Commutator of creation/annihilation operators (continuum limit)

1. Jul 29, 2015

soviet1100

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, $[a^{\dagger}_\textbf{p}, a_\textbf{p'}] = \delta^{(3)}(\textbf{p} - \textbf{p'})$. 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 $\langle\textbf{p}|\textbf{p'}\rangle$ as follows:

$\langle\textbf{p}|\textbf{p'}\rangle = \langle0| a_\textbf{p}a^{\dagger}_\textbf{p'} |0\rangle$

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

$\hspace{12mm} = \delta^{(3)}(\textbf{p} - \textbf{p'})$

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. Jul 29, 2015

ShayanJ

$a_{\mathbf p}$ 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 $\langle 0|\delta^{(3)}(\mathbf{p-p'}) |0\rangle=\delta^{(3)}(\mathbf{p-p'}) \langle 0|0\rangle=\delta^{(3)}(\mathbf{p-p'})$.

3. Jul 29, 2015

soviet1100

Ah, I get it now. Thanks a ton for the help.