Using the following as an example:

Show that the combinations $$-\frac{1}{\sqrt{2}}\left(Y_{11}-Y_{1-1}\right)\quad\text{&}\quad\frac{i}{\sqrt{2}}\left(Y_{11}+Y_{1-1}\right)$$ are real and normalized.

Where $$Y_{1\pm 1}=\mp\sqrt{\frac{3}{8\pi}}\sin\theta e^{\pm i\phi}$$

So for the first combination we have $$\begin{align}-\frac{1}{\sqrt{2}}\left(Y_{11}-Y_{1-1}\right)&=-\frac{1}{\sqrt{2}}\left[-\sqrt{\frac{3}{8\pi}}\sin\theta \,e^{i\phi}-\sqrt{\frac{3}{8\pi}}\sin\theta \,e^{-i\phi}\right]\\&=\frac{1}{\sqrt{2}}\left[\sqrt{\frac{3}{8\pi}}\sin\theta \left(e^{i\phi}+e^{-i\phi}\right)\right]\\&=\sqrt{\frac{3}{4\pi}}\sin\theta\cos\phi\\&=\sqrt{\frac{3}{4\pi}}\frac{x}{r}\end{align}$$

where in the last step I used the fact that $x=r\sin\theta\cos\phi$ in spherical polar coordinates.

For the second combination

$$\begin{align}\frac{i}{\sqrt{2}}\left(Y_{11}+Y_{1-1}\right)&=\frac{i}{\sqrt{2}}\left[-\sqrt{\frac{3}{8\pi}}\sin\theta \,e^{i\phi}+\sqrt{\frac{3}{8\pi}}\sin\theta \,e^{-i\phi}\right]\\&=-\frac{i}{\sqrt{2}}\left[\sqrt{\frac{3}{8\pi}}\sin\theta \left(e^{i\phi}-e^{-i\phi}\right)\right]\\&=\sqrt{\frac{3}{4\pi}}\sin\theta\sin\phi\\&=\sqrt{\frac{3}{4\pi}}\frac{y}{r}\end{align}$$

where in the last step I used the fact that $y=r\sin\theta\sin\phi$ in spherical polar coordinates.

The calculation is obviously not the problem here. It is the meaning of the result that concerns me. The author of this question mentions in the solution that:

They are obviously real. The spherical harmonics are orthogonal and normalized, $\color{blue}{\text{so the square integral of the two new functions will just give }}\color{blue}{\frac12(1+1)=1}$.

I know that the word 'normalize' is quite ubiquitous as it seems to have different meanings depending on the context.

For example:

The wavefunction for a particle confined to $0\le x \le a$ in the ground state is $$\Psi(x)=A\sin\left(\frac{\pi\, x}{a}\right)$$ where $A$ is the normalization constant. Find $A$ and hence the normalized wavefunction $\psi(x)$.

Omitting all calculations using the 'Normalization Condition':


The Normalized wavefunction is $$\psi(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{\pi\,x}{a}\right)$$

The words "Normalized wavefunction" make perfect sense to me as it has been normalized such that the condition $(1)$ is satisfied.

But what does it mean to say that $$-\frac{1}{\sqrt{2}}\left(Y_{11}-Y_{1-1}\right)=\sqrt{\frac{3}{4\pi}}\frac{x}{r}$$ is normalized?

Or put in another way; Where is the 'Normalization Condition' for this?

At a guess, I think the answer lies somewhere in what the Professor wrote in the quote above which I have marked blue. But I don't fully understand the logic behind what the professor is saying.


1 Answer 1


There are several choices for the normalization of spherical harmonics, but in your case it seems to be that $Y_{l,m} (\theta,\phi)$ is normalized if

$$ \int_0^\pi \mathrm{d}\theta \int_0^{2\pi} \mathrm{d}\phi \sin(\theta) \left|Y_{l,m} (\theta,\phi) \right|^2 = 1 $$

In problems like these it is often much easier to use the orthonormality of the spherical harmonic functions,

$$ \int_0^\pi \mathrm{d}\theta \int_0^{2\pi} \mathrm{d}\phi \sin(\theta) Y_{l,m} (\theta,\phi) Y_{l',m'}^{*} (\theta,\phi) = \delta_{l,l'}\delta_{m,m'} $$

to test whether linear combinations of spherical harmonic functions are normalized.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.