Free particle on a Möbius strip: boundary condition I've just see this video ( https://youtu.be/np6_1k99oRA ) and this guy tried to calculate the eigenvalues of a free particles on a Möbius strip using the boundary condition on the rectangle $[-\infty,\infty]\times [-w/2,w/2] $, $\psi(x+L,y)=\psi(x,-y)$ and $\psi(x,\pm w/2)=0$ where $L$ is the length of the strip. 
This makes sense to me.
Now let's consider the situation in which the particle can be on one face of the strip or on the opposite face. Here problems arise concerning orientation of the strip but, despite the fact that the strip is not orientable, it is however locally orientable (each point has an orientable neighborhood).
In other words now the particle can go to the other face of the strip in two ways.


*

*run across the entire strip 

*go to the other face of the strip by tunnel effect


In other words I'm thinking about 3 sheets with the Möbius strip boundary condition: in 2 of this the particle is free to move, the third is placed between the others and on this sheet is placed a delta potential. 
So I would schematize the situation considering a domain $ [-\infty,\infty] \times [-w/2,w/2] \times[-\epsilon,\epsilon]$ with the boundary condition $\psi(x,\pm w/2,z)=\psi(x,y,\pm \epsilon)=0$ and $\psi(x+L,y,z)=\psi(x,-y,-z)$. I would define a potential like $V(z)=\delta(z)$ 
The questions are:


*

*If this reasoning is right how can I bring back to the case of unions of 2 dimensional spaces (the 3 sheets)? when and how I can do the limit $\lim_{\epsilon \to 0^+}\psi$? 

*If we think about the strip as "union of 3 sheets" we are considering a non-connected space and a particle that can "jump" between non-connected components. This is so strange. Has it any sense? Is there a way to bypass this problem? 

*If our purpose is to calculate the probability that the particle travels along a length $L$ without go to the other face  ($z \to -z$) by tunnel effect, what we have to do? 

*Is there a way to solve the problem without introduce the third dimension? (with appropriate boundary condition on the rectangle $[-\infty,\infty]\times [-w/2,w/2] $ (or on other domains)

*If the answer to the 4. is "yes", what are the appropriate boundary condition?

*Physically, what non-orientability implies?

 A: 
This condition means that the particle is free to flip on the other side of the strip and this sounds weird to me.

I have no idea what you mean by this, but it seems to me that if you're talking about "sides" of a Möbius strip then you're already going wrong. The conditions you're talking about are obviously describing a particle on the 'flat', topological Möbius strip, which you get by identifying two opposing edges of the unit square.
Frankly, the important part of the Möbius-strip qualifier is the requirement that
$$
\psi(x+L,y) = \psi(x,-y),
$$
i.e. that a translation in one direction completely flips space in the orthogonal dimension. What you do along that dimension is frankly pretty irrelevant, but your elaborations (particularly your first proposal) sound pretty pointless to me.

Edit: your updated question, in the formulation

In other words now the particle can go to the other face of the strip in two ways.
  
  
*
  
*run across the entire strip 
  
*go to the other face of the strip by tunnel effect
  

does make enough sense to be answerable, though your attempts to do so are going in the wrong direction. The place where you go wrong is in your proposed domain,

i would schematize the situation considering a domain $ [-\infty,\infty] \times [-w/2,w/2] \times[-\epsilon,\epsilon]$

which is a three-dimensional domain that's attempting to model a fundamentally two-dimensional object.
What you need, ultimately, is two different copies of the unit square, to model the two different faces of the unit square ($\cong$ rectangle) that you twist and glue to make the Möbius strip. The answer is the obvious one: just make two copies explicitly! The right domain is thus
$$
\mathcal D =  [-\infty,\infty] \times [-w/2,w/2] \times \{1,-1\},
$$
a continuous domain in direct product with a discrete set of cardinality $2$. Thus, your wavefunction is an object of the form $\psi_i(x,y)$, where $i=1,2$, $x \mathbb R/L$ and $-w/2\leq y\leq w/2$.
To finish up, you need two things:


*

*The boundary conditions, which are given by
$$
\psi_i(x+L,y) = \psi_{-i}(x,-y),
\tag 1
$$
i.e., a translation by $L$ flips the orientation of the $y$ axis, and it also changes which face of the strip you're on. (Want to convince yourself that this is the right approach? Cut up a long strip of paper and draw a set of $x,y$ axes on one face and label it $1$. Then turn it around, label the other face $2$, and draw a set of axes directly tracing the ones on the other side. Then form the Möbius strip, and you'll see how the two coordinate patches connect as in $(1)$.)

*The hamiltonian, which will account for the free-particle motion as well as the tunnelling from one face to the other. This is dead simple:
$$
\hat H \psi_i(x,y) = -\frac{\hbar^2}{2m} \nabla^2 \psi_i(x,y) +t\,\psi_{-i}(x,y),
\tag 2
$$
where $t$ is a tunnelling constant with dimensions of energy.
That's all you need. I'll leave it to you to work out the eigenfunctions of this problem.
A: It makes no sense to talk about a free particle on a Möbius strip because a Möbius strip has a boundary that constrains the “free” particle. This is an oxymoron. For example, a particle in a box is not “free”; it is trapped by an infinite potential at the walls. (“Free inside”, which the OP did not state, is not the same as “free”, which the OP did state.) What makes sense is a free particle in a Klein bottle, which has no boundary.
The boundary conditions for a free particle in a Klein bottle with coordinates in $[0,W]\times[0,H]$ should be
$$\psi(0,y)=\psi(W,H-y),$$
$$\psi(x,0)=\psi(x,H).$$
You want to think of a Klein bottle as a “twisted” torus where two of the opposite edges are identified, running in the same direction, and the other two are identified, but running in the opposite direction.
