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Entanglement is simply a particular kind of quantum multiparticle state: it happens to be the "most common" kind of state in the sense that if you choose a random quantum superposition from a multiparticle state space, it will almost surelybe (in the measure-theoretic sense) entangled, so it's a little curious why entanglement takes some effort to observe in ...

7

To quote Stephen Gasciorowicz, Before evaluating these quantities to obtain an idea of their magnitude, we will introduce some notations that will be very useful. First, it is $h/2\pi$ rather than $h$ that appears in most formulas in quantum mechanics. We therefore define $$\hbar=\frac{h}{2\pi}=1.0546\times10^{-34}\,{\rm J\cdot s}$$ So basically it's ...

6

Quantum mechanical postulates So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms? The mathematical expression for each individual operator is sort of a postulate, but it should not be listed. The postulates define a (more or less) complete theory in that I can ...

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What's the mathematical process and physical logic? The Fourier transform of position space ($\vec x$ domain) is wave number space ($\vec k$ domain). This is an unambiguous, well understood mathematical result. By the De Broglie hypothesis, the momentum is $\vec p = \hbar \vec k$. This is physical hypothesis with experimental confirmation. Although ...

5

Let us consider a real, scalar field theory for simplicity (and metric signature $+ - - -$). In the free theory, one can use the mode expansions of the field $\phi(x)$ and its canonical conjugate momentum $\pi(x)$ to derive the following expressions for the creation and annihilation operators: $$a(p)=i\int d^3x\ ... 5 This is called "resolution of the identity", in case you wanted more information. The identity depends on the eigenvectors of H forming a complete orthonormal basis. Thus,$$\langle k | k'\rangle = \delta_{k,k'}$$Now consider the matrix$$O = \sum_q |q\rangle \langle q |,$$where the ket \left|q\right\rangle represents the wavefunction in the following ... 5 I don't intend to repeat the points already made in the other answers, consider this as a small addition to those, with the intention to give a more practical description (reminding some of the basic ideas) without getting into observer-philosophies. Quite clearly the act of observing, i.e. measuring a quantum system can be done via many different ... 5 I will give you some rigor, since it seems what you are looking for. The rigor takes the form of the very famous spectral theorem for self-adjoint operators. First of all, a simple example to elucidate: a purely discrete spectrum, that forms an orthonormal eigenbasis of the hilbert space (i.e. the operator H is either compact or with compact resolvent). ... 4 Even in special relativity, the claim that light always travels at a constant speed of c is only true in inertial reference frames, in non-inertial coordinate systems like Rindler coordinates, the coordinate velocity (change in position coordinate with respect to coordinate time) may vary. In general relativity, all coordinate systems covering large regions ... 4 Perhaps some additional information is in order to shed additional light... The whole discussion begs the question: If \hbar is so convenient, why do we have h around? As usual, "historical reasons". Planck originally invented h as a proportionality constant. The problem he was solving was blackbody radiation, for which the experimental data came ... 4 When does an interaction drop the system into an eigenstate? (i.e. when is a measurement=) This is an ill-posed question because, first of all, the system S doesn't drop into any state but each observer O has a state about it, as a state \rho is nothing but the coding of past measurements (so it should be named with reference to being dependent on ... 4 What are phonons? Phonons aren't particles like electrons or protons are, phonons are quasi particles, these type of particles are just used to describe excitations of a field: in phonons case, phonons are used to describe elementary lattice vibrations which have certain frequency. Electron-Phonon Interaction: Basically Cooper pairs are just pairs of ... 3 You are quite correct. If you have a wavefunction \Psi for the water molecule then swapping the two protons will multiply the wavefunction by -1. However the wavefunction is not observable, by which I mean that there is no experiment we could do that will measure the value of the wavefunction. Typically to measure some property Q there will be an ... 3 No such algorithm is known. The natural language description of experimental setups is far too informal to be turned into precise quantum mechanical statements. Therefore, we will in the following suppose that a quantum mechanical description of the measurement apparatus in spe has been provided. In the von Neumann measurement scheme, it is not subjective ... 3 The uncertainty principle may be stated more generally for two observables A and B as: $$\Delta A \Delta B \geq \dfrac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right|,$$ where \langle \hat{C}\rangle is the expected value of the observable C and [\cdot\,,\cdot] is the commutator (see here for details). From ... 