The problem is like this:

Let $$L_x = yp_z - zp_y, L_y = zp_x - xp_z, L_z = xp_y- yp_x, \\ L^2 = {L_x}^2 + {L_y}^2 + {L_z}^2 \\ L_\pm \equiv L_x \pm iL_y $$

We wish to find a "top rung" $f_t$ and a "bottom rung" $f_b$ $$ L_+ f_t = 0 \qquad L_-f_b = 0 $$

So he let $\hbar l$ be the eigenvalue of $L_z$ at this top rung: $$ L_z f_t = \hbar l f_t \qquad L^2f_t = \lambda f_t $$

so he found $ \lambda = \hbar^2 l(l+1)$

To solve the "bottom rung", he let $\hbar \bar l$ be the eigenvalue of $L_z$ at this bottom rung: $$ L_z f_b = \hbar \bar l f_b \qquad L^2f_b = \lambda f_b $$

so he found $ \lambda = \hbar^2 \bar l(\bar l+1)$

This allows us to show $\bar l = -l $

So questions here:

why do we assume the same $\lambda$ value when $L^2$ is applied on two different functions $f_t$ and $f_b$?


The eigenvalue of $L^2$ cannot depend on the state $f_t$ or $f_b$ since $L^2$ commutes with any $L_z, L_\pm$. In other words, since $$ L^2 L_+\,f= L_+ L^2 f \tag{1} $$ and since by definition $L^2 f=\hbar^2 l(l+1) f$, it follows from (1) that $L_+\,f$ will have the same eigenvalue for $L^2$ as $f$, as will all states of the ladder.

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