# Tag Info

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In addition to what @DavePhD says, the Schrodinger model also calculates the angular momentum correctly and shows the angular momentum degeneracy of energy states. A similarity between the results is that the Bohr model orbital radii are equal to the mean radius, $<\psi|r|\psi>$, values of some of the angular momentum states.

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A double well with a high or wide barrier will have a smaller $\Delta E=E_2−E_1$ than one with a low or narrow barrier. (Less coupling.) I think we can understand this intuitively as follows but first it has to be said that rob is right: the energies $E_i$ are NOT inputs but the eigenvalues of the Schrödinger equation. Width, height and potential of the ...

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We don't control the allowed energies $E_i$ independently of the potential: the energies must be the eigenvalues of the Hamiltonian. The "inputs" are the shape and height of the barrier between the two wells. You can kinda sorta think of the energy difference between the symmetric state (with energy $E_1$ in your diagram) and the antisymmetric state (with ...

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Electrons are Fermions so they are forbidden from being in exactly the same state. Two identical quantum wells placed an infinite distance apart will be identical because the wave functions do not overlap. However at the quantum wells are moved closer together the wave functions begin to overlap and the exclusion principle forces the energy levels to split ...

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Supersymmetry Well, for fixed $l$, the degeneracy of $m$ is because of SO(3) symmetry, we are just seeing a full representation of this group. The big question is why all the radial hamiltonians $H_l$ for different angular momenta have the same spectrum except a discrete number of eigenvalues. Note that particularly the tower-spectrum for $l$ and the ...

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Let me do a draft of the "Bohr model" answer that I think is failing in some O(1) factor, but still clarifies further the question. It is unclear how to find a bound for the existence of a bound state, but at least we can try for onset of the existence of a local minimum in the potential. In Bohr's way, model states are simply the equilibrium between force ...

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The field and the wavefunction look similar, but they don't really have much to do with each other. The main point of the field is to group the creation and annihilation operators in a convenient way, which we can use to construct observables. As usual I will start with the free theory. If we want to find a connection to non-relativistic QM, the field ...

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I'm not sure how you got that objective into your head: almost certainly that is not what you introductory second quantization text is telling you to do. Wikipedia has decent summaries of the bridge between QFT and QM, namely real scalar field theory ; multidimensional quantum oscillator; lattice phonon oscillators. The 1+1 quantum field $\psi$ you wrote is ...

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I'm as confused as you by the boxed equation. At best the author is making that all-too-common mistake of reordering the expressions in a transitive equals relation, making the equation nonsensical when read left to right. However, it is not quite a tautology to prove what I think this is trying to prove: If $E$ and $\psi$ satisfy $H \psi = E \psi$, ...

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It's the time independent Schrödinger equation written using Dirac notation.

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I would like to begin by your last question. I am of the opinion that this is correct. I am not versed in systems engineering but I see the equivalence in the mathematical treatments. And QM is very much about the mathematics, is more a descriptive body of knowledge than it is an explanatory. And proof of this is the many interpretations it has, all of ...

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Your Transformation is wrong and it is easy to see that. Your transformed derivative of $\partial_{r_2}$ does not depend on the $r = r_1-r_2$ part of the transformation at all. In other words with your logic I could equally write $\partial_{r_2} = \frac{\partial r}{\partial r_2} \partial_{r} = - \partial_{r}$. The correct transformation is $$\partial_{r_2} = ... 1 You seem to be thinking that the electron has a Coulomb potential energy and the nucleus has a separate Coulomb potential energy. That's not how you should think. Potential energy belongs to the whole system interacting with other parts of the system. Here we have an electron interacting with a proton. That system has potential energy. If we were to add ... 4 Consider two charges q_1 and q_2 kept at some separation. Suppose we want to calculate the potential energy of the system. By definition, potential energy is the work done to assemble such distribution. We can assemble the system in two ways: Bring q_1 to its place; no work done during this as there is no field present. Then bring q_2 to its ... 2 The matrix element \langle x |\psi\rangle where x is allowed to vary is just the wavefunction$$\psi(x) = \langle x |\psi\rangle.$$So the matrix element \langle x |p|\psi\rangle is just the wavefunction representation of the state p|\psi\rangle. We know that the momentum operator in the wavefunction representation acts as a derivative, so ... 2 The "time-independent Schrödinger equation" is just an equation for the eigenvalues and eigenvectors of the Hamiltonian operator on the Hilbert space of states (typically L^2(\mathbb{R}^3,\mathrm{d}x), the "space of wavefunctions") The spectral theorem tells us that the eigenvectors of any self-adjoint operator form a basis for the space the operator ... 2 It is a mathematical theorem that self-adjoint operators in Hilbert space have a complete spectrum. Note that "self-adjoint" has a special mathematical meaning. Not every Hermitian symmetric operator is self-adjoint. For example, the 1D free Schrodinger Hamiltonian on an open interval without boundary conditions is not self-adjoint. The reason is that we ... 2 Who says kets are not functions? A ket can be a function of time: |\psi(t)\rangle. Since they are elements of a vector space (with a complete inner product), you can define their derivative:$$\frac{d|\psi\rangle}{dt} =\lim_{h\rightarrow 0} \frac{|\psi(t+h)\rangle-|\psi(t)\rangle}{h}$$And given a Hamiltonian H : \mathcal{H}\to\mathcal{H}, which is a ... 3 The Schrödinger equation is not limited to any particular kind of Hilbert space. There's no problem with abstract kets. Given a space of states \mathcal{H}, a Schrödinger (or time-dependent) state is given by a (smooth) map$$ \lvert\psi(\dot{})\rangle : \mathbb{R} \to \mathcal{H}, t \mapsto \lvert \psi(t) \rangle so the Schrödinger state ...

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As you have discovered, the H$_2^+$ ion is separable but it is not exactly solvable. Spheroidal coordinates allow you to separate the three-dimensional time-independent Schrödinger equation into three separate one-dimensional Schrödinger problems, one of which is trivial, but that's as far as you can go. On the analytical side, you can do some simple ...

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So far, the only thing that I can be sure of is that there is no solution of the form $\psi(x,y)=g(y)f(x)$, which would lead to an obvious contradiction: If we have a solution $\psi(x,y)==f(x)g(y)$, denote $\frac{f''}{f}=F(x),\frac{g''}{g}=G(y)$, For any fixed x<-a, F(x)+G(y)=constant requires G(y) is constant, similarly, fixed y<-b, requires F(x) is ...

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After searching for a while I finally found a detailed solution to this general Legendre equation. So, I guess this question is answered! The source for the solution is this: http://www.physicspages.com/2011/03/22/associated-legendre-functions/

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To put it another way: typically with this sort of setup we want there to be an incoming wave from the left, $\psi_0 = e^{-i \omega t + i k x},$ which is partially "reflected" into an outgoing wave to the left $\psi_\ell = r e^{- i \omega t - i k x},$ and partly "transmitted" to an outgoing wave to the right, $\psi_r = t e^{- i \omega t + i k x}.$ The ...

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The idea is the following. You have three zones. One before the barrier, $x<0$, with $V(x)=0$ (zone 1), one inside (or above) the barrier $0\leq x\leq a$ with $V(x)=V_0>0$ (zone 2), and one after the barrier with $V(x)=0$ (zone 3). Let's say that a particle comes from the left, with energy $0<E<V_0$, so it will go "inside" the barrier. Then, ...

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