Is the product rule on operators different from traditional calculus?

  • Thread starter Thread starter Abigale
  • Start date Start date
  • Tags Tags
    Operators
Abigale
Messages
53
Reaction score
0
Hey Guys,

I regard two operators \Psi , \Phi, that don't commute.

Does the product-rule, looks like that?

$$\nabla (\Phi \Psi)

= \Psi (\nabla \Phi)

+\Phi (\nabla \Psi)

$$

THX
 
Physics news on Phys.org
How do you define the different items in this equation?
For example if Ψ = ∇, how do you define ∇Ψ ?
And in general?
 
Abigale said:
Does the product-rule, looks like that?
$$\nabla (\Phi \Psi) = \Psi (\nabla \Phi) +\Phi (\nabla \Psi) $$
You need to keep the original order:

$$\nabla (\Phi \Psi) = (\nabla \Phi) \Psi+\Phi (\nabla \Psi) $$
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top