# Tag Info

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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 ...

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`"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 ...

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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 ...

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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.

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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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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 ...

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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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Actually the $\sigma_x$ exponential is not so hard. Let's start by expanding the evolution: $$\exp\left[ -i \left( \|x_i\||0\rangle\langle 0| + \|x_j\||1\rangle\langle 1| \right)\otimes \sigma_x \;t\right] = \sum_{n=0}^{\infty}{\frac{(-it)^n}{n!}\left[\left( \|x_i\||0\rangle\langle 0| + \|x_j\||1\rangle\langle 1| \right)\otimes \sigma_x\right]^n} =\\ = ... 1 You can represent \rho_{AB} by tr_B(\rho_{AB})\otimes tr_A(\rho_{AB}) only if the two subsystems are not entangled, i.e. \rho_{AB}=\rho_A \otimes \rho_B. 1 In order to implement quantum cryptography, you need a link which allows to send quantum states, i.e., some kind of object/particle which carries a quantum degree of freedom. These particles should travel as freely as possible along their way. This is most easily accomplished using a single photon, thus the optical link, but in principle, sending e.g. ... 1 The A operators and the Bs all commute with each other because they always share an even number of sites and therefore an even number of Pauli matrices. Therefore, these are all conserved quantities and can be replaced by their expectation value. The ground state is the state with eigenvalues A = 1 = B in units where \sigma is a Pauli matrix. In ... 1 The simplest way to resolve this paradox is to require$$\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 ... 1 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 ... 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 ... 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 ... 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 ... 1 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 ...

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