The idea goes like this: silicon detectors (like modern silicon pixel detectors) can reach a better spatial resolution than gas detectors, and combining several layers of either technology improves the spatial resolution further, as one can average over several measurements. At the same time, scattering in the material is larger in silicon detectors than in gas detectors and the more layers are installed in a detector the further increases the scattering. Scattering in the material can thus have a larger influence on momentum measurements than the spatial resolution of detectors after a magnetic field. (Gluckstern went into details on this, I think at least, I could have a look at the original paper and add a reference but … it’s not open access, so apparently not intended to be accessed … means I won’t access it for reading and will spare you the struggle to access it.)

Now, any smart physicist (and we know that Physics is all about knowing what’s relevant and correctly assessing which effects need to be considered or can be neglected) will recognise the Heisenberg uncertainty principle here.

\[ \sigma_x \sigma_p \geq \frac{\hbar}{2} \]

Also, we all know that \(\hbar=c=1\) makes everything simpler:

$$ \sigma_x \sigma_p \geq \frac{1}{2} $$

I boldly claim that this is BS and plug in some numbers from LHCb Figure 5 and 17 I get:

$$ \sigma_x = 5\mu m$$

$$ \sigma_p = 0.005 \cdot p = 0.005 \cdot 10 \mathrm{GeV}/c = 0.05 \mathrm{GeV} (\text{using }c=1)$$

Plugging things together

$$ \sigma_x\sigma_p = 0.25 \mathrm{\mu m GeV} \geq 0.5 $$

The great joy of using natural units in real world computations. Let’s revisit:

$$ 1 = c \approx 3\cdot 10^8 \frac{\mathrm{m}}{\mathrm{s}} $$

Lets bring the numbers and the second on the other side of the equation

$$ \mathrm{m} = \frac{1 \mathrm{s}}{3\cdot 10^8} $$$$\Rightarrow \sigma_x\sigma_p = \frac{0.25 \mathrm{\mu s GeV}}{3\cdot 10^8} $$

Wikipedia to the rescue!

$$ 1 = \hbar \approx 6.6\cdot 10^{-16} \mathrm{eV s} $$

$$ \Rightarrow \mathrm{eVs} = \frac{1}{6.6\cdot 10^{-16}} $$

$$\Rightarrow \sigma_x\sigma_p = \frac{0.25 \mathrm{\mu G}}{3\cdot 10^8 \cdot 6.6 \cdot 10^{-16}} $$

now resolve these remaining \(\mathrm{G}\) and \(\mathrm{\mu}\) and put the numbers into digit.digits * 10^{exponent}

$$\sigma_x\sigma_p = \frac{2.5 \cdot 10^{-1} 10^{-6} 10^{9}}{2 \cdot 10^1 10^8 10^{-16}} $$

$$\approx 10^{-1-6+9-1-8+16} = 10^9 \geq 0.5$$

Well … let’s say … Heisenberg was right, the product of position and momentum uncertainty is larger than \(\hbar/2\) … 8 orders of magnitude!

So … certainly the interaction of charged particles with the detector material goes down to the fundamental interactions which are quantum effects, but on the actual measurement level, the resolutions we see, do not correspond to the Heisenberg limit anymore.

PS

It is kind of frustrating how in your first term studying Physics you get told how awesome the SI unit system is (which, if you grew up in a part of the world that uses metric units, is hard to appreciate, as you don’t know anything else) and then any other lecture you attend uses atomic units, cgs, natural units … and looking at HEP, we’re not even using natural units the way we claim. Using natural units would mean lengths are measured in \(1/\mathrm{MeV}\) (right? I honestly never got along with natural units), except … nuclear cross sections get measured in barn (which is no conversion factor in \(\mathrm{MeV}\)), and … inverse cross sections divided by time are neither \(1/\mathrm{barn}/\mathrm{s}\) nor \(\mathrm{MeV}^3\) (at this point I doubt I got the power of \(\mathrm{MeV}\) right), but rather \(1/\mathrm{cm}^2/\mathrm{s}\). And, okay, \(\mathrm{cm}^2\) and barn differ by powers of 10, but not by a factor of \(1000\). And once you go to detectors, you’re back to “proper” meters, where one doesn’t measure areas in absurd decimal powers of \(\mathrm{cm}^2\) but appropriate metric prefixed lengths (i.e. square micro meter or so).

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