Tag Info

5

When you have a bunch of interrelated phenomena in physics, trying to figure out which one is the "reason" for the other ones is often just a recipe for confusion. Different people will start from different postulates, so they will disagree on which results are trivial and which aren't, but hopefully everyone agrees on what's true. In a first course on ...

1

Yes, this is not the "true" reason. The direct reason is that they are not commuting operators. See the Robertson-Schrodinger Relation. You end up getting that the product of the uncertainties is bounded by 1/2 of the absolute value of the failure to commute. In the case of position and momentum this is $\hbar/2$.

1

The basics of atomic structure are studied in the non-relativistic limit (mass not changing with velocity). Relativistic corrections are then added as perturbations to the non-relativistic states & energies (this gives you the fine structure https://en.wikipedia.org/wiki/Fine_structure). For a full quantum treatment of relativistic effects you need to go ...

0

One can indeed see the reason behind the Heisenberg uncertainty principle as a mathematical propeties. The link between the physics and the mathematics is provided by the foundational work of Planck and de Broglie. They established the link between energy/momentum and frequency/k-vector. The general textbook on quantum mechanics therefore always starts by ...

0

It probably depends what interpretation of quantum physics you subscribe to. That sounds approximately right for the Copenhagen interpretation, in which you aren't allowed to analyze where the wave function comes from. For those who appreciate more what de Broglie, Einstein, Bell and others have put into quantum physics, there's always the interpretation ...

0

Your statements about quantum mechanics are correct, but there's more that needs to be said to get to the Heisenberg uncertainty principle. Specifically, we need a lower bound on products of variances. Fix a state $\left|\psi\right\rangle$. In this state, any linear operator $\hat{\mathcal{O}}$ on the Hilbert space (i.e. square matrix) has mean $\left\... 0 Short answer: position momentum uncertainty exists because their operators do not commute. Likewise the time and energy operators do not commute. Longer answer: First: Physics is an empirical science so "proof" must be an experiment. Sometimes thought experiments will be allowed because they are useful for building and testing models. Mathematical proofs ... 24 The main problem is, as you say, that time is no operator in quantum mechanics. Hence there is no expectation value and no variance, which implies that you need to state what$\Delta t$is supposed to mean, before you can write something like$\Delta E \Delta t\geq \hbar$or similar. Once you define what you mean by$\Delta t$, relations that look similar ... 4 This is the case. The uncertainty relationship with energy and time is a matter of Fourier analysis. In fact the relationship$\Delta\omega\Delta t \simeq 1\$ was know in classical EM and electrical engineering before quantum physics. The use of Fourier analysis in electrical engineering had much the same uncertainty relationship as the reciprocal ...

Top 50 recent answers are included