When do two state functions represent the same quantum state?

Question

According to the standard quantum mechanics, quantum states are one-dimensional subspaces of a separable Hilbert space. In practice, this Hilbert space is $L^2(M)$ where $M$ is the classical configuration space of the system. From this follows that for any state function $\psi_1 \in L^2(M)$ and for any $z\in\mathbb C\setminus \{0\}$, if $\psi_2\in L^2(M)$ differs from $z\psi_1$ at most on a zero-measure subset of $M$, then $\psi_1$ and $\psi_2$ represent the same quantum state.

But, as I saw in the Aharonov-Bohm effect, the converse of this implication is not true. If $\psi_2$ represents the same quantum state as $\psi_1$, from this does not follow that there is a complex number $z\in\mathbb C\setminus \{0\}$ so that $\psi_2$ differs from $z\psi_1$ at most on a zero-measure set.

For example, if $S$ is a simply connected proper subset of $M$ with nonzero measure and $\chi_S$ is the characteristic function of $S$ then for any $z\in\mathbb C\setminus \{0\}$, $\psi_2=(1+\chi_S(z-1))\psi_1$ is clearly in a different one-dimensional subspace of $L^2(M)$ than $\psi_1$, but it describes the same quantum state as $\psi_1$.

My question: What is the subset of $L^2(M)$ that contains all elements of $L^2(M)$ that describe the same quantum state as a given $\psi\in L^2(M)$?

Answer 1

The set of wavefunctions which corresponds to the same state as $\psi$ is just the set of multiples of $\psi$ by a non-zero complex number (respectively, a complex number of absolute value $1$, if you consider only normalized wavefunctions).

The Aharonov-Bohm effect does not change this. What it changes is that the Hilbert space is no longer the space of $L^2$ functions on $M$, but rather the space of $L^2$ sections of a hermitian line bundle, whose connection is the magnetic field.

Concretely, this means that you can only identify a wavefunction with a complex valued function after choosing a gauge on a simply connected patch of $M$. If you do that on two different patches $U $ and $V$, one wavefunction $\psi$ will yield two complex valued functions which differ from a phase on the overlap $U\cap V$ (even though they are still restrictions of the same element of the Hilbert space). This phase is a transition function which is part of defining the quantum system (it is part of the data defining the Hilbert space).

Edit : Wave-functions as section of a fiber bundle

I don't have a specific reference at hand right now, but I would be searching for key-word like "geometric phases", "geometry of the Aharonov-Bohm effect". In geometric quantization, wave-function appear naturally as section of a line bundle (see the discussion here for example), but this is a mathematics thing so it doesn't really address the physical reasoning behind this.

Let me try do give some explanation : in the presence of a magnetic field, the Hamiltonian and the Schrödinger equation are written in terms of the vector potential $\vec A$ : $$H = \frac{1}{2m}(p-qA)^2$$ To preserve gauge invariance, when we change vector potential $\vec A\to \vec A +\nabla \lambda$, we need to also change the wavefunction by a phase : $\psi \to e^{-iq\lambda }\psi$. Hence : we can only hope for the wave-function to be a complex valued function once a gauge is specified. To see the right framework to describe the wavefunction, we need to look into the geometry of gauge theory.

In the case of the Aharonov-Bohm effect, we can consider the whole of space, with non-vanishing magnetic field inside the flux tube. We can choose a vector potential defined over the whole $\mathbb R^3$ and solve the Schrödinger equation as usual, with wave-functions being just complex-valued functions.

We may also consider the flux tube to be infinitely thin and remove it from our space, which becomes $M = \mathbb R \times (\mathbb R^2 \backslash \{ 0\})$. Here, the magnetic field vanishes uniformly. Again, we can find a gauge defined over the whole $M$ (for example by taking the previous realistic solution and taking the width of the tube to $0$). We cannot choose $A = 0$ however, because the integral of $A$ along a path encircling the flux tube is the magnetic flux $\Phi\neq 0$.

We can also divide $M$ into two simply connected (open) subsets $U$ and $V$ (the left and right side of the tube). On both $U$ and $V$, we can take $A = 0$ (which makes solving the Schrödinger equation easier !). Formally, we do this by starting with the previous vector potential and performing a gauge transformation with $\nabla \lambda_{U,V} = -A$ on $U$ (resp. on $V$). The transformed wavefunctions are $\psi_{U,V} = e^{-iq\lambda_{U,V}}\psi$. On the overlap $U\cap V$, those two wavefunctions satisfy $\psi_V = e^{iq(\lambda_U-\lambda_V)} \psi_U$.

Mathematically, this is the formula relating two different trivialization of a hermitian line bundle : at each point on $M$, the wave-function takes it value on a different copy of $\mathbb C$; the vector potential is the connection which allows us to differentiate the wave function; on a small enough open subsets, you can identify the wavefunctions with complex valued functions, but you have to change gauge and multiply by a phase when switching from one of those small patch to another.