# Why does the quantum number n limit the value of l, and why does l, likewise, limit the value of m?

Any quantum number that represents the size of an orbital is a positive integer, $n$. Any quantum number that represents the shape of an orbital is a non-negative integer, $l : l < n$. Any quantum number that represents the orientation of an orbital, $m$, is an integer such that $|m| \leq l$.

Why do $n$ and $l$ restrict the ranges of $l$ and $m$, respectively?

To be clearer, I ask why the quantum numbers limit each other, not what they signify, which is already explained in the related question: What do the quantum numbers actually signify?

• – LordStryker May 28 '15 at 17:31
• @LordStryker, that question explains how to interpret quantum numbers, but does not explain why the ranges of allowable values are what they are. Why for example can $l=0,1,...,n-1$ but never $l=n$? – Ben Norris May 28 '15 at 22:51
• @BenNorris Because $l=n$ simply isn't a solution to the Schrodinger equation for hydrogen. I know you know this but for completeness, simply look at the periodic table. When $n=1$, that means that there can only be one subshell (i.e. orientation) of the $s$ orbital (which there is). Therefore $l=0$. When $n=2$, then there are two subshells, $l=0,1$ which corresponds to $s$ and $p$ orbitals. Furthermore, $m_l=-1,0,1$ since there are 3 orientations of $p$ orbitals when $n=2$ and $l=1$. – LordStryker May 29 '15 at 1:57
• Extremely instructive walktrough: hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html – ssavec May 29 '15 at 12:13

While it may seem that quantum numbers were invented out of the blue, that isn’t the case. They derive directly from solving the Schrödinger equation:

$$\hat{H}\Psi = E \Psi$$

One of the simplest Schrödinger equations is that for a particle in a box — the one most quantum chemistry or quantum mechanics courses start with and therefore found in every textbook for quantum chemistry which is why I won’t repeat it here. But in a nutshell: You need a function that you can differentiate twice and which will then reproduce itself with a constant factor E. That can be achieved by sine waves. You cannot say a priori which frequency the sine function will have, though: If the box has size l, it works for sine waves with the wavelengths $2l,\ l,\ \frac{2l}{3},\ \frac{l}{2},\ \dots$ and the corresponding wavenumbers $\frac{1}{2l},\ \frac{1}{l},\ \frac{3}{2l},\ \frac{2}{l},\ \dots$ or $\frac{n}{2l}$ with n being a natural number. n is therefore the simplest case of a quatum number.

If you solve this equation for an electron orbiting a proton, the solutions are much more complicated (and there are intermediate steps often presented in the quantum chemistry courses to make the transition easier). However, n still turns up as a variable, corresponding to quanted amounts of energy. For certain values of n, there are additional variables whose values cannot be derived a priori but which can only assume specific values — they correspond to the other quantum numbers.

This in itself is just a solution to a mathematical equation. The real beauty lies in the fact that these quantum numbers can be used to explain experimental observations such as hydrogen’s flame colours. All quantum numbers can be used to explain some property one has deduced by experiments.

So tl;dr: It’s just what maths tells us will be the case.

The shapes of the orbitals, i.e. the solutions to the Schrödinger equation for the hydrogen atom $\hat{H}\Psi_i = E_i\Psi_i$, are primarily due to the requirement that all orbitals are orthogonal: $\langle\Psi_i|\Psi_j\rangle=0$. Orthogonality is made possible by the fact that the orbitals have "positive" and "negative" regions separated by nodes.

One what to think about $n$ is that $n-1$ is the number of nodes in $\Psi_i$. So there can be no $1p$ orbitals since $p$-shaped orbitals have at least one node. Similarly, there can be no $2d$ orbitals, and so on.

Because $s$ orbitals have a spherical shape, there is only one way to orient them, i.e. $m=0$ if $l=0$. $p$-shaped orbitals lie along one axis and there are three Cartesian axes, so there are three possible - orthogonal - orientations (I'm sacrificing a bit of accuracy here for clearness).

It's a bit harder to visualize why there should be 5 and only 5 $d$-shaped orbitals, but the answer is that there are only 5 ways to construct 3-dimensional shapes with (a minimum of) 2 nodes that are orthogonal to the $s$- and $p$-shaped solution as well as the other $d$-shaped solutions.