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Concerning point 2: Operators do not always come through each other cleanly, but there are some very basic rules that always apply, which can be turned into less tedious rules that apply in special cases. Often the latter are taught first, causing mass confusion. General rule: Operators can be expressed as (Sum over a in the set of eigenvectors ) |a > ...

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Probably the best place to start classically is with integrable systems. A crude physicist definition is that these are systems that have, in the words of Nandkishore et al, "an infinite set of extensive conserved quantities that are sums of local operators" (1). Roughly speaking, such systems will never approach an equilibrium because none of these ...

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The Dirac matrices are invertible because of the defining Dirac anticommutators $$\gamma^\mu\gamma^\nu+\gamma^\nu \gamma^\mu = 2g^{\mu\nu}\cdot 1$$ For $\mu=\nu$, one simply obtains $(\gamma^\mu)^2 = g^{\mu\mu}\cdot 1$ (no summation) which means that the inverse of $\gamma^\mu$ is $\gamma^\mu g^{\mu\mu}$ i.e. basically $\gamma_\mu$ if $g_{\mu\nu}$ is ...

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I'll assume the question refers to calculating coefficients $c_i(t)$, not transition probabilities under external interaction. Since in Heisenberg representation all relevant quantities are calculated as observable/operator averages, the problem becomes one of choosing the correct observable to represent the final state. The simplest option is its ...

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It is a matter of definition of "same". Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the ...

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Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same. Even in your situation with photon ...

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Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such ...

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The inflationary theory you mention is probably eternal inflation. In this theory there is just one universe but different parts of it are causally disconnected i.e. the different parts cannot affect each other in any way. Whether these constitute a multiverse comes down to terminology. In principle there is a continuous spacelike straight line that links ...

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Does it go from one state to another via straight line or it makes smaller and smaller orbits till it reaches the next orbit The electron is an elementary particle and as such, at the level of atoms, it can only be mathematically modeled by a quantum mechanical probability distribution. These spatial probability distributions for the electron of the ...

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The left hand: rotate the state $|JM\rangle$ by applying a rotation $R$ on it. Right hand side: insert completeness condition $\sum_{M'} |JM'\rangle\langle JM'|$ $D$ is the matrix representation of rotation matrix $R$ in basis ${|JM\rangle}$. The rotated state is expanded in terms of basis ${|JM\rangle}$ with coefficient $D$ in terms of rotation matrix.

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You can define velocity as $v=\frac{\sqrt{\langle \hat{p}^2 \rangle}}{m_e}$. For ground state $v=\alpha$.

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"Perhaps a sagging potential is one of ratios rather than sizes." Your conclusion is correct, but the idea is not restricted to the sagging well. This is a neat exercise in scaling, and it all comes from the choice of parametrization of the potential for given length $L$. To see this better in the present case, just rewrite the hamiltonian in terms of ...

