How Does a Shifted Harmonic Oscillator Decompose into Eigenfunctions?

[prairiedogj]

I have a wave function which is the ground state of a harmonic oscillator (potential centered at x=0)... but shifted by a constant along the position axis (ie. (x-b) instead of x in the exponential).

How does this decompose into eigenfunctions?? I know it's an infinite sum... but I can't nail the coefficients.

...and...
without knowing the decomposition, how can I get the time development of the average position and momentum?

Help.

Thanks.

 

[dextercioby]

What does \exp\left[-(x-a)^{2}\right] instead of \exp\left(-x^{2}\right) have to do with time dependence of \langle \hat{x}\rangle...?


Daniel.

 

[prairiedogj]

Well, most intuitively, I would assume that shifting the initial wavefunction along the x-axis would affect average position.

Less intuitively, a shifted initial wavefunction in a zero-centered harmonic potential MUST be composed of eigenfunctions... ones centered at zero. Each of these have their own energies and own time development.

As an example... go to THIS applet http://groups.physics.umn.edu/demo/applets/qm1d/index.html
set it to harmonic oscillator, choose the ground state eigenfunction, and then adjust the offset.