Quantum: Spherical Harmonics Not Including r^2 term?

[laser1]

Homework Statement

N/A

Relevant Equations

N/A

In the solutions, I must integrate $Y_1^0 Y_1^{\pm 1}$. But my concern is when it states that the differential element (I think area?) is $d\theta d\phi$. I know that as

then $dA=r^2 \sin \theta d\theta d\phi$. But the solution only states $\sin \theta d\theta d\phi$. Why is this? Also, the units are not even area, so I have some problems with what this differential element is.

 

[pines-demon]

Is not the differential of area but the solid angle. I am not sure what you want to understand here, the spherical harmonics only are in radial coordinates and pertain only to the angular part.

 

[laser1]

Is not the differential of area but the solid angle. I am not sure what you want to understand here, the spherical harmonics only are in radial coordinates and pertain only to the angular part.

Where does the $\sin \theta d\theta d\phi$ come from? I had thought it came from the fact that the jacobian in spherical coordinates is $r^2 \sin \theta$ but maybe I was wrong.

 

[D H]

In the solutions, I must integrate $Y_1^0 Y_1^{\pm 1}$.

Do that and you should get zero, and you will get zero if you do the integration correctly. They are orthogonal functions, after all. What you need to integrate are $\left(Y_1^0\right)^2$ and $Y_1^{\pm1}\,\left(Y_1^{\pm1}\right)^\ast$.

But my concern is when it states that the differential element (I think area?) is $d\theta d\phi$.
then $dA=r^2 \sin \theta d\theta d\phi$. But the solution only states $\sin \theta d\theta d\phi$. Why is this? Also, the units are not even area, so I have some problems with what this differential element is.

You are integrating over the surface of the unit sphere, so $r=1$.

 

[laser1]

You are integrating over the surface of the unit sphere,

why unit sphere rather than an arbitrary sphere?

 

[kuruman]

In conventional spherical coordinates an area element $dA$ on the surface of a sphere of radius $r$ is the product of two arc lengths, $ds_{\theta}=r~d\theta$ and $ds_{\phi}=r\sin\!\theta~d\phi$ so that $$dA=ds_{\theta}\times ds_{\phi}=r^2~\sin\!\theta d\theta~d\phi.$$By definition this area subtends a solid angle $$d\Omega=\frac{dA}{r^2}=\sin\!\theta d\theta~d\phi.$$ The radius is irrelevant.

The orthonormality condition for the spherical harmonics is $$\int Y^*_{l'm'}(\theta,\phi)Y_{lm}(\theta,\phi)~d\Omega=\int_0^{2\pi}d\phi\int_0^{\pi}Y^*_{l'm'}(\theta,\phi)Y_{lm}(\theta,\phi)\sin\!\theta d\theta=\delta_{ll'}\delta_{mm'}.$$There is no $r$.

 

[pines-demon]

Where does the $\sin \theta d\theta d\phi$ come from? I had thought it came from the fact that the jacobian in spherical coordinates is $r^2 \sin \theta$ but maybe I was wrong.

Solid angle

 

[laser1]

Solid angle

Thanks, I guess the justification is "a unit sphere by definition" then!