In particle phyisics four-momentum is used and De Broglie relation is used to understand what lenghts can be "seen" in an experiment.

Here (page 6) it is claimed

The key factor for investigating the proton substructure is the wavelength of the probing photon, which is related to the transferred momentum by $$\lambda\sim \frac{1}{\sqrt{Q^2}}$$

Where $Q^2$ is not actually "momentum" but its the square of the four momentum transferred by the photon.

I can't understand why this is used, since the square root of square of four momentum is

$$\sqrt{p\cdot p}=\sqrt{|\textbf{p}|^2-E^2/c^2}$$

While De Broglie relation usually involves the three momentum $\textbf{p}$ $$\lambda\sim \frac{1}{|\textbf{p}|}$$

So is in this case $\sqrt{Q^2}$ approximately the three momentum? Or is the De Broglie relation in the relativistic case to be written using $\sqrt{Q^2}$ instead of three momentum?


1 Answer 1


In scattering experiments, specifically electron scattering, the four momentum of the virtual photon is:

$$ q_{\mu} = k_{\mu}-k'_{\mu}$$

where $k_{\mu}$ and $k'_{\mu}$ are the (on shell) initial and final electron momenta, respectively.

It is traditional to look at the process in the Breit Frame (also called the brick wall frame): this is frame where the electron bounces 180 degrees elastically off the target, so that:

$$ k_{\mu} = (E, \vec p) $$


$$ k'_{\mu} = (E, -\vec p) $$


$$ q_{\mu} = (0, \vec q) = (0, 2\vec p) $$

There is no energy transfer, and $q_{\mu}$ is entirely space-like.

In some sense, this is the frame where the photon can be seen as "most virtual", as it has lots of 3 momentum and no energy. Nevertheless, $Q^2$ is invariant, so you can evaluate it:

$$ -Q^2 = 4p^2 \approx 4EE'\sin^2{\frac{\theta} 2}$$

where the last expression is evaluated in the lab with $E>>m_e$.

It is in the Breit frame the $-\sqrt{Q^2}=||\vec q|| \propto 1/\lambda \ $ clearly defines the wavelength; moreover, the initial and final state plane waves are coupled by the transition element:

$$ T(q) \propto \int{(e^{-ipx}\ )^*\rho(x)e^{+ipx}\ d^3x} = \int{e^{iqx}\ \rho(x)d^3x}$$

which is the Fourier transform of the charge distribution--thus a scattering event with $Q^2$ is sensitive to structure on the scale of $1/\sqrt{-Q^2}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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