On page 102,103 of Peskin and Shroeder.
Let's denote a wavepacket $|\phi\rangle$ as
$$ |\phi\rangle = \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{k}}} \phi(\mathbf{k})\,|\mathbf{k}\rangle \tag{4.65} $$
where $\phi(\mathbf{k})$ is the Fourier transform of the spatial wavefunction.
And
It is important, however, to take into account the transverse displacement of wavepacket $\phi_{\cal{B}}$ relative to $\phi_{\cal{A}}$ in position space (see Fig. 4.2). Although we could leave this implicit in the form of $\phi_{\cal{B}}(\mathbf{k}_{\cal{B}})$, we instead adopt the convention that our reference momentum-space wavefunctions are collinear (that is, have impact parameter $\mathbf{b}=0$), and write $\phi_{\cal{B}}(\mathbf{k}_{\cal{B}})$ with an explicit factor $e^{-i\mathbf{b}\cdot\mathbf{k}_{\cal{B}}}$ to account for the spacial translation. Then, since $\phi_{\cal{A}}$ and $\phi_{\cal{B}}$ are constructed independently at different locations, we can write the initial state as $$ |\phi_{\cal{A}}\phi_{\cal{B}}\rangle_{\text{in}} = \int \frac{d^3k_{\cal{A}}}{(2\pi)^3}\int \frac{d^3k_{\cal{B}}}{(2\pi)^3} \frac{\phi_{\cal{A}}(\mathbf{k}_{\cal{A}})\phi_{\cal{B}}(\mathbf{k}_{\cal{B}})\, e^{-i\mathbf{b}\cdot\mathbf{k}_{\cal{B}}}}{\sqrt{(2E_\cal{A})(2E_\cal{B})}} |\mathbf{k}_{\cal{A}}\mathbf{k}_{\cal{B}}\rangle_{\text{in}} \tag{4.68} $$
And at the caption of Figure 4.2,
Incident wavepackets are uniformly distributed in impact parameter $\mathbf{b}$.
Why does the factor $e^{-i\mathbf{b}\cdot\mathbf{k}_{\cal{B}}}$ denote the uniform distribution in impact parameter $\mathbf{b}$?
Thanks.
