# Solving Gaussian Potential - Analyzing Energy Differences

• aaaa202
In summary: So you get a better approximation for larger values of V0.In summary, the physicist has solved the schrödinger equation numerically, and has approximated it on a finite interval to solve for the wave function of different potentials. He is then asked to show how he did it analytically, but is confused about how to do it. He can either go back to formulate it all in terms of x, or extract the scale of V from that prefactor.
aaaa202
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 = -V0exp(-x2)
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 V0. On the attached graph I have done this for V0=1..2..3...10
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) = -V0 + V0x2
Plugging this into my dimensionless Schrödinger equation I get:
½∂2ψ/∂x2 + (-V0 + V0x2)ψ = Eψ
I thought aha. The x2-term can just be put in the harmonic oscillator form if we pick k=2V0 and the -V0 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)=½x2 for the harmonic oscillator. Now I have 2V0 in front of that. How will this constant effect my energies?
And is all this even the right procedure?

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The difference in energy levels scales with the square root of the prefactor in a harmonic potential - in the dimensionless shape, this is hidden in the parameter transformations. In a region where this harmonic approximation is good (probably large V0), I would expect that the difference grows with the square root of V0, and your graph roughly looks like that. For small V0, you probably get additional effects from higher orders of the potential.

okay but I am meant to show this analytically. How can I do that?

With the standard formulas for a harmonic oscillator. In your dimensionless version, you should have the conversion factors somewhere.
Or with the simple sqrt-dependence if you like.

well the conversion formula to dimensionless units is x' = x * √(mω/hbar)
So do I go back to formulate it all in terms of x? :S I am very confused sorry.

That is possible. Alternatively, you can extract the scale of V from that prefactor.

okay I did the problem and did indeed find that the difference went like √(2V0) - I am just curious - how is it you can see that the harmonic approximation is better for bigger v0?

Intuition. Deeper wells of the same size tend to have more bound states, so the lowest states are "deeper" in the well and smaller in terms of their size in x.

## What is a Gaussian potential?

A Gaussian potential is a mathematical model used in physics to describe the behavior of particles in a system. It is a type of potential energy function that is based on the Gaussian distribution, also known as the bell curve. In this model, the potential energy of a particle is determined by its distance from a central point.

## How is a Gaussian potential solved?

To solve a Gaussian potential, the potential energy function is usually normalized and then expanded in a Taylor series. This allows for the calculation of various properties of the system, such as the energy differences between different particle configurations. The solutions to the Gaussian potential can also be obtained numerically using computer simulations.

## What is the significance of analyzing energy differences in a Gaussian potential?

Analyzing energy differences in a Gaussian potential allows for the understanding and prediction of the behavior of particles in a system. By comparing the potential energy at different points, scientists can determine the stability of a particular configuration and make predictions about the overall behavior of the system.

## How is the Gaussian potential used in real-world applications?

The Gaussian potential is used in a variety of fields, including chemistry, physics, and materials science. It is often used to model the behavior of molecules in chemical reactions, the properties of materials, and the behavior of particles in physical systems.

## What are some limitations of using a Gaussian potential?

While the Gaussian potential is a useful model, it does have some limitations. It assumes that the system is in equilibrium and does not take into account the effects of external factors, such as temperature and pressure. Additionally, it may not accurately describe the behavior of particles in more complex systems.

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