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I Creation operator and Wavefunction relationship

  1. Jul 4, 2017 #1
    Hello,

    I've noticed that some professors will jump between second quantized creation/annihilation operators and wavefunctions rather easily. For instance [itex] \Psi_k \propto c_k + ac_k^{\dagger}[/itex] with "a" some constant (complex possibly). I'm fairly familiar with the second quantized notation, and from my understanding the concept of wavefunctions is essentially abandoned. The creation/annihilation operators simply add a particle into a state in the number basis. So what would be the way to "translate" some single particle wavefunciton [itex] \Psi_k [/itex] into second quantization, and what would the method/reasoning be?
     
    Last edited: Jul 4, 2017
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  3. Jul 4, 2017 #2

    strangerep

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    We'd have to see the full context that you're quoting. Sometimes a ##\Psi## like that is indeed an operator, not a state.
     
  4. Jul 4, 2017 #3
    I see what you're saying; in this instance the [itex]\Psi(x) [/itex] is not a field operator, but a single particle wavefunction, like in the case [itex] H\Psi = \Psi E[/itex]. Is it legitimate to say that [itex]\Psi(x) [/itex] corresponds to a superposition of [itex] c_k[/itex]'s and [itex]c^{\dagger}_k [/itex]'s?
     
  5. Jul 4, 2017 #4

    strangerep

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    ... in which case writing ##\Psi_k \propto c_k + ac_k^\dagger## is nonsense (assuming the ##c##'s are a/c operators).

    Please give a reference, or post a link, showing the context of what you're trying to ask about.
     
  6. Jul 5, 2017 #5

    stevendaryl

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    Quantum field theory can be interpreted as many-particle quantum mechanics where the number of particles is allowed to change. But what's a little confusing is the fact that the expressions [itex]\psi(\vec{r})[/itex] and [itex]\psi^\dagger(\vec{r})[/itex] don't refer to wave functions, but to operators. There is a "wave function" associated with any state with a definite number of particles, although most treatments of quantum field theory don't talk about wave functions much. But it's definitely not correct to associate a wave function with a linear combination of creation and annihilation operators; a wave function is a complex number, not an operator.
     
  7. Jul 5, 2017 #6

    vanhees71

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    Well, that's why one should mark what's an operator and what's a c-number wave function, i.e., ##\hat{\psi}(\vec{r})## (field operator) vs. ##\psi(\vec{r})## (complex-valued function).
     
  8. Jul 5, 2017 #7
    I would see it as a shorthand that omits the vacuum state from the RHS. A bit sloppy but understandable.
     
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