Associated vs. Non-associated Laguerre Polynomials

[terp.asessed]

Homework Statement

Could someone pls clarify if the value of x changes from just Laguerre polynomial to associated one? I am confused about the role of variable x.

Homework Equations

From what I have learned in the class, I understand that L¹_n(x) = d/dx Ln(x), n = 1, 2, 3...

The Attempt at a Solution

Because L₁(x) = 1 - x and L₂(x) = 2 - 4x + x2
I did:
L₁¹(x) = d/dx L1(x) = d/dx (1 - x) = -1
L¹₂(x) = d/dx L2(x) = d/dx (2-4x+x2) = 2x - 4 = 2(x - 2)...I wonder if x is a different function of radius in L₁¹(x) (as in 1s orbital) and L¹₂(x) (as in 2s orbital)? I am assuming the orbital polynomial on the basis of node...as in:
L₁¹(x)= 1 because of 0 nodes...hence 1s
L¹₂(x) = 2x -4 because of 1 node...hence 2s

 

[DrClaude]

I wonder if x is a different function of radius in L₁¹(x) (as in 1s orbital) and L¹₂(x) (as in 2s orbital)?

Yes. The associated Laguerre polynomial found in hydrogenic wave functions is expressed as
$$
L_{n+l}^{2l+1}(\rho)
$$
with
$$
\rho \equiv \frac{2 Z}{n a} r
$$
where $n$ the principal quantum number and $l$ the angular momentum quantum number. You see that $n$ enters as a scaling factor in the argument of the associated Laguerre polynomial. Note also that the association between $n$ and $l$ and the associated Laguerre polynomial is a bit more complicated than your simple assumption based on the number of nodes, as angular nodes have to be considered also.