# Normalization of states and bracket notation

In Peskin & Schroeder's QFT, if we set $$|p\rangle = \sqrt{2E_p} a^{\dagger}_p|0\rangle \tag{2.35}$$ as in equation 2.35, then how do I get to the next equation 2.36:

$$\langle p|q\rangle = 2E_p (2\pi)^3 \delta^3(p-q) \tag{2.36}$$ Where I seem to be stuck is: $$\langle p|q\rangle =\sqrt{2E_p 2E_q}\langle0|a_pa_q^{\dagger}|0\rangle$$ I don't understand why $\langle0|a_pa_q^{\dagger}|0\rangle$ is prop. to the delta function.

## 1 Answer

Try commuting $a_p$ and $a_q^\dagger$ and keep in mind that $a_n|0\rangle = 0$.

• so I get $\sqrt{2E_p2E_q} \langle 0 |a_p a_q^{\dagger}|0\rangle = \sqrt{2E_p2E_q} \langle 0 |[a_p ,a_q^{\dagger}]|0\rangle + \sqrt{2E_p2E_q} \langle 0 |a_q^{\dagger} a_p |0\rangle$. The second term is zero since $a_n |0\rangle = 0 \forall n$. So we get, using $[a_p, a_q^{\dagger}] = 1$, $\sqrt{2E_p2E_q} \langle 0 |0\rangle$. But what is $\langle 0 |0\rangle$ equal to? Thank you – laguna Oct 23 '17 at 7:59
• Depending on whether you have Bosonic or Fermionic operators, the bracket inside the first term after the $=$ sign is either a commutator or an anti-commutator. But, regardless, that bracket is a the delta function. So you have: $\sqrt{4E_pE_q}\langle 0|\delta^3(q - p)(2\pi)^3|0\rangle = \sqrt{4E_pE_q}\delta^3(q - p)(2\pi)^3\langle 0|0\rangle= \sqrt{4E_pE_q}\delta^3(q - p)(2\pi)^3 =$ the desired answer. – IcyOtter Oct 23 '17 at 8:09
• Thank you. And by definition $\langle s | s \rangle = 1$ for all states $s$, right? – laguna Oct 23 '17 at 8:20
• No problem! And yes, for any normalised state, we have $\langle s|s\rangle = 1$. In other words, a state projects onto itself 100%. – IcyOtter Oct 23 '17 at 10:12