# Hydrogen radial wave function in Feynman

I'm reading the classical Feynman's lectures on hydrogen atom, and I want to calculate the radial component of the wave function with the formula (19.53)

$$F_{n,l}(\rho)=\frac{e^{-\alpha\rho}}{\rho}\sum_{k=l+1}^n a_k \rho^k$$ for $$n=2$$ and $$l=1$$. So I have $$\alpha=1/n=1/2$$ and $$F_{2,1}(\rho)=\frac{e^{-\rho/2}}{\rho}a_2 \rho^2$$ where $$a_2$$ should be given by the formula (19.50): $$a_{k+1} =\frac{2(\alpha k-1)}{k(k+1)-l(l+1)} a_k$$ but, for $$k+1=2$$ and $$l=1$$ the denominator becomes $$0$$.

It seems that I have some stupid mistake, but I don't see where. Someone can help me?

• The lower limit of the sum in your first equation requires... Dec 20, 2020 at 23:13
• It's k = l + 1, so k = 2, not k + 1 = 2 (which is the forbidden k = l case). Dec 21, 2020 at 4:26

• But, since $\alpha=1/2$, for $k=2$ the recursion formula gives $a_3=0$. And it's wrong. Dec 21, 2020 at 14:11
• @EmilioNovati why is it wrong? $k$ is supposed to span $\{l+1,...,n\}$, i.e. $\{2\}$. I.e., there's only one term in the whole sum. The values of $a_k$ other than $a_2$ are irrelevant. Or, well, since you're using an upwards recurrence relation, then the values of $a_k$ for $k>n$ are irrelevant. Dec 21, 2020 at 14:23
• I see. But my problem is: if I take $k+1=2$ the recurrence gives $a_2=-1/(1\cdot 2-1\cdot 2)$ and if i take $k=2$ the recurrence gives $a_2=0$. And I don't see where is the mistake. Dec 21, 2020 at 15:46
• Now I understand!! (I hope). We have to choose $a_2$, that is the first step in the recursion, say $a_2=1$. Then we can use the recursive formula. Correct? Dec 22, 2020 at 9:42