Undergrad Why is separation constant l(l+1) instead of +-l^2?

[Rupul Chandna]

While separating variables in the Schrödinger Equation for hydrogen atom, why are we taking separation constant to be l(l+1) instead of just l^2 or -l^2, is it just to make the angular equation in the form of Associated Legendre Equation or is there a deeper meaning to it?

 

[Orodruin]

Because those are the eigenvalues to the angular operator. You do not ”choose” the eigenvalues, they are what they are based on the differential equation and boundary conditions. In the case of the angular part of the Laplace operator, the eigenfunctions have those eigenvalues.

 

[strangerep]

While separating variables in the Schrödinger Equation for hydrogen atom, why are we taking separation constant to be l(l+1) instead of just l^2 or -l^2, is it just to make the angular equation in the form of Associated Legendre Equation or is there a deeper meaning to it?

Take a look at Ballentine section 7.1, where he derives the quantum angular momentum spectrum very directly using operators on an abstract Hilbert space, without all the extra baggage of differential equations.

He starts of using a symbol $\beta$ as the eigenvalue of ${\mathbf J}^2$, but after further analysis using $J_z$ and $J_\pm$ to determine $J_z$'s range of eigenvalues he derives a constraint on $\beta$ such that $\beta = j(j+1)$, where $|j|$ bounds the possible eigenvalues of $J_z$.

When you see an author simply choosing the value $j(j+1)$ (or $\ell(\ell+1)$ in your text), it just means that author is lazily skipping over the necessary extra analysis, as given in Ballentine's textbook.