Normalizing the wave function of the electron in hydrogen

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Homework Help Overview

The discussion revolves around the normalization of the wave function for an electron in a hydrogen atom, specifically addressing part d of a homework problem. Participants are exploring the conditions under which the wave function is considered normalized in a three-dimensional context using spherical coordinates.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration limits and the necessity of squaring the wave function. There are attempts to apply spherical coordinates, with some questioning the logic behind their integration limits. Others express confusion about constants and the results of their integrations.

Discussion Status

The discussion is active, with participants providing insights and guidance on the normalization process. Some have identified potential errors in their calculations, while others are clarifying the integration process. There is an ongoing exploration of the probability calculations related to the wave function.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may impose specific methods or approaches to solving the problem. There is mention of integration limits and the use of spherical coordinates, indicating assumptions about the setup of the problem.

Cocoleia
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Homework Statement


upload_2017-4-1_19-3-55.png

I am having trouble with part d, where they ask me to prove that the wave function is already normalized

The Attempt at a Solution


upload_2017-4-1_19-5-18.png

But that clearly doesn't give me 1. I tried to use spherical coordinates since it is in 3D? Not really sure how to proceed.
EDIT: I realize that I didn't square the wave function, so
upload_2017-4-1_19-8-5.png

Which still doesn't give me 1
 
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The integration limit should be from ##0## to ##\infty##.
 
blue_leaf77 said:
The integration limit should be from ##0## to ##\infty##.
Ok. I was thinking 0 to r since the most it could go was to the radius. I guess the logic is wrong
upload_2017-4-1_19-47-25.png

multiplied by 4pi, so the 4's will cancel out, leaving me with (a_0^3)(pi)
Which still doesn't = 1. What am I missing
 
It pays to be methodical
##\psi(r,\theta,\phi) =\frac{1}{\sqrt{\pi}} \left( \frac{1}{a_0} \right )^{3/2} e^{-r/a_0}##
##\psi^*(r,\theta,\phi) \psi(r,\theta,\phi) =\frac{1}{{\pi}} \left( \frac{1}{a_0} \right )^{3} e^{-2r/a_0}##
Now do the integrals.
 
kuruman said:
It pays to be methodical
##\psi(r,\theta,\phi) =\frac{1}{\sqrt{\pi}} \left( \frac{1}{a_0} \right )^{3/2} e^{-r/a_0}##
##\psi^*(r,\theta,\phi) \psi(r,\theta,\phi) =\frac{1}{{\pi}} \left( \frac{1}{a_0} \right )^{3} e^{-2r/a_0}##
Now do the integrals.
If I use the spherical coordinates I still get (a^3)π. I say that 0<Θ<π and 0<Φ<2π
 
Cocoleia said:
If I use the spherical coordinates I still get (a^3)π. I say that 0<Θ<π and 0<Φ<2π
Can you show the expression that you integrated to get this result?
 
kuruman said:
Can you show the expression that you integrated to get this result?
upload_2017-4-2_19-0-21.png

I took out the constant and then used spherical coordinates
 
Cocoleia said:
View attachment 117362
I took out the constant and then used spherical coordinates
Oh wait. I think I figured it out. I lost my constant along the way.
When I integrate to find the probability of finding the electron in a certain place will I use spherical coordinates again ?
 
Yes.
 
  • #10
Is it true that the probability of finding electron betwee r1= ao/2 and r2= 3a0/2 is 49,65%? For this problem
 
  • #11
rb120134 said:
Is it true that the probability of finding electron betwee r1= ao/2 and r2= 3a0/2 is 49,65%? For this problem
Yes.
 
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