I'm doing a course in Nuclear Physics but I think it focuses more on the experimental side so not everything is that rigorous. In the notes, lecturer gives an argument for why we should use electrons to study the structure of a nucleus. It goes as follows:

Electrons are the most accurate and precise probe of nuclear sizes:

 fundamental point-like particles

 leptons - family of particles that do not experience the strong force

Therefore, they predominantly interact via the well understood electromagnetic force. If Δp is the momentum transferred to the recoiling nucleus then the spatial precision we can study the nucleus, Δr is given by Heisenberg's Uncertainty Relation: ∆𝒑∆𝒓~ħ hence ∆𝒑~ħ/∆𝒓.

I understand the two points he makes about point particles and leptons. Where I get lost is how he can relate the momentum transferred to the uncertainty in the momentum and then also the relation between the uncertainty in the position of the nucleus (because that's what delta r actually is) to the "structure" of the nucleus (which I'm guessing he means what it's made of i.e protons and neutrons).

Furthermore, we had a question in our tutorial where we were asked to "calculate the minimum beam energy required to study the sub-structure of the proton down to a spatial resolution of 0.05 fm using an electron beam". He works this out in a similar fashion:

"We need to use the position-momentum uncertainty principle ΔpΔx~ℏ where Δp ~pmin and Δx to be the spatial resolution required

p_min~ℏ/Δx~ℏc/0.05c~(197 "MeV" ⁄c)/0.05

Now since electron is highly relativistic:E≃pc~200⁄0.05 MeV~4.0GeV".

I don't see how delta p could be the minimum momentum.

  • $\begingroup$ Please correct the Nobel laureate's name in the title. $\endgroup$
    – my2cts
    Oct 29, 2018 at 21:45
  • $\begingroup$ sorry, was a typo $\endgroup$
    – nic
    Oct 29, 2018 at 22:01
  • $\begingroup$ Consider the De Broglie wavelength available to you given a momentum (~energy) of your probe, and compare to the wavelength of light required to probe spatial structures of a given size. Often, loose invocation of the uncertainty principle is just a reminder of the conversion factor in the de Broglie relation. It all started with the slippery intuition of "Heisenberg's microscope" ... a logical appeasement construction against his thesis exam misadventure... $\endgroup$ Oct 29, 2018 at 22:24
  • $\begingroup$ In point of fact, said microscope was meant to illustrate the UP, not use it. $\endgroup$ Oct 29, 2018 at 22:29
  • $\begingroup$ Do you mean that for an electron "wave" of wavelength lambda (which we'd get from De Broglie) we'd need it to be of a wavelength comparable to the size of what we are probing so that it will diffract? If so, I tried asking my lecturer if this is another way of looking at it and he said no because delta p shouldn't be related to an uncertainty in energy (which is what would happen if there was an uncertainty in lambda) but and uncertainty in the direction of p. $\endgroup$
    – nic
    Oct 29, 2018 at 22:33

2 Answers 2


Notice how you are using approximation symbols. Also notice a lack of equal signs, as well as an estimate on the uncertainty principle (technically $\Delta x\Delta p\geq\frac{\hbar}{2\pi}$). These are just back of the envelope calculations assuming we are somewhat near the lower bound of he uncertainty relation. If your momentum is smaller than your uncertainty, then you would probably be in trouble experimentally.


I don't find the uncertainty principle that helpful in this context. Rather, when using and electron probe, there is a well defined initial state:

$$ |k\rangle $$

and a well defined final state:

$$ |k'\rangle $$

where $k_{\mu}$ could be the 4-momentum or the 4-wavenumber (depending on whether you have set $\hbar=1$ or not).

To lowest order, $k$ got to be $k'$, by exchanging a virtual photon with the target:

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

is the (space-like) photon 4-momentum that probes the target.

Rather than talking about the photon energy with respect to the uncertainty principle, think about the Breit Frame; it is the frame in which

$$ q_0 = 0 $$


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

If you work out the scattering probability connecting the final state to the initial state it looks something like

$$ \sigma \approx \langle k'|\rho|k\rangle $$

where $\rho$ is the charge density. If you go ahead and plug in plane waves for the bra and ket, you'll see that the right hand side is the Fourier transform of the spatial charge density with conjugate variable $\vec q/\hbar c$.

So that's how I interpret $p_{min} = \hbar c/x$ being the minimum momentum required to probe length scales as small as $x$. It comes for the matrix element connecting initial to final states being the Fourier transform of the charge density.

Meanwhile, from a math perspective, the uncertainty relation arises in the Fourier transform from the spatial domain to momentum space of plane old wave functions.

Note that in the lab frame, at relativistic energies:

$$ Q^2 \equiv -||q^2|| \approx 4EE'\sin^2{\theta/2} $$

so you don't just consider the beam energy $E$ when finding the smallest $E$ required to achieve a certain length scale.


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.