Is "Particle in a box" actually a misnomer? In the usual statement of the Particle in a Box problem, we assume two infinite potential barriers, to hold its wavefunction constrained, so it goes to zero on both ends:

But instead of invoking some unphysical infinite barrier, could be the case the wave just return to the same point. Using a simple warp drive(two paperclips):

That was for the X coordinate. So our one-dimensional box became a ring. Doing the same with the Y coordinate would get us a torus. With Z we would get a torus in the fourth dimension, and our box would be a single particle, inhabiting a boundless tiny universe.
So, our particle in a box is the actually a particle in a hypertorus?
At first I posted this question in the recently created materials modeling stackexchange, but it wasn't well received, quickly getting -6 votes and being deemed offtopic and deleted. People there pointed the physics stackexchange as a more adequate forum, so I'm giving it a try here.
 A: I'd like to address the OP's comment to one of the answers:

... I can see mathematics says they are different physical systems. The odd thing is, my intuition keeps saying they should be the same. I just can't grasp why, it just doesn't sit right with me.

Consider what happens with a classical particle in both cases.
In the particle in a box problem the particle goes to the right, hits the barrier and reflects, goes to the left, hits another barrier and reflects, and repeats this motion. If the particle is composite, these hits may break it down or at least deform and heat up.
In the case of a particle on a ring, the particle goes on and on without any reflections, like on a merry-go-round. It never hits anything, and the only force it could feel is the centrifugal force (if we view the "ring" literally), which is constant in magnitude.
What might be confusing you is the way you warped your piece of paper: you left the potential "walls" in place, just putting them together instead of erasing, so effectively your picture is now illustrating a different problem: particle on a ring with an impenetrable wall at one point of the ring. This problem (if we ignore centrifugal effects) is indeed equivalent to the original particle-in-a-box problem.
A: The ring scenario you call a "hypertorus" is an important one. For example, it's used to model the behavior of an electron in an effectively infinite crystal lattice.
But the two problems are not equivalent. 
For one thing, $\lambda=2l$ is a solution to one of these situations but not the other.
A: That's an interesting thought but no, we are talking about a particle in a box when we talk about a particle in a box and we can (and do) separately talk about a particle on a ring (or a torus). The differences between considering an infinite potential boundary (the case of boxes) and a periodic boundary (the case of rings/torii) are absolutely physical and the two are different physical systems.
In particular, the boundary condition for an infinite potential boundary is that the wave-function should vanish at the boundary, i.e., $\psi(0)=\psi(L)=0$. On the other hand, the boundary condition for the case of a periodic domain such as a ring is that the wave function should be periodic, i.e., $\psi(x)=\psi(x+L)$. You can verify that the former boundary condition implies that eigenfunctions be of the type $C(e^{ip_nx}-e^{-ip_nx})$ along with the condition that $p_n=n\pi/L$. Whereas the second case is simply a free particle on a periodic boundary, so the eigenfunctions would simply be $C \exp(ip_nx)$ but to satisfy the periodicity, $p_n$ would have to be of the form $2n\pi/L$.
So, there is a measurable physical difference between imposing infinite potentials and imposing periodic boundary conditions.
