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The total angular momentum of a closed shell is zero because for fixed $l$, we have the possible states labeled by eigenvalues of $L_i$ as $m_{l,i} = l,\dots,0,\dots,-l$ in integer steps. The sum over all $m_l$ inside a shell is always zero, so total angular momentum of a shell is zero. This is just the generalization of the argument with "up/down" for ...

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Quantum mechanics, wave function and all that, was not invented out of thin air and imposed on nature. It was experimental results that could not be fitted with the classical theories of the end of the 19th century that created the need for a new mathematical theory/model to explain observations. With classical mechanics we can solve the newtonian ...

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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 $\{\vert ... 4 It must not be greater than 1. To find the probability function you must integrate the probability density,$\psi^* \psi$, over the region in which you want to calculate the probability: $$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$ 3 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 ... 3 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). 2 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 ... 2 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 ... 2 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 ... 2 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 ... 2 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 ... 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 ... 2 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 $$A: V\otimes\cdots ... 2 Note: This is a brief summary. Wikipedia is helpful, if you can't look anywhere else at the moment. It notes that$$\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}$$... 2 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 ... 2 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 ... 2 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 ... 2 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 ... 2 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 ... 1 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 ... 1 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 ... 1 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 ... 1 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 an ... 1 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 ... 1 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 ... 1 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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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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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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Let me first correct you on a point that you make: Wave function collapses into an eigenstate of the measurement operator. The term "classical state" is not standard terminology. Now let me try to answer your question. It seems like you're asking whether there exists a physical reality separate from our observation (measurement). The answer is that we do ...

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However it did pass within Δx of the electron. The Δx is not the difference in space with the electron, as the electron is bound to a nucleus with a potential simulated by "an infinite potential well" . The Δx is related to the whole system, from the center of its mass as a possible location to start with. So the problem is : "photon + atom" as a ...

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