Suppose you have an electron in the infinite square well

aaaa202
Messages
1,144
Reaction score
2
Suppose you have an electron in the infinite square well. The system is completely isolated from the rest of the world and has been its entire lifetime.
Do we then know that the wave function describing the electron is an eigenstate of the Hamiltonian? The question arose because I was given a problem where the intial wave function was a linear combination of eigenstates, which I guess is perfectly possible. But then I thought: Did God create this state or was it created because we had some potential from outside which put it there and was then turned off.
It seems that for all practical calculations you always assume that the electrons are described by their corresponding eigenstates (free electron gas etc. etc.)
 
Physics news on Phys.org
aaaa202 said:
It seems that for all practical calculations you always assume that the electrons are described by their corresponding eigenstates (free electron gas etc. etc.)
Certainly not. The new Nobel prize in chemistry is a good counterexample for relevant work which is not based on this assumption.
 
aaaa202 said:
Suppose you have an electron in the infinite square well. The system is completely isolated from the rest of the world and has been its entire lifetime.
Do we then know that the wave function describing the electron is an eigenstate of the Hamiltonian? The question arose because I was given a problem where the intial wave function was a linear combination of eigenstates, which I guess is perfectly possible. But then I thought: Did God create this state or was it created because we had some potential from outside which put it there and was then turned off.
It seems that for all practical calculations you always assume that the electrons are described by their corresponding eigenstates (free electron gas etc. etc.)

Hamiltonian eigenstates have no physical time dependence whatsoever (there's a time dependent phase, but cancels out of all physical calculations). If everything were always in such an eigenstate, nothing would ever happen. So it's a good thing that's not the case!
 
But do you agree with me that for oscillations in solids, free electrons in solid you generally take the approach that the electrons are in eigenstates of the Hamiltonian for a start - it's sort of an equilibrium assumption.
Actually: Suppose an electron is in an eigenstate of some weird potential and you suddenly turn the potential off. Will the electron state then seek towards an eigenstate of the free electrons.
 
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 looking at the Laughlin wave function and it contains the term $$\prod_{j<k}^{N}\left(z_j-z_k\right)^q$$ In Wikipedia the lower index on ##\Pi## is ##1\le i<j\le N## and there is no upper index. I'm not sure what either mean. For example would $$\prod_{j<k}^{N}\left(z_j-z_k\right)^q=\prod_{k=2}^{N}\left[\prod_{j=1}^{k-1}\left(z_j-z_k\right)^q\right]?$$ (It seems that ##k## cannot be 1 because there would be nothing to multiply in the second product). I'm not sure what else the...
Back
Top