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Physics
Quantum Physics
Question about excited Helium states
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[QUOTE="king vitamin, post: 6108263, member: 134222"] Yes, that is correct - any linear combination of the four eigenstates will also be an eigenstate with the same energy. But the particular choice you've made in your post is a rather special one. You are ignoring electron-electron interactions and fine structure (effects due to electron spin), but if you were to take those into account, you would in principle need to use degenerate perturbation theory, which involves diagonalizing your degenerate subspace with the perturbed Hamiltonian. But the perturbing Hamiltonian conserves total angular momentum (the sum of the electron spins in this case, since these are [itex]/ell=0[/itex] states), so you know that the correct basis is the one which are eigenstates of [itex](\mathbf{S}_1 + \mathbf{S}_2)^2[/itex] and [itex]S^z_1 + S^z_2[/itex]. The is precisely the basis you've written down, so to leading order we expect the actual Helium atom with electron-electron interactions will actually split these four states into a single nondegenerate state (the first one you've written down) and a threefold degenerate state. [/QUOTE]
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Question about excited Helium states
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