Heisenberg's uncertainty in nuclear physics

Question

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.

Answer 1

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.

Answer 2

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

$$ q_0 = 0 $$

so

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

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.