3 If they are much far away will they have relative velocity of separation greater than speed of light and if so how can we even detect such galaxies. We can't detect such galaxies. Redshift goes to infinity at the cosmic horizon, and we cannot see beyond. Note that the cosmic horizon is different from the Hubble sphere: At the former, relative velocities ... 3 Such galaxies cannot be detected. and Quantum entanglement is just a correlation between two bits of information and it no way does, and no way can be used to transfer information faster than the speed of light. Any light sent from such galaxies towards ours would show a wavelength of infinity after considering Doppler shift. In short these galaxies can ... 2 These galaxies are not out of reach. The Hubble Sphere is the volume of space surrounding an observer where everything inside the sphere moves away from the observer with a speed less than c. Following this logic it immediately follows that the Hubble Sphere is equivalent to the cosmological event horizon that @Hritik is discussing in his answer. ... 2 The way you wrote it, they are distinguishable (unless a=b of course). For the particles to be indistinguishable their wavefunction must be of the form$$ \psi(r_1,r_2) = \frac{1}{\sqrt{2}} [ \psi_a(r_1)\psi_b(r_2) \pm \psi_a(r_2)\psi_b(r_1) ]$$where the sign depends on the fermionic/bosonic nature of the particles. If the particles are described by a ... 2 We don't know we only assume that it was a wave, and our assumption works well in some cases, and works badly in other cases. For instance if we put on the way of the particle a beam-splitter, we believe that we get a splitting of the wave, into a reflected wave and a transmitted wave. I.e. although we speak of one particle, we believe that we get two ... 2 As an experimental particle physicist I stick to observables . The complicated mathematical functions which have been established as necessary to describe the quantum mechanical state function of the particles under consideration are not observable. By the postulates of quantum mechanics the complex functions of space time or energy momentum are not ... 2 Quantum mechanics tells us the probability for an electron in a certain orbital to be in a certain position or momentum state. Inverting this to find the probability for the electron to be in a certain orbital given its position or momentum state requires some a priori idea of what orbitals are the most probable. This is stated succinctly by Bayes' theorem: ... 2 Then, there is the case that such an operator is defined on the full interval I assume that by "full interval" you mean the whole real line. First question: Do we then need any boundary conditions? Yes, as noted by Sam Bader, boundary conditions are part of the Hamiltonian. In my physics lecture we used so-called Born von Karmann boundary ... 2 The Schroedinger equation is non-relativistic and it propagates effects at an infinite velocity to begin with. It is thereof nonsensical to even talk about "locality". Schroedinger's equation doesn't describe local physics any more than a first order diffusion equation describes the speed of sound. There is no technical issue here, at all, you are simply ... 2 The interaction between two charged partices occurs through a change of momentum. All very well, but the next question is how we calculate the momentum exchange, and this is where quantum field theory comes in. The interaction between two electrons is described as a disturbance in the quantum fields involved. Quantum field theory gives us an expression for ... 2 Given that the \mathcal E_i are linear operators, i.e., \mathcal E_i(\alpha A + B)=\alpha \mathcal E_i(A) + \mathcal E_i(B) for all A and B, it is true that \mathcal E_1(\rho)=\mathcal E_2(\rho) for all \rho\ge0 implies that \mathcal E_1=\mathcal E_2. The argument goes as follows: Every operator A can be written as A=X+iY, with ... 2 It's actually easier to see if you write them as Dirac braket form. Equation (1) is$$\langle k\lvert k'\rangle = \delta(k,k') = \int \mathrm{d}r\, \langle k\lvert r\rangle\langle r\lvert k'\rangle = \int \mathrm{d}r\, \psi^*(r;k) \psi (r;k')$$Equation (2) is$$\langle r\lvert r'\rangle = \delta(r,r') = \int \mathrm{d}k\,\langle r\lvert k\rangle\langle ...

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You can try easily what is the good answer by calculating the difference in energy between the two participating levels. $E_{upper} - E_{lower} = -Rhc [\frac {1}{(n+1)^2} - \frac {1} {n^2}]$ $= Rhc [\frac {2n + 1}{n^4 + 2n^3 +n^2}]$ So, its obvious that the smaller n, the bigger $E_{upper} - E_{lower}$.

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The difference between the whole system properties and its constituents can be explained on two-particle system. Consider positronium (electron+positron) in the state $l=1$. The quantum number $l$ describes the relative motion of constituents. This state has a magnetic moment, which belongs to the whole system. But when you consider the angular momentum ...

1

Of course linear polarization is an observable, we measure it. Linear polarization along an axis x, and perpendicular to x, are the eigenstates of the observable "polarization along x". In fact, by measuring polarization along x, we measure the projection of the electric field along the axis x.

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