Tag Info

New answers tagged

1

The half life can in principle be determined using Fermi's Golden Rule. Well, this calculates transition probabilities per unit time, but the half life is simply derived from the transition probability. So if you know the initial and final wavefunctions and the appropriate operator then yes you can calculate the half life. However in practice nuclei are far ...


3

If we were able to have a quantum mechanical model for nuclei that needed no experimental input, then the half lives of unstable nuclei would be computed utilizing the fully known wavefunctions. An approximation to this ideal is the shell model, and there are papers in the literature using the wavefunctions of the model to calculate lifetimes that fit the ...


-1

You are labouring under some misconceptions. Quantum mechanics is deterministic. It describes each system in terms of observables that evolve deterministically according to an equation of motion. However, these observables describe multiple versions of that object that can interact with one another in quantum interference phenomena. Objects interact with ...


0

Since it is possible to build a macroscopic device whose state changes according to something that happens at the quantum level, the obvious answer is "Yes". By macroscopic determinism we usually mean that quantum effects are below a level that is measureable. Not that they don't exist. For example, quite large molecules such as fullerenes can show quantum ...


2

Hopping and tunneling are often used as synonyms, but they are really very different terms with a fundamentally different basis. Tunneling is an inherently quantum-mechanical feature which means that a particle wave-function tends to overlap into it's energetically disallowed area which leads to a non-zero probability of finding it "where it should not be". ...


0

When "hopping", the particle has enough energy to surmount the potential barrier. Its like water molecules passing from liquid state to gas: only those who happen to have enough kinetic energy KE to escape the average bounding of the other water molecules. This can happen even in room temperature, since their KE follows a Boltzman distribution and it can be ...


0

I will answer your question why the transmission coefficient depend on velocity in a very naive way. Consider a potential barrier like below, the electron is incident from left with energy $E=\hbar^2k^2/2m$, the barrier width is $a$ and height is $V$. Define $\kappa\equiv\sqrt{2m(V-E)/\hbar}$. The transmission coefficient $D$ can be easily ...



Top 50 recent answers are included