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Virtual particles are not real. They come, as I've said in many answers on this site, from a naive interpretation of Feynman diagrams which should not be taken as an actual, exact description of how the physics works. The actual description of an interaction in the quantum field theory is more complicated than "photons are exchanged". In particular, ...


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One way to facilitate this discussion is to think of what's classically forbidden but quantumly permissible. My favorite so far is a game that I call Betrayal. Let me explain that in this answer. Betrayal: Game Rules The players are a cooperative three-person team, they will either all win or they will all lose. They will be put through some number $N \gg ...


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So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $c_+^2 + c_0^2 + c_- ^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are ...


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Any one of operators Lx, Ly, or Lz can be called quantized. There exists a set of states which are eigenstates of Lz; the matrix Lz is diagonal but Ly and Lx are not. There exists a set of states which are eigenstates of Ly; the matrix Ly is diagonal but Lx and Lz are not. There exists a set of states which are eigenstates of Lx; the matrix Lx is diagonal ...


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I still don't quite understand the reasoning behind the conclusion that entangled particles somehow can communicate their state to each other instantaneously, even though they are separated by a substantial distance This isn't correct, they occupy a joint state. From what I gather [...] upon observation of one of the particles, it immediately (and ...


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Air tight and 'matter' tight are two separate issues. A container can be considered air tight even if it leaks slightly, and this can be tested via a simple pressure gauge and time in a vacuum chamber. Being 'air tight' would mean, by definition, 'air' consisting of mostly $\text{N}_2$, $\text{O}_2$, $\text{CO}_2$ and a few others cannot escape. This is ...


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I suppose that what is meant is the following: We can consider the neutron and proton as 2 states of the same particle, the nucleon N (regarding the strong interaction, not electromagnetism). Since neutrons and protons are fermions, the wave function of 2 identical particles (here 2 nucleons) must be anti-symetric because of Pauli principle. If the 2 ...


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Some time ago I have asked a similar question; although I have accepted one of the answers, It did not satisfy my main interest concerning the physics of the case when the diagonal $\Delta$ is not removed. Now I have some more information that I can share with you. Most of the authors give two reasons for the removal of the diagonal: If we include the ...


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The atom has some charge distribution $\rho(r)$. We don't don't know what form the function $\rho(r)$ has, but we do know it depends only on $r$ because an atom is spherically symmetric. When you have a spherical charge distribution the potential at a distance $r$ is simply due to the total charge inside the distance $r$: $$ V(r) = ...


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this is a description of an interaction between the electron and photons, which would collapse the wavefunction (right?). No this isn't right. As long as the system stays isolated, the interaction simply means that there are cross terms in the relevant Hamiltonian and that you have a two-particle quantum system, whose state space is the tensor product ...


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We usually assume that the screen, $L$, is much further away than the distance $d$ between the lattice planes. Then the lines which converge on a point are NEARLY parallel, although they are not quite parallel. Then we make the approximation that they ARE parallel. If you work through the math carefully, you'll see this approximation only changes the length ...


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Let's look at where the electromagnetic interaction comes from in hydrogen. At first quantization you have a multiparticle system so the wavefunction is defined as $\psi=\psi(x_1,y_1,z_1,x_2,y_2,z_2,t)$ and the point is to write the Hamiltonian. And the Hamiltonian comes from the Lagrangian. For a single particle of charge $q$ in an external ...


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Your integration is wrong. The probability density function measures the probability of finding the particle between $x$ and $x+dx$. If you integrate over $0 \leq x \leq L$ you don't get a function of $x$ but a number instead (one for this interval). If you state that $P_n$ is the integrated density, then it should depend on the interval in which you did the ...


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The discreteness of the allowed orbitals is a consequence of quantum mechanics, which was conceived precisely to explain this observation, among other things. However, the orbitals are not orbits - there is no "motion" in the classical sense going on, and an electron in an orbital does not have a fixed distance to the nucleus (it may even have non-zero ...


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Imagine you stuck a detector in front if the target and it went off a certain number of times per second (on average) per square meter of cross section. If you then placed it somewhere else at an angle from the forward direction it (the detector) would also go off a certain number of times per second (on average) per square meter of cross section. Super. ...


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Given two observables $A$ and $B$ such that $[A,B] = iC$, the most general form of the uncertainty principle is $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega(C)|,$$ where $\omega$ is any state of the algebra of observables. By the Riesz-Markov theorem, there is a regular probability measure such that $$\omega(f(A)) = \int_{\sigma(A)}f(\lambda)\ \text ...


