Normalize the wave function and more Please help

In Summary, you would expect the expectation value of x^2 to be zero if a < 0, but it is not. The function would increase to infinity as x goes to infinity.
  • #1
s_gunn
34
0
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!
 
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  • #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


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


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


Ok! I redid the integration for the wave function and got [tex]N=\sqrt{\frac{a}{2}}[/tex] now, which looks a lot 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


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:
 

1. What does it mean to "normalize" a wave function?

Normalizing a wave function means to adjust its magnitude so that the total probability of finding the particle in any location is equal to 1. In other words, it ensures that the probability of finding the particle somewhere in space is 100%.

2. Why is it important to normalize a wave function?

Normalizing a wave function is important because it allows us to accurately calculate the probability of finding a particle in a specific location. If the wave function is not normalized, the probability values will be incorrect and could lead to erroneous predictions.

3. How do you normalize a wave function?

To normalize a wave function, you need to square the amplitude of the wave function at each point in space, add up all the values, and then divide the original wave function by the square root of the total. This ensures that the total probability of finding the particle is equal to 1.

4. Can a wave function be normalized to a value other than 1?

No, a wave function must always be normalized to a value of 1. This is because the total probability of finding the particle must always be equal to 100%.

5. What does it mean if a wave function is not normalized?

If a wave function is not normalized, it means that the total probability of finding the particle is not equal to 1. This could indicate an error in the calculation or an incorrect wave function, and would need to be adjusted in order to accurately predict the behavior of the particle.

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