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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?

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

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I think this is just a simple arithmetic error. It's k = l + 1 so k = 2 (for the l = 1 case) , not k + 1 = 2 (which is the forbidden k = l).

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  • $\begingroup$ But, since $\alpha=1/2$, for $k=2$ the recursion formula gives $a_3=0$. And it's wrong. $\endgroup$ Commented Dec 21, 2020 at 14:11
  • $\begingroup$ @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. $\endgroup$
    – Ruslan
    Commented Dec 21, 2020 at 14:23
  • $\begingroup$ 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. $\endgroup$ Commented Dec 21, 2020 at 15:46
  • $\begingroup$ @EmilioNovati You can't take k + 1 = 2 for the l = 1 case. It's the l = 1, k=1 case which the text explicitly addresses in the line "The index k cannot be equal to l, the denominator becomes zero and al+1 is infinite." If you get a2 = 0 and all later terms are also zero this is perfectly legitimate, not a sign that anything has gone wrong. $\endgroup$ Commented Dec 21, 2020 at 21:54
  • $\begingroup$ 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? $\endgroup$ Commented Dec 22, 2020 at 9:42

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