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It must not be greater than 1. To find the probability function you must integrate the probability density, $\psi^* \psi$, over the region in which you want to calculate the probability: $$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$


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The terminology "continuous variable system" is non-standard, but likely refers to the fact that any canonical quantization of a classical Hamiltonian system (i.e. a system described by a continuous phase space) must have an infinite-dimensional Hilbert space since the canonical commutation relation $$ [x,p] = \mathrm{i}\mathbf{1}$$ cannot be realized on ...


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The linkage isn't particularly spooky. Record all of the flashes from the 2-slit pattern, and store the polarization for each one in some long-lived qubit. Peek at the flashes you see, and you'll see two overlapping bell curves, the "no interference" case looking like $|f(z - z_1)|^2 + |f(z - z_2)|^2$. If you measure the qubits in one basis -- let's say ...


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There is no resistance is this scenario. You need a scattering mechanism like electron-phonon scattering or electron-defect scattering to obtain a non-infinite conductivity.


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If $[A,B]=0$ then there is a unitary transformation $U_{A,B}$ that diagonalises both $A$ and $B$ simultaneously. This transformation depends on the pair of commuting operators $(A,B)$, so that for a different pair there could be a different unitary. Assume that all the eigenvalues of $H$ have multiplicity 1. Then there exists a unique (up to a phase factor) ...


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If $H$ commutes with $A_1$, then it will indeed share an eigenbasis with it. Your mistake is in supposing that it will share the same eigenbasis with both $A_i$s. Examples are easy to provide: On the trivial side, if $H=E_0\mathbf 1$ is trivial, then it shares an eigenbasis with $A_1=x$ and it shares an eigenbasis $A_2=p$, but it cannot share an ...


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There's kind of an indirect meaning to it. Suppose you have a photon flying along. Photons seem to experience infinite time dilation and so there's no oscillating thing that the photon is "carrying with it" to have its particular frequency: rather that frequency comes via some sort of interaction with its surroundings. Phase factors are how that sort of ...


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There is no physical relevance of a phase in front of a phase vector, as this is unobservable, hence unphysical. In the geometric approach to quantum mechanics this can be viewed as a gauge freedom that can be used to reduce the total Hilbert space to the quantum phase space, i.e. the projective Hilbert space (equipped with a natural Kähler structure).


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A 1-particle Hilbert space (neglecting spin for simplicity) is usually modelled as $L^2(\mathbb R^3)$ which is a function space, and the Hamiltonian is a differential operator. N-particle Hilbert spaces are usually constructed as tensor products of this 1-particle Hilbert space, but there exists an isomorphism such that you can again interpret them as ...


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The expression $$ \hat{H}=\sum_j \varepsilon_j\,a_j^\dagger a_j $$ is not the most general expression for free particles hamiltonian because it implies that you already found the eigenvalues $\varepsilon_j$ and diagonalized $\hat{H}$, i.e. already solved the Schrödinger equation. Maybe you should look at the problem in a different basis. Let say $\{\vert ...


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The second-quantization hamiltonian you wrote above is the hamiltonian for a collection of independent harmonic oscillators, where the quantities oscillating are now the amplitudes of the normal modes of the field being quantized. So, yes, the hamiltonian is a differential operator on the space complex functions of $N$ variables, one for each normal mode. ...


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Strangely enough, when you do a quantum mechanics experiment, you get a result that says something about what you already know. If you place a detector at one (or both) of the slits, you watch individual particles go through the slits, and you get a particle result from your detector (e.g., no interference pattern). If you don't know which slit the ...


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No, if you observe which slit they traveled through then there is NOT an interference pattern. The act of observing, or more accurately, the need for the location of the electron to be resolved causes it to take on a definite position and then continue on from that position as a particle. If it is not observed or interacted with in some way that would make ...


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Atoms do in fact have a sort of wave behavior you might say. Everything with mass does, even you! When the mass is small enough, like that of an electron or an atom, this behavior becomes more important to take into consideration. For example, when we go to look for an atom by shining light of a small wavelength on it, we can only say with a certain ...


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This is just a short expansion of Ernies comment (answer really) above, same reference, and the only thing I want to add is the size of the molecules, not just atoms but 58- and 114-atom molecules, made of links of carbon, hydrogen and nitrogen. $\mathrm{C}_{60}$ Fullerene Double Slit experiment and Neutron Interference Pattern both provide details of ...


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"What aspect of quantum mechanics makes it natural to say that probability and 'non-determinism' should take a central role?" The fundamental reason is philosophical and comes from the Teleportation thought experiment, which raises the question of what happens to subjective experience after an hypothetical teleportation device avoids deleting the ...


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Electron as a standing wave Yes, the electron is a standing wave. See atomic orbitals on Wikipedia: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". I couldn't understand how come Bohr who interpreted electron as a particle, formulated an equation for electron's angular ...


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I now come to my point, why one restricts a particle's motion to some discrete set of distances? Is it to provide a theory on the particle's stability? An attempt at an answer to your first question anyway. The electrons surrounding an atom need to obey energy level (and other) rules. As you mention distance, if you imagine that the further away the ...


