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In Weinberg's Lecture Notes on Quantum Mechanics, we find the spatial probability distribution of a scattering process is the following $$\left|\left(\Phi_{\mathbf{x}}, \Psi_{g}(t)\right)\right|^{2} \rightarrow \frac{\mu}{4 \pi^{2} \hbar^{4} t}\left|\int d^{2} k_{\perp} g\left(\mathbf{k}_{\perp}, \mu x_{3} / \hbar t\right) \exp \left(i \mathbf{k}_{\perp} \cdot \mathbf{x}_{\perp}\right)\right|^{2},\\\tag{7.115}$$ where $g$ is a smooth function that makes the initial momentum peaked at $k_3=\mu x_3/\hbar t$, which is not necessarily real, and $\mathbf{k}_{\perp}$ is the wave vector perpendicular to momentum of the direction of propagation ($p_3=\hbar k_3$). To verify the conservation of probability, we have $$\begin{aligned} \int d^{3} x\left|\left(\Phi_{\mathbf{x}}, \Psi_{g}(t)\right)\right|^{2} & \rightarrow \frac{\mu}{\hbar^{4} t} \int d^{2} k_{\perp} \int_{-\infty}^{\infty} d x_{3}\left|g\left(\mathbf{k}_{\perp}, \mu x_{3} / \hbar t\right)\right|^{2} \\ &=\hbar^{-3} \int d^{2} k_{\perp} \int_{-\infty}^{\infty} d k_{3}\left|g\left(\mathbf{k}_{\perp}, k_{3}\right)\right|^{2}=1 \end{aligned}\\\tag{7.116}$$


I had some difficulty realizing the step denoted by "$\rightarrow$" in $(7.116)$. As $$\int dx_1 dx_2\left|\int d^{2} k_{\perp} g(\mathbf k) \exp \left(i \mathbf{k}_{\perp} \cdot \mathbf{x}_{\perp}\right)\right|^2 = \int dx_1 dx_2 \left(\int d^{2} k_{\perp} g(\mathbf k) \exp \left(i \mathbf{k}_{\perp} \cdot \mathbf{x}_{\perp}\right)\right)\left(\int d^{2} k_{\perp} g^*(\mathbf k) \exp \left(-i \mathbf{k}_{\perp} \cdot \mathbf{x}_{\perp}\right)\right)$$ This is as far as I can get, I wonder how do we proceed from here.

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Just change the order of integration. After evaluating the integration over $x_1$ and $x_2$, you'll end up with a Dirac delta that will set the perpendicular parts of the k-vectors equal. (According to what you've given from the notes all the k-vectors are only the perpendicular parts.) Then you'll end up with one integration over the k-vector, which gives the expression you are looking for.

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