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Inspired by this other question.

The second is defined such that the electromagnetic radiation whose energy equals the hyperfine splitting of the ground-state of the Cs-133 atom has a frequency of exactly 9,192,631,770 Hz. However, any such transition has a "natural width" due to the fact that excited states that are subject to spontaneous emission are not eigenstates of the full interaction Hamiltonian; this implies that they don't have an exact energy, and even the most perfect measurements of the frequency of the radiation would therefore exhibit random variation.

How much is this inherent uncertainty in the case of the definition of the second?

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    $\begingroup$ This Wikipedia page mentions an atomic clock with an uncertainty of one second in 30 million years. I'm not sure if this uncertainty is only due to the energy width of the transition, but anyways it serves as an upper bound. $\endgroup$ Sep 30, 2021 at 18:10
  • $\begingroup$ see also this en.wikipedia.org/wiki/Atomic_clock#Evaluated_accuracy $\endgroup$
    – anna v
    Sep 30, 2021 at 18:34

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tl;dr: It is not currently a factor limiting the ultimate precision of how closely you can make a system oscillate to that defined frequency. To address your question specifically: each atom will absorb or emit photons at a random energy due to what you stated, but an ensemble of such atoms will average out to a distribution with a well-defined centre energy.

The linewidth of a transition only limits how narrow the spectrum you measure can be, it does not place a limit on how well you can determine the centre of that transition's spectrum. However a narrower line makes it easier to determine the centre: if you make $N$ measurements (or replace $N$ with an integration time $\tau$ if you want), the uncertainty is $\sigma/\sqrt{N}$ where $\sigma$ is the uncertainty of a single measurement. Thus if you want to get your uncertainty down by a factor of 10 you need to make 100 times as many measurements. If you can just switch to a transition with a linewidth that's 10 times smaller instead, your job is significantly easier.

At the risk of insulting the reader's intelligence, an illustration: the shape of an atomic transition is usually well-modelled by a Lorentzian (aka Cauchy distribution). Try to guess where the centre of this one (with width parameter $\gamma = 1$) is:

Lorentzian curve

Probably 0, right? Now a curve with the same centre but $\gamma = 0.05$ and a narrower range of x values:

Narrower Lorentzian

Ah! It's not centred at zero, but we can only tell easily because it has a narrower linewidth. In a real measurement where there's noise perturbing this curve, we could have found the centre of either to the same precision but only by taking a lot more data in the first case. As described in Alwin's answer there are also other effects which can shift the centre of a transition rather than just broadening it (often called systematic effects, or just systematics), which are usually bigger problems than the width of the line since you have to correct for them, and it's hard to know exactly how strong the perturbing effect is in the region where the atoms are. Look up co-magnetometers in neutron EDM measurements for a clever example of solving a similar problem.

There are also various methods used to narrow the spectrum you measure below the natural linewidth, the most prominent of which is known as Ramsey spectroscopy (or Ramsey interferometry, or the separated oscillatory fields method, or...), which got Mr. Ramsey a Nobel, to give you an idea of how important it is. It's difficult to give a short explanation, but in essence you drive the transition with two pulses separated in time, and in two types of experiment: with the radiation in the pulses being in phase, and 180 degrees out of phase, and subtract the measured spectrum from each case. The width of the resultant lineshape is then determined completely by the time separation between the pulses. This is due to some of the atoms being excited in the first pulse, some in the second, and acquiring a different quantum-mechanical phase of $\Delta E T / \hbar\space$ (where $\Delta E$ is the energy difference between the two states and T is the time separation of the pulses), which then causes interference when you measure the population of the atoms which have undergone a transition (image of such a lineshape from here):

enter image description here

However, you're still limited by the lifetime of the state $\tau$, as making $T$ larger decreases the amplitude of the signal you measure by a factor of $e^{-T/\tau}$. To toot my own horn a little, I've used a variation of this method to measure a transition in helium with a natural linewidth of 3.6 MHz to a precision of 25 Hz, where the uncertainty was dominated by how well we could control these perturbing effects - if we ignored those and just used the statistical uncertainty, it would have been 2 Hz. For another comparison, hydrogen masers use the 21 cm hydrogen line which has a lifetime of about 10 million years, and a natural linewidth of ~$10^{-15} \space \text{Hz}$, but when realised as a system they usually have a linewidth of several Hz due to all the broadening effects and finite cavity quality factor.

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  • $\begingroup$ In the case of the caesium-133 hyperfine transition, do you know the approximate width of the distribution that would be observed if all sources of error, and effects that would broaden the distribution, could be eliminated, and only the quantum-mechanical uncertainty were left? $\endgroup$
    – Brian Bi
    Oct 4, 2021 at 18:57
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    $\begingroup$ @BrianBi I can't seem to find the lifetime of the relevant state from a quick search, but I would guess it's pretty long since it's a spin-flip transition, so I think the natural linewidth is probably under a Hz. In this old paper they get a 44 Hz width, but that seems to be due to effectively having 20 ms microwave pulses, so it must be less than 44 Hz. osapublishing.org/ol/fulltext.cfm?uri=ol-14-5-269&id=9578 $\endgroup$
    – llama
    Oct 4, 2021 at 22:23
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This NIST publication is a great source for understanding these errors.

In practice, the clock emits radiation of some frequency to try to excite as many Caesium atoms as possible. The maximum number are excited when the clock's emitted radiation matches the peak resonance frequency of the transition. So, even though there is a quantum mechanical linewidth, its effect just introduces an uncertainty which can be reduced by observing a larger number of atoms (millions) for a longer time (large compared to the period). This is the benefit of spatially longer clocks and the fountain design.

From the NIST pub., the Q factor of NIST-F1 is $Q = \frac{f_{0}}{\Delta{}f_{a}}\sim Tf_{0}$ where $f_{0}$ is the transition frequency, $T$ is the observation time, and $\Delta{}f_{a}$ is the resonance width. Observation times of ~1 second provide a Q of $10^{10}$ (where larger is better). The effective line width in this case is ~1 Hz.

Other large effects come about from biased errors (as opposed to symmetric increases in width). These include a blackbody shift and a density shift. The uncertainty usually comes from the uncertainty in trying to correct for these biases.

You can see here to find a laundry list of other effects they correct for, such as gravitational redshift, second-order Zeeman, and microwave amplitude shift. There are several other effects with either small bias or small added uncertainty in the correction.

To summarize, quantum mechanical line width does not directly place an uncertainty on clocks, as its effect on uncertainty can be reduced with longer times and more atoms. Other relevant effects currently limit the precision of Caesium clocks.

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  • $\begingroup$ Are you basically saying that there's a way to interpret the definition as exact? In other words, whichever frequency yields the maximum transition amplitude is the frequency that is designated at 9,192,631,770 Hz? $\endgroup$
    – Brian Bi
    Sep 30, 2021 at 21:58
  • $\begingroup$ I think I'm providing a brief/simple description of various errors and how they are dealt with. Aside from my answer, I believe it's true that that is the definition. The maximum transition amplitude occurs at the frequency of the line by definition, which is related to the second by definition. $\endgroup$
    – Alwin
    Sep 30, 2021 at 22:01
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    $\begingroup$ @BrianBi small pedantic emphasis to add to Alwin's comment: The maximum transition amplitude would occur at that frequency, by definition, if there were no asymmetric perturbations to the line. This is an ideal which can't be reached in a real measurement, since there will always be some small perturbation which shifts the frequency of maximum transition amplitude that you can't perfectly correct for. $\endgroup$
    – llama
    Oct 4, 2021 at 17:41

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