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Okay, I have solved the schrödinger equation numerically by making it dimensionless (though I am still confused about this proces). And then approximating it on a finite interval and solving the resulting eigenvalue equation. This allows me to solve for the wave function of different potentials.

I started with the harmonic oscillator but have no reached the Gaussian one:

V = -V

In one simulation I am asked to find the difference in energy between the ground state and the first excited state as a function of V

Does it look right?

I am then asked the following: Solve the problem analytically by taylorexpanding the potential. So I taylor expand around x=0 to second order and find:

V(x) = -V

Plugging this into my dimensionless Schrödinger equation I get:

½∂

I thought aha. The x

But in thinking it over again there are some problems. With my dimensionless equation I just had V(x)=½x

And is all this even the right procedure?

I started with the harmonic oscillator but have no reached the Gaussian one:

V = -V

_{0}exp(-x^{2})In one simulation I am asked to find the difference in energy between the ground state and the first excited state as a function of V

_{0}. On the attached graph I have done this for V_{0}=1..2..3...10Does it look right?

I am then asked the following: Solve the problem analytically by taylorexpanding the potential. So I taylor expand around x=0 to second order and find:

V(x) = -V

_{0}+ V_{0}x^{2}Plugging this into my dimensionless Schrödinger equation I get:

½∂

^{2}ψ/∂x^{2}+ (-V_{0}+ V_{0}x^{2})ψ = EψI thought aha. The x

^{2}-term can just be put in the harmonic oscillator form if we pick k=2V_{0}and the -V_{0}term will just shift the energy of the oscillator, not alter the difference between E1 and E0.But in thinking it over again there are some problems. With my dimensionless equation I just had V(x)=½x

^{2}for the harmonic oscillator. Now I have 2V_{0}in front of that. How will this constant effect my energies?And is all this even the right procedure?