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I am trying to understand the derivation of the radial solution for the hydrogen atom in quantum mechanics. When solving the radial Schrödinger equation using a power series method, the solution for the radial function is expressed as a power series:

$$v(\rho)=\sum_{j=0}^\infty c_j\rho^j,$$

To ensure that this solution is physically meaningful and normalizable, the power series must terminate after a finite number of terms. This requirement leads to a recursion relation for the coefficients:

$$c_{j+1}=\frac{2(l+1+j)-\rho_0}{(j+1)(j+2l+2)}c_j,$$

where $l$ is the orbital angular momentum quantum number, and ρ0 is a parameter related to the energy. For the series to terminate, there must exist an integer $N = j + 1$ such that:

$$c_N\ne 0, \text{ but }c_{N+1}=0.$$

This leads to the condition:

$$2(N+l)=\rho_0.$$

At this point, the principal quantum number $n$ is defined as:

$$n=N+l,$$

My questions are:

  • Why is the quantum number $n$ specifically defined as $n = N + l$?
  • How could we know in advance to define $n = N + l$ this way, without prior knowledge of the hydrogen atom's energy levels or their dependence on the quantum number $n$?
  • Is the definition of $n$ chosen purely to connect the mathematical expressions to the known physical energy levels, or is there a deeper reason behind this choice?
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    $\begingroup$ Related: physics.stackexchange.com/q/724784/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 13 at 16:46
  • $\begingroup$ It is simply a fact that, if you do not make this choice, then the derived energy levels would look very ugly compared to the simplicity of the Rydberg formula (and Bohr model). i.e. that choice is solely to make connection with the all-important Rydberg formula. $\endgroup$ Commented Sep 13 at 19:24
  • $\begingroup$ Simply put: $E_n = -\frac{R_H}{n^2}$ is a pretty formula for the energies. $\endgroup$ Commented Sep 14 at 12:27

1 Answer 1

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It's the 'accidental' degeneration of the spectrum of the Coulomb-Schrödinger operator.

For a general form of the spherical potential $V(r)$ one solves $$-R''(r) - \frac{2}{r} R' + \frac{l(l+1)}{r^2} \ R + V(r) R' = n_{r,l} \ R.$$ In the cases $V(r)=r^k, k=-1,0, 2$ there exists an additional constant of motion, the classical Runge-Lenz vector pointing along the constant perihelion vector and derived from $$p\times L = r p^2- (r\cdot p) \ p$$.

It follows, that the spectral sequences for knot lines on the unit sphere from Legendre polynomials and knots on the radius from the radial equation appear as a linear combination in the radial energy eigenvalue equation.

As a principle, the spectrum is ordered along the energy scale by the value of $n_r+l$ and the subspaces of degeneration are ordered by the eigenvalue of the Laplacian on the unit sphere $-\Delta_{S^2} = L^2$ with eigenvalues $l(l+1)$.

The fact that $l$ and not a linear form of the eigenvalue like $\sqrt{l(l+1)}$ appears in the main quantum number is again a fact of symmetry: All subspaces of states on the unit sphere with equal $L^2$, $l,m=-l\dots l$ are degenerate and yield the same spherical energy contribution as the state with maximal $L_z=\pm l$ with no knots along meridians and kinettic energy in rotational form as opposed to the oscillating waves between the poles.

From the last sentence its clear whats the reason of the complicated representation of the symmetry group: The choice of the axis and the poles is arbitrary. A rotation of the sphere mixes all states of fixed quantum number $l$. The picture of static orbitals with fixed charge and current density has no absolute meaning; except if one breaks the symmetry by a weak magnetic field or considers the system in a fixed external lattice.

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