# Hydrogen atom

1. Dec 4, 2007

### ehrenfest

[SOLVED] hydrogen atom

1. The problem statement, all variables and given/known data
My book says that

$$R_{1,0}(r) = 2(1/a_0)^{3/2} e^{-r/a_0}$$

is a normalized wavefunction. But if you integrate R_{1,0}(r)^2 over r from 0 to infinity, you do not get 1. What's wrong here?

2. Relevant equations

3. The attempt at a solution

2. Dec 5, 2007

### Avodyne

You need a factor of r^2 that comes from the dx dy dz in spherical coords.

3. Dec 5, 2007

### Gokul43201

Staff Emeritus
That's not it - he's got the factor of r^2. LATE EDIT: No, you're right. I misread his (r)^2 as the r^2 from the volume element.

Ehrenfest: You should not get 1 for that integral. Write down the (general) normalization condition.

Last edited: Dec 5, 2007
4. Dec 5, 2007

### ehrenfest

Now, I am very confused. I did not have the factor of r^2. But I only need that when I am integrating over dtheta dphi.

Isn't the normalization condition

$$\int_{r=0}^{\infty}|R_{1,0}|^2dr = 1$$

Obviously that's wrong but why?

Are you implying that it should really be

$$\int_{all space}|\Psi|^2drd\thetad\phi = 1$$

But $$\Psi = R(r)Y_{l,m}(\theta,\phi)$$ and the spherical harmonics are already normalized. So the normalization condition is still $$\int_{r=0}^{\infty}|R_{1,0}(r)|^2dr = 1$$, right?

Last edited: Dec 5, 2007
5. Dec 5, 2007

### ehrenfest

I edited my last post before I saw yours but here it is again

Are you implying that it should really be

$$\int_{all space}|\Psi|^2drd\thetad\phi = 1$$

But $$\Psi = R(r)Y_{l,m}(\theta,\phi)$$ and the spherical harmonics are already normalized. So the normalization condition is still $$\int_{r=0}^{\infty}|R_{1,0}(r)|^2dr = 1$$, right?

6. Dec 5, 2007

### Gokul43201

Staff Emeritus
Only for a 1D problem.

That's not it either.

And that's important. Over what variables are the spherical harmonics normalized? Write down this expression.

Next:

$$\int d\Omega |\Psi(r,\theta ,\phi )|^2 = 1$$

How do you write the volume element $d\Omega$ in spherical co-ordinates?

Last edited: Dec 5, 2007
7. Dec 5, 2007

### malawi_glenn

What do you get? You said in your first post that you did not get 1, what did u get?

What happens if you insert r^2 and then integrate?

I have now checked, and I think you are missing the r^2

$$R(r) = \frac{u(r)}{r}$$ where u(r) solves the radial shrödringer eq.

And now the nowmalization becomes:

$$\int_{r=0}^{\infty}|R(r)|^2r^2dr = \int_{r=0}^{\infty}|u(r)|^2dr =1$$

EDIT: Gokul43201 saw that he made a misstake, now three of us are telling you that you missed the r^2 from the volume element of integration =)

8. Dec 5, 2007

### ehrenfest

I think I see the problem.

We know that

$$\int_{all space}\Psi(r,\phi,\theta)dr r^2 sin(\theta) d\theta d\phi = 1$$

$$\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)$$

So, $$\int_{\Omega}Y(\theta,\phi)d\Omega = \int_{0}^{\pi}\int_{0}^{2\pi} Y(\theta,\phi)d\phi d\theta = 1$$.

Wait, is the first equality in the last equation true, or do I need a factor of sine in there?

Last edited: Dec 5, 2007
9. Dec 5, 2007

### malawi_glenn

need or need.. you are forgetting to square the wavefunction..

the only problem was that you forgot was the r^2 from the volume element of integration.

$$\int Y^*Y sin(\theta )d\theta d\phi = 1$$ is the normalization condition for the angular part.

10. Dec 5, 2007

### ehrenfest

Here is what my last post should have been:

We know that

$$\int_{all space}\Psi(r,\phi,\theta)*\Psi(r,\phi,\theta)dr r^2 sin(\theta) d\theta d\phi = 1$$

$$\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)$$

So, $$\int_{\Omega}Y(\theta,\phi)*Y(\theta,\phi)d\Omega = \int_{0}^{\pi}\int_{0}^{2\pi} Y(\theta,\phi)*Y(\theta,\phi) sin\theta d\phi d\theta = 1$$.

I have no idea why the first equality in the last equation needs that sin \theta? I have solids angles, I have no idea what they mean any more?

11. Dec 5, 2007

### ehrenfest

Never mind. I see. Thanks all. :)

12. Dec 5, 2007

### ehrenfest

EDIT: Do not even read this post. It is all wrong. I am not confused any more.

Now I am really confused. Apparently the normalization condition is really

$$4\pi\int_{r=0}^{\infty}|R(r)|^2r^2dr = 1$$

If you try this for the wavefunction I have in my first post, you need that factor of 4\pi.

Which means that $$\int \int Y^*Y sin(\theta )d\theta d\phi = 1$$ was not correct!

But that makes no sense and I am so confused.

Last edited: Dec 5, 2007
13. Dec 5, 2007

### malawi_glenn

$$d\Omega = sin\theta d\phi d\theta$$

=)

Last edited: Dec 5, 2007