How do you show that a state is a simultaneous eigenstate of $\hat L_z$ and $\hat L^2$? what is the general process for showing that a given state is a simultaneous eigenstate of the angular momentum operators $\hat L_z$ and $\hat L^2$? I've searched for a while but I'm not really getting anywhere.
 A: I'll elaborate on the answer through a different approach.
Suppose that A and B are hermitian operators, then the so-called simultaneous diagonalization theorem holds.
The thereom states that $[A,B] = 0 \:$ if and only if there exists an orthonormal basis such that both $A$ and $B$ are diagonal with respect to that basis, that is you can express $A=\sum_i a_i |i\rangle \langle i|$ and $B=\sum_i b_i |i\rangle \langle i|$, where $\{|i\rangle\}$ is some common orthonormal set of eigenvectors for A and B.
Since $L^{2}$ and $L_z$ are hermitian operators, if we can prove that they commute, i.e. that $[L^{2},L_z] = 0$, then, since the previous theorem holds, they share some common orthonormal set of eigenvectors. Then, as pointed out by Nihar Karve, we can find out if a given state is a simultaneous eigenstate of $L_z$ and $L^{2}$.
So let's see that $L^{2}$ and $L_z$ commute:
\begin{align}
[L^{2},L_z] & = [L_x^2 + L_y^2  +L_z^2,L_z] \\
 & = [L_x^2,L_z] + [L_y^2,L_z]  +[L_z^2,L_z] \\
& = [L_xL_x,L_z] + [L_yL_y,L_z]  +[L_zL_z,L_z] \\
& = Lx[Lx,Lz]+[Lx,Lz]Lx + L_y[L_y,L_z]+[L_y,L_z]L_y  +L_z[L_z,L_z]+[L_z,L_z]L_z \\
& = L_x(-i\hslash L_y)+(-i\hslash L_y)L_x + L_y(i\hslash L_x)+(i\hslash L_x)L_y  +0+0 \\
& = 0 \\
\end{align}
Then, if $ |\psi\rangle$ is a given eigenvector of $L_z$, we have:
$$ L_z|\psi\rangle = \lambda |\psi\rangle $$
Applying $L^{2}$ to both side of this equation we obtain:
$$L^{2}L_z|\psi\rangle = L^{2}\lambda |\psi\rangle =  \lambda L^{2} |\psi\rangle  $$
Since the operators commute, we have:
$$L_z \bigl( L^{2}|\psi\rangle \bigr) =  \lambda \bigl( L^{2} |\psi\rangle \bigr)  $$
This equation expresses the fact that $L^{2} |\psi\rangle$ is an eigenvector of $L_z$ with eigenvalue $\lambda$. Now, there can be two cases:

*

*if $\lambda$ is a non-degenerate eigenvalue, then $L^{2} |\psi\rangle$ is necessarily proportional to $|\psi\rangle$, thus $|\psi\rangle$ is also eigenvector of  $L^{2}$;

*if $\lambda$ is a degenerate eigenvalue, it can only be said that $L^{2} |\psi\rangle$ belongs to the eigensubspace $E_\lambda$ of $L_z$ and that $E_\lambda$ is invariant under the action of $L^{2}$.

Hope it can help.
A: You can show this by considering commutable operators (just like in the question)
$$[A,B]=0$$
let
$$A|\alpha\rangle=a|\alpha\rangle$$
now apply operator $B$ then name $B|\alpha\rangle=|\beta \rangle$
$$BA|\alpha \rangle = Ba|\alpha\rangle=a|\beta \rangle$$
based on the commutation relation, you get
$$BA|\alpha\rangle =AB|\alpha \rangle=A|\beta\rangle$$
From these last two equations it's clear that $|\beta\rangle$ is an eigenvector of $A$. This is true if $|\beta\rangle=b|\alpha\rangle$, check out
$$B|\alpha\rangle = |\beta\rangle =b|\alpha\rangle$$
you know $|\alpha\rangle$ is an eigenket of $A$ with eigenvalue "$a$" and now it's an eigenket of $B$ with eigenvalue "b". So commuting operators have simultaneous eigenkets.
Note: $[L^2, L_z]=0$
Edit
It was my bad, I misread the question. Above answer is to show how commutable operators like $L^2$ and $L_z$ have simultaneous eigenkets. It's easier to show if a state is eigenstate or not, you need to apply the operators to the given state and see if your eigenstates are rescaled.
A: If you are already given the state $| \psi \rangle$, it is very easy to see if it is the eigenstate of both $L^2$ and $L_z$. We just need to show that there exist numbers $\alpha,\beta$ such that:
\begin{align}
L^2 | \psi \rangle &= \alpha | \psi \rangle \\
L_z | \psi \rangle &= \beta | \psi \rangle
\end{align}
In practice we would probably use coordinate representation to actually see if those relations hold or not.
