# Questions about finding “top rung” and “bottom rung” of angular momentum operator (Proof in Griffiths)

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.