Estimate the ground state energy of a particle of mass m moving in the potential

V(x) = lambda *(x)^4

by two different methods.

a. Using the Heisenberg Uncertainty Principle;
b. Using the trial function

psi(x)=N*e^{[- abs(x)]/(2a)}

where a is determined by minimizing (E)

*Note abs = absolute value

Solution Preview

To estimate the ground state energy using the uncertainty principle, you can use that the x^4 potential effectively confines the particle in a small interval around x = 0. Suppose that the particle has an energy of E. According to classical mechanics, the particle can be anywhere in the range - (E/lambda)^¼ to (E/lambda)^¼. At the boundaries of this interval, the potential energy equals the total energy and thus the kinetic energy is zero. Classically, the particle moves inside this region, the velocity reverses at the boundaries and at x = 0 it has a maximum momentum of (2 m E)^½. According to quantum mechanics, the wave function of the particle is not zero for x outside the interval, but it does decay exponentially to zero as you move to infinity.

So, the interval in which the particle is confined as a length of order Delta x = 2 (E/lambda)^¼. The momentum of the particle ranges from - (2 m E)^½ to (2 m E)^½, so
Delta P = 2(2 m E)^½

Inserting this in the uncertainty relation gives:

4(2 m)^½ E^¾ (lambda)^(-¼)>h-bar/2 (2)

From this equation you can see that if you make the energy low enough you will violate the uncertainty relation! But note that ...

Solution Summary

We explain how the ground state energy of a particle in a V(x) = lambda *(x)^4 potential can be estimated using the uncertainty principle. We also estimate the ground state energy using the variational method by using the trial function
psi(x)=N*e^{[- abs(x)]/(2a)}

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