Time derivative of creation/annhilation operators

pleasehelpmeno
Messages
154
Reaction score
0
Basically is it possible to take a time derivative of a creation/annhilation operator?
 
Physics news on Phys.org
I think that's not possible. If you are talking about creation and annhilation operators usually found in QFT, they live in the momentum space, so they don't depend on x, but on p. However, I am not sure if in some other circumstances this could be possible.
 
yeah so for eample \frac{d \hat{a}}{dt} would =0 if a was the standard QFT annihilation op
 
Saying you cannot take the "time derivative" does NOT mean that derivative is 0. It means that the derivative does not exist at all.
 
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
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
I am reading WHAT IS A QUANTUM FIELD THEORY?" A First Introduction for Mathematicians. The author states (2.4 Finite versus Continuous Models) that the use of continuity causes the infinities in QFT: 'Mathematicians are trained to think of physical space as R3. But our continuous model of physical space as R3 is of course an idealization, both at the scale of the very large and at the scale of the very small. This idealization has proved to be very powerful, but in the case of Quantum...
Back
Top