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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?

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# Commutator of creation/annihilation operators (continuum limit)

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