How Do You Normalize the Wavefunction ψ=Ae^(-λχ)e^(-iδt)?

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To normalize the wavefunction ψ=Ae^(-λχ)e^(-iδt), one must integrate ψ^2, specifically ψ multiplied by its complex conjugate. The confusion arises regarding the factor of 2 in front of |A|^2 e^(-2λχ), which may relate to the integration limits or symmetry considerations. The wavefunction as initially defined is not valid for negative x, as it diverges; it may need to be redefined to include the absolute value of x. It's essential to verify the integration limits used in the solutions manual, as they could clarify the presence of the factor of 2. Understanding these aspects will help in correctly normalizing the wavefunction.
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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 don't 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
 
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baldywaldy said:

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 don't 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,

\Psi = Ae^{- \lambda x}e^{-i\delta t}

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,
\Psi = Ae^{- \lambda |x|}e^{-i\delta t}?

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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