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3

All $\mathrm{SU}(2)_k$ with $k>2, k\neq 4$ are universal. For a proof see http://arxiv.org/abs/math/0103200.


3

Vectors (which kets are) don't have adjoints, they have duals. Whether the dual of $\lvert n_1,\dots,n_n\rangle$ is denoted by $\langle n_1,\dots,n_n\rvert$ or $\langle n_n,\dots,n_1 \rvert$ is entirely conventional.


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There are two things to consider: What does the potential look like? Is the wave function of the qubit narrow in the flux or charge basis? Potential shape The Hamiltonian of the transmon (a junction in parallel with a capacitor) is $$H_{\text{charge qubit}} = - E_J \cos(\phi) + \frac{(-2en)^2}{2C}$$ where $E_J\equiv I_c \Phi_0 / 2\pi$, $I_c$ is the ...


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I'm not an expert (mathematical physicist here), but as far as I know, the answer is no over really long distances and yes over shorter distances. There are inherently two problems that your question calls attention to: entangling particles over long distance maintaining this entanglement over longer time You only address the second one, but the first ...


2

It all goes back to EPR experiment. In a paper published by Einstein, Podolsky and Rosen the authors argued that quantum mechanics is "incomplete". They argued that principles (such as the principle of locality) needed to be restored in order for it be a complete theory. The problem is, ultimately quantum theory is a "nonlocal" theory. What this means is ...


2

Energy is not a Lorentz invariant quantity, it is the zero-th component of the four-vector. Only proper orthochronous Lorentz transformations preserve the sign of the zeroth component, so if the energy is positive in one frame, a non-orthochronous Lorentz transformation would yield a frame in which the energy is negative. But we usually only allow the ...


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There can be no blanket answer about what properties are shared between two entangled particles because it depends entirely on the type of entanglement. Moreover, the entanglement of a single property does not necessitate the entanglement of the others (see this question). My understanding is that if particles A and B are say, spin-entangled, a kick applied ...


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You have fallen into a very subtle trap, essentially the one I discussed here, but it's probably worth going into the specifics. Lets say that after measurement the combined state was $|00\rangle\langle00|$, so $B$'s actual reduced density operator is $|0\rangle \langle0|$ after measurement The problem with this is that you cannot consider the ...


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Firstly, $m$ does not have to be an integer, it is entirely possible for $m$ to be 1/2 for instance. Your points ,1-3 are fine. There are is a maximal and a minimal value of $m$. Call the maximal value $M$ (we have to call it something). Now we can apply the lower operator any number of times, each time it lowers the value of $m$ by a full integer ...


2

The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ...


2

Let's start from $$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$ with $\{f, f^\dagger\}=1$, $\{f, f\} = 0$ and define fermionic position and momentum coordinates by $$ \psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\ \psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right) $$ with the following anticommutation relations: $$ ...


1

If you try to analyze the power radiated by a black body and assume a continuous spectrum of energy, you find that the black body must emit an infinite amount of energy. This is known as the ultraviolet catastrophe and was solved by Max Planck assuming that energy is emitted in discrete units of $hv$ and not in a continuous spectrum.


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Discreteness of energy reproduces experimental results in a satisfying manner. If you assume discreet spectrum you get to preform summation instead of integration. To be precise, you get to sum geometric series which converges. After that, you get the factor of integration of the form 1/(exp (hν/kT) − 1). This now is OK. If you dont have discreteness ...


1

This is the spectrum of black body radiation, classical, and quantum calculations. As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans. So as you see the wavelengths are in the x axis ...


1

I think the confusion is due to a lack of mathematically precise definitions of what is quantum field theory? what is one trying to construct and how? etc. There are a lot of vague notions used in the physics literature: the partition function (which does not make much sense in infinite volume), the effective action,...but the bottom line is the collection ...


1

Protons and electrons both obey the de Broglie hypothesis: wavelength = Planck constant / momentum. But protons do not act within the atom the same as electrons - they move in a tighter radius at higher speed. They have to be accelerated to reveal their wave nature, and as the momentum of a proton would be much greater than that of an electron at typical ...


1

Note: There is a short summary at the bottom. This is actually also described in Nielsen&Chuang: You don't learn about general measurements, because they are completely equivalent to projective measurements + unitary time evolution + ancillary systems, all of which is described in your usual QM formalism. The Measurement Postulate Let's start from ...


1

That a Hamiltonian preserves a symmetry means $$ [H, C] = 0 \Rightarrow CHC^\dagger = CHC^{-1} = H$$ For a unitary symmetry operator $C$ (or anti-unitary if it is time reversal). The Hamiltonian of a crystalline condensed matter system written in terms of the Bloch matrix is: $$ H = \sum_{\vec k} \psi^\dagger(\vec k) H(\vec k) \psi(\vec k) $$ Where ...


1

If the light was emitted after the recombination, it can't have traveled over 13.7 billion years. The EGS-zs8-1 galaxy is located 13.1 billion light years away , which is close to the maximum for a plausible picture. More details on the wiki http://en.wikipedia.org/wiki/EGS-zs8-1 . If the light is not absorbed by an obstacle, it will probably travel to the ...


1

First off, there are some very misleading answers given above. Introductory quantum courses fail to properly discuss "time." It is a parameter, not an observable. E(operator)=ih(bar) d/dt has no meaning. That operator simply discribes the time evolution of a wavefuntion that is complex. So it does not describe a physical observable. I know this might be hard ...


1

Assuming that $X=X^\dagger$, $P=P^\dagger$ and $[X,P] = i\hbar$, let me try $$f = \sqrt{\frac{m\omega}{2\hbar}}\left( \alpha X + \frac{\beta}{m\ \omega } P \right) $$ where $\alpha$ and $\beta$ are complex numbers of modulus one. From this follows that $$ \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + ...


1

The previous answer is completely right but I want to add some historical comments, since from a historical point of view they have clearly different origins. The Correspondence Principle used to be one of the two fundamental principles of Bohr's theory (one of the main formulations of quantum theory between 1913 and 1923). Originally this principle didn't ...


1

You can use the teleportation protocol to teleport and part of a larger quantum state (which can be arbitrarily entangled), and it will work the way it should: I.e., if initially A+C hold $\vert\psi\rangle_{AC}$, after the protocol B+C hold $\vert\psi\rangle_{BC}$. The same is true if the initially shared state is mixed. This follows from the linearity of ...


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SUMMARY OF EDITED VERSION: You cannot place any conditions on $V(x)$ and $E$ that guarantee that solutions to the time-independent Schrödinger equation are normalizable, for something of a silly reason. Initial, partial answer: If the potential is bounded below by some value $V_\text{min}$, then a solution to the time-independent Schrödinger equation ...


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You are essentially correct. If you start on the real line with the Schrödinger equation $$ \left(-\frac12 \frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) = E\,\psi(x),$$ then at every point $x_0$ where $V(x)$ is analytic you are guaranteed that $\psi(x)$ will be analytic in a neighbourhood of that point. However, if $V(x)$ is not analytic then you ...


1

My answer to the question is first a question. How do you photograph a photon? You can measure a photon by detecting it's presence on photographic film, or by using some sort of photomultiplier and digital detector, or by a handful of other ways. My point is that these all require the photon to be absorbed by the detector, thus they must be localized, and ...


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The point of this thought experiment has been widely taken out of context and misused by new age science supporters. Schrodinger initially considered this experiment to show the RIDICULOUSNESS of the situation, not because it was physically what is happening. Additionally, anyone who says that there is a line between the quantum and classical regime is ...



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