# Normalize Hydrogen Wavefunction

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1. Feb 28, 2017

### Nacho Verdugo

• 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?

2. Feb 28, 2017

### BvU

Does your hydrogen live in 1 dimension or in 3 ?
In other words: what are you integrating over ?

3. Feb 28, 2017

### Jilang

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.

4. Feb 28, 2017

### Nacho Verdugo

1 dimension!

5. Feb 28, 2017

### PeroK

You need to revise integration in three dimensions using spherical coordinates.

6. Mar 2, 2017

### Nacho Verdugo

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.

7. Mar 2, 2017

### PeroK

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.

8. Mar 4, 2017

### vela

Staff Emeritus
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.