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This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that do not yet have commentary. Try "How to teach Physics to your dog", ...

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Is this concept of relativistic mass increase, related to the concept of Doppler effect of matter waves? No. Doppler's effect also happens for classical waves, including "classical matter wave", by which I meant Schroedinger's wave function. The effect is in fact trivial. When you change the reference frame, the momentum of the particle changes. By de ...

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1) Are we able to perform perturbative analysis and use diagrammatic expansion, Green function etc. – all these field-theoretical stuff [for bosons]? In general, the field-theoretic methods (at finite or zero temperature) can be applied to both bosons and fermions with slight differences which originate from the Fermi-Dirac and Bose-Einstein ...

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Perhaps you don't want a quantum superposition, but just a statistical mixture: $$\rho = \begin{pmatrix}1/2 & 0 \\ 0 & 1/2\end{pmatrix}$$ Although I'm not 100% sure that this will describe your situation any better...

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You don't specify the form of your Hamiltonian, but I'll venture that it might be something basic, like $$H = \sum_{\bf k}{\omega_{\bf k}{\hat a}^\dagger_{\bf k}{\hat a}_{\bf k}}$$ In this case a very simple and convenient toy model for a non-equilibrium state may be a displaced (or coherent) thermal state of the form $${\hat \rho}\left(\{\alpha_{\bf k} ... 0 I was re-reading von Neumann's tome in which he recapitulates his views on his own invention, the density matrix. As well as Dirac's inclusion of this in his second edition of his Principles, and the explanation of it in Landau--Lifschitz (second ed)...now, Landau independently had invented it, too. Von Neumann's rationale is that if our information ... 8 The signs are important for fixing an out of order machine. Define the states |\pm\rangle as:$$|\pm\rangle = \frac{1}{\sqrt{2}}\left[\left |\text{Working}\right\rangle\pm \left |\text{Down}\right\rangle\right]$$And we define the observable O as:$$O = |+\rangle\langle + |\ - \ |-\rangle\, \langle -|$$Suppose then that coffee machine is out of ... 0 If you want to declare indeterminacy and a probability of being either working or down you should use the vector notation: (working ) Status> = ( down) Status being the column vector analogous to the column state vector of the wavefunction in a matrix representation The user would be the operator :) -4 Do particles also have intrinsic linear momentum (linear analogue of spin)? Some do. A photon moves linearly at c, and its momentum is p=hf/c. The linear momentum is an aspect of energy-momentum, which is denoted by the Poynting vector as per this Blaze labs picture: In addition neutrinos travel at a speed which is so close to c that we can't tell ... 1 The state -|\text{Down}\rangle is the same as |\text{Down}\rangle just multiplied by a phase factor (of \pi). So they are the same physical state and it doesn't matter which sign you use. I wonder if you are getting mixed up with the LCAO method for approximating molecular orbitals, for example for Hydrogen, where we write:$$ \Psi = \psi_A \pm ...

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Quantum decoherence happens inside rocks and rocks are not conscious (although I know some people who would argue about that). Many people believe that consciousness is simply an emergent property of a sufficiently complicated process and needs no further explanation.

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It's true that "the perturbative series is valid only when the perturbed state is qualitatively similar to the unperturbed state". Generally perturbation theory is acceptable when the coupling is weak, in which case the coupling can be treated as a small perturbation of the free field theory at all energies (for example Yukawa theory and $\phi^{4}$ theory. ...

