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0

It is more difficult to think about the free particle case, because it is unnatural to put it on a lattice and not gap the Brillioun Zone edges. However, if you do want to do such a thing, the dispersion would look like (b) in the figure below for an arbitrary choice of lattice spacing $a$: Why is this so? It is because you have made something in real ...


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Bell's argument makes very weak assumptions about the behavior of the two particles (which is why it's interesting). In effect, the particles are black boxes that take an angle as input and produce a spin direction as output. There is no restriction on how they choose the spin direction; there could be a source of true randomness in there, or a human being ...


1

I think you misunderstood the significance of could for a classical theory. The text below the picture you took from Wikipedia says: "Many other possibilities exist for the classical correlation subject to these side conditions", so classicality does not imply linearity. It does, however, rule out the cosine, by the following (slightly heuristic) argument: ...


1

The formula $$ E = \frac{\hbar^{2}}{2m} |k|^{2}$$ is valid in the vicinity of $k = 0$, this is called parabolic band approximation, which is - as the name suggests - only an approximated formula. Regarding the actual calculation of energy, you need to calculate the eigenvalues of the Hamiltionian of the periodic system, then the theorem is trying to say that ...


0

Your question is really tricky. I am going to try to discuss two of the issues in the first two paragraphs. The rest of the paragraphs will speak only purely formally about how "old" classical cosmological equations would work in the case you describe and what a physically intuitive interpretation of the singularity could be. First, without quantum ...


1

The question is very speculative :-) Physics is based on models out of our experience of a world extended in space and time together with mass and energy. One example: The concept of point is already a Euclidean idealisation. It lives in the world of mathematics - not of Physics (otherwise, you fall into Democritus and QM) Indeed, in Maths, we still struggle ...


0

I would bet that consciousness is an emergent property of our neural network and not a quantum mechanical effect. Quantum effects are most probably washed away very quickly by the thermal bath in our brain.


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The Free will theorem of Conway and Kochen is simply an unfortunately titled theorem that, just like Bell's theorem, rules out a certain kind of hidden variable theory, i.e. shows that measurement results cannot be ultimately determined (if it is not somehow determined what measurement will be made, that's the "free will" of the experimenters), but are ...


2

(Disclaimer: The more rigourously inclined individual may be better suited by looking at the Stone-von Neumann theorem, as Qmechanic notes) One can deduce that the momentum operator takes the form $\hat p = -\mathrm{i}\hbar\partial_x$ in the position representation from the fact that the momentum operator generates the infinitesimal translations as ...


2

I) Comment to the question (v1): The Schrödinger position representation $$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{\partial }{\partial x^k}, \qquad \hat{x}^j ~=~x^j,$$ correctly reproduces the canonical commutation relations $$ [\hat{x}^j,\hat{p}_k ]~=~i\hbar ~\delta^j_k ~{\bf 1}, $$ while the proposal $$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{1}{x^k}, ...


-1

From the askers perspective, the explanatory powers of most of these answers seem pretty bad. I prefer Emilio Pisanty's answer here: Why isn't Hydrogen's electron pulled into the nucleus? because it explains exactly how the uncertanity principle dictates the facts of this atomic reality. The summarized problem is that, if the charged and attracted ...


0

Here's the Feynman response I read in his opening paragraphs in his Feynman lectures: The reason a proton and electron simply don't crash into each other is that if they did, we would exactly know their position-assuming one of them is stable, which one is (the proton). If we knew their position, we would be highly unaware of the momentum, meaning it could ...


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May I understand that, since the domain is 2D, the surface integral is actually the following: $\vec{E}\cdot d\vec{\sigma}' = \vec{E}\cdot\hat{n}dl$, where $\vec{E}$ is an arbitrary 2D vector, $\hat{n}$ denotes the normal of the domain boundary and $dl$ is the line element ?


0

Let us consider for a moment just a classical scalar field before quantization. For this field we have a Fourier decomposition: $$\phi = \int \frac{d^3 \!p}{...} (a_{\vec{p}}e^{ipx} + b^*_{\vec{p}} e^{-ipx})$$ Where $a_{\vec{p}}, b^*_{\vec{p}}$ are just numbers and the complex conjugation of $ b_{\vec{p}}$ is just a convention. In this description, the ...


0

Electrons are not waves. Neither are they little dots of matter. The wave/particle duality means that some properties of electrons are as if they are a wave, and some are as if they are particles. They are neither, and they "travel" exactly as the hybrid quantum object they are: Unless they interact with something, they do not take on any definite ...


0

I don't know of polarizing x-ray beam splitters, and I suspect they don't exist, but I'm always very interested to be shown wrong. It appears from a cursory search that working x-ray interferometers have been built, and are conceptually more similar to neutron interferometers than to optical interferometers. For neutrons there are polarizing supermirrors, ...


