# Tag Info

7

It's not completely clear what you're asking, but I can make one thing clear: quantum mechanics was not developed with the specific aim of correctly describing the precise amounts to which CHSH inequalities can be violated. Quantum mechanics was developed in the 1920s and early 30s. Bell published his first paper on Bell inequalities in 1964. The first ...

5

Quantum "teleportation" is a really strange phenomenon, but most popular-science descriptions fall a little flat, indeed. Here's the basic setup for every quantum teleportation experiment: Two particles A and B are entangled and separated. Two more particles, A2 and B2, are put into two special states: A2 is put into some complicated wavefunction state, ...

3

Unitary dynamics means that the evolution of quantum states are describen by unitary operators. Physically this means that there is no dissipation in the system. This is important, because inherently quantum phenomena needed for quantum computation eg. entanglement and quantum superposition are only observable for a long period if no dissipation or other ...

3

http://arxiv.org/abs/quant-ph/9607007 discusses necessary conditions on $T$ (more precisely, on its singular values) for $\rho$ to be positive. They don't seem to derive sufficient conditions, however. The basic idea is that one can perform a rotation $U_A$ and $U_B$ on the two qubits, respectively, which correspondingly transforms $r\mapsto O_Ar$, ...

3

please don't blame simplicity Sending information means : 1) identify the information you want to send, ie up or down 2) send it But , in EPR context , the best you can do is in reverse order 2) send an ( unknow yet ) information by doing a measure ( with a detector ) 1) identify the measure sent ( by reading the detector ) You cannot send what ...

3

This paper talks about factoring with 1.5n qubits, so you'd need 3072 qubits to factor a 2048 bit semiprime. It says you might be able to do it faster. I'm willing to bet you can't do it with less than n qubits since you need to store the number. You might be able to do it with one or two fewer qubits, since the first and last bits will always be one, but ...

3

Since state $\sigma$ is not in thermal equilibrium I don't think one can use your definition of "thermodynamical" entropy. In fact, one should instead use Von Neumann entropy, which is a correct measure of statistical (so not quantum!) uncertainty. There is no other "classical" or "thermodynamical" entropy in quantum systems. As you mentioned, for a thermal ...

2

The conditional probability talks about things that have happened, so it's an objective reality (ignoring the questions of the objective reality's existence). It's unimportant if you know it's happened or not. This of course changes a bit in experimental reality, where you (presumably) order your events $a$ based on your knowledge of the preconditions ...

2

In general, the integral $$V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$ is the integration of the identity over the space the measure $\mu$ is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.) Normalizing the measure means sending $\mu \mapsto ... 2 The antiunitary operator is an operator on the Hilbert space. Thus, it is nonsense to say that "the operator$K$apply on complex number". Nevertheless, it can be shown that$U\alpha |\phi\rangle=\alpha^*U |\phi\rangle$, where$U$is an antiunitary operator.$U|\phi\rangle$should not be confused with$\langle\phi|$, the complex conjugate of$|\phi\rangle$. 2 It is the misapplication of QM to problems that may or may not benefit from such application, but which produces no experimentally testable theory. In many cases not even a mathematical model is produced, but "all is handwaving explanation and no proof". 1 There is a problem with your statement about the condition of no signalling is$p = \frac{1}{2}$. We can expand your problem into three parts: how to understand the condition conceptually, how to use the condition in your problem and how the condition is reflected in the density matrix formalism. For the sake of convenience, I will denote$|\psi\rangle = ...

1

Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation $U$. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we ...

1

The following calculations are in the basis of $(|00\rangle,|01\rangle,|10\rangle,|11\rangle$) CNOT1 - when the first qubit is control - $CNOT1|00\rangle = |00\rangle$ $CNOT1|01\rangle = |01\rangle$ $CNOT1|10\rangle = |11\rangle$ $CNOT1|11\rangle = |10\rangle$ thus it's matrix representation is $$CNOT1 = \begin{pmatrix} 1 & 0 & 0 & 0\\ ... 1 Here is a reference for the continuous versions: http://arxiv.org/abs/1402.2966 1 If we consider local operators M_A and N_B acting on Alice's and Bob's part, respectively, then it holds that [M_A,N_B]=0, i.e., we can use this property in proofs involving local operations. Note that conversely, however, commutativity need not imply locality (to start with, there need not even be a tensor product structure). 1 In proper mathematical terms, the action on the left is actually an action on the dual of the Hilbert space. The most natural way of defining this map is just by a simple transposition of the operator. For instance one could define the left action of K to be$$K(\phi,\ \cdot\ ) := (K^*\phi,\ \cdot\ ).$$If one views this as a map on the Hilbert space to ... 1 No. Consider a unitary acting on the tensor product vector space that leaves that special elementary tensor fixed. Then the unitary has a direct sum decomposition 1\oplus U where U is a unitary on a vector space of codimension 1 (i.e. the orthogonal complement of the fixed tensor). For simplicity assume that the same finite dimensional vector spaces, of ... 1 I'd say that "device independence" is only interesting for cryptographic purposes (at least at present). These are my reasons: If you want to study a physical experiment, you'll always need to consider the device. You can compare different devices, but you can't do an experiment without specifying the device unless you want to run into philosophical ... 1 It is not impossible to send information using quantum entanglement. What is impossible is to send information faster than light, violating causality. The quantum teleportation protocol includes a classical step (sending information about one's measurament, if I'm not mistaken), where things are restricted by the usual laws of relativity. 1 In this answer we will stick totally to the Copenhagen interpretation of quantum mechanics. Before the system is measured, there is nothing physical (a consequence of realism not playing a part under this interpretation). There is only our mathematical, non-physical description of the possible measurement outcome called the wavefunction. On measurement ... 1 Alas, it was very simple!$$\delta U_j =-i\Delta t \delta u_k(j)\mathcal{H}_kU_j \, . \tag{1}\frac{\delta U_j}{\delta u_k(j)} =-i\Delta t \mathcal{H}_kU_j \, . \tag{2} Now, looking at $\Phi_o$, we see the only terms that actually have a $u_k(j)$ dependence are $U_j$ and $U_j^{\dagger}$, hence we have the following via the product rule: ...

1

Nielsen and Chuang are referring to a scheme known as measurement-based quantum computation, for which many good learning resources are a short google search away. The idea is that you prepare a large, highly entangled state involving a large number of qubits at the start of the experiment. You then proceed to make measurements on the state, potentially ...

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