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I think it is easy to do some matrix calculations to check your idea. In fact , the tensor product is non-commute, so your idea may be wrong.


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A crucial hypothesis is missed in your construction. Each $\phi_i$ must also satisfy $\phi_i \not \perp \psi_i$, otherwise $\langle \psi_i |E_i \psi_i\rangle >0$ is false. This point provides an answer to your last question as well. If $\psi$ is an added further vector, linearly dependent on the vectors $\psi_i$, the construction you made cannot be ...


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From a classical standpoint, it seems pretty clear that information can be easily lost. Nope, classical mechanics is reversible. It's so reversible that you can't even squeeze the state space: states that start out "very different" stay "very different". Those vague terms are formalized by Liouville's theorem. Because classical mechanics is reversible, ...


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Feynman never showed that quantum simulation has an exponential cost. He probably mentioned that it seemed to be the case. In complexity theory, it's surprisingly hard to prove lower bounds on the cost of solving a problem or to prove separations between complexity classes. The behavior of computer programs is wildly diverse, and eliminating all of them as ...


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The Gibbs form $\rho\sim \mathrm{e}^{-\beta H}$ is just a fancy way of writing the standard Boltzmann distribution. A quantum (mixed) state is written in general as $$ \rho = \sum_n p_n \lvert \phi_n\rangle \langle \phi_n\rvert, \qquad (1)$$ where $p_n$ is the probability to find the system in the pure state $\lvert \phi_n\rangle$. The thermal equilibrium ...


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It's a case of bad labeling: the $i$,$j$ labels in Fig.1 and Eqs.(4-5) have different meaning. In addition, subscript 1 was dropped on all $B$'s in Eq.(5). Other than that, it's straightforward algebra: Start by rewriting the final result of Fig.(1) in the familiar operator-product form, expand, and rearrange: $$ \overline{\left[ E \cos(B_1\tau) - i {\hat ...


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A CNOT gate can't be smaller than 4x4, because it operates on two qubits and $2^2=4$. The closest thing to an analogous gate, that applies to a single qubit, is the Z gate $\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$. You can think of Z, in the computational basis, as a negation of the global phase controlled by the target qubit. A ...


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You need to be careful about how you go from the full system to the subsystem $A$. You define $\rho^\text{eq}(T) = Z^{-1} \exp(-H/T)$ as the thermal state of the whole system, but then you use $\rho_A^\text{eq}(T)$ without defining how you are reducing the density matrix of the whole system onto just the subsystem. There are two reasonable ways to do so: ...


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I'll try to add a rather intuitive take on the difference between pure and mixed states. Take first the simple example of a single spin-1/2 particle. Its pure states may always be written as superpositions of spin-up and spin-down states measured along some particular direction $z$. That is, we write $|\psi\rangle = a |\uparrow_z\rangle + ...


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A pure state is a linear combination of basis states $|\psi\rangle = \sum_k c_k |b_k\rangle$. A pure state has unit 2-norm; pure states care about squared weight $\sum_k |c_k|^2=1$. Meaning the weights are amplitudes. A mixed state is a linear combination of adjoint-squared pure states $\rho = \sum_k p_k |\psi_k\rangle\langle\psi_k|$. A mixed state has ...


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In short, you cannot have stimulated emission without spontaneous emission (hermiticity requirement) . The spontaneous part will destroy the fidelity of your copying


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Okay, let me elaborate on my comment to show how you would calculate the Schmidt decomposition in general. This might also answer your second question. As I said in my comment, the Schmidt decomposition requires you to subdivide your system in two parts, A and B. It is then the (unique) decomposition $|\psi \rangle = \sum_\alpha \lambda_\alpha ...


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The simplest way to do this is to apply CNOT gate, so you're then looking at $\langle01|\rho|00\rangle$ instead of $\langle11|\rho|00\rangle$. Then you can post-select on the first qubit being $|0\rangle$ (i.e. keep redoing the experiment and throwing out runs where the first qubit measures as $|1\rangle$) and go about measuring the off-diagonal term of the ...


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Assuming $(X + zY)^*$ is the conjugate of $X + zY$, a counter-example is: $$z=0, Y=\begin{bmatrix} 0&0\\0&0 \end{bmatrix}, X=\begin{bmatrix} 0&-i\\i&0 \end{bmatrix}, \rho = \begin{bmatrix} 1/2&0\\0&1/2 \end{bmatrix}$$ Because: $tr[\rho(X+zY)^*(X+zY)]$ $= \text{Tr}\left[I/2\left(\begin{bmatrix} 0&-i\\i&0 ...


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Your definition of Holevo information is wrong. It corresponds to $C_{ea} $, the entanglement assisted capacity of the channel. See equation (5) of the paper. The Holevo information is defined for a probabilistic mixture of density matrices, or for a cq-state (cq = classical quantum state).


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Yes it is. Use the fact that the operator in the trace different from $\rho $ is positive and compute the trace using a basis of eigenvectors of $\rho $, whose eigenvalues are also positive. (By positive I actually mean nonnegative). Here is the proof (assuming the first version of the question regarding $tr(\rho (X+zY)^\dagger(X+zY))$). As $\rho$ is ...