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Your spin projection matrix $S_u$ actually looks like a rotation, not a projection: A projection has only degenerate $+1$ eigenvalues, but your $S_u$ has eigenvalues $\pm 1$. Also, I'm not sure the problem is meant to require a general solution for arbitrary $\alpha$. In any case, it's easier, even though less elegant, to go case by case: For $\alpha = ... -1 I got my answer :). In simple way electron also have both wave and particle nature and waves possess both kinetic energy and potential energy and potential energy increases with distance. Thus higher orbits have higher energy electrons. 0 To do Hartree-Fock to Helium atom, you just need to calculate one orbital, which for helium is spherically symmetric. The Hartree-Fock 'integro-differential' equation for spherically symmetric atom with one eigenstate can be written as $$\left(-\frac{1}{2}\nabla^2 - \frac{2}{r} + V_{Hx}(r) \right) u(r) = \epsilon u(r),$$ where$u(r) = \psi(r) r$and ... 0 I feel like this question is a little too open ended and I'm sorry for that...but I did find a WONDERFUL HF walk through http://www.phys.sinica.edu.tw/TIGP-NANO/Course/2011_Spring/classnotes/CMS_20110511.pdf that clears everything up. -1 My Chemistry book has a very witty answer to this exact question. Electrons are present in "clouds" around the nucleas, the orbital just gives a mathematical probability of finding an electron at a specific distance away from the nucleas. When a seemingly lower energy, thus a lower orbital electron gets a specific amount of energy, it goes a little further ... -1 To lift the electron into a higher orbit one has to transfer energy to the photon (in the form of a photon absorption). In the case the electron has a free space in a lower orbit, it will emit photon(s). Hence it is a lower energy electron. This is, what we get from experiments. QM gives a mathematical formulation about this phenomenon. That higher the ... 0 Because this suggests a force that depends on the orientation of the spins of the nucleons with regard to the vector joining the two nucleons and hence can't be called central force. This force is basically called tensor force. 0 This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature? No, it does not; it means that it is not a valid$\psi$function to use in a theory based on the Born interpretation of$|\psi(x)|^2|$as density of probability for configuration$x$. You're right there are no free particles in nature, but the reason is in ... 2 I henceforth assume$\hbar =1$. There is no reason to introduce Dirac deltas here, everything is elementary. Moreover as the function$\psi$is not differentiable, one cannot use the form of the momentum operator$P$as derivative which is valid only on smooth functions. Forcing this way would introduce unnecessary difficulties as the derivative must be ... 0 That's the covariance of X and Y. It tells you, in a way, "how much" X and Y are correlated, with covariance 0 meaning uncorrelated (Not to be confused with independent). Because$\langle A\rangle$and$\langle B\rangle$can be quite large, it is customary to define the correlation coefficient: ($cov(A,B)$is the covariance of A and B) ... 4 I) One problem is that the momentum operator$\hat{p}$is an unbounded operator, which means that it is only defined on a domain$D(\hat{p}) \subsetneq {\cal H}$of the Hilbert space${\cal H}=L^2(\mathbb{R})$. When we apply the differentiation operator$\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$to OP's wave function $$\tag{1} \psi(x)~=~A(a-x)\theta(a-|x|), ... 3 Well, you can conclude that something is wrong by the following logic: momentum is an observable, which means its allowed values must be things that you could read off a measuring device (assuming you had one that measures momentum). These are necessarily real values, and since the expectation value is some linear combination of possible measurements, it ... 10 The wavefunction has a discontinuity at x=-a, which gives a term -2aA i \hbar \delta(x+a) when you act with p. The contribution from this to the expectation value of momentum exactly cancels the imaginary value you have calculated. Two more-general points: The momentum operator is hermitian, which means its expectation value must be real (provided ... 2 We usually say that if two operators, \hat{A} and \hat{B} commute, then they have a simultaneous set of eigenstates. Saying that the eigenstates are the same isn't really correct. For example, let operator \hat{A} be hermitian and act on elements of the Hilbert Space \mathcal{H}_A and let operator \hat{B} also be hermitian and act on elements ... 4 Assumptions: I will be talking about Hermitian (more generally self-adjoint) operators only. This means that I will assume that the operators in question have a set of eigenvectors that span the Hilbert space. As mentioned by tomasz in a comment, this is not exactly necessary, since more general statements can be made, but since we are dealing with basic QM, ... 0 The quantity A|\psi\rangle, for general Hermitian operators A, is mostly meaningless. The eigenvalues and eigenvectors of A are meaningful, as is the quantity \langle \psi | A | \psi \rangle, but A|\psi\rangle by itself is not. This is a sort of confusing point when first learning QM because it feels like the most important thing about operators ... 0 I am going to focus on the precise meaning of the density matrix and the probability distribution w in Landau's notation, which could be called the quantum statistical distribution. Almost all texts and wikipedia mess this up. If the usual axioms of quantum mechanics are accepted, every system is in a pure state, given by a wavefunction. Sometimes this ... 1 \lvert \psi_{n'm'l'} \rangle is the state you start out with. A {\mid} \psi_{n'm'l'} \rangle is the new state you get when you apply A to the original state. \langle \psi_{nml} {\mid} A {\mid} \psi_{n'm'l'} \rangle is the projection of this new state onto \lvert \psi_{nml} \rangle; that is, it measures the overlap between the unprimed state and the ... 3 The wavefunction:$$ \Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)} $$is an infinite plane wave. So it describes a particle that has an infinite extent in both time and space. That is, it exists for -\infty \le x \le \infty and for -\infty \le t \le \infty. Unsurprisingly, if the particle has an infinite extent then it's amplitude is everywhere zero and ... 1 That is the transition dipole moment integral. It is basically the probability that an electric dipole (i.e. a photon) can cause a transition between the states \psi_{nml} and \psi_{n'm'l'}. 4 The von Neumann entropy, written in terms of the quantum mechanical density operator, is a constant of the motion if you keep track of everything (including entanglement with the environment) and don't have any collapse events (which, depending on your favorite interpretation of quantum mechanics, might not exist anyway). The thing is that this fact already ... 0 There exists a very simple and concise semi-classical explanation of the electron's spin angular momentum, without the notion of rotation of any material object: Qualitatively speaking, the electron’s spin angular momentum is the electromagnetic field's angular momentum resulting from the combined electromagnetic field surrounding an electron just in such a ... -1 It's not the same photon but the same energy. After the electron absorbs the energy of a photon it later releases a new photon in a random direction. 0 Firstly this equations applicable for full rotational invariance where potential dependence of r^2 only. Factor 2M can be written. It is a typo. The solution of free Schrodinger equation in 3d is R_{kl}^0=\sqrt{\frac{\pi}{2 k r}}J_{l+1/2}(kr) where J is a Bessel function also for x\gg1 asymptotic of Bessel function has the following form ... 0 No, the only thing you can conclude is that \langle\psi|[H,A]|\psi \rangle =0. Example, for some real constants a,b and for a particle described in L^2(\mathbb R^3) A= aL_x, H=bL_z,$$|\psi\rangle = |\phi(r)\rangle \otimes|l=0,m_z=0\rangle\:.$$In this case$[H,A] \neq 0$but$\langle \psi(t)|A|\psi(t) \rangle =0$for every$t \in \mathbb R\$ since ...

