Calculating Averages in a Unidimensional Quantum System

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Homework Help Overview

The discussion revolves around calculating averages in a unidimensional quantum system, specifically focusing on the Hamiltonian operator and the wave function provided. The original poster seeks to determine the average values of position and momentum at a later time, given the initial wave function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the Hamiltonian operator and its components, questioning the form of the momentum operator. There are attempts to express the time evolution of the wave function and concerns about the complexity of finding eigenvalues.

Discussion Status

The conversation is ongoing, with some participants suggesting the use of the Ehrenfest theorem as a potential approach. There is recognition of the challenges involved in finding eigenvalues, and the discussion reflects a mix of interpretations regarding the need for discrete eigenvalues.

Contextual Notes

Participants note the absence of a lower bound for energy eigenvalues in the context of the problem, which may affect the approach to calculating averages.

dirac68
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Homework Statement



Hi, i would to resolve this problem of quantum mechanics.

I have hamiltonian operator of a unidimensional system:

[itex]\hat{H}={\hat{p}^2 \over 2 m}-F\hat{x}[/itex]

where m and F are costant; the state is described by the function wave at t=0

[itex]\psi (x, t=0)=A e ^{-x^2-x}[/itex]

where A is a costant.

How can I calculate the the avarage of x and p at time t after t=0 ( so [itex]<x>_t[/itex] and [itex]<p>_t[/itex] )?

what is the fast procedure to solve it?

Homework Equations


[itex]\hat{H}={\hat{p}\over 2 m}-F\hat{x}[/itex]

[itex]\psi (x, t=0)=A e ^{-x^2-x}[/itex]

The Attempt at a Solution



I found a solution but it seems very long and boring...
 
Last edited:
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[tex]\psi(x,t)=\psi(x,0)e^{-iEt/\hbar}[/tex]... where E is given by: [tex]\hat{H}\psi=E\psi[/tex]

note: shouldn't the momentum operator appear squared in that hamiltonian?
 
Simon Bridge said:
[tex]\psi(x,t)=\psi(x,0)e^{-iEt/\hbar}[/tex]... where E is given by: [tex]\hat{H}\psi=E\psi[/tex]

note: shouldn't the momentum operator appear squared in that hamiltonian?

oh yes it's p2/2m... but find eigenvalue E is too hard!
 
Use the Ehrenfest theorem.
 
avarage of x and p
Ahhh yes - that's easier.

You don't have discrete E eigenvalues because you don't have a lower bound - but you don't need them. Sorry, my bad.
 

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