In example 4.1 of Lancaster and Blundell's "Quantum field theory for the gifted amateur", we verify that a field operator creates a particle as follow:
Let $|\Psi\rangle=\hat{\psi}^{\dagger}(\boldsymbol{x})|0\rangle$ and check that it does indeed correspond to a single particle at a particular location. Note that $$ |\Psi\rangle=\hat{\psi}^{\dagger}(\boldsymbol{x})|0\rangle=\frac{1}{\sqrt{\mathcal{V}}} \sum_{\boldsymbol{p}} \mathrm{e}^{-\mathrm{i} \boldsymbol{p} \cdot \boldsymbol{x}} \hat{a}_{\boldsymbol{p}}^{\dagger}|0\rangle . $$ To calculate the total number of particles in this state we can use the number operator $\hat{n}_{p}=\hat{a}_{p}^{\dagger} \hat{a}_{p}$ (which measures the number of particles in state $\boldsymbol{p}$ ) and then sum over all momentum states. Consider $$ \sum_{\boldsymbol{q}} \hat{n}_{\boldsymbol{q}}|\Psi\rangle=\frac{1}{\sqrt{\mathcal{V}}} \sum_{\boldsymbol{q} \boldsymbol{p}} \hat{a}_{\boldsymbol{q}}^{\dagger} \hat{a}_{\boldsymbol{q}} \hat{a}_{\boldsymbol{p}}^{\dagger}|0\rangle \mathrm{e}^{-\mathrm{i} \boldsymbol{p} \cdot \boldsymbol{x}} $$ and again using $\left\langle 0\left|\hat{a}_{\boldsymbol{q}} \hat{a}_{\boldsymbol{p}}^{\dagger}\right| 0\right\rangle=\delta_{\boldsymbol{p q}}$ we deduce that $$ \sum_{\boldsymbol{q}} \hat{n}_{\boldsymbol{q}}|\Psi\rangle=|\Psi\rangle $$ ...
I can't figure out why using the identity leads to the last equation.