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Do ladder operators give integer multiples of ћ?

  1. Jan 28, 2016 #1
    Say I apply a raising operator to the spin state |2,-1>, then by using the the equation

    S+|s,ms> = ћ*sqrt(s(s+1) - ms(ms+1))|s,ms+1>

    I get,

    S+|2,-1> = sqrt(6)ћ|2,0>

    Does this correspond to a physical eigenvalue or should I disregard it and only take states with integer multiples of ћ as eigenvalues? It makes sense that non-integer multiples of ћ wouldn't correspond to physical eigenvalues, but then again, ladder operators don't really correspond to observable quantities so should it even matter?
  2. jcsd
  3. Jan 28, 2016 #2

    Simon Bridge

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    Compare with the definition of an eigenvalue.
  4. Jan 28, 2016 #3
    "Any number such that a given matrix minus that number times the identity matrix has a zero determinant."

    I don't see how this answers the question. should non-integer multiples of ћ be omitted or not?
  5. Jan 28, 2016 #4


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    @baouba The linear operator whose eigenvalues are of interest here is not the raising operator S+. I imagine it is H+C where H is the Hamiltonian operator and C is a constant, and that operator certainly does correspond to a real-world quantity. You can see from your formulas in the OP that |2,-1> is not an eigenstate of S+. But it may be an eigenstate of H+C.
  6. Jan 28, 2016 #5

    Simon Bridge

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    Well, is that the situation you have here? What does it mean to be a physical eigenvalue?

    A|a> = a|a> satisfies the definition, does B|x> = a|y> satisfy the definition?
    If |x> were an eigenvector of operator B, wouldn't operation by B leave it unchanged?

    However - short answer: you should not just ignore a value just because you don't like it.
  7. Jan 29, 2016 #6


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    Think about the matrix form of the ladder operators. They're all sparse upper or lower triangular matrices with all zeros on the main diagonal, so the only eigenvalue of the ladder operators is 0. The eigenvectors are the top and bottom rungs.
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