That's just what comes out of quantizing a field. You find that the excited states can be interpreted as particles with a mass corresponding to the coefficient of a quadratic term with no derivatives. I don't know of any good heuristic explanation for this; sometimes you just have to do the calculation and see what comes out.
#3
PRB147
122
0
I think that it is determined by Einstein Mass-Energy Relation.
#4
BruceG
40
0
Yes - start with Einstein, since QFT is just the quantization of special relativity.
So with c=1, start with E^2 - p^2 = m^2
In terms of 4-momentum this is p^2 = m^2
In quantum theory with h= 1, p -> id/dx where x = (x,t) p = (p,E)
So we get the wave equation, [(d/dx)^2 + m^2] A(x) = 0
The Lagrangian L which produces this when we do DL/DA = 0 is then
L = A(x)[(d/dx)^2 + m^2]A(x)
So in summary we can say: quadratic Lagrangians give linear equations of motion.
At this point you should consider the advice we were given in our undergraduate math course: "the world is Linear because that is the only type of equations mathematicians know how to solve".
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!
The wavefunction of an atomic orbital like ##p_x##-orbital is generally in the form ##f(\theta)e^{i\phi}## so the probability of the presence of particle is identical at all the directional angles ##\phi##. However, it is dumbbell-shape along the x direction which shows ##\phi##-dependence!