Variational method: Why do parameters differ for two trial functions (optimization)?

Question

Below the potential and trial functions:

$$V(x)=(x^2-1)^2-x^2$$
Use the variational method with the two trial wave functions:
$$psi_{pm}(x)=Aleft(e^{-frac{(x-x_0)^2}{2sigma^2}}pm e^{-frac{(x+x_0)^2}{2sigma^2}}right)$$

Where $A$ is the normalisation constant and $x_0$ and $sigma$ are the variational parameters in the wave function.
Using Mathematica I found that:
for $psi_+$ -> $sigma = 0.612$ and $x_0 = 0.923$ with corresponding energy $= 0.136358$
for $psi_-$ -> $sigma = 0.528$ and $x_0 = 1.095$ with corresponding energy $= 0.470369$
I think the math I did is fine. So my question basically boils down to why the two parameters and corresponding energies differ for $psi_+$ and $psi_-$? What is the underlying physics reason? (Symmetry reasons?)

John Rennie · Accepted Answer

Let's graph the potential and your two wavefunctions (I have normalised the wavefunctions):

And this all makes sense. We have either the sum or the difference of the two gaussians. Now let's draw the same graph this time of the wavefunctions squared i.e. the probability density:

The total energy is the sum of the particle potential and kinetic energy. Of the two wavefunctions $psi^+$ has a higher probability density around the origin than $psi^-$ does. That is, $psi^+$ puts more of the particle in a higher energy region so you would expect it to have a higher potential energy.
The kinetic energy is related to the second derivative of the wavefunction, and I don't think it's easy to tell just by looking at the graphs how this will turn out. However we can wave our arms around and say that looking at the probability density $psi^+$ is less localised than $psi^-$, i.e. $psi^+$ is more spread out while $psi^-$ is more localised around the minima of the potential. And as a general rule the more localised a wavefunction is the higher its energy. So $psi^+$ is likely to have a lower kinetic energy than $psi^-$.
How the differences in the kinetic and potential energy will play out is hard to tell, though what we can say is we expect the two total energies to be different.
The small differences in the values of $x_0$ and $sigma$ are just the calculation getting as much of the particle as possible in the deepest parts of the well.

ZeroTheHero · Answer

A nice analogy can be made with the low-lying states of the ammonia molecule, where the potential and lowest energy levels are illustrated below.

The barrier in the middle splits the energy of the low-lying states and the symmetric combination has lower energy than the antisymmetric one.  Roughly speaking, the antisymmetric solution must go through $0$ and so must have greater curvature than the symmetric solution.  Since the kinetic energy $T$ is necessarily positive and represented by the operator $Tto -frac{hbar^2}{2m}frac{d^2}{dx^2}$, the average $int psi^* left(-frac{hbar^2}{2m}frac{d^2}{dx^2}right) psi $, which is related the curvature of $psi$, is greater for the antisymmetric case than the symmetric case.

If you accept that the two energies are going to be different then surely you can expect that the other variational parameters will also change.

Variational method: Why do parameters differ for two trial functions (optimization)?

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