Australian quantum researchers from the Centre for Quantum Computation and Communication Technology have (finally) been able to pinpoint the exact location of a single atom in a silicon crystal, a feat that used to be practically impossible without knocking atoms out of their original location. They were able to obtain high-resolution images of individual phosphorus and arsenic donors below the silicon surface. The researchers placed individual phosphorous atoms in a silicon crystal, as these atoms are great candidates for being used as quantum bits, or qubits. And the relation to quantum bits is precisely why the work is so significant, as accurately locating qubits would help us produce super fast, accurate computing power due to the multiple states that qubits can be in (compared to the binary system our current computers use). http://futurism.com/for-the-first-t...ere-able-to-pinpoint-the-location-of-an-atom/ Paper: http://arxiv.org/pdf/1601.02326v1.pdf
It won't. I'm sure the product of the uncertainty in location and momentum due to thermal motion will still be >= h-bar/2. In fact given the thermal vibration of the atoms in the lattice, there is no "exact" location anyway: it will be a slightly fuzzy blur, but close enough for their present purpose. I assume "pinpoint" is an inexact journo term Please Register or Log in to view the hidden image!.
Since nuclei are massive, localizing atoms to specific lattice points in a crystal is not a violation of the HUP.
Sorry, can you elaborate? I thought HUP told us that we could not know the exact location of an atom if we also held it still; it should smear out, should it not?
Nope -- (most terrestrially-encountered) atoms are fairly chunky compared to the grain of HUP. So if you know the position of a typical atom in a typical lattice to some not-too-small fraction of a lattice spacing constant, the uncertainty of the momentum should not force the atom far out of place. Planck's constant is about \(39.9 \; \times \; 1 \, \textrm{amu} \; \times \; 0.1 \textrm{nm} \; \times \; 100 \, \textrm{m} \cdot \textrm{s}^{-1}\). In Diamond, carbon-12 atoms of mass \(m = 1.99 \times 10^{-26}\, \textrm{kg}\) each in a lattice which puts the closest points \(L = 1.54 \times 10^{-10}\, \textrm{m}\) apart fitting 8 atoms per cube \(\frac{4 L}{\sqrt{3}} = 3.56 \times 10^{-10}\, \textrm{m}\) on a side. Check: The density of diamond is calculated to be \(\frac{8 \; \times \; 1.99 \times 10^{-26}\, \textrm{kg}}{\left( 3.56 \times 10^{-10}\, \textrm{m} \right)^3} \approx 3.53 \, \textrm{g} \cdot \textrm{cm}^{-3}\) which compares well with references. So if \(\Delta x = \alpha \times L\) what is the lower bound on \(\Delta p\) ? \(\Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{1}{\alpha} \frac{h}{4 \pi L} \approx \frac{3.42 \times 10^{-25} \, \textrm{m} \cdot \textrm{kg} \cdot \textrm{s}^{-1}}{\alpha} \) Err, OK -- what does that mean. Oh, wait. For low velocity atoms, we know that the momentum is related by \(\Delta p = m \Delta v\), so \( \Delta v \geq \frac{1}{\alpha} \frac{h}{4 \pi m L} \approx \frac{17.2 \, \textrm{m} \cdot \textrm{s}^{-1}}{\alpha} \) Since air molecules have typical speeds in hundreds of meters per second, localizing carbon atoms down to \(\alpha \approx \frac{1}{20} \) is not a crazy idea if diamond is a strong enough material not to erode due to kinetic forces caused by exposure to air.
To paraphrase from "The Right Stuff", what rpenner is sayin'.................. is that for objects with relatively large mass the uncertainties, though still there of course - QM still applies in principle - have such little effect as to be practically irrelevant. A small uncertainty in momentum for a massive object only requires a very small uncertainty in velocity. It's possibly worth recalling that most examples of the HUP talk about QM objects such as electrons, which are leptons, i.e. very light (1/1837 of the mass of a single proton). Atoms such as phosphorus, with an atomic mass equivalent to 31 protons, are not going to exhibit much uncertainty in practice.
