How are quantum numbers used to classify eigenvectors of a Hamiltonian?

[samblohm]

What exactly are quantum numbers? I read the Wikipedia page on them but it is just way to complicated for me.

 

[Matterwave]

They are basically just numbers denoting the state of the particle.

For example, if you're talking about the Bohr atom, then the quantum number is "n" which denotes which orbit the electron is orbiting at. For the Bohr atom, that's all that matters.

As you get to more formal QM, there are more numbers which are needed to denote the full state of the system.

 

[turin]

It depends whether you are discussing "ordinary" (i.e. nonrelativistic) QM or QFT.

In QM, quantum numbers are eigenvalues of observables. In particular, their values depend on the particular state, which is essentially random.

In QFT, it is more confusing. There are properties that define particle species, and there are properties that define states (configurations of various fields/particles). Only some of the particular properties that define states are regarded as "quantum numbers" (just as is the case in QM). However, things like energy and momentum (and helicity, actually) are not so-called ("good") quantum numbers in QFT. This is in stark contrast to (nonrelativistic) QM, where things like energy, momentum, and angular momentum are discretized into quantum numbers. Things like spin, charge, and chirality are so-called ("good") quantum numbers in QFT. This is again in stark contrast to (nonrelativistic) QM, where such quantities are just taken as part of the problem setup.

 

[peteratcam]

Given a Hamiltonian, H, it will have a complete othonormal set of eigenvectors and associated real eigenvalues.

Quantum numbers (which can be symbols as well as numbers) are a classification scheme for the eigenvectors.

The idea is that for any given operator, there is a unique way of turning a quantum number into the eigenvalue of that operator. Eg, for total angular momentum, if the quantum number is \ell, experienced physicists automatically recognise that the eigenvalue is \ell(\ell+1)\hbar^2. So associated with an observable is a set of quantum numbers which have a one-to-one mapping onto the set of eigenvalues, which themselves have a one-to-many mapping onto eigenstates. (Due to the possibility of degeneracy). A "good" quantum number means the operator commutes with the hamiltonian, ie. is a conserved quantity.

How do we classify eigenvectors of H using good quantum numbers then?
Choose some operator (eg, A) which commutes with H: it follows that eigenvectors of A are also eigenvectors of H.
Label the eigenvectors of H by the A quantum numbers.
If there are eigenvectors of H with the same label, find another operator B, which commutes both with A, and with H.
In addition to the A-quantum number, add a B-quantum number label.
Keep going with C,D, etc until no two states have the same set of quantum numbers.

eg Hydrogen:
1)Choose A as the hamiltonian itself.
States are labelled by 'n'. Still some degeneracy.
2)Choose B as total orbital angular momentum.
States are labelled by n\ell. Still some degeneracy.
3)Choose C as azimuthal angular momentum
States are labelled by n\ell m_\ell. Still some degeneracy.
4)Choose D as azimuthal spin angular momentum.
States are labelled by n\ell m_\ell m_s. No degeneracy.

The procedure is not unique, it boils down to identifying a maximally commuting set of observables and in general there are many of these.

In the presence of perturbations, the states might no longer be eigenstates of H, but they provide a useful basis to do perturbative calculations with.