Squaring the wavefunction

  • Thread starter baldywaldy
  • Start date
  • #1
20
0

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
Gold Member
2,933
1,360

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?
 
Last edited:

Related Threads on Squaring the wavefunction

  • Last Post
Replies
8
Views
921
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
593
  • Last Post
Replies
3
Views
671
Replies
1
Views
3K
Replies
2
Views
589
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
18
Views
772
  • Last Post
Replies
4
Views
860
  • Last Post
Replies
11
Views
2K
Top