1

Your book is correct. Remember $ \psi $ is the wave function that represents the state of the system and is the solution of the Schrodinger's equation. Now, the Schrodinger's Equation is a linear partial differential equation (PDE) and the solution has several interesting properties: It has infinitely many particular solutions and we consider only those ...


0

While both the answers given in some sense are correct, the true reason has to do with energetic considerations. It is a matter of what is stronger and can be phrased as the following question: Will the wavefunction alter itself to accommodate the flux, or will the flux quantize itself because the wavefunction is trying to remain single valued? As an ...


2

The adiabatic theorem is required to derive the Berry phase equation in quantum mechanics. Therefore the adiabatic theorem and the Berry phase must be compatible with one another. (Though geometric derivations are possible, they usually don't employ quantum mechanics. And while illuminating what is going on mathematically, they obscure what is going on ...


2

As I showed in my article http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.), in a general case you can use one real function instead of the four complex components of the Dirac spinor, as 3 out of four components can be algebraically eliminated from the Dirac equation, and the remaining component can be made ...


3

Here is some reasoning for representation both of particles and antiparticles by one (complex in general, not only in usual sense but also in a sense of the irreducible representation of the Lorentz group) field. All of QFT processes are described by S-matrix, which can be written in a form $$ S_{\alpha \beta} = \langle \beta | \hat{S} | \alpha \rangle , ...


2

This Question is Beyond the Scope of This Site This subject takes much, much more room to precisely answer than a single post can do here. People get degrees to answer this question. A single post here will not answer your question with exactness. A few years studying physics and then going on to study material science- that will answer your question as ...


0

Silicon is interesting because it is a semi-conductor. That is, it can be used to build transistors and other devices. Moreover, with the appropriate techniques the same silicon wafer can be used to build transistors, diodes, resistances, capacitors and electrical connections. Now, transistors, diodes and resistances can be connected to build logical gates. ...


0

The TWIN axiom just says that entanglement exists (at any distance), so it is trivial in proper quantum mechanics. The relevant entangled state is of the form $$ |\Psi \rangle = \sum_{m=-j}^j |m\rangle \otimes |-m\rangle $$ or something of this form (phases, sign flips). The SPIN axiom is also simple in proper quantum mechanics. It says that the eigenvalues ...


3

Such operators are ill-defined in an interacting theory because whatever counterterms we try to subtract, their expectation value in any finite-energy state will diverge. The closest operators that are well-defined are densities of charge – number operators with signs labeling antiparticles – because the divergent contributions naturally cancel for them. ...


0

In fact, your question is not so clear. I try my best. The Yukawa potential is an exchange potential, so it is based on the particle which is exchanged between 2 interacting particles. So if that particle (a boson) is coupled to the Higgs boson, it will get mass and the potential between the 2 interacting particles should change from U(r)~1/r to U(r)~ ...


1

In your previous question, you say that these two images (a)  (b)  have been taken of the same area of your sample but with a different lens and lighting configuration. I think that I believe this: in this first image, there are two dark areas (the left shaped sort of like New York State, and the right shaped sort of like a heart), and in the second ...


2

Feynman said the following about QM: The difficulty really is psychological and exists in the perpetual torment that results from your saying to yourself, "But how can it be like that?" which is a reflection of uncontrolled but utterly vain desire to see it in terms of something familiar. I will not describe it in terms of an analogy with something ...


0

I'm not sure what you're asking, but it's not really true to say that particles interact with the Higgs field. A quantum field like the electron field interacts with the Higgs field and the result is that the electron field is massive i.e. its excitations (electrons) have a mass. If you consider the Higgs boson, rather than the Higgs field, then the ...


1

Consider a system in the state $$|\psi\rangle=\sum_kc_k|\psi_k\rangle$$ Suppose that this system couples to another system. The coupling will leave unchanged some set S of commuting observables of the original system and will change the others. In general, unless the system happens to be in an eigenstate of that observable this will change the expectation ...


1

You can find derivation of these operators in most standard quantum mechanics textbooks. For your convenience, see https://en.wikipedia.org/wiki/Momentum_operator and https://en.wikipedia.org/wiki/Position_operator. For the second question, Paul Dirac said in his classic The Principles of Quantum Mechanics: A measurement always causes the system to jump ...


2

The plane wave – the wave function for a particle moving in a clear direction and with a sharp momentum, $\vec p$, is $$ \psi (\vec x) = C \exp (i\vec p\cdot \vec x / \hbar ) $$ where $\exp(ia)=\cos a + i \sin a$. However, the dependence of an $s$-wave wave function on the angular direction is trivial (constant), $$\psi_s (r,\theta,\phi) = \psi(r). $$ ...


1

Let $A$ hermitian operator corresponding to an observable. If $\psi$ is an eigenfunction of A, then $$ A\psi = \lambda\psi $$ We say: $A$ has the value $\lambda$ on $\psi$. If it's not an eigenfunction, then $$ A\psi = \lambda_1\psi_1 + \lambda_2\psi_2+\dots $$ That is, after the measurement the state changes and you cannot associate a definite value ...


