I Uniqueness of quantum numbers

[ecurbian]

The states of the Schroedinger atom are indicated by several quantum numbers - principle, azimuthal, magnetic, and spin. From the point of view of differential equations, the first three can be derived by using radius and two angles and solving for seperable solutions. One then has what amounts to the same thing as harmonics on a string, that only discrete wavelengths can occur - for the function to fit around the loops of the coordinates.

What would happen if, instead, one were to solve the Schroedinger equation for inverse radius potential using cartesian coordinates. By one means or another and never separating the variables. One would still have the same set of solutions, but it is not clear that the same set of numbers to characterise them would occur.

Are the quantum numbers in some sense locked into the the choice of coordinates and measurements - rather than being objectively forced. An example of the thought is the simple observation that if you have two numbers (a,b) then one can store instead (a+b,a-b). There being nothing deep about the original choice of basis.

 

[pines-demon]

Quantum numbers are basis dependent. What this means is that once you specify a set of commuting observables (more specifically a complete set of commuting observables) that you think are important, you get a given set of quantum numbers.

Position in the Cartesian basis, more specifically $x,y,z$ are valid quantum numbers to label position space but (1) position usually does not commute with the Hamiltonian, (2) it is not a complete set of observables, you still need to account for spin somehow.

In the hydrogen atom the Hamiltonian commutes with $L_z,L^2,S^2,S_z$ so all you can find a basis, where all basis states are eigenstates of these operators, and their eigenvalues are the quantum numbers. Note that you could choose instead the set $H_{\mathrm H},J_z,J^2,S^2,S_z$ where $\mathbf J=\mathbf L + \mathbf S$ instead. For a Hamiltonian that is not rotational symmetric neither of these two sets would work.

 

[Paul Colby]

What would happen if, instead, one were to solve the Schroedinger equation for inverse radius potential using cartesian coordinates.

Even for a single particle, the Schrodinger equation describes many different possible situations. The hydrogen atom has a unique ground state, however, the Schrodinger equation allows for a ground state hydrogen atoms in any state with any velocity. Typically one is interested in atoms at rest. For hydrogen as a two body problem, one separates coordinates. The set that describes the center of mass motion isn't very interesting. The second set describes the relative motion of the particles and is of more interest. The separation is usually done in Cartesian coordinates. It's only after this is done, spherical coordinates are introduced for the relative coordinates. One doesn't need to do this, it's just much easier. One may continue to use Cartesian coordinates. All the observables listed in post #2 remain the same and so do their Eigenvalues independent of coordinates chosen.

 

[Nugatory]

One would still have the same set of solutions, but it is not clear that the same set of numbers to characterise them would occur.

Why not? It is precisely because “there being nothing deep about the original choice of basis” that we expect that the eigenvalues will be the same.

We can write $\hat{H}$, $\hat{L^2}$, $\hat{L_z}$ in Cartesian coordinates and then (in principle - in practice only a serious masochist would attempt this) solve Schrodinger’s equation to find the eigenfunctions written in these coordinates. These will of course be our original polar-coordinate eigenfunctions rewritten in Cartesian coordinates (which is why only a masochist would go through the exercise - we already know the answer).
So we’ve been writing down equations in the form $\hat{O}\psi=\lambda\psi$ in polar coordinates, then making a whole bunch of substitutions ($r$, $\theta$, $\phi$ are replaced with functions of $x$, $y$, $z$) on both sides. $\lambda$ is a constant so its value isn’t affected by these substitutions.