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Quantum angular momentum proportionality constant

  1. Oct 7, 2015 #1
    Hi, not a homework problem as such, but I am studying introductory quantum physics and having some trouble understanding how applying the rising or lowering operator for angular momentum implies:

    L+/-|l,m> ∝ |l, m+/-1>
    Basically, my question is the same as the first part (Q1) as described in the following link.
    http://www.thestudentroom.co.uk/showthread.php?t=2552515

    Regarding the answer to that post, I can get to the two expressions quite easily by using the commutation relations, but even after reading it multiple times, I dont understand how the proportionality is implied. I know this is probably trivial and I'm probably just missing some obvious fact due to the time being 1.30am. Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 7, 2015 #2

    blue_leaf77

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    Because the state ##L_{+/-} |l,m\rangle## is not necessarily normalized, while all ##|l,m\rangle## must be normalized.
     
  4. Oct 7, 2015 #3
    blue_leaf77, thanks. It was one of those lack of sleep things. I just woke up and realised this. Also, mathematically speaking, there is an infinite amount of choices for our eigenvector based on a given eigenstate and scale factor. We must obviously use the one that fits our operator, ie it must be normalised.
     
  5. Oct 7, 2015 #4

    blue_leaf77

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    The vectors (including eigenvectors of an operator) in QM are agreed to be normalized because of the probabilistic interpretation assigned to the inner product between any two vectors. The probability of finding a given state to be in that same state must be equal to the maximum allowed value for a probability, which is unity.
     
  6. Oct 7, 2015 #5
    okay thanks I understand my mistake and (hopefully) will not make it again :smile:
     
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