Physical meaning of a complete set of compatible observables

First of all, the common orthormal system is not unique: you can always change every vector in the set with a complex factor with absolute value 1. What is unique is the set of pure states associated to these families of orthonormal vectors (which, in fact, are vectors up to phases).

I do not know what type of physical meaning you are interested in. However, a physical meaning is that if you simultaneously measure such a set of observables on any given initial state, you define a pure post-measurement state. This state does not depend on the initial one, but only depends on the set of outcomes of the observables. A complete set of compatible observables (with point spectrum) together with a filter (on the oucomes) is therefore used to prepare the quantum system in a desired pure state.

Regarding maximal complete sets of (independent) compatible observables, they always contain a number of elements identical to the dimension of the Hilbert space: a complete set of this type is nothing but the set of orthogonal projectors onto the one-dimensional spaces of a Hilbert basis. So, minimal complete sets of compatible observables may seem more appealing. However, mathematically speaking, a set of this sort is always made of a single observable: it is sufficient to assign different eigenvalues to a chosen Hilbert basis and defining the corresponding selfadjoint operator. However, physically speaking this observable has no physical meaning in general. In summary, minimality or maximality are not physically meaningful guides for evaluating complete sets of compatible observables.

Most of quantum mechanics (at least initially) is about labelling states, i.e. giving them names. We hope of course these would be meaningful names, like energy or angular momentum, so we can talk about a state with this energy or this angular momentum.

The point of a complete set of commuting observables is to provide sufficiently many names to completely identify any state. Suppose for instance your operator is “what is your first name”. The eigenvalue of this operator on the name “Zero the Hero” is “Zero”. It could be that there is more than one user with this first name (I can’t verify because we only get a list of the users but I’m pretty sure there were other usernames starting with “Zero”). In other words, the eigenvalues of the operators “what is your first name” are not always enough to uniquely identify all users so you need a second operator, like “what is your last name”. That may be enough to uniquely identify the users.

We want eigenstates because

1. if I ask “What if your first name” I don’t want you to be ashamed and change your first name after (i.e. the action of the operator changes the first name to something else) and
2. we want the answer to be the same all the time, i.e. 0 variance. These are precisely properties of eigenvectors.

In this view,

1. The eigenvectors “react nicely” to being asked their first names, and
2. the answer is always the same because there is one and only one possible answer to that question (as opposed to — say — a username which is a superposition of names, i.e. not an eigenstates of “what is your first name”.)

The fact that the operators are commuting is related to quantum mechanics because you want to be able to simultaneously label users by their first and last name.

Coming back to physics, of course energy is something we care about so eigenstates of the energy operator (the Hamiltonian) and its labels (the energy of an eigenstate of $H$) are really useful. Moreover, because eigenstates of the Hamiltonian maintain their energy under time evolution, you know that if you ask “what is the energy of this state” at the start of the experiment, it will be the same at the end of the experiment if the system has evolved under the Hamiltonian. Energy may not be enough, as in hydrogen, where there are multiple states with the same energy. In this case, you use angular momentum $\ell$ and the projection $m_z$ so as to completely label each state using $E_n, \ell,m_z$.

There is no way of knowing if a set of observables is complete because experiment may reveal unknown properties of the system which are not accounted for by the naming $E_n,\ell,m_z$. Spin is an example of this. One then accommodates spin by adding it “by hand” as an observable.

Not all commuting observables are useful. If you have a 1d symmetric well, then parity commutes with the Hamiltonian but parity is not enough to uniquely label states, so you still need the eigenvalues of $H$ to do the complete job.

For a single particle in 1d, the only quantum number you need (i.e. the only observable you need) is $H$ and its eigenvalue $E$.

From @TobiasFünke

There is no real "physical meaning" a priori. It is just that a CSCO can yield a nice way to label an orthonormal basis. Think of the hydrogen atom for example.--And, depending on the definition one employs, one calls a CSCO a set which is minimal; that is, if one removes one operator from that set, the remaining set does not yield a unique orthonormal basis, cf. Galindo and Pascual, QM book Vol I.

