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Does it still have rotational symmetry if the elecron has n>1? Doesn't this lead to wavefunction shapes that aren't spherically symmetric? Thank you. (I'm coinc
by the__alchemist 4y ago
Does it still have rotational symmetry if the elecron has n>1? Doesn't this lead to wavefunction shapes that aren't spherically symmetric? Thank you. (I'm coincidentally running into this conundrum while trying to build a chemistry visualizer. Have only attempted for n=1 with the potential being a single proton.)
What about an electron in more complicated potentials, like the ones you'd see in real life vice textbook examples?
- Elpis1 4y agoGood question! What's important is that the Hamiltonian or Lagrangian has these symmetries. Particular solutions having rotational invariance may signify some nice properties of that solution, but it's quite irrelevant to Noether's theorem.
- wnoise 4y agoSymmetry of boundary conditions cashes out in symmetry of solutions in two ways. The obvious one is solutions that are themselves symmetrical. The less-obvious one is families of solutions, where the symmetry maps one solution to another. All of the "s" states are spherically symmetric. The "p" (and higher) states aren't spherically symmetric, but are instead a basis for an entire family of states that are related to each other through rotations.