Is This Wave Function Normalization Correct?

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Homework Help Overview

The discussion revolves around the normalization of a wave function in quantum mechanics, specifically addressing the setup and evaluation of integrals related to the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the inclusion of the parameter \alpha in the normalization process and question the correctness of their integral setup. There are inquiries about the cancellation of terms and the implications of signs in the calculations.

Discussion Status

There is an ongoing examination of the integral setup and the signs of terms involved. Some participants have pointed out potential errors in reasoning, while others are reflecting on their understanding and seeking clarification.

Contextual Notes

Participants are considering the implications of their integral limits and the behavior of terms as they approach infinity, indicating a focus on the mathematical details of wave function normalization.

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Homework Statement


http://www.ph.qmul.ac.uk/~phy319/problems/problems1.doc"
Question 2)b

The Attempt at a Solution



http://img685.imageshack.us/img685/9033/p270210111001.jpg

Is this correct?

Thanks!
 
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No. The parameter [itex]\alpha[/itex] should appear in the answer. Your set-up looks fine, though.
 
vela said:
No. The parameter [itex]\alpha[/itex] should appear in the answer. Your set-up looks fine, though.

But once we take out the infinity terms after the integral, we're left with the [itex]\alpha[/itex] terms, one is + and one is -, so they cancel.

What is it I'm doing wrong here?

Thanks!
 
They can't cancel or else you'd be left with N2x0=1, which has no solution. Recheck the sign on each term.
 
I can't seem to find the problem, anyone care to take a look?

The only reason I can think of is the original way I set up the integrals, and whether I add them or subtract them!
 
[tex]\left[\frac{1}{-2\alpha}e^{-2\alpha x}\right]^\infty_0 \ne -\frac{1}{2\alpha}[/tex]
 
vela said:
[tex]\left[\frac{1}{-2\alpha}e^{-2\alpha x}\right]^\infty_0 \ne -\frac{1}{2\alpha}[/tex]

Ahhhhhh, I see now, thank you VERY much! :) Let the studying continue!
 

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