Why is the ground state energy of particle in a box not zero? 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?
 A: 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. 
A: 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 $. In addition, it generates the same time evolution (apart from a non-physical phase factor). Therefore it is, from the physical standpoint, indistinguishable from the original one.
A: I am sharing my opinion$ \ :\ \ \ $
As we know that quantum mechanical discrete energy states appears only when we consider a bound state problem . Generally it's told that ground state energy can never be zero quantum mechanically .Now think that you have a bound state problem . That means the particle is bounded in some region . Then if we get a zero energy state (means energy is zero : as $n=0  \ in \  E_n = \frac{n^2\pi^2\hslash^2}{2ma^2} \ $ particle in the box ) then the particle will have no energy in that state . Means it  will be resting at some definite position . Then you can predict it's position perfectly by measuring as it is resting and you can also tell the momentum = $0$ as it's in rest . But Heisenberg uncertainty Principle clearly tells us that $ \ \Delta x.\Delta p \geq \frac{\hslash}{2} \ $  . But then this product will be zero . In this way zero energy violates the H.U.P . And so it is eliminated .
