Hydrogen probability distribution

[unscientific]

I have found that:

For l = 1:

\sum_{m=-l}^l |Y_l^m|^2 = \frac{3}{4\pi}

For l = 2:

\sum_{m=-l}^l |Y_l^m|^2 = \frac{5}{4\pi}

What significance does this have for the probability distribution in an hydrogen atom?

 

[jfizzix]

the significance is that there are five-thirds as many states of orbital angular momentum l=2, than for l=1. This makes sense because there are five orbitals for l=2, and three orbitals for l=1, for all the values of m.

 

[Simon Bridge]

Do you know what those terms mean?
Do you know why you did the sum in each case?

 

[unscientific]

Do you know what those terms mean?
Do you know why you did the sum in each case?

I think the numerator gives the degeneracy for a certain l?

since -l ≤ m ≤ l, m can take on 2l+1 possible values.

Like possible states for |n,l,m>:

For l = 1:

Possible states are |n,1,-1> and |n,1,0> and |n,1,1>

 

[Simon Bridge]

I think the numerator gives the degeneracy for a certain l?

Each spin/orbit state contributes a factor of $1/4\pi$ to the overall thingy being calculated.

But I meant the terms on the RHS.

For a particular |n,l,m> state, what does |Y_l^m|² mean?

Now - what was your question?