Re: Quantum question?

[DEVARAKONDA VENKATA NARAYANA SARMA <narayana@HD1.VSNL.NET.IN>]

(psi)(psi)* is function only of position cordinates
and not of energy. The time-dependent parts which are
functions of energy are eliminated in the process of
taking (psi)(psi)*. Hence there appears to be no question
of probabilities being given for different energy states.
Energy occurs as the eigenvalue and Sch.Eq has single valued,
continuous and finite (square-integrable) solutions
which are physically meaningful, only for those values.

But energy states do have a minimum width fixed by
Heisenberg's Uncertainity relation (delta E)(delta T) = h.
This does not come from Schrodinger equation.

regards,

sarma.

At 05:25 PM 1/26/00 -0600, you 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?