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Re: Quantum question?



I am not happy with the characterization of the Schroedinger
Equation as telling us "the probability of an electron being in any
particular energy state."
If I prepare an atom in a particular state at the beginning
of an experiment, I can use the S-Eq to calculate the transition
probabilities to other states. The intensity of radiation from
the different transitions will, for an assembly of atoms, be proportional
to the transition probabilities.
Key to such calculation, of course, is the interaction of the
atom with the radiation field, a highly non-trivial subject.
Regards,
Jack

Adam was by constitution and proclivity a scientist; I was the same, and
we loved to call ourselves by that great name...Our first memorable
scientific discovery was the law that water and like fluids run downhill,
not up.
Mark Twain, <Extract from Eve's Autobiography>

On Wed, 26 Jan 2000, Cliff Parker wrote:

Today a student asked a question that I had not considered
before. I think I know the answer but am not 100% sure. Please
let me know if I am on the right track.

The question... If Schrodinger's equations only tell us the
probability of an electron being in any particular energy state
and therefore a non zero probability exists that it will be in
various unusual states, why do we not see the entire visible
spectrum when we look at the emission spectrum of excited atoms?
I thought for awhile and said I would give my best guess but
warned it was only my best guess. I told her that in effect we
could see the entire spectrum because any particular electron
could possibly be changed from any energy level to any other
level. However the probability for the energy changes
represented by the dark space between spectral lines is extremely
low. Therefore any light generated with such energy would be
very infrequent and not detectable with normal means. How did I
do?