# Wave function and impact parameter

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.

Let $\Phi_B(\vec x)$ be the spatial wave function corresponding to a collinear incident particle. The Fourier transform is $$\Phi_B(\vec k) = \int \mathrm d^3 \vec x\, \Phi_B(\vec x)\, \mathrm e^{-\mathrm i \vec k \cdot \vec x} .$$
The spatial wave function corresponding to an incident particle with impact parameter $\vec b$ is then $\tilde \Phi_B(\vec x) = \Phi_B(\vec x - \vec b)$, its Fourier transform $$\int \mathrm d^3 \vec x\, \Phi_B(\vec x - \vec b)\, \mathrm e^{-\mathrm i \vec k \cdot \vec x} = \Phi_B(\vec k)\, \mathrm e^{-\mathrm i \vec b \cdot \vec k}$$ by substitution.
So far (equation (4.68)) we do not describe a "uniform distribution", but a single packet with impact parameter $\vec b$. If you now continue reading, in (4.75) they perform that average: $$\int \mathrm d^2 \vec b\, \mathcal P(\vec b) .$$