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Angular momentum eigenvalues

  1. Oct 31, 2005 #1
    Hi,

    How can you infer from these equations,

    [tex]a = b_{max}(b_{max}+\hbar) \quad \text{and} \quad a = b_{min}(b_{min}-\hbar),[/tex]​

    that [tex]b_{max} = -b_{min}[/tex]? It is used in the derivation of the angular momentum eigenvalues...
     
  2. jcsd
  3. Oct 31, 2005 #2

    Physics Monkey

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    Set them equal. You should find [tex] b^2_{max} + \hbar b_{max} = b^2_{min} - \hbar b_{min} [/tex]. This implies that [tex] b^2_{max} - b^2_{min} = - (b_{max} + b_{min} )\hbar [/tex]. Try factoring the left hand side ...
     
  4. Oct 31, 2005 #3
    Aarh, very clever indeed :rolleyes:

    Thanks!
     
  5. Nov 2, 2005 #4
    Why, you little.... :mad:

    That doesn't work since:


    [tex] b^2_{max} - b^2_{min} = - (b_{max} + b_{min} )\hbar \quad \Leftrightarrow \quad b_{max} - b_{min} = -\hbar \quad \Leftrightarrow \quad \text{nonsense} [/tex]

    What am I missing here??
     
  6. Nov 2, 2005 #5

    George Jones

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    Now, Now.

    Physics Monkey's hints were good ones.

    All kinds of nonsense can be "proved" by dividing by zero.

    Regards,
    George
     
  7. Nov 3, 2005 #6

    George Jones

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    You have used a reductio ad absurdum argument to prove the required result, i.e., assume that (b_max + b_min) is not equal to zero, divide by (b_max + b_min), arrive at nonsense.

    Thus the assumption is false and (b_max + b_min) = 0.

    Regards,
    George
     
  8. Nov 3, 2005 #7

    vanesch

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    That's true. One day, my math teacher told us how we could convince our parents to double our pocket money by showing that A = 2 A (using again some division by zero). My dad's comment was: since A = 2 A, I'll give you half of what you get normally, that shouldn't then make a difference for you :redface:
     
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