# Tag Info

Heisenberg's uncertainty principle is in fact not a principle but a consequence of the operator formalism of QM. If we associate to the operator $X$ the standard deviation $$\Delta_X = \sqrt{ \langle{X^2}\rangle -\langle X \rangle^2}$$ it can be then shown that, given two operators $A,B$ $$\Delta_A \Delta_B \geq \frac{1}{2} \left| \langle ... 2 The point is that the domain D(P) of P must be such that P is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that D(X)= L^2([0,L],dx) instead, where X is automatically self-adjoint. The vector \psi you use to prove Heisenberg inequality has to belong to D(PX) \cap D(XP) as you see by direct ... 1 For some detailed analysis of the limits for electron microscopy, see: Viewpoint: What Are the Resolution Limits in Electron Microscopes? This is a brief review of the technology, and they summarize the recent improvements in resolution with: "The authors estimate that the resulting resolution limit is in the range 0.50–0.8Å, which is consistent with the ... 1 Is momentum of a particle "random" because it is uncertain, or is it uncertain in addition to being random? In quantum mechanics systems are represented using wave-functions (wave-vectors). The momentum of a particle is completely uncertain if it's position is certain (a localized particle) . But it is also possible to create wave-functions that have a ... 1 A particle is a wave. It's wave function (consider non relativistic Quantum Mechanics), when absolute valued squared, is a probability density function. The particle's momentum is a multiple of the gradient of the wave function, with h, Planck's constant, one of the proportionality constants. That is then the probability density function for momentum. It is ... 1 For a neutron of that speed, the uncertainty in the momentum is expected to be less than the momentum magnitude. Using the actual momentum will be an upper bound on the momentum uncertainty. That correlates to a lower bound on the position uncertainty. So, \Delta x is lower bounded by \hbar/(2p):$$\Delta x \ge \frac{\hbar}{2m_nv}. $\Delta x$ could ...