I'm trying to demonstrate the small effect of Natural Broadening as compared to other types of broadening (Doppler, Stark, van der Waals, etc.) and my calculations don't match the accepted values. My understanding is that the calculation for Natural Broadening is:

$$ \Delta \lambda = \frac{ \lambda^{2}}{2 \pi c} \big( \sum_{n<u} A_{un} + \sum_{n<l} A_{nl}\big) $$

Where $\Delta \lambda$ is the Full Width at Half Maximum (FWHM), and $A_{un}$ and $A_{nl}$ being the Einstein coefficients for the upper and lower levels. The sums are equal to the radiative lifetime of the state $n$ in question.

For instance, for the $H_{ \beta }$ line from $n=4$ to $n=2$, I interpret the equation to mean that the necessary Einstein coefficients are $A_{43}, A_{42}, A_{41},$ and $A_{21}$. I found the values for these on the NIST Atomic Spectral Database (I can list them here if no one wants to look them up), however, when I perform the calculation, the result is not correct. The accepted value from what I have found in the literature for $H_{ \beta }$ is $ 6.2 \times 10^{-5} {\buildrel _{\circ} \over {\mathrm{A}}}$, and my result isn't even close ($ 7.9 \times 10^{-3} {\buildrel _{\circ} \over {\mathrm{A}}}$).

Am I using the correct Einstein coefficients for this?

  • 1
    $\begingroup$ It seems this pdf has all the gory details. zuserver2.star.ucl.ac.uk/~idh/PHAS2112/Lectures/Current/… $\endgroup$
    – MaxW
    Apr 12, 2016 at 1:57
  • 2
    $\begingroup$ @MaxW That document doesn't do it. I don't see where it explains why there a line width in the first place (but I did peruse it quickly, so I may have missed something). The natural line width arises from coupling of an atomic state to an infinity of degenerate electromagnetic modes. As with coupled oscillators, the result is a broadening of the natural frequency. $\endgroup$
    – garyp
    Apr 12, 2016 at 3:06
  • $\begingroup$ What you mean by 'isn't even close'? What value do you get? $\endgroup$
    – gigacyan
    Apr 13, 2016 at 9:39
  • $\begingroup$ @gigacyan More than 2 orders of magnitude. Like I said, I can provide the actual values of the Einstein A coefficients that I used, but I am wondering if I chose the correct levels for those coefficients. The value I got was 7.9*10^-3 angstroms. $\endgroup$
    – iwantmyphd
    Apr 13, 2016 at 13:55
  • $\begingroup$ @garyp Yes, you're right, that do simply presents the same formula as the OP is using (modulo some $\pi$ factors that I haven't checked in detail). However, the results used by the OP are consistent with the mechanism you name, because if you do the calculation with a constant coupling co-efficient per unit frequency interval to the EM modes, you end up with a probability amplitude for transition to happen in time interval $[t,\,t+\mathrm{d}t)$ that varies like $\exp(-(t/(2\tau)) + i\,\Delta\omega\,t)$, where the transition lifetime $\tau = 1/(2\,\pi\,|\kappa^2|)$ is related to the ... $\endgroup$ Jul 17, 2016 at 23:47

1 Answer 1


I am not sure whether you have solved this question already. This problem occurred to me recently as well, and I think leaving what I got might be helpful to people that need help with this in the future.

Your understanding of adding up the Einstein A values of A$_{41}$, A$_{43}$, A$_{42}$, A$_{21}$ is correct. I took values from Wiese W L, Smith M W and Glennon B M 1996 Atomic Transition Probabilities. Vol. 1. Hydrogen through Neon (US National Bureau of Standards National Standard Reference Series, Washington, DC NSRDS-NBS), with

A$_{41}$ = 1.278e7, A$_{42}$ = 8.419e6, A$_{43}$ = 8.986e6, A$_{21}$ = 4.699e8.

Inserting these into the formula you gave I got 6.3$\times 10^{-5}$ angstrom.

A further comment on the possible confusions of the Einstein A values is that, those "A" I adopted above are the "average" transition probabilities that are for the transitions between the lower state of principal quantum number n$_l$ and the upper state, n$_u$. Einstein A of different orbital ($l$) quantum numbers degenerate with the same principal, $n$ (see Section E of the book I referred to above).

  • $\begingroup$ Your Einstein A coefficients produce the right result, so I must have read them wrong off of NIST. I should have listed the ones I used in the question so someone could have pointed that out, but after checking it again, I found the same Einstein A coefficients that you have here. $\endgroup$
    – iwantmyphd
    Nov 6, 2021 at 21:56

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