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Another, better way, (because imo, the wiki line you quote is a bit obscure) of dealing with your question is based on John Baez's Notes: The Compton wavelength of a particle, roughly speaking, is the length scale at which relativistic quantum field theory becomes crucial for its accurate description. A simple way to think of it is this. Trying ...


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Rather than using a magnetic field, you are better off using a strong electric field to separate them - since the initial direction of the electron / positron is somewhat random, a magnetic field will deflect but not separate in a meaningful way. An electric field can pull the positrons one way, and the electrons the other way - regardless of their initial ...


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Note: This is a brief summary. Wikipedia is helpful, if you can't look anywhere else at the moment. It notes that $$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$ because of the de Broglie relation, $$p=\hbar k$$ where $p$ is momentum and $k$ is the wave vector. de Broglie's equations, in turn, relate to the de Broglie wavelength, $$\lambda = \frac{h}{p}$$ ...


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Consider a $N$-dimensional vector space $V$ and let $\{\chi_1, \cdots, \chi_N\}$ be basis of $V$. Next focus attention on the anti symmetric space $(V\otimes\cdots \otimes V)_A$ where $V$ occurs $M\leq N$ times. A basis of $(V\otimes\cdots \otimes V)_A$ can be constructed out of $\{\chi_1, \cdots, \chi_N\}$ making use of the projector $$A: V\otimes\cdots ...


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Light changes wavelength in the presence of gravity I'm sorry Alex, but actually, it doesn't. You know how if you're moving away from a light source you see a redshift? The photon energy appears to reduce and the wavelength appears to increase? Well the photons coming from that light source haven't changed one iota. Instead you've changed. You also ...


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In general taking $B_-=0$ (or some similar aproach) does not solve this problem because you already have 2 equations and you have to add also the normalization condition, therefore you have 3 equations and you need all the constants. So rewritting the wave function in terms of only one constant in the right path.


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To answer your last 2 questions: "Are there bounds on optimal encoding, whose purpose would be to delay a release of a finite fraction of the information?" and "Can there be encoding which releases a finite fraction of the information only at the very last stages of the decoding?" It's not a potential purpose of optimal encoding, but a required ...


2

A wavefunction of a quantum system is the system's state written in a particular form - no more nor any less than that. Here the word state has an exactly analogous meaning to the state of a classical system insofar that the state at any time uniquely defines the state at any other time and contrariwise. The state's evolution with time in either a quantum or ...


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The "wave function of an orbital" is just referring to "the wave function for electrons the corresponding state". When speaking of electron levels in an atom, it is commonly accepted the Copenhagen interpretation of Quantum Mechanics, where a wave function of an isolated system, contains all the information relevant for its full description. Rigorously you ...


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In quantum mechanics we use a wave function to describe the quantum state that an electron is in. It basically tells you where you are likely to find a particle. Notice how it is typically introduced as a function of position. That is because by taking the value at a certain location and squaring it we get the probability of finding it there.


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After alpha decay, the alpha particle can be thought of as a doubly positively ionized helium atom, and the parent atom is now a doubly negatively charged ion. Under normal circumstances, the two ions will eventually neutralize their charges. In a near vacuum, this may take some time. In a crystalline solid, the nuclei share an electron cloud, so the two ...


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It is complicated and we ignore it, but your intuition is right. When the nucleus loses an alpha particle its charge decreases by two. The atomic physicists now claim their job is done and don't care. The solid state physicists don't consider radioactivity, so they don't care either. If it is an atom floating freely in space, two electrons will move off ...


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You're missing the very last step: averaging the intensity over time. Since there's a $t$ inside the argument of the sinusoid, the contribution of that term is zero. Averaging over time is always done in these kinds of problems, though often implicitly. For example, if you just have one slit, the electric field might be proportional to $\cos(\omega t)$, in ...


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As noted in some of the comments already, the answer to your question depends on what properties of quantum eigenstates you want your classical analogues to have. If you're thinking of eigenstates of arbitrary observables, quantum eigenstates have the property that the value of that observable is precisely defined. In classical mechanics, individual points ...


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This is because usually a electron can ONLY stay in certain energy states in a given atom, this is because of the quantum mechanical forces such as strong force, and Heisenberg uncertainty principles. Let me explain further, as electron gets closer to an atom the electrostatic & other quantum mechanical forces between the electron starts pulling it, and ...


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If the pencil were at absolute zero it would necessarily assume its lowest energy state, which is not the vertical state. If the pencil were modeled as a quantum rotor with an infinite potential barrier covering half its solid angle space (i.e. the table) then there are certainly excited but stable states where the pencil remains in a more or less vertical ...



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