# Hydrogen probability distribution

## Main Question or Discussion Point

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?

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jfizzix
Gold Member
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.

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Simon Bridge
Homework Helper
Do you know what those terms mean?
Do you know why you did the sum in each case?

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
Homework Helper
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 |Ylm|2 mean?

Now - what was your question?