Effect of the raising operator on an l=1 m=1 state

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SUMMARY

The raising operator L+ acting on the quantum state |1,1> results in zero, as the maximum value of m (M_max) cannot exceed l. Specifically, applying the operator yields L+|1,1> = 0, confirming that the eigenstate does not change when m is at its maximum. This aligns with the established principle that L+|l,m> = ħ√(l(l+1) - m(m+1))|l,m+1>, which results in zero when m equals l.

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neish88
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Pretty basic question but I was wondering what the result of acting the raising operator on an l=1, m=1 quantum state for a hydrogen wavefunction would be.

Specifically,

L+|1,1> = ?

I know that normally L+|l,m>=hbar(l(l+1)-m(m+1))1/2|l,m+1> but I wasn't sure if the eigenstate remained the same (since m cannot be larger than l) or if it results in zero, like if you were to act the lowering operator on the lowest possible state.

Thanks in advance!
 
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it will be zero

L+ on |L,L> = 0

since M_max = L

Same with L- on |L,-L>

since M_min = -L
 
Much appreciated thank you!
 

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