Radial Wave Function: Solving Equation for Normalized Radial Wave Function

[Shackleford]

Radial Wave Function; Normalized Radial Wave Function

5. Is the brute-force method the only way to show the Equation is satisfied? By that I mean differentiating R20 once, twice, and substituting it in? I tried that earlier, but it was very nasty.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18100956.jpg?t=1284823040

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18100944.jpg?t=1284823023

7. I'm not quite sure what I'm doing wrongly. I think I'm kind of close to the correct procedure.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18114302.jpg?t=1284828367

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18114318.jpg?t=1284828434

 

[vela]

5. Is the brute-force method the only way to show the Equation is satisfied? By that I mean differentiating R20 once, twice, and substituting it in? I tried that earlier, but it was very nasty.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18100956.jpg?t=1284823040

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18100944.jpg?t=1284823023

I'm afraid you're going to have to revise your notion of nasty. The function is of the form c(1-x)e^-x. It's pretty straightforward to show it satisfies the radial equation by direct substitution.

7. I'm not quite sure what I'm doing wrongly. I think I'm kind of close to the correct procedure.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18114302.jpg?t=1284828367

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-18114318.jpg?t=1284828434

You need a factor of r² in your normalization integral. It comes from the volume element in spherical coordinates.

 

[Shackleford]

I'm afraid you're going to have to revise your notion of nasty. The function is of the form c(1-x)e^-x. It's pretty straightforward to show it satisfies the radial equation by direct substitution.

You need a factor of r² in your normalization integral. It comes from the volume element in spherical coordinates.

I'll look at it again. I must not have simplified it enough. But having to do the product rule so many times made it "nasty."

Also, I'm only integrating with respect to r, not theta and phi. Do I still need r^2? I'm a bit rusty on my coordinate transformations.

 

[vela]

Also, I'm only integrating with respect to r, not theta and phi. Do I still need r^2? I'm a bit rusty on my coordinate transformations.

Yes, you need the factor of r². The complete wave function is ψ_nlm=R_nl(r)Y_lm(θ,φ), and its normalization condition is

1=\int\psi_{nlm}^*\psi_{nlm}r^2drd\Omega=\int R_{nl}^*R_{nl}r^2dr\int Y_{lm}^*Y_{lm}d\Omega

By convention, the individual pieces are normalized so that the radial and angular integrals are equal to 1 individually.

 

[Shackleford]

Yes, you need the factor of r². The complete wave function is ψ_nlm=R_nl(r)Y_lm(θ,φ), and its normalization condition is

1=\int\psi_{nlm}^*\psi_{nlm}r^2drd\Omega=\int R_{nl}^*R_{nl}r^2dr\int Y_{lm}^*Y_{lm}d\Omega

By convention, the individual pieces are normalized so that the radial and angular integrals are equal to 1 individually.

So, d-omega is just d-theta, d-phi, right?

For this problem I only need to show the radial function is normalized. How do I show the angular function is similarly normalized? Or is that even possible given that I do not know the magnetic quantum number?

Of course, showing the radial function is normalized is quite easy in this case.

 

[vela]

So, d-omega is just d-theta, d-phi, right?

Not exactly. It's sin θ dθ dφ, which is sometimes written d(cos θ) dφ.

For this problem I only need to show the radial function is normalized. How do I show the angular function is similarly normalized? Or is that even possible given that I do not know the magnetic quantum number?

You can't. Without knowing m because you don't know what Y_lm is (though there's only three possibilities), but you're not asked about the angular part anyway. I was just showing you why the factor of r² appears in the integral for normalizing the radial wave function.

Of course, showing the radial function is normalized is quite easy in this case.

 

[Shackleford]

Not exactly. It's sin θ dθ dφ, which is sometimes written d(cos θ) dφ.

Ah. I forgot about those extra terms that appear when you transform to polar spherical coordinates. I think we covered how they got there in my vector analysis class last semester, but I forgot how to do that. And I'm not talking about simply using the little formulas that transform from Cartesian to polar cylindrical/spherical. We used tensor notation for all that stuff.

You can't. Without knowing m because you don't know what Y_lm is (though there's only three possibilities), but you're not asked about the angular part anyway. I was just showing you why the factor of r² appears in the integral for normalizing the radial wave function.

That's what I figured. I always like to know the general case or context, so I'm glad you wrote everything out.