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Given that the "state" of a single qubit cannot be independently factored and considered, how does the hadamard gate now affect the state of system? To apply a single-qubit operation $M$ to an n-qubit system you hit the system with $I \otimes M \otimes I \otimes ... \otimes I$. The position of $M$ within that tensor product determines which qubit you ...


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How can I write a single Qubit operator $O$ (given as a $2\times 2$ matrix) as a multiQubit operator $O'$ (given as a $2^k \times 2^k$ matrix) that acts as the identity on all inputs except the first where it acts as $O$ traditionally would. This has a very direct answer: $$O' = O\otimes \underbrace{I_2 \otimes \cdots \otimes I_2}_{k-1\text{ ...


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As I stated in the comments the post by Arnold Neumaier in reply to this question answer mine too. In particular the papers he links (arXiv papers cond-mat/0102428 and cond-mat/0203460). In my mind the two papers are essentially a continuation of what Leggett and Caldeira showed (that the density matrix becomes diagonal when coupling to certain statistical ...


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Quantum teleportation works on any qubit. It isn't restricted to only working on entangled qubits. It works on unentangled qubits. Quantum teleportation does require the sender and receiver to have shared an EPR pair $P_{A,B}$ in order to send your qubit $|\psi\rangle$, and the teleportation process will use up $P_{A,B}$, but there is no restriction on ...


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You need to use a more precise notion of the cloning process, in order to understand the general statement and its repercussions. I will give you some outline here (mainly following the explanations of B. Schumacher and M. Westmoreland given in the reference), with an emphasis on the most important aspects of it, but to fully appreciate the importance of the ...


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The No-Cloning Theorem means that if you have an unknown state then it is not possible to make an identical copy. The original reference is to Wooters, A single quantum cannot be cloned. Of course, if you know the state, you can manufacture duplicates; or if you have many identical copies of the unknown state, provided by some quantum machine, you could ...


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Take a pair of qubits in a maximally entangled state $\phi=A\otimes B+B\otimes A$. (Here $A$ and $B$ are a basis for the state space of a single qubit.) One stays with you, the other resides at the intended location. You've therefore got two qubits --- the one you want to transport (in state $\psi$) and one of the entangled pair. Now make an observation ...


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If a superposed gate is equivalent to a choice of gates controlled by some pre-initialized ancilla qubits, then you can get the exact same effect with a normal gate. Just have the appropriately initialized ancilla be passed in, instead of hidden inside. I don't think hiding the ancilla inside will give any polynomial benefit in gate count or other metrics. ...


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The question, as I understand it, asks the following: Consider a state $|\psi\rangle$ which is uniquely determined among all other pure states by its K-party reduced density matrices. Is it guaranteed that this state is the ground state of a K-local Hamiltonian? First, we should note that not only is it the case that any unique pure ground state of a ...


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An ensemble of interacting particles will, over time, develop entanglement between widely separated parts*, so this is similar to asking whether an interacting system can still be a BEC. The short answer is yes, but a subtlety is that various authors define BEC in slightly different ways. One way of defining BEC, as I mention in a recent answer, is the ...


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The answer to your first question is interpretation-dependent. In many-worlds, the other state is unaffected. In the Copenhagen interpretation, the other state is affected. Since interpretations all give the same predictions, and they disagree about the answer to your first question, the answer can't be determined experimentally or mathematically. What is ...


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You are right. You can only (repeatedly) simulate a system where you have perfect knowledge of the quantum state. If you don't have perfect knowledge of the quantum state, a "perfect" simulation in the sense you have in mind is not possible. If the system is very small, you can try to projectively measure all quantum states. Then you have destroyed all ...


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If I understand this correctly, the setting is the following: We have a Hilbert space $\mathcal{H}$ and $n$ states $|\psi_i\rangle\in \mathcal{H}$ that we want to distinguish. We want to distinguish "correctly" i.e. we want a protocol with the following properties: We know that the state we are given is one of the $n$ states specified before with some ...


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A universal NOT gate would allow deterministic remote state preparation with one classical communication bit. See http://arxiv.org/abs/1505.05615 and references.


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You already do this to some extent when your power supply is an electrolyte-based battery (a, b) --- there you have motion of both negative and positive ions in the electrolyte. Electrons are free to move in conductors when the Fermi energy falls within a band of energy levels rather than in a gap when there are no allowed energies. However, the band/gap ...


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It looks like your mistake is doubling-up the Us instead of doubling-up the controls. You computed the matrix for applying a U to wires 2 and 3 controlled by wire 1, instead of the matrix for applying a U to just wire 3 controlled by wires 1 and 2. Quirk used to show the matrices of operations you hovered over. You can play around with that old version on ...


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A Universal-Not allows FTL communication I added a Universal-Not gate to my toy circuit simulator and played around with it, trying to find interesting things. I was experimenting with "measuring along the UniversalNot axis" when I stumbled onto this: The left half of the circuit is only varying the global phase, but by the end an observable is varying. ...


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You can interpret $1-Tr[\rho^2]$ as a kind of entropy. If you think of $\rho$ as a classical mixture of quantum states, then $Tr[\rho^2] = \sum_i p_i^2$, where $p_i$ is probability of being found in eigenstate $i$. The more dispersed the states, the smaller the quantity. So for a completely mixed state you have $1/n$, whereas for a pure state you have 1.



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