Normalize the wave function and more Please help

  • Thread starter s_gunn
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  • #1
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normalize the wave function and more!! Please help!!!

Homework Statement



i) Normalize the wave function

ii) Calculate <x>

iii) Calculate [tex]<x^{2}>[/tex]

iv) What would happen if a < 0?



Homework Equations



[tex]\psi\left(x\right) = N\left(1+i\right)exp\left(-a|x|\right)[/tex], for -inf < x < inf and a > 0

The Attempt at a Solution



It would take ages for me to work out the latex for all my steps (I'm new to latex!!) so I'll do what i need to and hope someone can help!!

First:

[tex]^{inf}_{-inf}\int 2N^{2}e^{-2a|x|}dx[/tex]
[tex]=N^{2}\left[-e^{-2a|x|}\right]^{inf}_{-inf}[/tex]

so: [tex]\frac{-N^{2}}{a}=1[/tex]
so: [tex]N=\sqrt{\frac{-1}{a}}[/tex]

therefore:
[tex]\psi\left(x\right) = \sqrt{\frac{-1}{a}}\left(1+i\right)exp\left(-a|x|\right)[/tex]


ii+iii) for the expectation values, I got both equalling zero

iv) and if a < 0, you'd get exponential growth as x approaches infinity (+ and -)

Is this right??!

I get so confused when the limits are infinity!!
 

Answers and Replies

  • #2
501
2


I think you lost a 2 when doing your initial normalization.

You should get an expectation value for position of zero. Think about physically why that is the case.

The expectation value of x^2 should not be zero. Check that integral.

Your completely correct on the a<0 regieme. The function would increase to infinity as x goes to infinity and you could not normalize the function. The solution becomes non-physical.
 
  • #3
nicksauce
Science Advisor
Homework Helper
1,272
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Your integration to normalize the WF looks wrong. You shouldn't get a negative answer when you're integrating a positive definite function. I'd try it again more carefully if I were you.
 
  • #4
501
2


Also, it is difficult to integrate an absolute value. You need to split the integral into two regions. One where the Abs(x) = x and one where Abs(x) = -x
 
  • #5
34
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Ok! I redid the integration for the wave function and got [tex]N=\sqrt{\frac{a}{2}}[/tex] now, which looks alot better and my expectation value of x is still zero so that's good too!! My problem now is that when i try to calculate the expectation value of x^2, (using integration by parts) I just keep getting deeper and deeper into it!!

[tex]<x^{2}>=^{inf}_{-inf}\int\psi^{*}x^{2}\psi\: dx[/tex]
[tex]=\sqrt{\frac{a}{2}}\left(1-i\right)e^{-a|x|}x^{2}\sqrt{\frac{a}{2}}\left(1-i\right)e^{-a|x|}\: dx[/tex]
[tex]=^{inf}_{-inf}\int ax^{2}e^{-2a|x|}[/tex]
[tex]=^{0}_{-inf}\int ax^{2}e^{2ax}+\;^{inf}_{0}\int ax^{2}e^{-2ax}[/tex]
[tex]=\left[\frac{x^{2}e^{2ax}}{2}\right]^{0}_{-inf}-\:^{0}_{-inf}\int \frac{e^{2ax}}{x}\:dx+\left[\frac{-x^{2}e^{-2ax}}{2}\right]^{inf}_{0}-\:^{inf}_{0}\int \frac{-e^{-2ax}}{x}\:dx[/tex]

=......

so far I have carried on another 2 integrals and the denominator in each new integral has an increased power of x (this one is x, next x^2, then x^3)! I can see no end!!!

Have I made yet another mistake?!
 
  • #6
34
0


I don't know how where i was going with that intgral!! I started again from scratch following your advice and every thing seemed to fall into place itself so thanks to both of you for your help!:smile:
 

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