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

AI Thread Summary
The discussion centers on proving that a wave function of a particle in a one-dimensional infinite double well returns to its original state after a time T=4ma²/(πħ). The approach involves analyzing a combination of the first two stationary states and factoring out the common exponential term to find the oscillation frequency. It is noted that regardless of the specific stationary states involved, the terms in the summation will yield multiples of 2π at time T. The connection to standing waves in a string illustrates that the fundamental frequency dictates the oscillation behavior of all harmonics. Ultimately, the conclusion emphasizes that the wave function's return to its original state is a general property applicable to any combination of states.
Javier141241
Messages
3
Reaction score
0

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.)
 
Physics news on Phys.org
The answer is staring at you and you almost had it. You have
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top