees with the ...


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Since $\lvert a_i\rangle$ and $\lvert b_i\rangle$ are both bases for the space $W$, there exists a unitary $U=\sum \lvert b_j\rangle \langle a_j\rvert$ which maps $\lvert b_i\rangle=U\lvert a_i\rangle$ for all $i$. This $U$ can be naturally embedded in $V$, i.e., we can think of it as an operator $U:V\rightarrow V$. Then, $$ P_2 = \sum \lvert ...


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In terms of the splitting $V=W\oplus W^{\perp}$, the projection operator $P_1=P=P_2$ and unitary operator $U$ are block diagonal $$ P ~=~ \begin{bmatrix} {\bf 1} & {\bf 0} \\ {\bf 0} & {\bf 0}\end{bmatrix} $$ and $$ U ~=~ \begin{bmatrix} U|_W & {\bf 0} \\ {\bf 0} & {\bf 1}\end{bmatrix}, $$ respectively. Clearly, the two operators ...


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This is an answer to the second version of the question, namely would it possible for the two agents to make contradictory predictions about other experiments The answer is no, assuming that your original distinction between the two observers holds. To be precise, this distinction is that observer 1 has access to the state of the detector after the ...


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QBism is much more than that: the second observer's subjective experience contains the two slits, the measuring device installed by the first observer, the first observer himself and the rest of the world. Only by taking into account all this data can he estimate probabilities properly. But even if their experiences disagree, there is no way they could know ...


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The lower bound $-\log\,d\le H(A|B)$ follows from strong subadditivity, $$ H(ABC)+H(B)\le H(AB)+H(BC)\ . $$ To this end, choose a purification $\lvert\psi_{ABC}\rangle$ of $\rho_{AB}$. Then, $H(ABC)=0$, and $H(BC)=H(A)$, and thus, we have $H(B)\le H(AB) + H(A)$ (this is also known as the Araki-Lieb inequality), which implies $$ -\log\,d\le -H(A)\le ...


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We will prove the following statement: Let $\rho=\sum p_i \rho_i$ ($p_i>0$) be a decomposition of a given state $\rho$ which minimizes the cost function $C(\{p_i,\rho_i\})=\sum p_i f(\rho_i)$ for that state. Then, $\rho'= \sum p_i' \rho_i$ is a decomposition which minimizes $C$ for the state $\rho'$. Note. In the case of entanglement of ...


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As @Phoenix87 said, you should claim what is the meaning of system in your question. If I have not missed something. the system is one of the states $|ψ_i⟩$ but we don't know which. Here the system means the system of a certain particle and a certain particle will be in a pure state $|ψ_i⟩$ but we don't know which. two systems $S_1$ and $S_2$ and ...


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The density operator of a quantum system summarises the expectation values of the observables of that system alone. If you have two systems $S_1$ and $S_2$ that are entangled with one another, then there is no pure Schrodinger picture state for either of the entangled systems. However, there are observables of $S_1$ (or $S_2$) and those observables have ...


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The outcome of measuring an observable $O$ on a system described by a density operator $\rho$ is given by $\mathrm{tr}[O\rho]$. Thus, if two systems are described by the same density operator, they cannot be distinguished by any measurement, i.e., they describe the same state. So if those density operators encode everything you know about the two states, ...


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You can find an explanation of how to obtain all ensemble decompositions here: Proving the unitary relation of ensemble decompositions. As Martin said in his comment, they are related by unitaries (or, more precisely, (partial) isometries), regardless of whether the decomposition consists of orthogonal states. Most conveniently, you would start from the ...


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Firstly, what is a state? A state gives you the complete description of a system. Let's label the state of a system $\lvert \psi \rangle$. This is a normalised state vector which belongs in the vector space of states. Keep in mind that we are talking about the full state; I haven't decomposed it into basis states, and I will not. This is not what the ...


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Check the form of your density operator it should be something like $$\rho=\sum_i p_i |\psi_i\rangle\langle\psi_i|$$ I find the most intuitive way to think of the density matrix as follows. Consider an experimenter in his lab who has a "machine" which produces the quantum state $|\psi\rangle$ but the machine doesn't work perfectly and produces some other ...


