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Difficulties with overlap rule

 
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Feb19-07, 10:34 AM   #1
 

Difficulties with overlap rule


First post and not used to the symbols that I have seen used in posting formulae, so please be patient!

The probability of obtaining the ith energy eigenvalue Ei is given by the square modulus of the integral psi_i*(x)PSI(x,t)dx where PSI(x,t) is the wave function of the system.

How can this be calculated since the wavefunction is a product of the superposition of all the possible energy eigenfunctions, psi_i(x)? I guess what I am trying to ask is - how can we know PSI(x,t)? Every text I have read typically starts problems in this area with "Suppose we have a wavefunction,...."

Thanks to all that help
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Feb19-07, 01:02 PM   #2
 
As noted above, I'm new to the board and just checked the forums to see my question moved to this Homework section. Just to note that this is not part of any assignment - I'm trying to clear up gaps in my understanding.

Thanks
Feb19-07, 01:06 PM   #3

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To actually calculate a probability, they'll need to give you some more or less explicit form for PSI. Right?
Feb19-07, 02:58 PM   #4
 

Difficulties with overlap rule


Dick,

Well, this is how I have always encountered the overlap integral - you're given a wavefunction and can use it to e.g. calculate the probability of finding the system in a particular eigenenergy. I'm trying to broaden my horizons though - where does PSI come from? I seem to remember something about the variational principle being used to produce a "best guess" but would like to hear from someone with a definitive answer here.

Best,

John
Feb19-07, 03:06 PM   #5

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Ok. A 'useful' example of this sort of problem would be the following. Suppose an electron is in a particular energy state for an atom. The atom undergoes radioactive decay so the charge on the nucleus suddenly changes. What is the probability the electron transitions to some given state in the new atom. Then PSI is the old wave function and the P_i are the new energy eigenfunctions.
Feb20-07, 02:32 AM   #6
 
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That PSI is the solution to the SE

[tex] \frac{d\left|\Psi\right\rangle}{dt}=\frac{1}{i\hbar}\hat{H}\left|\Psi\r ight\rangle [/tex]
Feb20-07, 09:54 AM   #7
 
Still not feeling enlightened as to my straightforward question - how can we know PSI so as to evaluate the overlap integral?
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