From @naturallyInconsistent

I have absolutely no idea what you mean by maximal, not least because Tobias above mentioned that it should be a minimal set. But consider the H atom that has the LRLP vector as an extra symmetry. Then you could have too many symmetries and their conserved quantities, i.e. too many labels, and that would make it annoying. In that case we usually just ignore some extra labels and make do with the minimal set, which is then nicer to work with.

There is physical meaning. But to get at it, one must first clean up the language to make it more precise. The Complete Set of Compatible Observables that is referred to is more commonly (and more helpfully) known as a Complete Set of Commuting Observables (CSCO). The collection of CSCO operators is used to define what is called the Universal Enveloping Algebra of the associated Lie algebra.

So far, this was just mathematics. The physical interpretation is that the CSCO allows one to find the Casimir invariant operators of the associated algebra. These Casimir operators are the collection of all the physical invariants of the quantum system defined through the Schrodinger Hamiltonian operator.

To help get a feel for how this works, it is best to use a concrete example. Consider the Hamiltonian for the hydrogen atom. For $x\in\mathbb{R}^3$, and ,$\nabla^2$, the Laplace operator, the Hamiltonian, \begin{equation} H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{kQ^2}{|x|}, \end{equation} gives the total energy operator on the Hilbert space. This operator has an associated collection of Casimir operators, which are the complete collection of the invariants of this system. One knows without calculation from the laws of physics what these should be, for the most part. The operators that are intuitively obvious are, the total charge, $Q$, the total energy, $H$, the total momentum, $P$, and the angular momentum, $L$. But the interesting thing is that these are not all the invariants of this system. The system is not quite rotationally invariant because the Laplace operator has non-radial pieces. But angular momentum is definitely conserved.

The CSCO then brings to light that the missing connection has two pieces. The first is that the relevant operator for angular momentum is $L^2$, and not $L$. This is because the action of $L$ on an eigenfunction is complicated but the action of $L^2$ is always $L^2\psi = \ell(\ell+1)\hbar^2\psi$. While the second piece is the little known Laplace-Runge-Lenz vector named $A$. This is because while the Hamiltonian above is not quite rotationally invariant, i.e. not invariant under the Lie group $\mathcal{SO}(3)$ of 3d rotations, it is, in fact, invariant under the Lie group $\mathcal{SO}(4)$. The Laplace-Runge-Lenz vector, $A$, helps identify this fact as it is related to the constants of the motion and hence an invariant of the system. Furthermore, one has that the two vector operators contracted on each other through the 3d dot product is $A \cdot L = 0$. It was not at all obvious by looking at $H$ above that the system was invariant under 4d rotations.

In summary, the CSCO allows one to find all the Casimir invariant operators by finding the collection of all the quantum operators that commute with each other. Refining this collection allows one to find all the symmetries and hence invariants of the system by constructing the Universal Enveloping Algebra. For the example system of the Hydrogen atom, the Casimir invariants were $C_1 = A^2 + L^2$, and $C_2 = A\cdot L$. The physical interpretation of the CSCO is then the collection of all physically invariant operators of the quantum system associated to $H$.

For more detail, see Arno Bohm's book Quantum Mechanics: Foundations and Applicaitons.

Edit: I should also add that for the specific system in the original question, one should start with the operators that one knows like the total energy $H$, and the momentum $P$. But in that particular case, I cannot tell how many dimensions the system is in. To make things murkier in the Hamiltonian of the original question, the "potential", $V(x)$, is not an operator of multiplication because it is not even a function of $x$, for any number of dimensions. Normally for some real-valued, semi-bounded, and differentiable scalar $V:\mathbb{R}^d\to\mathbb{R}$ the commutation relation with the momentum operators is given by $[P,\,V(X)] = -i\hbar\,\nabla V(X)$. But when $V(x)$ is not a function, things get tricky very quickly.