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Consider the unit torus $\mathbb T:=\mathbb R /\mathbb Z$. As far as I understand, the points of $\mathbb T$ are equivalence classes $[x]$ defined through the relation $x\sim x+k$, with $k \in \mathbb Z$. So for $y\in \mathbb R$ we have $y\in [x]$ if $x=y+k$ for some $k$.

I want to define functions on $\mathbb T$. Do all functions have to be periodic with period $1$? I guess so because I'd define functions by the following relation:

$$f([x]) :=f(x) \quad , \tag{1}$$

where $x\in [x] \in \mathbb T$ and $f$ on the RHS is a function defined on $\mathbb R$ (abusing the notation a bit). So the function evaluated at an equivalence class is defined through the function value of a representative of that class and thus

$$f([x])=f(x)=f([x+1])=f(x+1) \quad . \tag{2}$$

Is my analysis correct? Does the definition make sense?

Background: I study a quantum mechanical system defined on $L^2(\mathbb T)$ and in particular a linear operator $V$ defined on $L^2([0,1])$ by $(V\psi)(x):= v(x)\psi(x)$ for some real-valued function $v$. I want to 'lift' this operator to $L^2(\mathbb T)$ and I guess this is only possible if I consider some sort of periodic extension of the function $v$. I think that for $\psi\in L^2(\mathbb T)$ we have that $(V\psi)([x]) = v(x) \psi(x)$ is well defined only if $v(x)$ is periodic, no? In other words: $V:\mathcal D(V) \subset L^2(\mathbb T)\longrightarrow L^2(\mathbb T)$ only if $v$ is periodic?!

Kleber
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    A function is by definition single-valued, so indeed a function defined on a torus must take the same value at equivalent points. It is worth noting, however, that some authors argue that the wavefunction is not directly observable and hence need not be single-valued - it can acquire a phase shift upon making a round trip through space (as long as its squared amplitude is the same). – Raad Shaikh Sep 18 '22 at 13:07
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    @RaadShaikh Wouldn't a phase shift mean a change in momentum, thus be an observable effect? I agree with you on a global constant phase shift, though, since wavefunctions are really elements of the projectivisation of the Hilbert space, not the space itself. – lisyarus Sep 18 '22 at 15:12
  • @lisyarus I must admit I'm not sure of the details, I just read this in the 'Quantization' section in here: https://en.m.wikipedia.org/wiki/Angular_momentum_operator – Raad Shaikh Sep 18 '22 at 15:23
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    @lisyarus Not a shift in the wavefunction $\psi \mapsto e^{ikx}\psi$, but rather a quasi-periodic boundary condition $\psi(1)=e^{i\theta} \psi(0)$. This is indeed observable; this is what happens e.g. to a particle on a ring which is threaded by some non-zero magnetic flux. The phase accumulated by going around the ring once is the Aharanov-Bohm phase. – Albatross Sep 18 '22 at 15:38
  • @J.Murray Sorry for being a bit late to the party here, but if you take a look to my recent question: https://physics.stackexchange.com/questions/777579/aharonov-bohm-effect-and-periodic-boundary-conditions-for-particle-on-a-ring you'll see that your comment is directly relevant. If we impose periodic BCs without a "phase shift", the wavefunction with an Aharonov-Bohm phase fails to satisfy the perioic BCs, except if we quantise the magnetic flux, which seems dissatisfying to me, since the magnetic flux is an external parameter. So, the boundary conditions are indeed the ones you mentioned? – TheQuantumMan Sep 20 '23 at 13:51

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Yes, that looks right to me. Note that the function need not be continuous, but it must be periodic.

More precisely, a function $f$ on $\mathbb T$ induces a function $\tilde f$ on $\mathbb R$ in the obvious way, namely $\tilde f = f\circ q$ where $q:x\mapsto [x]$ is the quotient map corresponding to $\sim$. In order to go in the other direction - which we might loosely write as $f = \tilde f \circ q^{-1}$ despite the fact that $q$ is not injective - we must have that $\tilde f (x)= \tilde f(x+k)$ for $k\in \mathbb Z$ (otherwise it would not be a well-defined function on the quotient space).

There would be nothing wrong with having a function $\tilde f:\mathbb R\ni x \mapsto x-\lfloor x\rfloor$, which returns the fractional part of $x$. This function is discontinuous, but it is periodic, and so $\tilde f(x) = \tilde f(x+k)$ for $k\in \mathbb Z$ and our induced function $f = \tilde f \circ q^{-1}$ is well-defined.

Albatross
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