Angular Momenta and Magnetic Dipole Moments — Question 10

Question 10

A normal Zeeman triplet follows transition energies rather than upper levels alone

Consider an ideal spinless electric-dipole transition from $l_u=1$ to $l_l=0$. Its zero-field photon energy is $E_0$. Use orbital shifts $m_l\mu_BB$ and allowed $\Delta m_l=0,\pm1$.

1. Find the possible emitted photon energies.

2. For $E_0=2.00\,\mathrm{eV}$ and $B=1.00\,\mathrm T$, estimate the magnitude of the wavelength shift of either outer component using $hc=1240\,\mathrm{eV\,nm}$ and $\mu_B=5.788\times10^{-5}\,\mathrm{eV/T}$.

Question 10 - Solution

A normal Zeeman triplet follows transition energies rather than upper levels alone

(a) Method and calculation

$$m_l^{\mathrm{lower}}=0,\quad m_l^{\mathrm{upper}}=-1,0,1.$$$$\boxed{E_\gamma=E_0-\mu_BB,\ E_0,\ E_0+\mu_BB}.$$

(b) Interpretation and further calculation

$$\lambda=hc/E\Rightarrow\delta\lambda\simeq-\frac{hc}{E_0^2}\delta E,\quad\lambda_0=620\,\mathrm{nm}.$$$$\boxed{|\delta\lambda|\simeq\frac{1240(5.788\times10^{-5})}{(2.00)^2}=.01794\,\mathrm{nm}}.$$

Physical check

Transition energies subtract both upper and lower shifts. Here the lower shift vanishes; the shorter-wavelength component has the larger photon energy.