Riemann surface











问题 1.

Give an example of a metastable state in hydrogen. Why is it metastable? Label this state as appropriate for $L S$ coupling (4)

Answer: The 2 s electron configuration of hydrogen is metastable. It is metastable because it is not the ground state (or lowest energy state) $1 \mathrm{~s}$ and given time it will decay to the ground state but at a rate much slower than say $2 \mathrm{p}$. The reason for this is that the selection rules for electric dipole radiation state that $\Delta \ell= \pm 1$. The transition from the $\ell=02 \mathrm{~s}$ state to the $\ell=0$ is state would have $\Delta \ell=0$ so it cannot decay by electric dipole radiation.
In the $L S$ coupling scheme this state would be labelled $2 \mathrm{~s}^2 \mathrm{~S}_{1 / 2}$.

问题 2. What is fine structure? Physically (not just the formula) why is the degree of fine stucture splitting so much bigger in lead than in hydrogen? Referring to Fig. 1.3 where is fine structure evident? How big is it in this case? (6)

Answer: Fine structure arises from an extension of the Hamiltonian to include relativistic effects. A proper solution would involve the Dirac equation but it possible to class them as the relativistic mass correction, spin-orbit interaction, and Darwin term (for $\ell=0$ states). (Note: the Lamb shift is not predicted by the Dirac equation; you need to quantize the electric and magnetic fields as well to calculate it).Our studies of relativity have told us that the first order relativistic corrections are on the order of $\frac{v^2}{c^2}$ (consider the $\gamma$ factor of special relativity). The inner electrons in lead move much faster than they do in hydrogen so we expect larger corrections. We also expect relativistic corrections to break the degeneracy with respect to $\ell$ (recall our discussion of Sommerfeld’s ellipitical orbits). That is exactly what we see in Fig. 1.3. The innermost or $K$ shell with $n=1$ only permits $\ell=0$. The $n=2$ or $L$ shell can have $\ell=0$ or $\ell=1$ and the degeneracy is broken. The overall splitting is between 3 and $4 \mathrm{keV}$. (I believe that the 3 levels correspond to a Lamb shifted $2 \mathrm{~S}{1 / 2}, 2 \mathrm{P}{1 / 2}$, and $2 \mathrm{P}_{3 / 2}$. Remember they are flipped upside down on this diagram.)

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