Ground state energy of particle in quantum harmonic oscillator.

[Lengalicious]

Homework Statement

Consider a quantum mechanical particle moving in a potential V(x) = 1/2mω²x². When this particle is in
the state of lowest energy,
A: it has zero energy
B: is located at x = 0
C: has a vanishing wavefunction
D: none of the above

Homework Equations

The Attempt at a Solution

Ok from what i know, all energy levels of the particle traveling through the harmonic potential are discrete, and as i understand, the lowest of these discrete energies is the ground state energy but i don't quite understand how this energy is determined? The ground state energy is not zero right? Because if the ground state energy was zero then there would be no wave function in the 1st place right? Also the ground state wave function of the particle when normalised forms a probability density in the form of an upside down porabola almost with exponential decay on both sides, from this you can conclude that the particle is most likely to be at the center or (x=0), so i think the correct answer is B, is my understanding and answer correct or not? If not can someone explain where I am going wrong please.

 

[Mindscrape]

Yes, ground state is the lowest energy. The energies are determined as solutions (observables) to the hamiltonian operator. So actually, the ground state is the first eigenvalue (the first energy), and I'm pretty sure anything that would not have the first eigenvalue be the lowest energy would be unphysical. In the case of the harmonic oscillator, you'd determine the eigenvalues (energies) with ladder operators (I remember there are some power series methods too, but those are messy/brute force).

You could have a zero ground state. I can't really think of an example where a system typically has a zero ground state, but you could. I mean, to some extent the actual 'value' of the energies is completely arbitrary, it's the separation between the energies, their form, that matters.

The ground state wave function is gaussian.