1
$\begingroup$

Wikipedia says that the (newly proposed, more rigorous) definition of a second is:

The second, symbol s, is the SI unit of time. It is defined by taking the fixed numerical value of the caesium frequency ΔνCs, the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom, to be 9192631770 when expressed in the unit Hz, which is equal to s−1.

My question: at such a tiny scale (the scale of atoms), do the probabalistic laws of quantum mechanics make this definition imprecise in any way?

$\endgroup$
  • 3
    $\begingroup$ No, in fact the laws of quantum mechanics are the only thing that let this work in the first place. They produce the discrete energy levels that are being transitioned between in the 'hyperfine transition frequency' they talk about. $\endgroup$ – knzhou Jan 19 '18 at 9:08
  • $\begingroup$ What do you mean by imprecise? You certainly can't measure a time interval with perfect precision under this definition, but you can't measure a time interval with perfect precision under any definition. $\endgroup$ – Chris Jan 19 '18 at 9:11
  • $\begingroup$ @Chris I'm obviously not talking about measuring time intervals with perfect precision in the real world. I am talking about the definition of a second, and that's it. $\endgroup$ – user182050 Jan 19 '18 at 9:17
  • $\begingroup$ @AlexAdamov It's not really clear what it means for a definition to be "precise," though. The only thing that comes to mind is "you get a different thing if you measure the same interval twice," but that's a real world precision thing. (And a sense in which this definition is very, very precise, for that matter) $\endgroup$ – Chris Jan 19 '18 at 9:22
  • 1
    $\begingroup$ @The_Sympathizer you are spot on, my friend. $\endgroup$ – user182050 Jan 19 '18 at 10:42
1
$\begingroup$

No, Quantum Mechanics actually makes this definition of a second really precise. The fact that energies of the different states of the Caesium atom are quantised makes it possible to define the exact frequency of this transition (remember $E = h\nu$) so that you can actually have an incredibly precise definition of a second in terms of an integer number times the period $1/\nu$. The hyperfine transition is actually a correction to 'usual Quantum Mechanics' coming from Relativity, so that the energy $E$ is tiny compared with the energies of other excited states. Why they use this particular transition? It must be related with the precision to which it can be measured, but I am not an expert in atomic physics.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy