I understand that we want to solve for non-zero values of wave function. I always thought that is to avoid the obvious answer to Schrodinger equation. But from physical standpoint, if we have a particle of mass $m$, is it really impossible for it to have energy of zero? From mathematical point of view shouldn't the ground state energy of every system be zero? If yes, what does that mean? Nothingness. Void as ground state?
As far as most textbooks on (non-relativistic) quantum mechanics go, we do not consider the solution for $n=0$ because it gives us a trivial solution (and we interpret it to mean that there is no particle inside the box/well).
However, if there were a ground state with zero energy for a square well potential, it would imply that (since the particle has zero energy), it will be at rest inside the square well, and this will clearly violate Heisenberg's uncertainty principle!
By confining a particle to a very small region in space, it acquires a small but finite momentum. So, if the particle is restricted to move in a region of width $\Delta x \sim a$ (i.e., the entire length of the well), we can calculate the minimum uncertainty in momentum (using the uncertainty principle) and it comes out to be $\Delta p \sim \hbar/a$. And, this in turn gives us the minimum kinetic energy of the order $\hbar^2/(2ma^2)$. This (qualitatively) agrees with the exact value of the ground state energy.
So physically, the existence of a zero-point energy is a necessary feature of a quantum mechanical system. It indicates that the particle should exhibit 'a minimum motion' due to localization. Classically, the lowest possible energy of system corresponds to the minimum value of the potential energy (with kinetic energy being zero). But in quantum mechanics,the lowest energy state corresponds to the minimum value of the sum of both potential and kinetic energy, and this leads to a finite ground state or zero point energy.
The zero of the energy is completely arbitrary, as the zero of time or space.
In fact, suppose that the ground state energy of $H $ is $a $, then $H-a I$, where $I $ is the identity operator, has ground state energy zero and the same eigenvectors of $H $. In addition, it generates the same time evolution (apart from a non-physical phase factor). Therefore it is, from the physical standpoint, indistinguishable from the original one.
My simple version of the answer:
- because for the energy to be zero, time must be stopped
I'm sorry I don't know how to state that in the maths you're looking for, perhaps someone else can, I suspect that's what noir1993 has provided.
I think in terms you can visualise, it's kind of like saying that yes it's "at rest" from a perspective of Newtonian physics, but nothing is "at rest" from the quantum perspective unless time actually stops.
I don't know if that helps, but I was just fascinated by the question and wanted to chime in ( yes I know this isn't facebook )