Spherical-Wave State: E, l & m Quantum Numbers Explained

[rbwang1225]

Why can spherical-wave state be described by the 3 quantum number, E, l, and m?

And, how can we assume the normalization convention $$<E', l', m'|E, l, m> = \delta\delta_{ll'}\delta_{mm'}\delta(E-E')$$?

 

[henry_m]

This is to do with the idea of a complete commuting set of observables. An observable (hermitian operator) has a basis of orthogonal eigenvectors. If you have two observables which commute, then they can be simultaneously diagonalised, which means that there is an orthogonal basis of simultaneous eigenvectors, i.e. states which are eigenvectors for both operators. This is useful because the eigenvalues of a single operator may not be enough to distinguish all of the eigenvectors: some eigenvalues may be degenerate. Adding an extra commuting operator helps to break up the degenerate subspace by providing an extra label. It's a bit like labelling by rank and then file to identify a square on a chess board.

A complete set of commuting observables (A,B,C,...) is a set whose eigenvalues (a,b,c,...) completely label the states, so there is a basis of simultaneous eigenvectors, and any two states of the basis must differ in at least one of the eigenvalues.

In the example of a spherically symmetric potential, the symmetry means that the energy alone is not enough to label every state uniquely. But spherical symmetry implies that the Hamiltonian commutes with the angular momentum operator, whose eigenvectors are labelled by the number l. This label adds more information, but for fixed energy and total angular momentum there are still degeneracies coming (roughly speaking) from the possible directions of the angular momentum. So, we introduce the operator of angular momentum in the z-direction, which commutes with both total angular momentum and Hamiltonian operators. This then may be enough to uniquely label a full basis of eigenstates (we may still need more though, for example if the particles have spin).

Now to the normalisation. The eigenstates as we mentioned above are orthogonal due to hermiticity, so we have
$$\langle E',l',m'|E,l,m\rangle=c_{E,l,m}\delta_{EE'}\delta_{ll'}\delta_{mm'}$$
where the c's are some positive constants depending on E,l,m. But we can always rescale the vectors to make these all 1. If E has a continuous spectrum, so we are talking about scattering states, the Kronecker delta must be replaced with a Dirac delta. (The technical mathematics of scattering states is a little sensitive, but you won't go far wrong if you just think of this as the continuum analogue of the ordinary orthonormality condition).