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## Hot answers tagged quantum-information

4

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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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 ...

3

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{ ... 3 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 ... 3 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 ... 3 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: ... 3 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 ... 3 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. ... 3 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 ... 3 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 ... 3 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)$$where p_n is the probability to find the system in the pure state \lvert \phi_n\rangle. The thermal equilibrium ... 3 Feynman never showed that quantum simulation has an exponential cost. He probably mentioned that it seemed to be the case. In complexity theory, it's surprisingly hard to prove lower bounds on the cost of solving a problem or to prove separations between complexity classes. The behavior of computer programs is wildly diverse, and eliminating all of them as ... 2 A Universal-Not allows FTL communication I added a Universal-Not gate to my toy circuit simulator and played around with it, trying to find interesting things. I was experimenting with "measuring along the UniversalNot axis" when I stumbled onto this: The left half of the circuit is only varying the global phase, but by the end an observable is varying. ... 2 Yes it is. Use the fact that the operator in the trace different from \rho  is positive and compute the trace using a basis of eigenvectors of \rho , whose eigenvalues are also positive. (By positive I actually mean nonnegative). Here is the proof (assuming the first version of the question regarding tr(\rho (X+zY)^\dagger(X+zY))). As \rho is ... 1 Your definition of Holevo information is wrong. It corresponds to C_{ea} , the entanglement assisted capacity of the channel. See equation (5) of the paper. The Holevo information is defined for a probabilistic mixture of density matrices, or for a cq-state (cq = classical quantum state). 1 It's a case of bad labeling: the i,j labels in Fig.1 and Eqs.(4-5) have different meaning. In addition, subscript 1 was dropped on all B's in Eq.(5). Other than that, it's straightforward algebra: Start by rewriting the final result of Fig.(1) in the familiar operator-product form, expand, and rearrange:$$ \overline{\left[ E \cos(B_1\tau) - i {\hat ...

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A 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 ...

1

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 $|\psi\rangle = a |\uparrow_z\rangle + ... 1 It looks like your mistake is doubling-up the Us instead of doubling-up the controls. You computed the matrix for applying a U to wires 2 and 3 controlled by wire 1, instead of the matrix for applying a U to just wire 3 controlled by wires 1 and 2. Quirk used to show the matrices of operations you hovered over. You can play around with that old version on ... 1 If I understand this correctly, the setting is the following: We have a Hilbert space$\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 ... 1 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 ... 1 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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