# 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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I wanted to complement the answers above. For (1) $so(4) = so(3) \times so(3)$, one $so(3)$ is from the geometric 3D symmetry of the Hamiltonian, and the other $so(3)$ is from the potential term of $\frac{k}{r}$. For (2). the second $so(3)$ symmetry is a dynamic symmetry and only holds when potential term is inversely proportional to $r$. One has to do ...

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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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Emilio Pisanty has already given a good answer. Here we offer a qualitative (as opposed to quantitative) proof of the angular momentum dependence. Recall first of all that the energy-levels $$\tag{2} E_n ~=~-\frac{R_{\mu}}{n^2}$$ in the non-relativistic hydrogen atom without spin-orbit interactions are linked to the principal quantum number ...

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The $E$ in your expression is the quantity calculated using the operator $i\hbar\frac{\partial}{\partial t}$, so it is the total energy. As $n \rightarrow \infty$ this energy $E$ goes to the rest energy of the electron $m_ec^2$ as we'd expect. For finite $n$ the energy is lower than $m_ec^2$ with the difference being the binding energy of the electron. To ...

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I have found the answer myself here Energy in Dirac model $E_d$ is related to energy in Bohr's model $E_b$ as $E_b \approx E_d - m_ec^2$ where $m_e$ is mass of electron and $c$ is speed of light. The answer above is not useful.

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This is a tricky bit of intuition to get right. In essence, having a lower angular momentum expands the radial range that the electron is allowed to span - the inner turning point moves inward and the outward turning point moves outward - but the electron is moving much slower at the outward turning point, which means that it spends more time there and ...

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The hydrogen atom has an infinite number or quantum mechanically allowed energy levels, as explained on this web page. Using that same link, scroll up the page a bit to better understand how transitions between these energy levels give rise to absorption or emission of photons of very specific frequencies. Then scroll further down to see how the hydrogen ...

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The most simple way of solving this is using Gauss' law and spherical symmetry. Then $$\int_{\delta \Omega} \mathbf{E}\; \text{d}\mathbf{F} = \int_{ \Omega} \nabla \mathbf{E} \; \text{d}V = \int_{ \Omega} \frac{\rho}{\varepsilon_0} \; \text{d}V$$ In the case of spherical symmetry this can be simplified to $$\int_{ \Omega} \mathbf{E}\; \text{d}\mathbf{F} = ... 1 Your Transformation is wrong and it is easy to see that. Your transformed derivative of \partial_{r_2} does not depend on the r = r_1-r_2 part of the transformation at all. In other words with your logic I could equally write \partial_{r_2} = \frac{\partial r}{\partial r_2} \partial_{r} = - \partial_{r}. The correct transformation is$$\partial_{r_2} = ...

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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 ...

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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 ...

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