Why do we consider the evolution (usually in time) of a wave function ?

rajesh_d
Messages
5
Reaction score
1
Why do we consider evolution of a wave function and why is the evolution parameter taken as time, in QM.

Look at a simple wave function $\psi(x,t) = e^{kx - \omega t}$. $x$ is a point in configuration space and $t$ is the evolution parameter. They both look the same in the equation, then why consider one as an evolution parameter and other as configuration of the system.

My question is why should we even consider the evolution of the wave function in some parameter (it is usually time)?. Why can't we just deal with $\psi(\boldsymbol{x})$, where $\boldsymbol{x}$ is the configuration of the system and that $|\psi(\boldsymbol{x})|^2$ gives the probability of finding the system in the configuration $\boldsymbol{x}$?

One may say, "How to deal with systems that vary with time?", and the answer could be, "consider time also as a part of the configuration space". Why wonder why this could not be possible.
 
Physics news on Phys.org
Why do we consider evolution of a wave function and why is the evolution parameter taken as time, in QM.
Because we want to know what happens next. Generally a system will change in time... so the square modulus of the wavefunction only give the correct probability-density at a particular point in time.

Yes we can work time into the configuration space - this is the point of relativistic quantum mechanics.

Even so, we have to do experiments in a laboratory which is mired in time - so the coordinate that is usually most convenient to evolve the wavefunction with respect to is still time.
 
Time in quantum theory is necessarily a parameter labelling the "causal sequence of events" or something like that, no matter whether you consider non-relativistic or relativistic quantum theory.

The reason has been given by Pauli very early in the development of quantum theory: If time would be taken as an observable than it would have the commutation relation
[\hat{t},\hat{H}]=\mathrm{i}\hbar.
This would mean, as for the case of the position and momentum operator in non-relativistic quantum theory, that the spectrum of both the time and the Hamilton operator would be whole \mathbb{R}, and this contradicts the stability of matter since for that the energy must have a lower boundary, so that a ground state of lowest energy exists.
 
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!

Similar threads

Replies
16
Views
2K
Replies
14
Views
698
Replies
1
Views
1K
Replies
9
Views
2K
Replies
61
Views
5K
Replies
3
Views
1K
Replies
15
Views
3K
Back
Top