2

The former and the latter are really the same: "$c_n=\psi(x)$". If you want to measure positions, then possible outcome states are $|x\rangle$, therefore you write $$ |\psi\rangle = \sum_x|x\rangle\langle x|\psi\rangle:=\sum_x\psi(x)|x\rangle :=\sum_xc_x|x\rangle $$ This tells you, the probability to find the particle at position $x$, i.e. to measure it in ...


4

It's impossible for an electron not to encounter some electromagnetic radiation. Even if we try and screen our experiment by putting it in a Faraday cage there will be some black body radiation from any parts of the equipment not at absolute zero. What matters is that the energy of the photons must be much less than the energy of the electron. In that case ...


0

As far as I know, local hidden variables have not been ruled out experimentally so far. Not without the so called loopholes. For what it's worth, Zeilinger called those loopholes "essential" (arXiv:quant-ph/9810080v1 (Phys.Rev.Lett. 81 (1998) 5039-5043)).


1

If you want to stick to a n-particle-interpretation, for the electron, your answer is the Dirac Equation. However, it breaks down as soon as you reach sufficiently large energies, a unified description is only possible within quantum field theory. For the photon, the situation is more difficult as it only appears as excited state of the electromagnetic ...


1

If you are looking for intuition about the subject, read Dirac's Principles of Quantum Mechanics. You need to read §2, §3 and the beginning of §10 until you see fit.


1

Here's a simple-minded answer: Let's just compute the momentum of a particle with a Bloch wave function $$\begin{eqnarray} \left.\langle x \right| \hat{p}\left|\Psi \rangle\right. &=& -i\hbar \left(\frac{d}{dx}\right) u_k(x) e^{i k x} \\ &=& -i \hbar \left( i k u_k(x) e^{ikx} + u_k'(x)e^{ikx}\right) \\ &=& \left( pu_k(x) - i\hbar ...


2

The idea of the question is to find the temperature at which the average interparticle spacing is equal to the average de Broglie wavelength. Both of these are averages because the atoms of the ideal gas are not evenly spaced and the velocity (and therefore de Broglie wavelength) of the ideal gas atoms follows the Maxwell-Boltzmann distribution. So this is ...


2

"The size of a single gas particle" is a bad term for what you calculate. A better term would be: "The volume one of the gas particles would occupy, if the total volume were distributed equally among all gas particles" And this translates loosely to "The volume, in which you find one gas particle on average" If you then imagine every particle sitting at ...


0

I will reply to this: why is then frequency defined for matter waves(in other words what is its use?)? Frequency is defined for electromagnetic waves. When the photon was discovered and the theory assigned to it an elementary particle identity, on par with electrons and protons (at the time ) it was found that the frequency of the electromagnetic wave ...


-3

I think that matter wave could be both longitudinal as well as transverse!!! Even if, by this time, any scientist anywhere proved that it is either longitudinal or transverse, may be corrected by somebody else in future with this change - it is both longitudinal as well as transverse. As we know generally, the concept of matter wave is just the idea of ...


0

There are numerous applications of quantum tunnelling. A few that pop in my mind right now are: Radioactive decay: Particles tunnel out of the nucleus of which they are bounded by a potential. Classically this is forbidden as the nucleus is very strongly being held together by strong nuclear forces. Scanning tunneling microscope: A scanning tunneling ...


1

Well, first of all, it is important to realize that the integrals (1) and (3) are not merely ordinary double integrals over a single $x$- and a single $p$-variable. Instead they are (Wick-rotated) path integrals containing, heuristically speaking, infinitely many integrations. The path integral derivation of the free particle and the harmonic oscillator ...


1

(Not really an answer, but as one should not state such things in comments, I'm putting it here) You commented: "This seems to boil down to the relationship between the phase space and the Hilbert space." That's a deep question. I recommend reading Urs Schreiber's excellent post on how one gets from the phase space to the operators on a Hilbert space in a ...


0

For very long times, a decay process starts to compete with the inverse process. For instance, right now you are bathed in an ocean of matter and antimatter neutrinos with lots of different energies. For a given beta-decaying nucleus, some fraction of these background neutrinos will have enough energy to drive the inverse decay process, transforming the ...


0

As for the derivation of the Fermi golden rule, there is a crystal clear one by myself: http://arxiv.org/abs/1404.4280 There is no hand-waving argument at all. It is completely rigorous in the mathematical sense. Of course, it is based on some assumptions on the continuum spectrum and the couplings. In talking about Fermi golden rule, one must keep in ...


0

So my question is, is it like future of an isolated system is already determined but is just not perfectly predictable by an observer because of limitations in observability? In quantum mechanics the outcome of measurements is in general not determined when given the exact physical state of the system in general. But apart from the uncertainty arising ...


-4

The universe cannot be predicted from a single data point about a moment in time because inertia does not exist in any one moment but is critical to how a system would develop.



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