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Félix Faisant's user avatar
Félix Faisant's user avatar
Félix Faisant's user avatar
Félix Faisant
  • Member for 6 years, 9 months
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Why are Alkali atoms used in many Cold Atom experiments?
There's an historical aspect too : the production of Rb BECs is now very well known, to the point that entire experimental apparatus are available commercially. Why use an other elements when Rb suits you well ? On the other hand, Sr or Yb are "newer" elements, although they are not particularly difficult to condense.
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Why are Alkali atoms used in many Cold Atom experiments?
I'd argue that the many other elements do have (almost) closed optical transitions, so that's not a very important point. Sr for example, which is increasingly used. Molecules are an other story however, and laser cooling molecules is severely hindered by the lack of closed optical transitions. As you pointed out, the availability of laser technologies at the cooling transition wavelength is very important. For example, very few labs use Mg because it's in the (near) UV (among other reasons), as for many other "light" elements.
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Why are Alkali atoms used in many Cold Atom experiments?
One could add two more important reasons. First, an alkali atom possesses a magnetic moment, allowing us to trap it in a magnetic trap. This greatly helps for forced evaporative cooling, combined with the RF knife technique. Alkali-earth elements do not, and the only option left is optical trapping, which is arguably less favorable for initial evaporative cooling. Second, the ground state manifold of alkali elements allows for sub-Doppler cooling (that was a surprise in the 80's), without which it would be very difficult to reach high enough phase space densities.
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Why is the position operator a vector operator?
I finally did not edit the question.
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Why is the position operator a vector operator?
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Why is the position operator a vector operator?
Did you mean $\Psi$ rather than $\Psi'$ in the two integrals ? If it is the case, maybe I wasn't clear, but I don't want to assume that $\Psi'(\mathbf{x})=\Psi(R^-1\mathbf{x})$ in the first place (without the phase). In fact, I want to prove that the phase $\alpha$ can be dropped, and I just realized I need something stronger : that $\hat{R}\hat{\mathbf{X}}\hat{R}^{-1}=R\hat{\mathbf{X}}$, so I edited my question.
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Potential energy and conservation law
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