 

[Shackleford]

Well, chief, it hasn't been pretty straightforward. Forgive my chicken scratch.

http://i111.photobucket.com/albums/n149/camarolt4z28/1-1.jpg?t=1284850615

http://i111.photobucket.com/albums/n149/camarolt4z28/2-1.jpg?t=1284850626

http://i111.photobucket.com/albums/n149/camarolt4z28/3.jpg?t=1284850640

 

[vela]

Well, I'm not terribly interested in checking your algebra, but I can make some suggestions to make the algebra a bit less tedious. First, pull a factor of 2 out so you have

R_{20}=\frac{1}{\sqrt{2}a_0^{3/2}}\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0}

Note when you do this, r always appears over 2a₀. Next, when you find R'₂₀, simplify before proceeding. You should find

R'_{20} = \frac{1}{\sqrt{2}a_0^{3/2}}\left[\frac{1}{2a_0}\left(\frac{r}{2a_0}-2\right)e^{-r/2a_0}\right]

Keep the original normalization constant separate because it's going to be common to every term so you can cancel it out. Same with the exponential factor.

Finally, the sum of the first two terms of the differential equation will be a polynomial multiplying an exponential and the normalization constant. That polynomial will factor. The product of one factor, the exponential times, and the normalization constant will be R₂₀.

 

[Shackleford]

Well, I'm not terribly interested in checking your algebra, but I can make some suggestions to make the algebra a bit less tedious. First, pull a factor of 2 out so you have

R_{20}=\frac{1}{\sqrt{2}a_0^{3/2}}\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0}

Note when you do this, r always appears over 2a₀. Next, when you find R'₂₀, simplify before proceeding. You should find

R'_{20} = \frac{1}{\sqrt{2}a_0^{3/2}}\left[\frac{1}{2a_0}\left(\frac{r}{2a_0}-2\right)e^{-r/2a_0}\right]

Keep the original normalization constant separate because it's going to be common to every term so you can cancel it out. Same with the exponential factor.

Finally, the sum of the first two terms of the differential equation will be a polynomial multiplying an exponential and the normalization constant. That polynomial will factor. The product of one factor, the exponential times, and the normalization constant will be R₂₀.

Well, I took out the alpha and only differentiate the non-constant terms. Quite a few of them canceled. I think my algebra is correct up until the last image I posted. That's probably the only work that needs review. But let me incorporate your simplifications and see what I get.

Okay. I got the same thing for R'20. I assume with R'', I can bring out the 1/2a0 to make things easier since it's just another constant.

 

[Shackleford]

Here's what I got for R''20.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-19135530.jpg?t=1284922635

This is pissing me off. Nothing canceled this time, assuming I did R'' correctly.

 

[vela]

That matches what I got, so now you have

R''(r)+\frac{2}{r}R'(r) = \frac{1}{\sqrt{2}a_0^{3/2}}e^{-r/2a_0}\left[-\frac{1}{(2a_0)^2}\left(\frac{r}{2a_0}-3\right)+\frac{2}{r}\frac{1}{2a_0}\left(\frac{r}{2a_0}-2\right)\right]

Since r/2a₀ appears to be a natural combination, I'd write the factor of 2/r in the differential equation in terms of it:

\frac{2}{r} = 2\left(\frac{1}{2a_0}\right)\left(\frac{2a_0}{r}\right)

then you'll have a common factor of 1/(2a₀)² that you can factor out. You should be left with some combination of constants and powers of r/2a₀.

I misspoke earlier when I said you'd end up with a polynomial multiplying the exponential. You actually have to pull out a factor of 2a₀/r before you get the polynomial you can factor.

 

[Shackleford]

That matches what I got, so now you have

R''(r)+\frac{2}{r}R'(r) = \frac{1}{\sqrt{2}a_0^{3/2}}e^{-r/2a_0}\left[-\frac{1}{(2a_0)^2}\left(\frac{r}{2a_0}-3\right)+\frac{2}{r}\frac{1}{2a_0}\left(\frac{r}{2a_0}-2\right)\right]

Since r/2a₀ appears to be a natural combination, I'd write the factor of 2/r in the differential equation in terms of it:

\frac{2}{r} = 2\left(\frac{1}{2a_0}\right)\left(\frac{2a_0}{r}\right)

then you'll have a common factor of 1/(2a₀)² that you can factor out. You should be left with some combination of constants and powers of r/2a₀.

I misspoke earlier when I said you'd end up with a polynomial multiplying the exponential. You actually have to pull out a factor of 2a₀/r before you get the polynomial you can factor.

Okay. That's making sense. I don't know why I didn't factor out the common terms. Let me see how far I can get now.

Besides the common terms, I'm getting -r/2a₀ + 7 -6a₀/r