I recently started learning about quantum mechanics and its applciations in atomic structure in chemistry.

In this inorganic textbook Inorganic Chemistry, it describes

"Each of the wavefunctions obtained by solving the Schrödinger equation for a hydrogenic atom is uniquely labelled by a set of three integers called quantum numbers."

What is the relationship between the quantum numbers and the wave function? If the above is saying that each wavefunction is uniquely labelled by 3 quantum numbers, then it sounds to me that the 3 quantum numbers can describe the wavefunction on their own. But how can just 3 numbers describe a function?

On this website, it says the following

"A complete solution to the Schrödinger equation, both the three-dimensional wavefunction and energy, includes a set of three quantum numbers ( n,l,ml )."

Is the complete solution to the Schrodinger's Equation the wavefunction in 3D? what is the "energy" mentioned in the last quote to do with solution of the equation?

Thank you.

  • $\begingroup$ Have you looked at the general solution of the S.E for a hydrogen atom ? Have you seen the quantum numbers there ? $\endgroup$
    – Poutnik
    Commented Feb 21, 2021 at 10:56
  • $\begingroup$ Imagine you have a differential equation leading to a general solution y=k+l.x+m.x^2, where k,l,m are integers. Does the knowledge of k,l,m completely describe the function y=f(x) as the particular solution of that implied equation ? In both cases, here and wave function case, the general function is already known but the parameter values. $\endgroup$
    – Poutnik
    Commented Feb 21, 2021 at 11:09
  • $\begingroup$ A number can not define a function. It is just part of the function, like the constant $k$ in trigonometric problems, where you find a solution with an angle $\alpha + 2k\pi$ $\endgroup$
    – Maurice
    Commented Feb 21, 2021 at 11:11
  • 4
    $\begingroup$ @Maurice A number can define a function, if the general form of function is a priori known. If y=n.x is given, than giving n defines the particular function. By other words the wave function is partially predefined, waiting just for particular values of allowed quantum numbers.( as allowed numbers are predefined as well ). $\endgroup$
    – Poutnik
    Commented Feb 21, 2021 at 11:12

2 Answers 2


The hydrogen atom

Indeed, three numbers cannot define a function. There are several more criteria, one of which you mentioned already: it must satisfy the time-independent Schrödinger equation. The Schrödinger equation is a nasty differential equation, which looks something like the following:

$$-\frac{\hbar^2}{2\mu}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) - \frac{e^2}{4\pi\varepsilon_0r}\psi = E\psi$$

On top of this, it must also satisfy the usual criteria for a wavefunction ("boundary conditions"): for example it must be square integrable (i.e. its integral over all space must be finite, meaning that it can be normalised):

$$\int |\psi|^2\,\mathrm{d}\tau < \infty$$

Once you put all of these conditions together, it turns out that not every plain old function $\psi(x,y,z)$ is a permissible atomic orbital. There is a family of functions $\psi$ which do, and they all have a certain form which is completely determined by the three quantum numbers $(n, l, m_l)$.

A simpler example

This is pretty abstract, so why don't we make it more concrete with a simpler example. (The hydrogen atom is not easy to solve explicitly.) Consider the one-dimensional function $f = f(\phi)$, where $\phi$ is an angle. You should recognise at this point that, given no other information, one quantum number is not sufficient to specify what $f$ is, as there is an infinite possible different $f$'s.*

However, assume now that $f$ obeys the differential equation

$$\frac{\mathrm{d}f}{\mathrm{d}\phi} = \mathrm{i}mf$$

The general solution of this differential equation is (after dropping the normalisation factor)

$$f = \exp(\mathrm{i}m\phi)$$

But note also that since $\phi$ is an angle, we have that for all $\phi$

$$f(\phi) = f(\phi + 2\pi)$$

meaning that

$$\begin{align} \exp(\mathrm{i}m\phi) &= \exp[\mathrm{i}m(\phi + 2\pi)] \\ &= \exp(\mathrm{i}m\phi) \exp(\mathrm{i}m2\pi) \\ 1 &= \exp(\mathrm{i}m2\pi) \end{align}$$

which means that $m$ must be an integer. Consequently, the allowed $f$'s are

$$f = \exp(\mathrm{i}m\phi) \quad\quad m \in \mathbb{Z}$$

and so we can now label the $f$'s by the value of $m$. The quantum numbers $m$ do not represent actual values of $f$ at any point in space, nor do they represent actual values of $\phi$: they instead serve to parameterise the $f$'s. That is to say, the value of $f$ at any value of $\phi$ is completely determined by the value of $m$. We do not write this as $f(m)$, but rather $f_m$, and thus we have:

$$\begin{align} &\,\,\,\vdots \\ f_{-2} &= \exp(-2\mathrm{i}\phi) \\ f_{-1} &= \exp(-\mathrm{i}\phi) \\ f_{0} &= 1 \\ f_{1} &= \exp(\mathrm{i}\phi) \\ f_{2} &= \exp(2\mathrm{i}\phi) \\ &\,\,\,\vdots \end{align}$$

Indeed, $m$ can be said to be a quantum number here. Notice how we've used a simplified problem: it has one instead of three dimensions, and it has a much simpler differential equation. However, it illustrates that by enforcing additional conditions on the wavefunction — namely that it obey a differential equation and boundary conditions — we can get a family of solutions which can be labelled with quantum numbers.†

What is the energy?

When you find the allowed $\psi$'s that solve the Schrödinger equation, you can plug it back into the differential equation and evaluate the value of $E$ inside it. That will give you the energy of the atomic orbital.


* Technically, there is an uncountably infinite number of $f$'s. If it were countably infinite, then we could establish a one-to-one mapping between the different $f$'s and the natural numbers, which would allow us to label the $f$'s as $f_1, f_2, f_3, \cdots$ which is exactly what a quantum number does.

† With regards to the previous footnote, this set of permissible $f$'s is still infinite, but it is countably infinite this time. Also, this actually happens to be the exact way in which the quantum number $m_l$ is derived, so my choice of $m$ as the quantum number was not arbitrary.


You need to add two quantum numbers for the electron: s and ms to take in account its Fermionic nature.

The concept of quantum number is quite general and can be applied both to monoelectronic wavefunctions both poliectronic ones. Given a physical system, a quantum number is an integer or a semi integer associated to an invariant. An invariant is a quantity that does not change during the dynamics. For an isolated system, the most important invariant is the total energy.

Why quantum numbers are so useful? Because they can label in a unique way the wavefunctions. That's it.

I would suggest you to read the masterpiece of Eyring, Walter and Kimball "Quantum Chemistry". It uses a very old notation, but if you want to understand the root of QM it is the best choice.


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