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$\newcommand{\ket}[1]{\left|{#1}\right\rangle}$ The $\tfrac{\theta}{2}$ can be seen on the picture given at the end of my answer. Maybe you are puzzled because you are tempting to interpret $$\Psi=\cos\frac{\theta}{2} \lvert 0\rangle + e^{i\phi}\sin\frac{\theta}{2} \lvert 1\rangle$$ as a linear combination of vectors in the 3D Euclidean space. Note ...

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Causality in special relativity implies that no signal can travel faster than light. In quantum mechanics, that does not translate in bounds on the speed of Dirac-delta wave-packets, not least because that would be in general an ill-defined condition (a localised wave-packet typically spreads out when evolving in time, so a delta at x=a would never evolve to ...

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You are essentially observing an interference effect. This is equivalent to what you would see in a single slit experiment. The fringes are caused as the light rays that passes between your fingers are travelling slightly different distance to reach you eye. This causes their phases to be somewhat different. As a result the light rays partially cancel, ...

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It took me a while to figure it out, even though it was obvious from the start that it should be ridiculously simple. Reminded me of the (in)famous question about the plane and the conveyor belt in that sense. The longest part was to figure out why we need to maximize time even though we're supposed to look for the minimum time. I guess that's because I'm ...

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Hint. ( I give absolutely no guarantee about not making calculational mistakes!) \begin{eqnarray*} L(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\left( 1-\frac{1}{ \sqrt{\pi }\sigma }\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{ \sigma ^{2}}]\right) e^{-iPy}\rho e^{+iPy}=L_{1}(\tau )-L_{2}(\tau ) \\ L_{1}(\tau ) ...

