Spherical Harmonics: Showing $\delta_{m,0}\sqrt{\frac{2\ell + 1}{4\pi}}$

[Dustinsfl]

I am trying to show that
\[
Y_{\ell}^m(0,\varphi) = \delta_{m,0}\sqrt{\frac{2\ell + 1}{4\pi}}.
\]
When \(m = 0\), I obtain \(\sqrt{\frac{2\ell + 1}{4\pi}}\).

However, I am not getting 0 for other \(m\). Plus, to show this is true, I can't methodically go through each \(m\).

How can I do this?

 

[topsquark]

I am trying to show that
\[
Y_{\ell}^m(0,\varphi) = \delta_{m,0}\sqrt{\frac{2\ell + 1}{4\pi}}.
\]
When \(m = 0\), I obtain \(\sqrt{\frac{2\ell + 1}{4\pi}}\).

However, I am not getting 0 for other \(m\). Plus, to show this is true, I can't methodically go through each \(m\).

How can I do this?

The \(\displaystyle \delta _{m, 0} \) forces the expression to be 0 unless m = 0. There is no other m to compute with. I'm not sure what you are trying to get at with the "other" m values?

-Dan

 

[Dustinsfl]

The \(\displaystyle \delta _{m, 0} \) forces the expression to be 0 unless m = 0. There is no other m to compute with. I'm not sure what you are trying to get at with the "other" m values?

-Dan

I understand the kronecker delta. I am trying to show the identity is true.

 

[topsquark]

I understand the kronecker delta. I am trying to show the identity is true.

Oh! I see the problem now. Yes, the expression is not correct.
\(\displaystyle Y_l^m( \theta, \phi ) = (-1)^m \sqrt{ \frac{2l + 1}{4 \pi} \frac{(l - m)!}{(l + m)!}} P_l^m( cos ( \theta ) ) e^{i m \phi }\)

Gah! I can't get the LaTeX to code the second line. Anyway, the only term that drops out for theta = 0 is the associated Legendre polynomial. Not much of a simplification. Are you perhaps adding some together?

-Dan

 

[Dustinsfl]

Oh! I see the problem now. Yes, the expression is not correct.
\(\displaystyle Y_l^m( \theta, \phi ) = (-1)^m \sqrt{ \frac{2l + 1}{4 \pi} \frac{(l - m)!}{(l + m)!}} P_l^m( cos ( \theta ) ) e^{i m \phi }\)

Gah! I can't get the LaTeX to code the second line. Anyway, the only term that drops out for theta = 0 is the associated Legendre polynomial. Not much of a simplification. Are you perhaps adding some together?

-Dan

No but I am pretty sure it is correct. I have Mathematica so I have entered in SphericalY[l,m,0,\phi] and tried different l's and m's, but every time m is nonzero, I do get zero back.

 

[topsquark]

No but I am pretty sure it is correct. I have Mathematica so I have entered in SphericalY[l,m,0,\phi] and tried different l's and m's, but every time m is nonzero, I do get zero back.

Okay, yes you are correct. I had been thinking that \(\displaystyle P_l^m(1) = 1\) but that's only true for m = 0. When theta = 0 \(\displaystyle P_l^m(1) = 0\) for non-zero m as all the non-zero m are proportional to sin(theta).

-Dan