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For a bipartite state $\vert\psi\rangle_{AB}$, the entropy of entanglement (EoE) is defined as $$ E(\vert\psi\rangle) = S(\mathrm{tr}_B\vert\psi\rangle\langle\psi\vert)\ , $$ with $S(\rho)=-\mathrm{tr}(\rho\log\rho)$ the von Neumann entropy. Why is the entropy of entanglement such a great measure for the entanglement in a pure bipartite state? The reason is ...


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No. Take, for instance, the fully depolarizing channel, where $A_a=\{I,X,Y,Z\}$. Since $\mathcal E(\rho)=\tfrac12 I$, your condition $(1)$ holds for all $H$. On the other hand, there is no operator which commutes with all $A_a$. (Let me take the opportunity to advertise my list of canonical counterexamples for quantum channels ;) ).


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Photons are elementary particles, part of the SM, they are traveling at speed c in vacuum, when measured locally. The world line (or worldline) of an object is the path that object traces in 4-dimensional spacetime. It is an important concept in modern physics, and particularly theoretical physics. https://en.wikipedia.org/wiki/World_line Now ...


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Writing the requirement explicitly $$\mathcal{E}(U\rho U^\dagger)=U\mathcal{E}(\rho)U^\dagger $$ in terms of Kraus operators $$\sum_a A_aU\rho U^\dagger A_a^\dagger=\sum_a U A_a\rho A_a^\dagger U^\dagger $$ Hence we want the channel with Kraus operators $ A'_a=A_aU $ and $A_a''=UA_a$ to be equal. We know that two channels are equal if and only if their ...


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To simulate means to mimic or model the characteristics or appearance of something. Cloning means to make an exact replica- another actual instance of the subject being cloned. The two actions are utterly different in essence. So yes, you could straightforwardly simulate the two slits experiment, and no that would not violate the no-cloning theorem.


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If it allows for faster-than-light travel, either the article, or your interpretation of it, or the scheme you build following from it, must be wrong. More concretely, entanglement swapping (which is what the protocol you describe sounds like) requires classical communication before the entanglement between 1 and 4 is actually built up. Just as in ...


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Yes, except for the specific case in which the state is in an eigenstate of the measurement operator. Measurements with deterministic outcomes More specifically, for any quantum (pure) state $|\psi\rangle$, there is just one class of measurements that can be performed without changing the state, and these are the measurements which ask questions of the ...


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I want to add a point: entanglement not only does not allow for faster than light communication, but entanglement alone does not allow for any communication at all! By the no communication theorem local operations on a quantum system cannot cause measurable consequences to a far away quantum system, entanglement notwithstading. Entanglement can be used to ...


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Hint: $A_1^{\uparrow} = \begin{pmatrix}\cos\left(\frac{\theta_1}{2}\right)\end{pmatrix}$, $A_1^{\downarrow} = \begin{pmatrix}e^{i\phi_1}\sin\left(\frac{\theta_1}{2}\right)\end{pmatrix}$


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In the proper time of the object falling into the black hole, it crosses the event horizon and hits the singularity in finite time. But the observer outside 'sees' the object never reaching the event horizon. So, for the observer, the object never reached the event horizon. Whatever the observer sees, are events that happened before the object crossed the ...


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It is possible to simulate quantum computers with classical computers. However, the time (and typically memory) needed for such a simulation will scale exponentially with the time a quantum computer needs. Thus, if a quantum computer needs $10$ times the computation time for a more complicated case, the simulation would need $2^{10}$ times more time, and so ...


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Many Hamiltonians will be good. In particular, if you have control over the individual couplings, any interaction, together with a local terms in two directions, will suffice. As an example, the Ising model with two fields, $$ \sum p_is_z^is_z^{i+1} + \sum p_i' s_z^i + \sum p_i'' s_x^i $$ will do: The local terms allow to implement arbitrary rotations ...


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Of course this is true, the point of a useful error correcting scheme is that the errors introduced with the additional complexity needed to implement the scheme are either themselves corrected or less likely than the one it's supposed to correct. Let's take the example of a simple classical repetition code: say you want to encode a single classical bit, ...


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Since your title asks "Is there a theoretical limit": No, there is no theoretical limit. (And even if our gates in practice have some noise, we can use quantum error correction to perform gates to any desired accuracy.)


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You are absolutely right that by adding extra qubits, and extra error correction operations, you also produce extra entropy. However, with more qubits it is also possible to move all the entropy onto a subset of the qubits, making some those qubits more noisy, while the remaining -- those which become error corrected -- become less noisy! By discarding the ...


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No, that is not a safe assumption. The class of experiments you refer to tend to have two measurements taking place at different times, in such a way that it is possible to interpret the later measurement as influencing the outcome of the earlier one. However, there is in each case a reciprocity that allows you equally to consider that the earlier ...


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There isn't a simple answer to this because it depends on what the light interacts with. If you start with two entangled photons then they form a single system described by a single wave function that isn't separable into the two photons. As long as this single system doesn't interact with anything else it remains unchanged. So if you manage to emit the ...


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Your title question and your question in the body is different. You are not specifically saying where the detector (and what type of detector it is) is positioned. Now in the body your the question you are asking what would happen if you would turn on the detector after the photon has already hit the screen and interacted with it (left a dot). In this case, ...


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If you could go back the way you describe it, nothing would prevent Alice from measuring $|\psi\rangle$ again, again, and again, in whatever basis she likes. At the same time, as per your protocol she would keep full record of the past measurement outcomes. This would therefore allow her to perform full tomography of $|\psi\rangle$, and thus obtain a full ...


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First, what is he calling the number of excitations for the 2-atom system? There are two atoms coupled to a radiation mode, i.e. to a harmonic oscillator. The states $|n\rangle$ refer to the harmonic oscillator, the states $|S, m_s\rangle$ refer to the two atoms. (Each atom is assumed to have only two energy levels and thus behaves like a spin-1/2. Two spin-...


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Let me argue by analogy: We know that we can draw a straight line on a sheet of paper by taking a pen and moving it from left to right. Now imagine the paper is rotating. If we try to follow the same procedure as before to draw a line, we will instead be left with a spiral-shaped curve. Following your reasoning, this would mean that we cannot draw a ...


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These questions are a bit hard to answer in full generality, as quantum architectures can differ quite a bit in the way the quantum information is encoded and processed. But generally speaking, yes, it is true that most quantum systems/qubits will have some dephasing time due to environmental noise etc. Similarly, very often operations will have to be timed ...


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No, the concept of a mixed state is different from what you have. It is not a state composed of states from different Hilbert spaces. A state that one can write as a tensor product of pure states (subsystems) is still a pure state. One can show that by using the $\rho^2=\rho$ test. (Each subsystem will only contract on subsystems from the same Hilbert space.)...


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The idea of the collapse of the state is not a fundamental part of quantum mechanics. It's a feature of the Copenhagen interpretation (CI). The CI is not the only way to think about quantum mechanics. Even within CI, it's not necessarily true that measurement must disturb the system, depending on what you mean by "measurement" and "disturb." In the Stern-...


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You want to prove that given an arbitrary matrix $A$, we can write $A$ as a linear combination of positive, unit-trace matrices. To do this, you start by writing $A$ in terms of its Hermitian and skew-Hermitian components (see also this post about this decomposition): $$A=\underbrace{\frac{A+A^\dagger}{2}}_{\equiv A_1}+i\underbrace{\frac{A-A^\dagger}{2i}}_{ ...


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Given an arbitrary state $\rho\in\mathcal B(\mathbb C^n)$ whose eigendecomposition reads $$\rho=\sum_{k=1}^{\mathrm{rank}(\rho)} p_k|\psi_k\rangle\!\langle \psi_k|,\quad p_k>0,\tag A$$ The set of its purifications is the set of vector states $|\Psi\rangle\in\mathbb C^{n}\otimes\mathbb C^{m}$ for $m\ge \mathrm{rank}(\rho)$ whose SVD (called Schmidt in this ...


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