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

  1. Mar 3, 2009 #1
    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!
     
  2. jcsd
  3. Mar 3, 2009 #2

    malawi_glenn

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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
     
  4. Mar 3, 2009 #3
    Much appreciated thank you!
     
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