0
$\begingroup$

Can we proof the relation that the principal quantum number $n$ and azimuthal quantum number $l$ have the relation $l=0,1,...n-1$ in any spherical symmetric potential $V(r)$ or this just apply to Coulomb potential?

I would appreciate reference to books or articles

$\endgroup$

2 Answers 2

1
$\begingroup$

This relationship does not hold for all spherically-symmetric potentials. For example, for the 3D harmonic oscillator the relationship between $n$ and $l$ is

$$l=0,2,4,...,n$$

for even $n$ and

$$l=1,3,5,...,n$$

for odd $n$.

$\endgroup$
2
  • $\begingroup$ So why in multi electrons atoms we have this relation,since in multi electron atoms we approximate the potential to be spherically-symmetric potentials? $\endgroup$ Dec 25, 2018 at 18:59
  • 1
    $\begingroup$ @amiltonmoreira But we do more than this. To first approximation (the orbital approximation) we consider each electron to have hydrogen-like orbitals created from an effective nuclear charge $Z_{eff}$ created from the nuclear charge and shielding of inner electrons. See chemicool.com/definition/orbital-approximation.html $\endgroup$
    – isometry
    Dec 25, 2018 at 20:29
0
$\begingroup$

The radial $P(r,\nu,l)$ part of the wave function in a central potential obey a one dimensional Schrodinger equation an so it is characterized by the number of nodes $\nu$, and since the equation contains $l$ it is also characterized by the quantum number $l$. Now we define $n=l+\nu+1$

So instead of characterizing the radial part by $l$ and $\nu$ we characterize it by $l$ and $n$ ,that is, the radial part of the wave function is of the form $P(r,n,l)$. Now since the number of nodes $\nu$ and angular number $l$ are positive integer we should have $l=0,1,...n-1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.