- #1

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**[SOLVED] hydrogen atom**

## Homework Statement

My book says that

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

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?

- Thread starter ehrenfest
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- #1

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My book says that

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

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

Avodyne

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

- #3

Gokul43201

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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.

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

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- #4

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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.That's not it - he's got the factor of r^2.

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

Isn't the normalization condition

[tex]\int_{r=0}^{\infty}|R_{1,0}|^2dr = 1[/tex]

Obviously that's wrong but why?

Are you implying that it should really be

[tex]\int_{all space}|\Psi|^2drd\thetad\phi = 1[/tex]

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

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- #5

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Are you implying that it should really be

[tex]\int_{all space}|\Psi|^2drd\thetad\phi = 1[/tex]

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

- #6

Gokul43201

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Only for a 1D problem.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

[tex]\int_{r=0}^{\infty}|R_{1,0}|^2dr = 1[/tex]

That's not it either.Obviously that's wrong but why?

Are you implying that it should really be

[tex]\int_{all space}|\Psi|^2drd\thetad\phi = 1[/tex]

And that's important. Over what variables are the spherical harmonics normalized? Write down this expression.But [tex] \Psi = R(r)Y_{l,m}(\theta,\phi)[/tex] and the spherical harmonics are already normalized.

Next:

[tex]\int d\Omega |\Psi(r,\theta ,\phi )|^2 = 1 [/tex]

How do you write the volume element [itex]d\Omega[/itex] in spherical co-ordinates?

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- #7

malawi_glenn

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

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

[tex] R(r) = \frac{u(r)}{r} [/tex] where u(r) solves the radial shrödringer eq.

And now the nowmalization becomes:

[tex]\int_{r=0}^{\infty}|R(r)|^2r^2dr = \int_{r=0}^{\infty}|u(r)|^2dr =1[/tex]

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

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I think I see the problem.

We know that

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

[tex]\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)[/tex]

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

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

We know that

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

[tex]\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)[/tex]

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

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

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- #9

malawi_glenn

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

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

- #10

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We know that

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

[tex]\Psi(r,\phi,\theta) = R(r)Y(\theta,\phi)[/tex]

So, [tex]\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[/tex].

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

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Never mind. I see. Thanks all. :)

- #12

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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

[tex]4\pi\int_{r=0}^{\infty}|R(r)|^2r^2dr = 1[/tex]

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

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

But that makes no sense and I am so confused.

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

[tex]4\pi\int_{r=0}^{\infty}|R(r)|^2r^2dr = 1[/tex]

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

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

But that makes no sense and I am so confused.

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- #13

malawi_glenn

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[tex] d\Omega = sin\theta d\phi d\theta [/tex]

=)

=)

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