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According to Richard Feynman, the charge is the probability of a particle interacting by the electro magnetic force. More specifically it describes the amplitude of the "probability arrow" of a certain electromagnetic interaction taking place. Much like @Asher has mentioned already, the standard model cannot provide an explanation for why certain particles ...

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There is one, and only one, electron, neutron, proton. If they are in rest to the observer and fields are not involved, the electron always has the identical energy, electric field and magnetic dipole moment. The same we say about the proton and the neutron too. Photons are never in rest, they all have the same velocity in vacuum. Furthermore they have an ...

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The first thing to point out is that there are two equivalent ways to describe a quantum statistical distribution: the density matrix, and a probability distribution on the results of measurements of a "complete" set of observables. (It is remarkable that one should think of givin such a probability distribution on the classical phase space of the quantum ...

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Charge means that the body experiences a force in an electric field. A charge generates an electric field, which generates a force on other charges particles. Two bodies are said to repel if they force each other away and two bodies are said to attract if they force each other closer together. Now, I'm not really answering your question here of "why," I ...

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To add to ArtforLife's answer you are speaking of the famous Quantum Measurement Problem. I disagree with his/her answer a little (even though I upvoted it) insofar that we're not sweeping the question under the rug so much as using abstraction to decouple descriptions from one another so that we can tackle and conceive of one description at a time. In other ...

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Nobody knows. In large part, this issue and question have been swept under the rug for most of the twentieth century physics. If you have ever heard the nostrum of "shut up and calculate" as applied to Quantum Mechanics, you can safely assume that you are being instructed not to ask questions like that. What is more, there is no such thing as a "collapse" ...

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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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1) I should note that most perturbative expansions that are of interest in physics are not formally convergent (and more often than not, not Borel-resummable either). 2) There are many examples of useful perturbative calculations for bosons. The oldest example (probably) in Many-Body physics is the calculation of the energy per particle of the weakly ...

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Just open any string text which has a discussion of the relativistic point particle. http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1 Punchlines: 1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge ...

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This is one of the open questions in Physics. J.S. Bell felt there was a fundamental clash in orientation between ordinary QM and relativity. I will try to explain his feeling. The whole fundamental orientation of Quantum Mechanics is non-relativistic. Even though, obviously, QM can be made relativistic, it goes against the grain to do so, because the ...

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You may be worried about the boundary conditions: the difference between the smooth condition or the discontinuous boundary condition. In general, if you are thinking about the Transmission and Reflection coefficients, a powerful method is so called the WKB (Wentzel–Kramers–Brillouin) approximation. For example, such a continuous potential is solved via ...

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Yes, you could set up a series of multiple SG experiments so that the electrons leaving one experiment entered the next. If you're absolutely determined to use just the one piece of SG kit you'd need some form of magnetic guide to route electrons round in a circle back to the starting point. A bit like putting the SG experiment in the beam line of a (very ...

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I love making small pieces of simulation. I would recommend octave/Matlab for such a simple simulations. To make that point, here is a small piece of code I wrote in 10 minutes with octave/Matlab. It simulates a double slit experiment by solving a 2D Schrödinger equation in a box with a "double-slit source term" using finite difference and Crank-Nicholson ...

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