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

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The discussion focuses on the action of angular momentum operators J_x, J_y, and their commutator [J_x, J_y] on the eigenstate |0,0⟩, which has eigenvalues of 0. The attempt to express J_x and J_y using ladder operators J_+ and J_- leads to the realization that these operators cannot change the m eigenvalue when j=0, as m must remain within the range defined by j. Consequently, applying J_x and J_y to |0,0⟩ results in zero, confirming that the outcome for all operations is indeed zero. This conclusion is deemed unsatisfactory but is consistent with the properties of the eigenstate. The final consensus is that all operations yield zero due to the constraints of the j=0 state.
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Homework Statement


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

2. The attempt at a solution
It seemed reasonable to write J_x and J_y in terms of ladder operators
<br /> J_{+}=J_x + iJ_y<br />
<br /> J_{-}=J_x -i J_y<br />
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
<br /> J_{+}|j,m\rangle = \hbar\sqrt{(j+m+1)(j-m)} |j,m+1\rangle<br />
<br /> J_{-}|j,m\rangle = \hbar \sqrt{(j-m+1)(j+m)}|j,m-1\rangle<br />
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,
 
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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.
 

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