1
$\begingroup$

This is a very simple question, yet I'm not sure how to approach it. I want to calculate the spherical harmonic:

$$ Y_{l m}^{*}(\theta = \pi/2, \phi) $$

I know the general formula:

$$ Y_{l m}^{*}(\theta, \phi)=\sqrt{\frac{2 l+1}{4 \pi} \frac{(l-m) !}{(l+m) !}} P_{l}^{m}(\cos \theta) e^{-i m \phi} $$

But for $\pi/2$ I need to calculate the associate Legendre polynomial of 0 ($P_{l}^{m}(\cos \pi/2)$), which I'm not sure how.

The Rodrigues' formula is not clear to me for the case of $x=0$:

$$ P_{l}^{m}(x)=\frac{(-1)^{m}}{2^{l} l !}\left(1-x^{2}\right)^{m / 2} \frac{d^{l+m}}{d x^{l+m}}\left(x^{2}-1\right)^{l} $$

Any guidance on how to calculate it for that special case would be appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

For each choice of $l$ and $m$, you can find a closed form solution for $P_l^m(x)$ as a function of $x$. Typically, I would start there and only then evaluate it at a particular value of $x$ such as $x = 0$.

That said, if you prefer we can find a formula for $P_l^m(0)$ in terms of $l$ and $m$

Instead of starting from the Rodrigues' formula, we could just use this closed form [1] of the associated Legendre polynomials:

$$P_l^m(x)=(-1)^{m} \cdot 2^{l} \cdot (1-x^2)^{m/2} \cdot \sum_{k=m}^l \frac{k!}{(k-m)!}\cdot x^{k-m} \cdot \binom{l}{k} \binom{\frac{l+k-1}{2}}{l}$$

where $\binom \alpha k$ is the generalized binomial coefficient: $$\binom \alpha k = \frac{1}{k!} \prod_{i=0}^{k-1} (\alpha-i)$$

In particular, for $x = 0$ and $k > m$ we have $x^{k-m} = 0$. Therefore, the only term in the sum of $k$ which may be non-zero is the $k = m$ term.

Now, if we naively plug in $x = 0$ and $k = m$, we'd get $x^{k-m} = 0^0$, which is indeterminate. But there are a few reasons I feel confident the value it takes here should be $1$. For one thing, I can see that this will give me the correct values for the first few associated Legendre polynomials, such as $P_0^0(x) = 1$. For another thing, I know that $P_l^m(x)$ for fixed $l$ and $m$ should be a continuous function in $x$, and clearly any small but non-zero $x$ to the $0$-th power is $1$.

But if we wanted to be rigorous, we could go back to the derivation of this closed form and see for ourselves that we could have instead produced a series over $k$ from $m + 1$ to $l$, and with a separate term in front (playing the same role as our $k = m$ term did previously). This would give us a closed-form solution like this:

$$P_l^m(x) = (-1)^{m} \cdot 2^{l} \cdot (1-x^2)^{m/2} \left[m! \binom{l}{m} \binom{\frac{l+m-1}{2}}{l} + \sum_{k=m+1}^l \frac{k!}{(k-m)!}\cdot x^{k-m} \cdot \binom{l}{k} \binom{\frac{l+k-1}{2}}{l} \right]$$

Now all terms of the sum are positive powers of $x$, so at $x = 0$ the whole summation goes away and we're left with this:

$$P_l^m(0) = (-1)^{m} 2^{l} m! \binom{l}{m} \binom{\frac{l+m-1}{2}}{l}$$

After replacing the binomial coefficients using the definition above and making some cancelations, we have: $$P_l^m(0) = \frac{(-1)^{m} 2^{l}}{(l-m)!} \prod_{i=0}^{l-1} \left(\frac{l+m-1}{2} - i\right)$$


We can check a few values just to be sure. For example: $$P_{3}^{1}(x)=-\tfrac{3}{2}(5x^{2}-1)(1-x^2)^{1/2} \\ P_{4}^{2}(x)=\frac{15}{2}(7x^2 - 1)(1-x^2)$$ so $P_{3}^{1}(0) = \frac{3}{2}$ and $P_{4}^{2}(0) = - \frac{15}{2}$

From the above formula, we have:

$$P_3^1(0) = \frac{(-1)^{1} 2^{3}}{(3-1)!} \prod_{i=0}^{3-1} \left(\frac{3+1-1}{2} - i\right) = \frac{-8}{2!} \prod_{i=0}^{2} \left(\frac{3}{2} - i\right) \\ = -4 \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{-1}{2} = \frac{3}{2}$$

and

$$P_4^2(0) = \frac{(-1)^{2} 2^{4}}{(4-2)!} \prod_{i=0}^{4-1} \left(\frac{4+2-1}{2} - i\right) = \frac{16}{2!} \prod_{i=0}^{3} \left(\frac{5}{2} - i\right) \\ = 8 \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{-1}{2} = - \frac{15}{2}$$

$\endgroup$
1
  • $\begingroup$ Beautiful answer! Thank you !! $\endgroup$ Mar 15, 2022 at 5:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .