Highly localized initial psi in harmonic well

  • #1

Main Question or Discussion Point

Say we start with a wavefunction inside a harmonic potential well, such that the initial ##\psi(x)## is confined to a central region much smaller than the ground state (hence ##V(x)\approx0##).. and the expectation Kinetic Energy is equal to an energy eignenvalue ##E_n## of the system.

Starting from here, will it ultimately converge over time into an energy eigenstate corresponding to ##E_n## ... OR.. will it slosh around forever in a complicated way?
 

Answers and Replies

  • #2
vanhees71
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It will slosh around forever in a complicated way. You can just solve the equation of motion by using the well-known energy-eigenstates. Given the wave function ##\psi(t,\vec{x})## at ##t=0## you define the corresponding coefficients
$$\psi_j=\int_{-\infty}^{\infty} u_j^*(x) \psi(0,\vec{x}),$$
where ##u_j(x)## is the energy eigenfunction with eigenvalue ##E_j=(j+1/2)\omega##, ##j \in \{0,1,2,\ldots \}##. Then the wave function at any later time is given by
$$\psi(t,x)=\sum_{j=0}^{\infty} \exp(-\mathrm{i} E_j t) \psi_j u_j(x).$$
This immediately shows that you never converge to an energy eigenfunction but that for any time all components of the initial wave function stay involved. This must be so, because only the energy eigenfunctions represent stationary states, i.e., if initially you don't have the system prepared in an energy eigenfunction the state can never become an energy eigenstate later.
 
  • #3
Thanks !
 
  • #4
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To paraphrase vanhees' answer more abstractly, the Schrödinger equation is linear and unitary (the eigenvalues are just phases) so any nontrivial linear combination of its eigenfunctions will never converge to a single eigenfunction.
 

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