When do two state functions represent the same quantum state?

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 $$\psi\in L^2(M)$$?

• Why is $\psi_2$ a the same state as $\psi_1$? If $\psi_1\neq \psi_2 e^{i\varphi}$ there always exists a (bounded) hermitian operator with different expectation values. Hence, if you really mean "state", you should restrict the notion of observables, i.e. not every hermitian operator is an observable anymore. Which then basically answers your questions. Two vectors (or density matrices) can be said to be equivalent if and only if they give the same expectation value for every bounded observable. Commented Mar 28 at 8:37
• To add: The answer to your question depends on how you define "state". I guess my definition above is the most natural (at least to me, lol) which can be put on completely rigorous ground, but you should find some things in the literature (say, e.g. regardin $C^*$-algebras, or good books on QM/QI). Commented Mar 28 at 8:42
• This reference may not directly be related to your specific question, but I thought I'd pass on anyway since it discusses the overlap idea. See the PBR Theorem (2011), figure 1, which discusses what kind of sets relate to a given quantum state. arxiv.org/pdf/1111.3328.pdf Commented Mar 28 at 16:28

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.

• Hi Soluble Fish, nice answer. Can you provide a (introductory) reference for this? Thanks in advance. Commented Mar 28 at 10:12
• I accepted this answer because it's clear and follows the nice "state=1-dimensional subspace" principle, but, on one hand, I don't know this "Hilbert space=the space of $L^2$ sections of a hermitian line bundle, whose connection is the magnetic field." theory (a reference would be greatly appreciated), on the other hand, to my other question I've got an argument for restricting the physically meaningful Hermitian operators to the compositions of kinetic momentum and coordinate, so saving the original $L^2(M)$ space.
– mma
Commented Mar 28 at 16:23
• (cont.) ...but losing the 1-dimensional spaces as states.
– mma
Commented Mar 28 at 16:48
• Thanks for the addendum! Your link contains also a reference for this topic, this is Quantization, Classical and Quantum Field Theory and Theta - Functions by Andrey N. Tyurin. Meanwhile I also found a chapter using Hermitian line bundles with connection: "1. Quantum Mechanics on a manifold with magnetic field" in Ruedi Seiler's Magnetic Bloch analysis and Bochner Laplacians. @TobiasFünke, I hope you also will find them useful. I
– mma
Commented Mar 30 at 5:12