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You seem to be thinking that if Alice reads the Z spin on one of the entangled pair that Bob would be able to detect that. Alice can't control what the value is that will be read. She might read an up or a down. Bob will read the complement, but since that is the complement of a "random" value, it appears random as well and cannot be differentiated.


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Result for one orthogonal and one non-orthogonal decomposition We will prove the following Theorem: Let $\rho=\sum_{i=1}^Np_i\lvert\phi_i\rangle\langle\phi_i\rvert$ be a eigenvalue decomposition and $\sigma=\sum_{i=1}^Mq_i\lvert\psi_i\rangle\langle\psi_i\rvert$ ($M\ge N$). Then, $\rho=\sigma$ if and only if \begin{equation} ...


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It is almost never useful to try to analyze these things with photons. The clorophyll in a plant cell is innundated by light which is properly analyzed as a classical oscillating electric field. All the clorophyll molecules in a cell are driven by the same field, so they are all excited to the same extent. There is no particular constraint whereby the amount ...


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There is a very easy, but highly unsatisfying way that your question could be answered: Given a noisy channel, a code is error-correcting, iff it increases the capacity. Normally, the idea is that the channel has zero-capacity and afterwards, it has capacity. Now, the various capacities (e.g. classical, quantum-classical, quantum, one-shot-quantum) are ...


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There are approaches to quantum systems, and photosynthesis from the viewpoint of quantum information systems too. The approach is to basically see photosynthesis as a process in which electrons involved in photosynthesis reactions "sample" different energy-level routes in much the same way quantum-computer algorithms can. To understand the process in such ...


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You certainly want to have $\langle\psi_i|\psi_j\rangle=\delta_{ij}$ and $\langle\phi_k|\phi_l\rangle=\delta_{kl}$, or some equivalent condition ensuring that both ensemble decompositions are optimal. Otherwise, it is easy to come up with many different totally unrelated decompositions. Because of the optimality condition, both decompositions are ...


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What can we say about the quantum state from the number of zero and non-zero eigenvalues of the corresponding density matrix? The number of zero eigenvalues has no significance, and is not really well defined anyway. If the number of non-zero eigenvalues is not one, then there are many different ways to write the density matrix $\rho$ as a coherent ...


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Measurement is not the same as wavefunction collapse. A measurement is an interaction that produces information about a system that can be copied to multiple other systems. Such an interaction does not copy all of the information in the state of a system, only the information instantiated in some set of orthogonal projectors, see ...


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The MWI (many worlds interpretation) doesn't preserve the unitarity, but it preserves something else. The apparatus that measures is reset after each detection to a state, it is not left in the state that reported $q_2$. Let's name the "ready" state of the apparatus, $|A>$, s.t. $|A>|q_1> \to |a_1>|q_1>$, $|A>|q_2> \to ...


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I think I can give a more technical and detailed explanation of "long range entanglement". I felt it puzzling some time before and still puzzle me for some situations. For generic topological states, the entanglement of ground states scales as $S=\alpha L-ln(D)$ ($D$ is the total quantum dimension of the system. For topologically nontrivial system, ...


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Quantum teleportation requires a "classical channel" of information to be communicated between the two experimenters, so it doesn't violate the no-communication theorem because that theorem only rules out the possibility that two experimenters could communicate purely by their choice of measurements on parts of an entangled system. Referring to the schematic ...


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Yes, a many-qubit state (even if it entangled) can be teleported by teleporting each qubit separately using one (perfect) Bell pair and two bits of classical communication. (This is the idea of quantum teleportation: The qubit, including all of its quantum correlations, is teleported.) This can be seen by observing that teleportation of a qubit implements ...


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1) The trace of the rank-one operator $|\psi_1\rangle\langle\psi_2|$ is $\operatorname{Tr}(|\psi_1\rangle\langle\psi_2|) = \langle\psi_2|\psi_1\rangle$ (think of the trace of the matrix $[v_iw_j]$ where $v_i$ and $w_j$ are the components of vectors $v$ and $w$ respectively in an orthonormal basis), so for the product of $\Pi_1\Pi_2$ you get ...



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