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The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ... 1 As phrased, this question suggests suggests a very Bohr-like image of electrons. This answer is intended to nudge you in the direction of a quantum model, but is by no means complete. Honestly, there's no way to understand QM without delving deep into the math, and even then its difficult to make a math <-> reality correspondence. Firstly: quantum ... 0 You can think, on the left is a short hand notation for a (2J+1) x (2J+1) matrix R applied to a (2J+1) component vector |J,M> with the components labelled by M. On the right, the matrix elements are explicitly shown, and the sum over M' is the matrix multiplication. Actually on the left is an abstract rotation operator R that will rotate any J. When ... 1 Write$H=H_0 + H'$where$H_0$is constant and$H'$is how it changes. Let's take a superficially very different situation: Assume$H'$is sinusoidally oscillating, e.g.$H' \propto \cos(\omega t)$. Then$H'$will be very good at exciting transitions between states whose energy differs by$\approx \hbar \omega$. For details on this, see any QM textbook ... 0 We have a state ket$\mid\alpha\rangle$. We can obtain any allowed information about the quantum mechanical system from this ket. All you have to do is to operate the ket by the corresponding operator of an observable, say energy. In that case, we define our state ket$\mid\alpha\rangle$spanned by the eigen kets of the observable energy. Otherwise these ... 2 The point is that the domain$D(P)$of$P$must be such that$P$is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that$D(X)= L^2([0,L],dx)$instead, where$X$is automatically self-adjoint. The vector$\psi$you use to prove Heisenberg inequality has to belong to$D(PX) \cap D(XP)$as you see by direct ... 1 As all the other commentators have said, both diffraction and interference are manifestations of the same thing: the fact that waves superpose. The two words are slightly different, but it's clunky to work with two concepts when you only need one. You never do a "diffraction calculation" or an "interference calculation", you just do wave mechanics and ... 1 These high reputation members aren't confused about the difference between diffraction and interference. The two terms are synonymous in the physics community (we're not arguing on the basis of linguistics. I can agree the terms are linguistically different just because people sometimes prefer one over another in certain contexts). The way the physics ... 1 Is momentum of a particle "random" because it is uncertain, or is it uncertain in addition to being random? In quantum mechanics systems are represented using wave-functions (wave-vectors). The momentum of a particle is completely uncertain if it's position is certain (a localized particle) . But it is also possible to create wave-functions that have a ... 1 A particle is a wave. It's wave function (consider non relativistic Quantum Mechanics), when absolute valued squared, is a probability density function. The particle's momentum is a multiple of the gradient of the wave function, with h, Planck's constant, one of the proportionality constants. That is then the probability density function for momentum. It is ... 0 Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric. 0 The work function doesn't depend on time because we're assuming a$\textbf{steady state solution}$. This means the differential equations which describe the total charge on the surface of the metal are at an equilibrium, where all the time derivatives are$0$. When you shine light on the metal, the system actually reaches this steady state almost ... 0 The more electrons we take out from the metal, the more ionized it becomes Looking at an experimental setup description (such as this one or this one taken at random from google), you should find that the target is not electrically isolated. Indeed the potential of the target can be directly controlled to change the behavior of the experiment. ... 0 No. The frequency of EM radiation such as visible light is many orders of magnitude greater than the frequency of mechanical vibrations. You would not be able to detect any difference in the spectrum even if there was a difference. 4 No,$|\phi\rangle$doesn't need to be orthogonal to$|0\rangle$. Neither does$|\psi\rangle$. They only need to be orthogonal to each other. Example Suppose you know that you will be given a state$|\omega\rangle$that will either equal$|\phi\rangle = \frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$or equal$|\psi\rangle = \frac{1}{\sqrt ...

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Cort and Ilmari have given good answers about the practical issue: the inverse square law is for point sources, and so a non-point source (like an emergency light) will only appear to have the same properties at some minimum distance that depends on the geometry of the real source. However, it seems nobody has mentioned a different "minimum distance" that ...

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The quote from the reference says it all: (I added caps) "The minimum test distance IN PHOTOMETRY of these sources is called the 'minimum inverse-square distance.'" The minimum distance is therefore a photometry issue, in other words, a measurement problem. The essence of the measurement problem is how far away you have to be before you can approximate the ...

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The solution that you wrote in your last (not numbered) equation is not a basis of a Hilbert space of sections because the phase factor: $(-1)^n e^{2i\pi A}$ depends on $n$. The phase factor should not depend on $n$. Please see your (correct) equation (2) defining the boundary conditions, in which the phase factor does not depend on $n$. Thus there is no ...

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That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be ...

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The inverse square law says that the intensity of incident light falls off in proportion to the inverse of the square of the distance from the light source. The important word here is "the distance" — the inverse square law implicitly assumes that all parts of the light source are at the same distance from the measurement point, or at least ...

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The way I like to understand this is in terms of generators of translation. A well known result from classical mechanics (see Goldstein) is that momentum is a generator of translation in the canonical coordinate conjugate to that momentum. For example, linear momentum generates space translations, and angular momentum generates rotations. In Hamiltonian ...

