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.


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 $\mathbf{b}$?


1 Answer 1


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: $$ \int \mathrm d^2 \vec b\, \mathcal P(\vec b) . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.