The quantum number n and its restrictions

1. Jun 3, 2015

pondzo

So in class we went through the process of solving the S.W.E for the hydrogen atom.

During the process a constant $\lambda_n=\frac{ze^2}{4\pi\epsilon_0\hbar}(\frac{\mu}{2|E_n|})^{\frac{1}{2}}$ is introduced, where mu represents the reduced mass of the electron.

Later this constant is put on the following restriction so that the power series has a finite power; $\lambda_n=l+k+1$ (where k is the index number for the power series).

It then follows that this constant must be an integer since l and k are. $\lambda_n=n\geq l+1$.

I would like to know the physical reason why the quantum number n arises and why it is put under the constraint $n\geq l+1$.

Last edited: Jun 3, 2015
2. Jun 3, 2015

DEvens

Maybe I am not remembering correctly But I think it is n that is the integer, not lambda. The constant lambda is a physical quantity with physical units.

Asking "why" in physics is a troublesome thing. It depends on what kind of answer you will be happy with.

At one level, the answer is what you have already done in class. It arises because of the various boundary conditions and the need to solve the equation.

At another level it is a physical continuity and symmetry kind of thing. In essence, the wave function has to fit into a "mode" of rotational symmetry, for example. And it has to satisfy certain restrictions in the radial direction in order to be finite and correspond to a finite probability of detecting an electron. So there are only certain possibilities. So you come to discover the atomic orbitals. And those are arranged by integers.

With only a little more sophistication you come to group theory and functional representations. To motivate that, think about a system that has mirror symmetry in the plane x=0. Such a system will naturally break into functions that are even, f(x) = f(-x), and functions that are odd, g(x) = -g(-x). Depending on the details of the system, it means you can have the system in either f or g, but not a mix. Similarly with a system with spherical symmetry. There are categories of functional characteristics that are possible, and they are characterized by a set of integers.

But fundamentally, the "why" is still because it has to satisfy the equation under the boundary conditions.

3. Jun 3, 2015

PeroK

The reason that $\lambda$ must be equal to an integer is so that the power series terminates. If $\lambda$ is not an integer, then you have an infinite power series and a non-normalisable wave function. There might be a bit of work to show this, but that's the reason.

4. Jun 3, 2015

Khashishi

Not really sure what your question means. The problem of the hydrogen atom can be separated into a radial and angular part. The angular momentum quantum number is quantized because the angular wavefunction has to be single valued at the poles. k is just the excitation level of the radial wavefunction for a fixed angular momentum l. As you increase the energy (fixing l), the number of radial nodes increases, and these states are labeled by k. Since the number of nodes has to be integral, k is an integer. And n is just l+k+1. We defined n this way because it just so happens that the different states with the same 1+l+k have about the same energy, given by Rydberg's formula. But really, k is perhaps more "fundamental" than n.