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Good quantum number for the weak field Zeeman effect

Physics Asked by yhmilan on January 7, 2021

To find the fine structure of hydrogen using nondegenerate perturbation theory, we choose the eigenstates of L2, S2, J2, and Jz. As stated in Griffiths Introduction to Quantum Mechanics, the good quantum numbers are n, l, s, j, and mj.

For the weak-field Zeeman effect, the fine structure dominates and the Zeeman effect becomes the perturbation. The book states that the good quantum numbers are n, l, j, and mj. What about s? Why s is explicitly mentioned for fine structure but not for Zeeman effect? Is s suppressed since s is always 1/2?

One Answer

In the presence of a magnetic field $vec B$, the Hamiltonian of this system is given by

$$H = -(vec mu_l + vec mu_s). vec B$$

where $vec mu_s$ and $vec mu_l$ are the magnetic dipole moments for spin and orbital angular momentum respectively.

A “good set” of unperturbed energy eigenstates is given by $| n l j m_j rangle$ and $m_l$ and $m_s$ are considered “not good” quantum numbers, since in the presence of spin-orbit coupling, the Hamiltonian $H$ does not commute with the operators $L$ and $S$. Operators that commute have computable simultaneous eigenvalues.

Note that in the strong-field limit, the unperturbed Hamiltonian does commute with $L^2$, $S^2$, $L_z$ and $S_z$, and so a “good set” of unperturbed energy eigenstates is given by $|n l m_l m_s rangle$.

Probably funny terminology, but that is probably what the author is getting at.

Answered by Dr jh on January 7, 2021

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