Calculating Averages in a Unidimensional Quantum System

AI Thread Summary
To calculate the averages of position <x>_t and momentum <p>_t in a unidimensional quantum system, the Hamiltonian operator is given as \hat{H}={\hat{p}^2 \over 2 m}-F\hat{x}. The wave function at time t=0 is \psi(x, 0)=A e^{-x^2-x}. A suggested approach involves using the Ehrenfest theorem, which simplifies the calculation of these averages without needing to find discrete energy eigenvalues. The discussion highlights the complexity of finding eigenvalues due to the lack of a lower bound in this system. Overall, the Ehrenfest theorem is recommended as a more efficient method for determining the averages.
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Homework Statement



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

I have hamiltonian operator of a unidimensional system:

\hat{H}={\hat{p}^2 \over 2 m}-F\hat{x}

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

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

where A is a costant.

How can I calculate the the avarage of x and p at time t after t=0 ( so &lt;x&gt;_t and &lt;p&gt;_t )?

what is the fast procedure to solve it?

Homework Equations


\hat{H}={\hat{p}\over 2 m}-F\hat{x}

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

The Attempt at a Solution



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

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

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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