You can expand the free, real scalar field in the following manner $$ \phi(x) = \int \frac{d^{3}\mathbf{k}}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{k}}}} \bigg[ e^{- i \omega_{\mathbf{k}}x^0+i \mathbf{k} \cdot \mathbf{x}} a_{\mathbf{k}} + e^{+ i \omega_{\mathbf{k}}x^0-i \mathbf{k} \cdot \mathbf{x}} a_{\mathbf{k}}^{\dagger} \bigg] $$ where $\omega_{\mathbf{k}} = \sqrt{ |\mathbf{k}|^2 + m^2 }$ and the operators $a_{\mathbf{k}}$ and $a_{\mathbf{k}}^{\dagger}$ obey the bosonic canonical commutation relations (CCRs): $$ [a_{\mathbf{k}},a_{\mathbf{p}}]=[a_{\mathbf{k}}^{\dagger},a_{\mathbf{p}}^{\dagger}]=0 \ \ \ \ \ \ \ , \ \ [a_{\mathbf{k}},a_{\mathbf{p}}^{\dagger}] = \delta(\mathbf{k}-\mathbf{p}) $$
From here you build a symmetric Fock space as follows: You define a vacuum state $|0\rangle$ such that: $$ a_{\mathbf{k}} |0\rangle = 0 $$ which is assumed to be normalized $\langle0|0\rangle=1$. Then a single-particle state can be defined as $$ | 1_{\mathbf{k}} \rangle \equiv a_{\mathbf{k}}^{\dagger}|0\rangle $$ which are orthonormal $\langle 1_{\mathbf{k}} | 1_{\mathbf{p}} \rangle = \delta(\mathbf{k} - \mathbf{p})$ on account of the CCRs. From here you can construct a state containing $\sum_{s=1}^{N} n_s$ particles: $$ | (n_{1})_{\mathbf{k}_1} (n_{2})_{\mathbf{k}_2} \ldots (n_{N})_{\mathbf{k}_N} \rangle \equiv \frac{ \big( a_{\mathbf{k}_1}^{\dagger} \big)^{n_{1}} }{\sqrt{n_1!}}\frac{ \big( a_{\mathbf{k}_2}^{\dagger} \big)^{n_{2}} }{\sqrt{n_2!}} \cdots \frac{ \big( a_{\mathbf{k}_N}^{\dagger} \big)^{n_{N}} }{\sqrt{n_N!}} | 0 \rangle $$ where there are $n_s$ particles with the momentum $\mathbf{k}_{s}$.
I am confused about the way the following states are normalized: $$ | n_{\mathbf{k}} \rangle = \frac{ \big( a_{\mathbf{k}}^{\dagger} \big)^{n} }{\sqrt{n!}} | 0 \rangle $$ What is the normalization property of $\langle m_{\mathbf{p}} | n_{\mathbf{k}}\rangle$? Or also $\langle n_{\mathbf{p}} | n_{\mathbf{k}}\rangle$? For example, I find that \begin{align} \langle 2_{\mathbf{p}} | 2_{\mathbf{k}} \rangle &= \tfrac{1}{2} \langle 0 | a_{\mathbf{p}} a_{\mathbf{p}} a^{\dagger}_{\mathbf{k}} a^{\dagger}_{\mathbf{k}} | 0 \rangle\\ &= \tfrac{1}{2} \langle 0 | a_{\mathbf{p}} \big[ a^{\dagger}_{\mathbf{k}} a_{\mathbf{p}} + \delta(\mathbf{p} - \mathbf{k}) \big] a^{\dagger}_{\mathbf{k}} | 0 \rangle\\ &= \tfrac{1}{2} \langle 0 | a_{\mathbf{p}} a^{\dagger}_{\mathbf{k}} a_{\mathbf{p}} a^{\dagger}_{\mathbf{k}} | 0 \rangle + \tfrac{1}{2} \delta(\mathbf{p} - \mathbf{k}) \langle 0 | a_{\mathbf{p}} a^{\dagger}_{\mathbf{k}} | 0 \rangle\\ &= \tfrac{1}{2} \langle 0 | a_{\mathbf{p}} a^{\dagger}_{\mathbf{k}} \big[ a^{\dagger}_{\mathbf{k}} a_{\mathbf{p}} + \delta(\mathbf{p} - \mathbf{k}) \big] | 0 \rangle + \tfrac{1}{2} \delta(\mathbf{p} - \mathbf{k}) \langle 1_{\mathbf{p}}| 1_{\mathbf{k}} \rangle\\ &= 0 + \tfrac{1}{2} \delta(\mathbf{p} - \mathbf{k}) \langle 1_{\mathbf{p}} | 1_{\mathbf{k}} \rangle + \tfrac{1}{2} \delta(\mathbf{p} - \mathbf{k}) \langle 1_{\mathbf{p}}| 1_{\mathbf{k}} \rangle\\ &= \delta(\mathbf{p} - \mathbf{k})^{2} \end{align}
This makes no sense, as the square of the Dirac delta function is ill-defined. What am I doing wrong? How does the normalization actually work here?
EDIT: I have been using definitions for particle states that correspond to discrete labels - it is wrong to use them for continuous state labels. You cannot have more than one particle with exactly the same momentum. As usual Weinberg (Chapter 4) says it the clearest, he calls an $N$-particle state; $$ | \mathbf{k}_1\mathbf{k}_2 \cdots \mathbf{k}_N \rangle = a_{\mathbf{k}_1}^{\dagger} a_{\mathbf{k}_2}^{\dagger} \cdots a_{\mathbf{k}_N}^{\dagger} | 0 \rangle $$ Which has the normalization condition $$ \langle \mathbf{p}_1\mathbf{p}_2 \cdots \mathbf{p}_N | \mathbf{k}_1\mathbf{k}_2 \cdots \mathbf{k}_M \rangle = \delta_{NM} \sum_{\mathscr{P}} \prod_{j=1}^{N} \delta(\mathbf{k}_{j} - \mathbf{p}_{\mathscr{P}(j)}) $$ where $\mathscr{P}$ is the permutation over the integers $1,\ldots, N$. This is symmetric under interchange of any of the labels and 0 if $N \neq M$.
