I'm studying the hydrogen atom from a quantum mechanics perspective, but I'm having troubles understanding a step.

Consider the stationary Schroedinger equation: $$\hat H \psi = E\psi$$

Let $M$ be the mass of the nucleus, and $m$ the mass of the electron. Then the reduced mass is $$\mu = \frac{mM}{m + M} \approx m$$

Hence Hamiltonian can be written as $$\hat H = -\frac{\hbar^2}{2m}\nabla^2 + P(\vec r) = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\varepsilon_0 r}$$ where $\nabla^2$ is the Laplace operator and $Z$ is the atomic number, so it should be $1$. The next step is the one I fail to understand. Since the potential has spherical symmetry, we change to spherical coordinates. The Hamiltonian is then $$\hat H = -\frac{\hbar^2}{2m}\underbrace{\left(\frac{\partial^2}{\partial r^2} + \frac2r\frac{\partial}{\partial r}\right)}_{\nabla^2} + \underbrace{\frac{\hat L^2}{2mr^2}}_{\text{??}} - \frac{Ze^2}{4\pi\varepsilon_0 r}$$ where $\hat L$ is the angular momentum operator. This passage comes with no explanation whatsoever. Here is what I have trouble understanding:

  1. How can I derive the new expression for $\nabla^2$ in spherical coordinates?
  2. Why is the angular momentum appearing?
  • 1
    $\begingroup$ Item #3 looks like a typo, but since you didn't give a reference for where the last equation came from there's absolutely no way for anyone reading this to tell. Whenever you ask a question, put yourself in the position of your audience and make sure the question can be answered without having to read your mind :-) $\endgroup$
    – DanielSank
    Jan 18, 2016 at 9:05
  • $\begingroup$ $4\pi \epsilon_0 = 1$ in atomic units. So that could explain it, except that $\hbar=1$ also in atomic units, so maybe the units are just wrong, or at least confusing. Your underbrace ?? is the angular part of kinetic energy operator. Your underbrace $\nabla^2$ is the radial part of the kinetic energy operator. That is convenient, because spherical harmonics are the eigenfunctions of L^2, and whole expression can be made angle independent. $\endgroup$ Jan 18, 2016 at 9:06
  • $\begingroup$ @DanielSank I crosschecked this and it is indeed a typo. I'm correcting the question. I don't have a title because it is material provided by the professor. $\endgroup$
    – rubik
    Jan 18, 2016 at 9:18
  • $\begingroup$ @MikaelKuisma But why is the angular momentum appearing only after the change in coordinates? Shouldn't the hamiltonian include that in cartesian coordinates as well? $\endgroup$
    – rubik
    Jan 18, 2016 at 9:20
  • $\begingroup$ You can find this in books about quantum mechanics. A. Messiah "Quantum Mechanics" is a nice reference. $\endgroup$
    – Urgje
    Jan 18, 2016 at 9:55

1 Answer 1

  1. Deriving the expression of the Laplace operator in spherical coordinates: http://digitalcommons.uconn.edu/cgi/viewcontent.cgi?article=1034&context=chem_educ

  2. The angular momentum operator comes from gathering the angular parts $ (\theta, \phi) $ of the Laplace Operator:
    $$ {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left(\sin \theta {\partial \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 \over \partial \phi^2}. $$

In addition if you assume spherical harmonic solutions, you find: $\hat{L}^2 = l (l+1) \hat{1} $ as the operator equality when acting on the spherical harmonics. (https://en.wikipedia.org/wiki/Particle_in_a_spherically_symmetric_potential).

  • 4
    $\begingroup$ The last statement is slightly misleading, $\hat{L}^2\neq\ell(\ell+1)$. However, $\hat{L}^2Y_{\ell,m}=\ell(\ell+1)Y_{\ell,m}$ (with $\hbar=1$ in both). $\endgroup$
    – Kyle Kanos
    Jan 18, 2016 at 12:04

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