Could someone please explain why the 1s2s3p excited state of helium is metastable.
Generally, metastable states (i assume you know it's definition) are determined by the electrons of the He gas. In the case of a Helium gas, electrons can be excited from the ground level E to a higher level E' which is stable. Now, once they are there, the electrons can make fast radiationless transitions to an energy level E'' which lies a bit beneath E'. From E'', the electrons fall down to the ground level E. The E'' is the metastable energylevel because the electrons don't stay there long. The transition from E' to E'' goes fast, the transition from E'' towards E takes much longer.
The reason of this transition behaviour is explained using QM (Fermi's Golden Rule, which calculates transitions rates) and depends on the electronic configuration of the He atoms making up the gas.
Some states are metastable typically because the transition from that state to the ground state manifold is a "forbidden" transition as it does not satisfy the transition rules of quantum mechanics. Some other mechanism must occur to trigger an optical transition such as a collision with another atom, or an incoming photon to trigger stimulated emission.
The way I was taught about those "forbidden transitions" was in terms of symmetry arguments. What symmetry breaking mechanism is at work with collision-allowed "forbidden transitions"?
keep in mind that your analytic solution (or symmetry arguments, in this case) applies to a single molecule in vacuum. as soon as you start coupling your vacuum system to some other molecules then your eigenstate may be very different.
also, for what are the transitions "allowed"? dipole allowed? spin allowed? just because the dipole transition is forbidden, doesn't mean that there isn't a small amount of spin coupling allowed.
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