# Tag Info

9

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 ...

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 ...

3

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 ...

2

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 ...

1

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 ...

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 ...

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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