# Hydrogen Atom Radial Wave Function (limit problem)

1. Dec 4, 2009

### Master J

The Hydrogen Atom wave function.

With the substitution u(r) = r.R(r)

p=kr

We get a simplified version: d^2u/dp^2 = [1 - (p_0)/p + l(l + 1)/(p^2) ].u

Im sure some of you have seen that before.

Now, in the limit, p goes to infinity, I understand that we get u = A.exp[-p], but in the limit that p goes to 0, the answer is:

d^2u/dp^2 = [l(l + 1)/(p^2)].u

I dont get where that comes from. Surely the terms with p at the bottom explode and become infinite??

Last edited: Dec 4, 2009
2. Dec 4, 2009

### Bob_for_short

Yes, it is the biggest term in the square bracket. It should be multiplied by u too.

3. Dec 4, 2009

### Master J

But I dont get it. Why do the 2 terms with p at the bottom not become infinity???

4. Dec 4, 2009

### George Jones

Staff Emeritus
As p becomes small, the two terms become large at different rates, with the second term getting large faster. So, for small enough (but non-zero) p, the second term will be much larger than the first, and the first term can be ignored.

As a simple example, consider

A = 1/x + 1/x^2

with x = 0.001. What does A equal? What does 1/x equal? What does 1/x^2 equal? Is 1/x^2 a good approximation to A?

5. Dec 6, 2009

### Master J

What happens to the one then?

Should the answer not be [1 +l(l + 1)/(p^2)].u

6. Dec 6, 2009

### George Jones

Staff Emeritus
Again, if p is small enough (but still nonzero), l(l + 1)/(p^2) will be much larger than one, so the one can be ignored.

7. Dec 7, 2009

### Master J

Thanks for the help, it is very much appreciated!!!

I figured out the last question anyway.

Now......sums!!!! Yay!

So its still the same ball game. 2 solutions have been found for the 2 limiting cases, as p goes to zero and infinity. Using these 2 regimes, we assume a solution that is the product of these to aforementioned solutions, and another function of p, which is to be determined, call it v(p) (while Im here, is this a general rule that one can assume that type of solution??)

Assuming a series solution:

v' = $$\Sigma$$ j.(c_j).(p)^(j-1)

the book equates this to $$\Sigma$$ (j+1).(c_j+1).(p)^j

where both sums are from zero to infinity, index of summation is j, and c is the coefficient. Im unsure about how the above equality is made, and why??

Then, we end up with a recursive formula for the coeffecients. But if this expansion is compatable with the previous asymptotic analsis, it must terminate after a finite j value, j_max. Why is this?

I understand there is alot there, any help is appreciated!

Last edited: Dec 7, 2009
8. Dec 7, 2009

### RedX

Is it a general rule to assume that type of solution? Well, it's mathematically legal. Usually what happens in these type of problems is that the asymptotic solution has two solutions, one of which blows up somewhere, e.g., $$e^{\pm r}$$ for the hydrogen atom. So you want to choose - and not + by writing $$R(r)=U(r)e^{-r}$$. When you solve U(r) in the new differential eqn by power series, you find that the solution of U(r) in general behaves like $$e^{2r}$$ at large r, which gives you back $$R(r)=e^{2r}e^{-r}=e^{r}$$, so the solution blows up for large r. This is to be expected in general, because you already established $$e^r$$ was a valid solution to the differential equation at large r. However, there are certain choices of the energy that shuts down the power series by making it a finite series, so that U(r) would be polynomial and not behave like an exponential at large r. This quantizes the energy.

The very same type of thing happens in the harmonic oscillator. You get two asymptotic solutions, one of which that blows up. You plug in the product of the one that doesn't blow up with an arbitrary function into the differential equation to get a new equation. In general, the solution to the new equation when solved in power series, multiplies with the well-behaved asymptotic solution to give the mis-behaved asymptotic solution. But certain values of energies cuts of the power series, so you avoid all this, and energy gets quantized.