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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?

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  • $\begingroup$ 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. $\endgroup$
    – march
    Commented Mar 16, 2021 at 20:40

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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.

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