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It is just a misleading concept that electron is particle or a wave ... the answer is neither particle nor wave but BOTH. How we came to know that electron is particle?? Suerly through experimental evidences. A man is affected by its very imediate surroundings .. and what is in surroundings .... ?? Most oftenly matter. Electron is particle or a wave is ...

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The most effective way to think about such problems is by means of operator algebras. The best way to treat systematically quantum observables (and their associated unitary operators) is to collect them in a C$^*$-algebra, roughly speaking a Banach algebra where observables have a norm, can be summed and multiplied (with some additional technical ...

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Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...

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Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $\psi_{a}$ and $\psi_{b}$, both of which correspond to energy $E_{0}$. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...

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As many have said, the inverse square law applies to point-sources. These are idealized light sources which are sufficiently small compared to the rest of the geometry that their size is of no importance. If a light source is larger, it is typically modeled as a collection of idealized light sources, potentially using integration. The exact definition of ...

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Yes. For example, consider the harmonic oscillator potential $V(x)$, where the ground state has energy $\hbar \omega / 2$. Then the ground state of the potential $V(x) - \hbar \omega / 2$ has zero energy. This works because in quantum mechanics, like in classical mechanics, absolute energies don't matter. You can always add or subtract constants. ...

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The inverse square law applies to point sources. For extended sources becomes accurate at distances that are large compared to the size of the source. At large distances the source looks like a point. What "large" means depend on the application. In the case of light fixtures, the Illuminating Engineering Society and other organizations have made ...

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Robert Griffiths is quite fond of Mach-Zehnder experiments as useful windows into interpreting quantum foundations, and he presents calculations for some toy models of M-Z interferometers in Chapter 13 of his book, Consistent Quantum Theory. As a caution, most of this book is written to parallel the consistent quantum histories formulation of quantum ...

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The inverse square law applies to point sources. A real emergency light is not a point source, and therefore the law appears to not apply at close distances, because any real point is at a varying distance from different parts of the emergency light.

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For a neutron of that speed, the uncertainty in the momentum is expected to be less than the momentum magnitude. Using the actual momentum will be an upper bound on the momentum uncertainty. That correlates to a lower bound on the position uncertainty. So, $\Delta x$ is lower bounded by $\hbar/(2p)$: $$\Delta x \ge \frac{\hbar}{2m_nv}.$$ $\Delta x$ could ...

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If we setup the camera to record like above but NEVER EVER EVER look at the result of what was recorded. Does the wave function still collapse? The answer is that we just don't know. We can tell that the wave function has collapsed (in Copenhagen terms) only when we humans look at the system -- in the canonical experiment that means looking at the ...

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Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently — instead, a quantum state must be described for the system as a whole. Measurements of physical properties such as position, momentum, spin, ...

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You can find an extensive treatment of the double-slit experiment with electrons in Feynman Path Integral approach to electron diffraction for one and two slits, analytical results (Beau, 2012). The paper discusses both Fraunhofer and Fresnel regimes. These regimes do hold for electrons. Interestingly it does not use the standard semi-classical ...

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Quantum entanglement is the property of two objects $A,B$ – more precisely two subsystems – or a relationship between these two objects whose quantities or observables aren't independent of each other. It means that there exist some quantities $a_j$ and $b_k$ describing $A,B$, respectively, such that the probability distribution for these observations ...

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Experimental determination of $c_i$ values starts with preparing multiple identical systems, then making measurements. From all the measurements, one determines the probabilities, which are the $|c_i|^2$. The square root of the probabilities will tell you the $c_i$ to within a phase factor of the form $e^{i\beta}$, where $\beta$ is real, and may or may not ...

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Once a measurement ( observation ) is made on a quantum system the system will be in an eigenstate of that property, so if the energy of an electron is measured the electron will afterwards remain in an energy eigenstate ( until some other measurement or interaction occurs ), but if the angular momentum is measured the electron system will afterwards remain ...

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The answer is no, and the details are clearly spelled out in Glauber's Les Houches lectures (circa 1964). Glauber introduces a "T-representation" which can represent any operator in the Fock space of harmonic oscillator states, a less general "R-representation" which can represent any density operator, and the still less general "P-representation" which can ...

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