Estimating Atomic Transition Strengths

Short Topical Videos

• Estimating the Strength of Atomic Transitions (Parsons, YouTube)

Estimating the Strength of Atomic/Molecular Lines

1 Einstein A’s for $Ly\alpha$

${A_{21}}$ is a measure of the probability of decay per unit time, so ${A_{21}}^{-1}\sim lifetime$. This should be about equal to the energy of the electron state divided by the average power radiated by an electron being accelerated:

${A_{21}}^{-1}\sim {E \over P}\sim {\hbar \omega _{0} \over {2 \over 3}{e^{2}{\ddot {x}}^{2} \over c^{3}}}\sim {3\hbar \omega _{0}c^{3} \over 2(e{\ddot {x}})^{2}}\,\!$

Now $e\cdot {\vec {x}}={\vec {d}}$ (the electric dipole moment) and ${\ddot {x}}\sim \omega _{0}^{2}x$ for a spring, so:

${A_{21}}^{-1}\sim {3\hbar \omega _{0}c^{3} \over 2d^{2}\omega _{0}^{4}}\,\!$ ${{A_{21}}\sim {2d^{2}\omega _{0}^{3} \over 3\hbar c^{3}}}\,\!$

For H: $d\sim ea_{0}$ and $\lambda _{L}y\alpha =1216\mathrm {\AA}$ so:

${A_{21}}\sim 5\cdot 10^{8}s^{-1}\,\!$

2 Magnetic Dipole for $Ly\alpha$

The magnetic dipole of an electron is:

$\mu _{e}={e\hbar \over m_{e}c}\,\!$

Thus we can estimate the ratio of ${A_{21}}$ for magnetic dipole transitions to that of electric dipole transitions:

${{A_{21}}{\big |}_{mag} \over {A_{21}}{\big |}_{elec}}\sim \left({\mu _{e} \over d}\right)^{2}\sim \left({e^{2} \over \hbar c}\right)^{2}\sim \alpha ^{2}\,\!$

This tells us that the magnetic dipole states (that is, fine and hyperfine states) are longer lived than electric dipole states by a factor of $\alpha ^{2}$.

${{A_{21}}{\big |}_{21cm} \over {A_{21}}{\big |}_{Ly\alpha }}\sim \alpha ^{2}\left({1216\mathrm {\AA} \over 21cm}\right)^{3}\,\!$ ${A_{21}}{\big |}_{21cm}\sim 6\cdot 10^{-15}s^{-1}\,\!$

The actual value is $2.876\cdot 10^{-15}s^{-1}$.

3 Electric Quadrupole

If one is nearby a rotating quadrupole, one sees the ${\mathfrak {E}}$ (electric) field rotating rigidly. However, from far away, there are kinks in the field, resulting in a retarded potential. The radiation nearby goes as $r_{near}\sim \lambda$. For a monopole, the electric field is ${\mathfrak {E}}\sim {q \over r^{2}}$. For a dipole, it is ${\mathfrak {E}}={q \over r^{2}}{s \over r}$, where s is the charge separation. For a quadrupole:

${\mathfrak {E}}={q \over r^{2}}\left({s \over r}\right)^{2}\,\!$

Since $P\propto {\mathfrak {E}}^{2}$, the ratio of the powers emitted by a quadrupole vs. a dipole should be:

${P_{quad} \over P_{di}}\sim \left({s \over r}\right)^{2}\sim \left({s \over \lambda }\right)^{2}\,\!$

An acoustic analogy: a kettle whistle is a monopole, a guitar string is a dipole, and a tuning fork (with its two out-of-phase prongs) is a quadrupole.\

Anyway, since ${A_{21}}\sim {P \over E}$,

${{{A_{21}}{\big |}_{quad} \over {A_{21}}{\big |}_{di}}\sim \left({s \over \lambda }\right)^{2}}\,\!$

Thus $28\mu m$, the lowest quadrupole rotational transition of $H_{2}$, should have an ${A_{21}}$ of about:

${A_{21}}{\big |}_{28\mu m}\sim {A_{21}}{\big |}_{Ly\alpha }\left({s \over \lambda _{H_{2}}}\right)^{2}\left({\lambda _{L}y\alpha \over \lambda _{H_{2}}}\right)\sim {A_{21}}{\big |}_{Ly\alpha }\left({a_{0} \over 28\mu m}\right)^{2}\left({1216\mathrm {\AA} \over 28\mu m}\right)^{3}\sim 7\cdot 10^{-11}s^{-1}\,\!$

The actual value is $3\cdot 10^{-11}s^{-1}$.

4 Radio Recombination Lines

In HI, the $n=110\to n=109$ transition has a wavelength of 6 cm. We can estimate its ${A_{21}}$:

${A_{21}}{\big |}_{6cm}\sim {A_{21}}{\big |}_{Ly\alpha }\underbrace {\left({1216\mathrm {\AA} \over 6cm}\right)^{3}} _{change \atop in\ \lambda }\underbrace {\left({a_{110} \over a_{0}}\right)^{2}} _{change \atop in\ atom\ size}\,\!$

This presents the question of which dipole moment to use. It turns out we must use $\langle i|{\vec {k}}\cdot {\vec {r}}|f\rangle$.

5 Back to $\sigma$

$\sigma _{12}{\big |}_{line \atop center}\sim {\lambda ^{2} \over 8\pi }{{A_{21}} \over \Delta \nu }\,\!$

Now $\Delta \nu \sim \nu$ for doppler broadening, and ${A_{21}}\sim \nu ^{3}$, so for electric and magnetic dipole transitions:

$\sigma _{12}\sim \lambda ^{0}\,\!$

So the cross-section for these transitions does not depend on wavelength.