Return time to its original state of a particle in the double infinite well

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Homework Help Overview

The discussion revolves around the behavior of a particle in a one-dimensional infinite double well potential, specifically focusing on the time it takes for a wave function to return to its original state. The original poster presents a mathematical expression for this time and explores the implications of their findings.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to analyze a simpler case involving a combination of the first two stationary states and questions how the result might generalize to combinations of multiple states. Other participants suggest considering the implications of factoring out common terms in the wave function and the behavior of the summation at specific time intervals.

Discussion Status

Participants are actively engaging with the mathematical framework and exploring different interpretations of the wave function's behavior over time. Some guidance has been offered regarding the treatment of the common factor in the wave function, and there is an acknowledgment of the periodic nature of the terms involved.

Contextual Notes

The original poster expresses uncertainty about the generalization of their findings to arbitrary combinations of states and reflects on the implications of their calculations. There is a mention of potential misconceptions regarding the simultaneous behavior of different states in relation to time intervals.

Javier141241
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Homework Statement


First sorry for the traduction mistakes.

Prove that any wave function of a particle in a 1 dimensional infinite double well of width a, returns to its original state in time T=4ma2/(πħ) .

Homework Equations


Ψ(x,t)=∑cnψn(x)·exp(-i·Ent/(ħ))

En=n2π2ħ2/(2ma2)

The Attempt at a Solution


I will explain my reasoning for a simpler case (combination of first two stationary states)
Ψ(x,t)=c1ψ1exp(-i·E1t/(ħ))+c2ψ2exp(-i·E2t/(ħ))
Since the global phase of Ψ doesn't matter ( |Ψ|2 ) you obtain the ω of oscillation taking common factor exp(-i·E1t/(ħ)) and obtaining ω=(E1-E2)/(ħ). Therefore for the wave function to return to the same state it was at t=0, t must be
t·(E1-E2)/(ħ)=2π
this leads to
t=4a2m/(2ħπ)·1/Q where Q its a term dependent of n

Since de statement says for any wave function and I get a similar result but depending of the Ψ in question, what I am missing? I wrote it for a combination of the two first stationary states, how would it be for a combination of n states? (since you can't take common factor the same way.)
 
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Javier141241 said:
Ψ(x,t)=∑cnψn(x)·exp(-i·Ent/(ħ))
Suppose you factor out exp(-i·E1t/ħ). What happens to the common factor exp(-i·E1t/ħ) at t = T? What happens to each term in the summation at t = T?
 
I just checked and it seems that doesn't matter which n's you take in (∑Ei)T/ħ, it just turns into multiples of 2π, diferent for each term. Should have checked it before more rigourous ( I was thinking that they couldn't all be 2π at the same time, but wasnt thinking of multiples of 2π).
Thanks for your time!
 
The wavefunctions of a particle in a box are mathematically the same as the standing waves in a string tied at both ends. The ground state (n=1) of the particle in a box corresponds to the fundamental frequency of the string. Any harmonic frequency is a multiple of the fundamental which means that in the time required for the string to complete one oscillation at the fundamental frequency, it has completed an integer number of oscillations at all harmonic frequencies.
 

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