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About ladder operator

  1. Apr 27, 2009 #1

    KFC

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    For harmonic oscillator, let |0> be the ground state, so which statement is correct?

    a|0> =|0>

    or

    a|0> = 0 (number)

    where a is the destroy operator
     
  2. jcsd
  3. Apr 27, 2009 #2

    clem

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    It is the number zero. If a were the identity operator, you would get |0> back.
     
  4. Apr 29, 2009 #3
    you sure about that? i know that the books say zero but those operators aren't hermitian anyway hence unphysical so what does it matter what it does to |0> ?
     
  5. Apr 29, 2009 #4

    dx

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    Neither. [tex] a|0\rangle = [/tex] the zero ket.

    EDIT: The zero ket and [tex] |0\rangle [/tex] are different things.
     
    Last edited: Apr 29, 2009
  6. Apr 29, 2009 #5

    KFC

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    zero ket? Do you mean [tex] a|0\rangle = |0\rangle[/tex] ?
     
  7. Apr 29, 2009 #6

    dx

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    No. The zero ket is the zero vector in the state space. [tex] |0\rangle [/tex] is not the zero vector, it's just the vector with eigenvalue [itex] \hbar \omega ( 0 + \frac{1}{2}) [/itex].
     
  8. Apr 29, 2009 #7

    KFC

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    Oh, got it :)
     
  9. Apr 29, 2009 #8

    Pengwuino

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    Or better yet, a|0> = null vector.
     
  10. Apr 29, 2009 #9

    Ben Niehoff

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    Yes. The thing to remember is that all physical states must be normalizable. So

    <0|0> = 1
     
  11. May 2, 2009 #10

    KFC

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    and what about [tex]\hat{N}|0\rangle[/tex], where [tex]\hat{N}=\hat{a}^\dagger\hat{a}[/tex] is the number operator? Should it be

    [tex]\hat{N}|0\rangle = 0|0\rangle = 0 [/tex] (number) ?
     
  12. May 2, 2009 #11

    Pengwuino

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    Technically speaking, it is the null vector.
     
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