Before measurement the wavefunction of a quantum particle, eg an electron, evolves in a superposition state. Can the following wavefunction $$\Psi = 1\,\,\lvert \uparrow \rangle + 0 \,\,\lvert \downarrow \rangle$$ in priciple also be considered as a superposition (conservation of probabilities), or does it describe already and only the "collapsed" eigenstate?
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Wavefunctions are solutions of specific wave equations, if there is a "colapse" it means there was an interaction, and after the interaction a new wavefunction describes the probabilities for the electron. my answer here https://physics.stackexchange.com/questions/151457/how-does-a-wavefunction-collapse/151522#151522 is extensive and the other answers too. – anna v Mar 10 '22 at 04:57
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$\Psi = 1,,\lvert \uparrow \rangle + 0 ,,\lvert \downarrow \rangle$ and $\Psi = \lvert \uparrow \rangle$ are identical. So yes, $\Psi$ describes the "already collapsed eigenstate." – Kurt G. Mar 10 '22 at 09:54