- #1

aliens123

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Suppose we have a hamiltonian $$H_0$$ and we know the eigenvectors/values:

$$H_0 |E_i \rangle = E_i|E_i \rangle $$

We then add to it another perturbing Hamiltonian:

$$H’$$ which commutes with $$H_0.$$ According to nondegenerate first order perturbation theory:

$$\langle H \rangle \approx \langle H_0 \rangle + \langle H’ \rangle.$$

Question: with the extra assumption that $$H’$$ commutes with $$H_0$$ is this an

If the answer is yes, then why is it that we need to consider different types of Zeeman splitting (weak, intermediate, strong)? Taking $$B_{ext}$$ to lie in the z direction gives:

$$H’_Z = \mu_B B_{ext}( L_Z + 2S_Z).$$

This commutes with the original Hamiltonian for the hydrogen atom, so we should be able to solve this new Hamiltonian exactly, and then add in fine-structure as a perturbation. But there would be no need to treat the Zeeman effect as a “perturbation.”

$$H_0 |E_i \rangle = E_i|E_i \rangle $$

We then add to it another perturbing Hamiltonian:

$$H’$$ which commutes with $$H_0.$$ According to nondegenerate first order perturbation theory:

$$\langle H \rangle \approx \langle H_0 \rangle + \langle H’ \rangle.$$

Question: with the extra assumption that $$H’$$ commutes with $$H_0$$ is this an

**exact**result?If the answer is yes, then why is it that we need to consider different types of Zeeman splitting (weak, intermediate, strong)? Taking $$B_{ext}$$ to lie in the z direction gives:

$$H’_Z = \mu_B B_{ext}( L_Z + 2S_Z).$$

This commutes with the original Hamiltonian for the hydrogen atom, so we should be able to solve this new Hamiltonian exactly, and then add in fine-structure as a perturbation. But there would be no need to treat the Zeeman effect as a “perturbation.”

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