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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:

Usual version

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):

Alternative version

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.

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    $\begingroup$ The particle in a box and the particle on a ring have different eigenvalues and eigenfunctions, so they’re definitely not the same system. Why don’t you try and solve for the eigenfunctions of the particle on a ring and see how they compare to the particle in a box? $\endgroup$ – Jahan Claes May 30 at 22:55
  • $\begingroup$ @JahanClaes, a ring exists embedded in our 3D space. I was thinking about bending space itself, so there is nothing outside the "box", not even the potential barrier. Just a single tailspinning particle, interacting with itself. $\endgroup$ – ksousa May 30 at 23:01
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    $\begingroup$ Your ring is like the compactified dimension in Kaluza-Klein & related theories. $\endgroup$ – PM 2Ring May 30 at 23:43
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    $\begingroup$ @ksousa it doesn’t matter whether you embed the ring or treat it like it’s own manifold, that doesn’t effect the spectrum or the eigenfunctions. You can find them, and see they are not equal to the eigenfunctions of the particle in a box. $\endgroup$ – Jahan Claes May 30 at 23:45
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    $\begingroup$ If you never consider normals to the hypersurface, it hardly makes a difference whether it is embedded in a larger space or self-standing. You are just like an ant on it. $\endgroup$ – Cosmas Zachos May 31 at 0:11
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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.

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    $\begingroup$ Isn’t it $2n\pi/L$? $\endgroup$ – G. Smith May 31 at 0:08
  • $\begingroup$ Thank you for the answer, @Dvij D.C. With your explanation, 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. $\endgroup$ – ksousa May 31 at 0:20
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    $\begingroup$ @ksousa The particle in a box is equivalent to the particle on a ring if you choose some privileged point on the ring and demand that all physically allowed wavefunctions must vanish there. Intuitively, does this feel like it should be equivalent to the unrestricted case (which has no such privileged point)? $\endgroup$ – J. Murray May 31 at 0:38
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    $\begingroup$ Imagine the difference between a circular hallway that you can run around in, and a circular hallway that is blocked at one point by a concrete wall. You can still run around the second hallway, but every time you get to the wall you'll have to turn around. You can run laps in the first but not the second. $\endgroup$ – Luke Pritchett May 31 at 3:46
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    $\begingroup$ @ksousa The wavefunction on the ring does not have to vanish anywhere. For example, $\psi(x)=e^{2\pi i x/L}$ is a good wavefunction on a ring of circumference L, and it doesn't vanish anywhere. The box, on the other hand, DOES have to have its wavefunctions vanish somewhere $\endgroup$ – Jahan Claes May 31 at 15:49
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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.

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  • $\begingroup$ Problem becomes similar to delta potential. $\endgroup$ – Kartik Chhajed May 31 at 9:47
  • $\begingroup$ @KartikChhajed it's very different from delta potential. Delta potential admits even solutions (in the form of $\exp(-|x|))$, while this problem doesn't, since it forces the wavefunction to zero at the barrier. $\endgroup$ – Ruslan May 31 at 9:51
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    $\begingroup$ @KartikChhajed The problem is similar to the delta potential in one sense that there is only one point where potential is non-zero and that potential is a delta potential. However, as @Ruslan points out, actually, the problems are different because of different boundary conditions. In particular, the delta potential problem for bound states requires the eigenstates to vanish at infinity but for the ring with a delta barrier, there is no infinity and there is a periodic boundary condition (which isn't there in the delta potential problem). $\endgroup$ – Dvij D.C. May 31 at 12:10
  • $\begingroup$ @DvijD.C. well, one can put a delta potential somewhere in the ring's configuration space, still imposing periodic boundary conditions. The difference of this setup from the OP's barrier is that in this setup we'll still have both even and odd solutions (although no longer degenerate), while in the OP's case the even solutions will be discarded as not satisfying the $\psi=0$ condition at the (more powerful than delta) potential. $\endgroup$ – Ruslan May 31 at 12:46
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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.

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