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Commutator of creation/annihilation operators (continuum limit)

  1. Jul 29, 2015 #1
    Hi,

    This is a question regarding Example 3.6 in Section 3.5 (p.35) of 'QFT for the Gifted Amateur' by Lancaster & Blundell.

    Given, [itex] [a^{\dagger}_\textbf{p}, a_\textbf{p'}] = \delta^{(3)}(\textbf{p} - \textbf{p'}) [/itex]. This I understand. The operators create/destroy particles in the momentum state p and p'.

    However, the authors use this commutator in example 3.6 to calculate [itex] \langle\textbf{p}|\textbf{p'}\rangle [/itex] as follows:

    [itex] \langle\textbf{p}|\textbf{p'}\rangle = \langle0| a_\textbf{p}a^{\dagger}_\textbf{p'} |0\rangle [/itex]

    [itex] \hspace{12mm} = \langle 0| [\delta^{(3)}(\textbf{p} - \textbf{p'}) \pm a^{\dagger}_\textbf{p'} a_\textbf{p}]| 0\rangle [/itex]

    [itex] \hspace{12mm} = \delta^{(3)}(\textbf{p} - \textbf{p'}) [/itex]

    I understand the second step too; +/- for bosons/fermions depending on whether a commutator or anticommutator is used. It's the third and last step that I don't understand. How does that follow from the second?
     
  2. jcsd
  3. Jul 29, 2015 #2

    ShayanJ

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    Gold Member

    ## a_{\mathbf p} ## is an annihilation operator, which means ## a_{\mathbf p}|0\rangle=0 ##, so the term ## \langle 0|a_{\mathbf p'}^\dagger a_{\mathbf p} |0\rangle ## is zero. The other point is that the Dirac delta there is representative of 0 or ## \infty ## which means its a number and you can take it out to gain ##\langle 0|\delta^{(3)}(\mathbf{p-p'}) |0\rangle=\delta^{(3)}(\mathbf{p-p'}) \langle 0|0\rangle=\delta^{(3)}(\mathbf{p-p'}) ##.
     
  4. Jul 29, 2015 #3
    Ah, I get it now. Thanks a ton for the help.
     
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