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

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SUMMARY

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⟩. The participants confirm that since |0,0⟩ is a j=0 state, applying the ladder operators J_+ and J_- results in zero, as the m eigenvalue cannot be raised or lowered. Consequently, the results for J_x|0,0⟩, J_y|0,0⟩, and [J_x,J_y]|0,0⟩ are all zero.

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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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