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I have a doubt. If Legendre equation has a polynomial solution, is the constant $l$ in $l(l+1)$ necessarily a integer number? Asked in another way, is possible $l(l+1)$ be a integer if $l$ is not an integer?

Thanks in advance.

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up vote 1 down vote accepted

Yes (although this is not really relevant for the Legendre polynomials): $$l = \frac{-1 + \sqrt{5}}{2}$$ $$l(l+1) = 1$$

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I think it is relevant because you can have physical relevant solutions in the polar angle when $l$ is not integer. I was reading a book saying that l must be integer to have physical meaning. What do you think? – TheStudent Nov 3 '12 at 19:48
It is not relevant for physics. Not considering half-integer intrinsic angular momentum (spin), the (orbital) angular momentum quantum number is always an integer. The reason that your text book says that non-integer $l$ is not relevant, is that in that case the resulting solution of the Schrodinger equation is not square-integrable - and hence the solution does not fulfill the probability postulate of quantum mechanics. – Maestro13 Apr 23 '14 at 6:19

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