Eigenvalues/functions of a mixed potential

[jdwood983]

Homework Statement

Find the first two energy eigenfunctions and eigenvalues for a particle in a potential

<br /> V(x)=\frac{1}{2}m\omega_0^2\left(x^2-2cx\right)<br />

Homework Equations

<br /> H=\frac{p^2}{2m}+V(x)=\frac{p^2}{2m}+\frac{1}{2}m\omega_0^2\left(x^2-2cx\right)<br />

<br /> -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega_0^2x^2\psi(x)-m\omega_0^2cx\psi(x)=E\psi(x)<br />

The Attempt at a Solution

I think what is confusing me is having the mixed potential, which is the whole point of the problem. I have tried using a completing the square and making the substitution of

<br /> \frac{1}{2}m\omega_0^2\left(x^2-2cx\right)\rightarrow\frac{1}{2}m\omega_0^2\left(u^2-c^2\right)<br />

but not really sure where this leads me (thought I might've ended up with a Bessel function or. I also tried solving it without substitution but end up with a quadratic in x differential equation

<br /> y&#039;&#039;+(ax^2-bx-c)y=0<br />

but unsure of where to start with this as wikipedia says this is the Weber-Hermite function, but I don't know what to do with this, at least not from what wiki says.

Any pointers as to where I should go from either starting point? (Or perhaps a starting point not considered that would be a better option)

 

[gabbagabbahey]

Hi jdwood 983, welcome to PF!

I haven't tackled the problem myself yet, but you might try solving it in the momentum basis instead and see if that leaves you with something less complicated...

 

[jdwood983]

The professor said he wants this done in coordinate space, so that's a no go :(

Thanks for the welcome!

 

[kuruman]

I have tried using a completing the square and making the substitution of

<br /> \frac{1}{2}m\omega_0^2\left(x^2-2cx\right)\rightarrow\frac{1}{2}m\omega_0^2\left(u^2-c^2\right)<br />

but not really sure where this leads me ...

This leads you to

-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(u)}{du^{2}}+\frac{1}{2}m\omega^{2}u^{2}\psi(u)=(E+\frac{1}{2}m\omega^{2}c^{2}})\psi(u)

It is ye olde harmonicke oscillator equation with the origin shifted by c and the "zero of energy" shifted by -(1/2)mω²c². A parabola is a parabola is a parabola.

 

[jdwood983]

This leads you to

-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(u)}{du^{2}}+\frac{1}{2}m\omega^{2}u^{2}\psi(u)=(E+\frac{1}{2}m\omega^{2}c^{2}})\psi(u)

It is ye olde harmonicke oscillator equation with the origin shifted by c and the "zero of energy" shifted by -(1/2)mω²c². A parabola is a parabola is a parabola.

Wow, I was too busy looking at special differential functions that I didn't even see that one. Thanks for your help!