# Question about the procedure to find eigenvalues for angular momentum in QM

I am reviewing griffiths QM explanation of Angular momentum, and when finding their eigenvalues he uses a ladder approach. He states that there has to be a "top rung", where $$L_+f_t=0$$, and a lower one where $$L_-f_b=0$$.

He also states that the total angular momentum for the top state is $$L^2f_t=\hbar^2l(l+1)f_t$$ and for the bottom one is $$L^2f_b=k(k-1)f_b$$

I understand how he got this far. Next he states that those values have to be equal, so: $$l(l+1)=k(k-1)$$ which means that $$k=-l$$ or $$k=l+1$$. I understand why only the first option is valid, but I don't know on what grounds one can compare these two expressions, or what logic is behind that.

Could someone explain this to me please?

• Sorry, I played a little fast and loose with the first version of the answer. The edit is actually correct, whereas the first was not. Commented Mar 16, 2021 at 20:40

You'll notice at the start of the derivation that he picks a particular eigenvector, corresponding to the "top rung of the ladder", that satisfies the eigenvalue equations, given by $$L_zf_t = \hbar l f_t;~~~~~~~~L^2f_t = \lambda f_t.$$ Since this is a particular eigenvector, $$\lambda$$--even though it's unknown--has a particular value. In the derivation that follows, he derives an expression for $$\lambda$$.
He then starts at the "bottom rung of the ladder", satisfying $$L_zf_b = \hbar \tilde{l} f_b;~~~~~~~~L^2f_b = \lambda f_b.$$ Crucially, this is the same $$\lambda$$ because of what he showed on the previous page: by stepping up and down using the ladder operators, we get to states that are still eigenvectors of both $$L_z$$ and $$L^2$$, but the eigenvalue of $$L_z$$ is changed while the eigenvalue of $$L^2$$ is not (this last point is the important one!).
Thus, he is able to derive a second expression for $$\lambda$$, and they must of course be equal.