Two-particle states of bosons I'm reading Advanced Quantum Mehcanics of Schwabl and I'm having a hard time trying to reproduce this calculation (page 57). Consider the following general two-particle state:
$$\displaystyle{|2\rangle=\int d^3x_1 d^3x_2 \;  \varphi(\bf{x_1},\bf{x_2})\psi^\dagger(\bf{x_1})\psi^\dagger(\bf{x_2})}|0\rangle$$
The claim is that the normalization condition $\langle2|2\rangle=1$ yields to the following:
$$\langle2|2\rangle=\int d^3x_1 d^3x_2 \, \varphi^*(\bf{x_1},\bf{x_1})(\varphi(\bf{x_1},\bf{x_2})+\varphi(\bf{x_2},\bf{x_1}))=1,$$
which is the part I don't get.
Any help would be very valuable for me, I'm following this book on my own and it is now very descriptive about the steps or the calculations it takes. I'm getting used to second-quantization formalism, so it is non-trivial for me to derive this.
 A: We can expand the inner product by using two additional integration variables:
\begin{align}
\langle 2|2\rangle&=\langle 0|\int d^3x_3 d^3x_4  \varphi^*(\mathbf{x_3},\mathbf{x_4})\psi(\mathbf{x_3})\psi(\mathbf{x_4})\int d^3x_1 d^3x_2  \varphi(\mathbf{x_1},\mathbf{x_2})\psi^\dagger(\mathbf{x_1})\psi^\dagger(\mathbf{x_2})|0\rangle\\
&=\int  d^3x_1 d^3x_2  d^3x_3 d^3x_4  \varphi^*(\mathbf{x_3},\mathbf{x_4}) \varphi(\mathbf{x_1},\mathbf{x_2})\langle 0|\psi(\mathbf{x_3})\psi(\mathbf{x_4})\psi^\dagger(\mathbf{x_1})\psi^\dagger(\mathbf{x_2})|0\rangle.
\end{align} Using reasoning like in your previous question, we can  evaluate the expectation value
\begin{aligned}
\langle 0|\psi(\mathbf{x_3})\psi(\mathbf{x_4})\psi^\dagger(\mathbf{x_1})\psi^\dagger(\mathbf{x_2})|0\rangle&=\langle 1_{\mathbf{x}_3},1_{\mathbf{x}_4}|1_{\mathbf{x}_1},1_{\mathbf{x}_2}\rangle \\
&=\left[\delta({\mathbf{x}_3}-{\mathbf{x}_1})\delta({\mathbf{x}_4}-{\mathbf{x}_2})+
\delta({\mathbf{x}_3}-{\mathbf{x}_2})\delta({\mathbf{x}_4}-{\mathbf{x}_1})\right],
\end{aligned} where we have used the second-quantized notation saying that, for example the state $|1_{\mathbf{x}_1},1_{\mathbf{x}_2}\rangle$ has $1$ boson with coordinates $\bf{x}_1$ and one with coordinates $\bf{x}_2$ (if both have the same coordinates, we get an extra factor of $\sqrt{2}$ on both the bra and the ket, which is accounted for by the sum over the two delta functions).
We can evaluate the integrals over $\mathbf{x}_1$ and $\bf{x}_2$ using the  delta functions (I guessed wrong about the initial labels so we have to adjust the labels) to find
\begin{align}
\langle 2|2\rangle
&=\int  d^3x_3 d^3x_4   \varphi^*(\mathbf{x_3},\mathbf{x_4}) \left[\varphi(\mathbf{x_3},\mathbf{x_4})+\varphi(\mathbf{x_4},\mathbf{x_3})\right]\\
&=\int  d^3x_1 d^3x_2   \varphi^*(\mathbf{x_1},\mathbf{x_2}) \left[\varphi(\mathbf{x_1},\mathbf{x_2})+\varphi(\mathbf{x_2},\mathbf{x_1})\right].
\end{align} This is exactly the desired expression, up to a typo in $\phi^*(\mathbf{x}_1,\mathbf{x}_2)$.
