Normalize Hydrogen Wavefunction

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Homework Help Overview

The discussion revolves around the normalization of the wave function for the hydrogen atom in its fundamental state, specifically the expression for the wave function in one dimension. Participants are examining the mathematical steps involved in proving that the integral of the squared wave function equals one.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to integrate the squared wave function over one dimension but encounters discrepancies in the results. Some participants question the dimensionality of the integration and suggest that the integration should be performed in three dimensions instead. Others point out potential errors in the integration process, such as squaring the exponential term incorrectly and missing a factor related to spherical coordinates.

Discussion Status

The discussion is ongoing, with participants providing guidance on the need to consider three-dimensional integration due to the nature of the hydrogen atom. There is a recognition that the problem statement may imply a one-dimensional integration, but the physical context suggests a need for a three-dimensional approach. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the normalization constant has units of 1/volume, indicating that the integration must account for the appropriate dimensionality to yield a unitless result. The original poster expresses confusion regarding the problem statement's requirement for one-dimensional integration.

Nacho Verdugo
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Moved from a technical forum, so homework template missing
I'm trying to prove that the wave function of Hydrogen for the fundamental state is normalized:

$$ \Psi_{1s}(r)=\frac{1}{\sqrt{\pi a^3}}e^{-\frac{r}{a}} $$

What I tried is this:

$$ I= \int_{-\infty}^{\infty} | \Psi^2(x) | dx = 1$$

$$ \int_{-\infty}^{\infty} \frac{1}{\pi a^3}e^{-\frac{r^2}{a^2}} dr $$

Which lead me to two differents results:

a) First, when integrating the next integral

$$ \frac{e^{1/a^2}}{\pi a^3} \int_{-\infty}^{\infty} e^{-r^2}dr$$

and using polar coordinates to integrate ## \int_{-\infty}^{\infty} e^{-r^2}dr ##, I finally obtain:

$$ \frac{e^{1/a^2}}{\pi a^3} \sqrt{\pi} $$

and I don't see a way that this expression equals to one.

b) for my second attempt, I used the next property

$$ \int_{-\infty}^{\infty} e^{-bx^2} dx = \sqrt{\frac{\pi}{b}} $$

So, considering ## b=\frac{1}{a^2} ##, I obtain

$$ I=\frac{a\sqrt{\pi}}{\pi a^2} $$

which as well, doesn't look like one.

did i miss something?
 
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Does your hydrogen live in 1 dimension or in 3 ?
In other words: what are you integrating over ?
 
The first issue is that you are squaring the term in the expontial, but you just need to double it. The second issue as Bvu alludes to is that you are missing a factor of 4 pi r^2.
 
BvU said:
Does your hydrogen live in 1 dimension or in 3 ?
In other words: what are you integrating over ?
1 dimension!
 
Nacho Verdugo said:
1 dimension!
You need to revise integration in three dimensions using spherical coordinates.
 
PeroK said:
You need to revise integration in three dimensions using spherical coordinates.
I thought about that, but the statement of the problem I'm trying to solve just says to integrate in one dimesion. I'll try again.
 
Nacho Verdugo said:
I thought about that, but the statement of the problem I'm trying to solve just says to integrate in one dimesion. I'll try again.
A normalised function of one Cartesian variable ##x## is different from a normalised function of the spherical variable ##r##.

You have a function of one variable here, sure, but it is not a Cartesian variable.
 
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Nacho Verdugo said:
I thought about that, but the statement of the problem I'm trying to solve just says to integrate in one dimesion. I'll try again.
A hydrogen atom lives in three dimensions, so you have to integrate across three-dimensional space. It turns out because of spherical symmetry, the integration over the angles is trivial, leaving you with only one variable to integrate over, which is probably what the problem statement was referring to.

Also, note that the normalization constant squared has units of 1/volume. The only way the integral is going to produce a unitless result is to multiply ##\psi^2## by a quantity with units of volume. ##dr## by itself isn't going to cut it.
 

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