About ladder operator

  • Thread starter KFC
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  • #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
 

Answers and Replies

  • #2
Meir Achuz
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It is the number zero. If a were the identity operator, you would get |0> back.
 
  • #3
1,707
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It is the number zero. If a were the identity operator, you would get |0> back.
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> ?
 
  • #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:
  • #5
KFC
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Neither. [tex] a|0\rangle = [/tex] the zero ket.
zero ket? Do you mean [tex] a|0\rangle = |0\rangle[/tex] ?
 
  • #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].
 
  • #7
KFC
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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].
Oh, got it :)
 
  • #8
Pengwuino
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Or better yet, a|0> = null vector.
 
  • #9
Ben Niehoff
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Yes. The thing to remember is that all physical states must be normalizable. So

<0|0> = 1
 
  • #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) ?
 
  • #11
Pengwuino
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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) ?
Technically speaking, it is the null vector.
 

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