The wavefunction I've been given for a 1s hydrogen orbital is:

$$ \Psi = A e^{-r} $$

And I need to normalize this to find the value of A. I understand to normalise this I would inset this wave function into:

$$\int_{-\infty} ^\infty \Psi^*\Psi dr = 1$$

When I do this, I get:

$$A^2 \int_{-\infty} ^\infty e^{-2r} dr = 1$$

From the literature, I believe that the normalisation constant of this wavefunction should be equal to $\frac{1}{\sqrt{\pi}}$ but I can't work out where the pi comes from? Or how to evaluate this integral since it is unlike any I have seen. Are my limits wrong? Is there a standard integral I don't know about?

Any help would be much appreciated!

  • $\begingroup$ Note that atomic orbitals are 3-dimensional, while your integral there is 1-dimensional. The wavefunction you've been given is actually in spherical coordinates: $\Psi(r,\phi,\theta)=Ae^{-r}$, so you should be using a 3-dimensional integral. $\endgroup$
    – Sandejo
    Jun 10, 2021 at 18:34
  • $\begingroup$ You are right, I should have added that the hydrogen orbital is spherically symmetric therefore I figured since $\theta$ and $\phi$ would be constant, $d\theta$ and $d\phi$ would equal 1.... Is that correct? So I wouldn't need to integrate over those coordinates? But I understand that means my limits of $\infty$ definitely aren't correct. $\endgroup$
    – Harry
    Jun 10, 2021 at 18:41
  • $\begingroup$ an infinitesimal element $d\theta$ is certainly never equal to $1$. $\endgroup$ Jun 10, 2021 at 19:15

2 Answers 2


Your limits are certainly wrong, since the variable $r$ extends from 0 to $\infty$, not from $-\infty$ to $\infty$.

The actual condition for normalisation of a 3D wavefunction is $\psi(r,\theta,\phi)$ that $$\int_\text{all space} \psi^*\psi \,\,\text{d}V = 1,$$ where the integral is done over all space.

In the case of the Hydrogen atom, the wavefunction can be separated into the radial part that you call $\Psi(r)$, and an angular part (call it $Y_l^m(\theta,\phi)$ ), so that: $$\psi(r,\theta,\phi) = \Psi(r) Y^m_l(\theta,\phi).$$

By shifting to an integral in 3D over the variables $r,\theta,$ and $\phi$, the normalisation condition now simply reduces to:

$$\int_{r=0}^\infty r^2 \text{d}r \,\,\Psi^*(r)\Psi(r) \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi}\sin(\theta) \text{d}\theta\, \text{d}\phi\,\, Y^{*m}_l(\theta,\phi) Y^m_l(\theta,\phi) = 1,\tag{1}\label{1}$$

where I have used the fact that $\text{d}x\text{d}y\text{d}z \rightarrow r^2\sin(\theta)\text{d}r\text{d}\theta\text{d}\phi$.

Now, the spherical harmonics $Y^m_l$ are defined such that they are already normalised, and so $$ \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi}\sin(\theta)\text{d}\theta\, \text{d}\phi\,\, Y^{*m}_l(\theta,\phi) Y^m_l(\theta,\phi) = 1,$$ which just leaves

$$\int_{0}^\infty r^2 \text{d}r\,\, \Psi^*(r)\Psi(r) = 1,$$

which should be easy enough to calculate by integration by parts or Feynman's Integral Trick.

EDIT: If you do this calculation, you will find that $A$ does not have a factor of $\pi$ in it. The question really is this: is $A$ supposed to represent the normalisation factor for the radial wavefunction, or for the overall wavefunction? This is a matter of convention. I have -- in the analysis above -- used the former definition. As a result, I have chosen the (standard) convention that the $Y^m_l$s are normalised to 1. If you choose that definition, then the 1$s$ wavefunction would be:

$$\psi(r,\theta,\phi) = A \Psi(r) Y^0_0(\theta,\phi),$$ where the $A$ is the term you calculated above, and $Y^0_0$ is the appropriate spherical harmonic $$Y^0_0 = \frac{1}{2\sqrt{\pi}}.$$

If, on the other hand, you don't want to use this convention because you don't know about spherical harmonics and so on, you can keep things simple by going back to Equation (\ref{1}) and choosing $\psi(r,\theta,\phi) = A e^{-r}$. In this case, $A$ is the normalisation constant for the entire wavefunction and we forget all about the spherical harmonics and so on. Equation (\ref{1}) becomes:

$$\int_{r=0}^\infty r^2 \text{d}r \,\,\Psi^*(r)\Psi(r) \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi}\sin(\theta) \text{d}\theta\, \text{d}\phi = 1.$$

Since $$\int_{\theta=0}^\pi \int_{\phi=0}^{2\pi}\sin(\theta) \text{d}\theta\, \text{d}\phi = 4\pi,$$ this just means that $$4\pi A^2 \int_{0}^\infty r^2 \text{d}r\,\, \Psi^*(r)\Psi(r) = 1,$$ which should lead to the factor you're looking for. Of course, the overall normalisation factor turns out to be the same in both cases, as it should.

  • 1
    $\begingroup$ Thank you! The second way was more what I was looking for (basic). This makes perfect sense, with this inserting my wavefunction I can use the standard integral $\int_0 ^\infty r^2 e^{-2r} dr = 2!$ And I end up with a factor of $\pi$ in the normalization constant that way! $\endgroup$
    – Harry
    Jun 10, 2021 at 20:28

In 3d spherical coordinates, the integration is $$ \int_0^\infty dr r^2 \int_0^{\pi} \sin\theta d\theta \int_0^{2\pi} d\phi \vert\psi(r,\theta,\phi)\vert^2 $$ since

\begin{align} \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy\int_{-\infty}^\infty dz \to \int_0^\infty dr r^2 \int_0^{\pi} \sin\theta d\theta \int_0^{2\pi} d\phi \end{align} in spherical, i.e. the volume element $$ dV=dx\,dy\,dz\to r^2\sin\theta dr\,d\theta\,d\phi $$ in spherical. The range of the integration is the standard one for the spherical coordinates, where $0\le r< \infty$, $\theta\le 0 < \pi$, and $0\le \phi < 2\pi$ in the definition of the coordinates.


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