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The discrepancy comes from the fact that there are different ways to define a quantum Fourier transform and the direct equivalence to the Hadamard transform holds for only one such definition. See for instance this Introduction to Quantum Computing. Consider the general case of $N$ qubits, let the $2^N$ computational basis states be $$|{\bf x}\rangle ... 0 A general 2-dimensional quantum state, that is to say a density matrix \rho admits a bloch sphere representation because it admits a decomposition into the form: \rho=\frac{\mathbb{I}}{2}+ \vec{r}.\vec{\sigma} Where \vec{\sigma} is a vector of the three pauli matrices and \vec{r} is a real vector. There are analogous constructions in higher ... 0 If I understand you correctly, you have a relatively large Hilbert space \mathcal H of dimension \dim(\mathcal H)>2, which may be finite or infinite, and you are interested in a subspace \mathcal S\leq \mathcal H of dimension two. In that case \mathcal S contains two linearly independent (and w.l.o.g. orthonormal) vectors |u⟩ and |v⟩, and ... 0 A great college-level quantum mechanics textbook is John S. Townshend's Introduction to Quantum Mechanics. It starts with spin and angular momentum, and includes an entire chapter on EPR and entanglement. It sounds like that will be your best bet. A few other options include: J. J. Sakurai, Modern Quantum Mechanics. This is a famous textbook praised for ... 0 Tl;dr: There is a theory of reference, it's called "experiments" or more precisely "how to do experiments and make sense of their results". Equivalently, it's any of the multitude of answer to "how does science infer something 'real' about nature". Long answer: This is slightly off-topic, as it is ultimately about the theory of science and not physics - but ... 1 After a while I realized that the heart of the poster's questions are very relevant, and difficult questions, not all of which have answers. Here I have rephrased/reorganized the questions to make for easier answering. 1) Does accessing quantum memory always take constant time? In both quantum and classical computers one could make a physical system that ... 0 Just relabel j,k\to \ell (note that j=k due to the \delta_{jk}) and i\to k, and you are done, since \sum_\ell \alpha_\ell=\langle \alpha\rangle. 0 BEC qubits are set up by splitting one harmonic condensate trap into a double-well trap by means of a suitable laser or microwave field. See for instance pg.6 on these LANL slides: Quantum dynamics, measurement and decoherence in Bose-Einstein condensates. Under strong enough separation the overall ground state becomes two-fold degenerate and the two ... 0 The initial state does not need to be one of eigenstates of the hamiltonian, it could be superposition. Therefore time evolution will change it. I don't think your first assumption is correct. 4 The most general description of a quantum system is given by a density matrix \rho. It has dimensions of N \times N, where N is the number of degrees of freedom of the system: 2 for a 2 level quantum system (qubit), 3 for 3-level etc. But often we deal with the systems that have infinite number of degrees of freedom. Such systems are quantum harmonic ... 1 Better calibrated response: I'm a little confused about your first argument, that decoherence should somehow reduce entropy. Decoherence is the coupling of a system to a much larger system and of course adds entropy. Another way of putting it: if one ignores the environmental degrees of freedom, decoherence essentially maps pure states (zero entropy) to ... 5 The terminology of a mode of a free quantum field \phi(x) comes from writing it as a Fourier transform, often also called mode expansion:$$ \phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec p)^\dagger\mathrm{e}^{-\mathrm{i}\vec x\cdot\vec p}\right)$$where for a ... 1 The quantum state is a vector in a Hilbert space. Each basis vector in the Hilbert space is a possible observation one can make. For example, when modelling the position of a free particle, we assign one basis vector to each point in space. The length of the shadow that the state casts upon each vector is the (square root of) the probability that a ... 2 The term mode is used to define a particular state of a system and may refer for instance to its spin, wavevector, polarisation, charge etc. If we wanted to create a boson at position x with an up-spin and with wavevector k, we may use the field operator \hat{a}^\dagger(x, k, \uparrow) on the vacuum state \vert0\rangle. The most clear distinction ... 0 At the risk of being too obvious, let me first point out: any state that changes with time is not an eigenstate of the Hamiltonian. So if you are describing a system that is at equilibrium for all time, then you may indeed assume that system A is in an eigenstate of the many-body Hamiltonian, but for any other situation (including anything in the real ... 4 `"qubits can be 1,0 or both" - the accurate statement is that a qubit is in a superposition of the two states \left|0\right> and \left|1\right>:$$\left|q\right> = \alpha\left|0\right> + \beta\left|1\right>$$where \alpha,\beta are two complex numbers with |\alpha|^2+|\beta|^2=1. The probabillity to measure either 0 (1) is given ... 0 Yes and no (for bipartite systems). For your first question, if you don't take it as a definiton (which is often done), you can prove it by noting that a maximally entangled (pure) state has the Schmidt decomposition \sum_{i=1}^d 1/\sqrt{d} |i\rangle\otimes |i\rangle is an appropriate basis. The partial trace is the identity. Your second question is ... 1 You can represent \rho_{AB} by tr_B(\rho_{AB})\otimes tr_A(\rho_{AB}) only if the two subsystems are not entangled, i.e. \rho_{AB}=\rho_A \otimes \rho_B. 4 A Gaussian state is a ground or thermal state of a (bosonic or fermionic) Hamiltonian which is quadratic in the creation and annihiliation operators. Those states are fully characterized by expectation values of quadratic operators, and thus 4N^2 parameters for N fermions or bosons. 2 Actually the \sigma_x exponential is not so hard. Let's start by expanding the evolution:$$ \exp\left[ -i \left( \|x_i\||0\rangle\langle 0| + \|x_j\||1\rangle\langle 1| \right)\otimes \sigma_x \;t\right] = \sum_{n=0}^{\infty}{\frac{(-it)^n}{n!}\left[\left( \|x_i\||0\rangle\langle 0| + \|x_j\||1\rangle\langle 1| \right)\otimes \sigma_x\right]^n} =\\ = ...

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The theorem isn't called a no-manufacturing theorem. You can indeed make thousands, or millions, of copies of a known state. The theorem is about making a device that takes an arbitrary and unknown state and then produces two of it. So the point is that you have to copy something without knowing what it is that you are copying. And since you don't know ...

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This is not my field (so I may be wrong), but since no-one else has tried to answer I will have a go... Bekenstein bound limits the amount of information that can be stored in a system of bounded size and mass. Does that imply that the number of possible states is finite? Since we are talking about a finite-volume system, its energy spectrum must ...

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The simplest way to resolve this paradox is to require $$\rho_B(t=0)\simeq \tilde{\rho}_B=\frac{1}{Z}e^{-\beta H_B}.$$ That is to say, you do not need any time evolution to reach thermal equilibrium, and this is the statement of ETH. This is very different from a classical chaotic system, where you need some time to explore the phase space and reach ...

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