What is the result of J_x, J_y, and [J_x,J_y] acting on a specific eigenstate?

[jfy4]

Homework Statement

Let $|0,0\rangle$ be the simultaneous eigenstate of $\mathbf{J}^2$ and $J_z$ with eigenvalues 0 and 0. Find
$$J_x|0,0\rangle \quad\quad J_y |0,0\rangle \quad\quad [J_x,J_y]|0,0\rangle$$

2. The attempt at a solution
It seemed reasonable to write $J_x$ and $J_y$ in terms of ladder operators
$$J_{+}=J_x + iJ_y$$
$$J_{-}=J_x -i J_y$$
and then to have them operate on the states (for the last one I would just use the x,y,z commutation relations). But I was looking at the normalization constant out front
$$J_{+}|j,m\rangle = \hbar\sqrt{(j+m+1)(j-m)} |j,m+1\rangle$$
$$J_{-}|j,m\rangle = \hbar \sqrt{(j-m+1)(j+m)}|j,m-1\rangle$$
but given the initial state, these seem to be zero... Please tell me I'm doing this wrong, zero is such a unsatisfactory answer...

Thanks,

 

[Matterwave]

Because you are in a j=0 state, you can't actually raise or lower the m eigenvalue since m ranges between -j and +j so in this case it must stay 0.

So, it would seem like indeed you get 0 for everything.