Where Did I Go Wrong in Determining the Normalization Constant?

Click For Summary

Homework Help Overview

The discussion revolves around determining the normalization constant for a wave function expressed as ψ(x) = c cos(kx) exp[(-1/2)(x/L)²]. The context involves improper integrals and the application of normalization conditions in quantum mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to evaluate the normalization condition by integrating the square of the wave function but expresses uncertainty about their approach to improper integrals. Some participants question the validity of breaking up the integral and suggest considering identities related to trigonometric functions.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's attempts and raising points about mathematical identities and integration techniques. There is acknowledgment of the challenges faced with the integration process, and some guidance has been offered regarding the structure of the integrals involved.

Contextual Notes

Participants are discussing the complexities of integrating the wave function, particularly focusing on the Gaussian integral and its relevance to the problem. The original poster indicates a lack of confidence in their mathematical skills, which may affect their approach to the problem.

atomicpedals
Messages
202
Reaction score
7

Homework Statement



Determine the normalization constant c in the wave function given by
[itex]\psi[/itex](x) = c cos(kx) exp[(-1/2)(x/L)2 ]

Homework Equations



1=[itex]\int[/itex] |[itex]\psi[/itex](x)|2 dx

limits of integration being -inf to inf.

The Attempt at a Solution



I'm very much sure that my math is wrong, I'm very rusty with improper integrals.

1= [itex]\int[/itex] |c cos(kx) exp[(-1/2)(x/L)2|2 dx

= [itex]\int[/itex] |c2 cos2(kx) exp[-(x2/L2)| dx

it's at this point I start getting into trouble

= c2 [itex]\int[/itex] |cos2(kx) exp[-(x2/L2)| dx

= c2 [itex]\int[/itex] cos2(kx)dx [itex]\int[/itex] exp[-(x2/L2)dx

= c2 (lim((2kx+sin(2kx))/4k)) ([itex]\pi[/itex])1/2/(1/L2)1/2

I think I'm pretty solidly wrong by this point... where did I go wrong?
 
Physics news on Phys.org
You can't break up an integral like that. Think about it.

[itex]\int x^2 dx = \int x * x dx = \int x dx \int x dx = x^4/4[/itex] ??

Also this identity might make it less painful for you:

[itex]cos(kx) = \frac{e^{ikx} + e^{-ikx}}{2}[/itex] (Euler's formula)
 
Ah, right... well at least I made it three steps in before totally going off the deep end. Still working on it though.
 
So my sticking point mathematically really seems to be the

e(-1/2)(x/L)2

This almost certainly simplifies down to something reasonably basic after being squared and/or integrated shouldn't it?
 
Last edited:
do you know what a gaussian integral is?
 
Yep, Arfken is a life-saver!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K