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In the radial equation of hydrogen atom the differential equation is described by enter image description here

But why is l taken to be integer. I know the principal quantum number n correspond to energy levels so that's why it's taken as integer. Why should azimuthal quantum number be taken as integer though

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  • $\begingroup$ Do you know what $l$ is and whether number $1$ is an integer? $\endgroup$
    – andselisk ♦
    Commented Oct 29, 2020 at 16:23
  • $\begingroup$ l is the angular momentum $\endgroup$
    – zeeman
    Commented Oct 29, 2020 at 16:27
  • $\begingroup$ But why is it an integer always when solving for hydrogen we use laguerre polynomials and solve the radial equation but then we substitute l as an integer in l(l+1) $\endgroup$
    – zeeman
    Commented Oct 29, 2020 at 16:28
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    $\begingroup$ When $\ell$ is not an integer, you'll get bad solutions. $\endgroup$
    – Ivan Neretin
    Commented Oct 29, 2020 at 16:40
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    $\begingroup$ If you want the maths, though, I honestly suggest checking out a QM book. There are lots of great ones out there. I'm sure many of them have the same derivation that show how $n, \ell, m$ have the conditions they do. $\endgroup$
    – orthocresol
    Commented Oct 29, 2020 at 18:21

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There are two parts to the answer to your question. (a) When solving a differential equation the boundary conditions must be specified (this is not always emphasised enough in textbooks), and (b) these conditions are determined by the physics of the problem via the postulates of quantum mechanics, including the nature of the wavefunction. Quantum mechanics (QM) has no derivation; we accept its postulates and QM survives only because experiment has so far always confirmed it.

As an example of quantisation consider the particle in a box. Here there are all sorts of solutions to the differential equation, i.e Schroedinger's equation. However, as the walls are infinitely high the wavefunction must have zero amplitude here and this means that the only solutions that remain and are physically realistic are when quantum number $n$ is quantised. The fact that the wavefunction must be zero at each wall are the boundary conditions.

A similar situation occurs on a particle on a ring where the angular momentum quantum number $l$ is quantised only because we insist, by applying the postulates of QM that the wavefunction repeats itself exactly each $2\pi$ round the circle. This is in effect the boundary condition.

In the H atom the equation is far more complicated but the general conditions imposed on the wavefunction remains the same; the imposition of boundary conditions, as determined by QM, results in the generation of integer quantum numbers so that theory matches experimental data.

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  • $\begingroup$ you said by applying the postulates of QM that the wavefunction repeats itself exactly each 2π round the circle but isnt this the boundary condition for n like 2×pi×r=n lambda. Can you explain a bit more on how it extends to azimuthal quantum number and imposition of boundary conditions that force integer quantum numbers $\endgroup$
    – zeeman
    Commented Oct 31, 2020 at 9:03
  • $\begingroup$ The boundary condition on $\psi$ is that $\psi (0)=\psi(2\pi)$ and that the gradient $d\psi/d(\theta) $is the same at $0$ and $2\pi$ $\endgroup$
    – porphyrin
    Commented Oct 31, 2020 at 9:45

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