A question about ladder operator in QM Taking angular momentum for example:since $L^2$ commute with $L_z$,we have simultaneous eigenfunctions $Y_{lm}$. Define $L_{+} =L_x+iL_y$,we know $[L^2,L_+] = 0$ so the operator $L_+$ will not change the eigenvalue (for $L^2$) of the previous eigenfunction. Since $[L_z,L_+] = i\hbar L_+$, we know it will increase the eigenvalue for $L_z$ by a step, since the expectation of $L^2$ is constant (due to the fact that expectation value at corresponding eigenstate is the eigenvalue), the eigenvalue for $L_z$ cannot be raised to infinty by $L_+$.
In this case $L_+Y_{lm+} =0$ for some $m_+$. Similarly for $L_-Y_{lm-} = 0$ for some $m_-$. Since otherwise there must exist another step up to higher/lower eigenvalue that would exceed the bound, the question is why are both ladder operators sending the state to the same eigenfunction 0?
 A: I tried to change some of the English to make your question have more sense.
What I think you are asking is:

Given that $L_+ |m_{\mathrm{max}}\rangle = 0$ and $L_-|-m_{\mathrm{max}}\rangle = 0$ > for some $m_\mathrm{max}$, how can both
operators $L_+$ and $L_-$ (whose action is to shift to higher and to
lower eigenstates respectively) result in the same eigenstate $0$?

$0$ is not an eigenstate. It's just the number $0$. It's the same that you get when you annihilate the vacuum $a|0\rangle  = 0 $. You can show that assuming that lowering the lower bound returns the same bound, $a|0\rangle = |0\rangle$, leads to a contradiction. The same reasoning can be applied here to your example.
Keep in mind though that there indeed are some eigenvalues of $L_z$ that are $0$. But this is a "different" zero from the one we got above. If the eigenstate $|m_0\rangle$ has eigenvalue $0$ when acted on with the operator $L_z$, then a measurement of $L_z$ returns $0$: $L_z |m_0\rangle = 0$. However, the state remains $|m_0\rangle$. Whereas in the case above the 'state' has now become $0$. Meaning essentially there is no state available.
