Squaring the wavefunction

  • Thread starter baldywaldy
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  • #1
baldywaldy
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Homework Statement


Normalise ψ=Ae^(-λχ)e^(-iδt)


Homework Equations


I know you have to intergrate ψ^2 i.e (ψxψ*)

The Attempt at a Solution


Im literally just stuck at the first bit , i can do the rest. I have the solutions manual and I dont understand how they get 2|A|^2 e^(-2λχ) from ψxψ*well only the 2 in front of A^2 really when i do it i get |A|^2 e^(-2λχ)

thanks for the help
 

Answers and Replies

  • #2
collinsmark
Homework Helper
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Homework Statement


Normalise ψ=Ae^(-λχ)e^(-iδt)


Homework Equations


I know you have to intergrate ψ^2 i.e (ψxψ*)

The Attempt at a Solution


Im literally just stuck at the first bit , i can do the rest. I have the solutions manual and I dont understand how they get 2|A|^2 e^(-2λχ) from ψxψ*well only the 2 in front of A^2 really when i do it i get |A|^2 e^(-2λχ)

thanks for the help
I'm not sure, but it might have something to do with the fact that the wavefunction you specified, as it is specified,

[tex] \Psi = Ae^{- \lambda x}e^{-i\delta t} [/tex]

is not a valid wavefunction for negative x. It blows up toward infinity as x becomes more negative.

Are you sure it's not something like,
[tex] \Psi = Ae^{- \lambda |x|}e^{-i\delta t} [/tex]?

Maybe the '2' in front of the A comes taking symmetry into account. Check the limits of integration used in the solutions manual. Are the limits from negative infinity to infinity, taking the entire wavefunction into account? Or is only half the wavefunction integrated (taking advantage of symmetry), and integrated from 0 to infinity?
 
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