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Normalizing wave function, factor of 2 out

  • Thread starter Frinkz
  • Start date
  • #1
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Homework Statement


Consider the wave function

[tex]\Psi(x, t) = Ae^{-\lambda|x|}e^{-i\omega t}[/tex]

where A, [tex]\lambda[/tex] and [tex]\omega[/tex] are positive real constants.

Normalize [tex]\Psi[/tex]


Homework Equations


[tex]\int |\Psi(x, t)|^{2} dx = 1[/tex]

[tex]|\Psi(x, t)|^{2} = \Psi^{*}\Psi[/tex]


The Attempt at a Solution


I have a model solution - with a step missing, I think my error is in complex conjugate math...

1, Finding [tex]|\Psi(x, t)|^{2}[/tex]

[tex]\Psi^{*}\Psi = (Ae^{-\lambda|x|}e^{i\omega t}) (Ae^{-\lambda|x|}e^{-i\omega t})[/tex]

[tex] = A^{2}e^{-2\lambda|x|}e^{i\omega t}e^{-i\omega t}[/tex]
[tex] = A^{2}e^{-2\lambda|x|}e^{0} = A^{2}e^{-2\lambda|x|}[/tex]

I think this is where my problem is, I am told that

[tex]|\Psi(x, t)|^{2} = 2|A|^{2}e^{-2\lambda|x|}[/tex]


So I am missing a factor of 2?

Is there a complex conjugate rule somewhere I am missing?
 

Answers and Replies

  • #2
286
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your steps looks fine to me, im not sure from where they got the factor of 2?.. and there is no other complex conjugate rule you miss ..

is their answer says that psi^2 is after obtaining the normalized wavefunction? or is it just the step that follows the one you did ?..
 
  • #3
8
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Thanks for checking.


Turns out, the model answers were a bit badly written.


They had put in the factor of two, because the integral was changed from -infinity to +infinity, to 0 to +infinity, so you can make use of a definite integral

[tex]\int_0^{\infty} \! e^{-\lambda x} = \frac{1}{2}\sqrt{\frac{\pi}{\lambda}}[/tex]

That step was just omitted from the solution I was trying to understand, but I get it now :)
 
  • #5
8
0
Sorry, that was a typo.

Should have been e^(-lambda x^2)
 

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