# Possible measurements of z-component of angular momentum

• I
I'm looking through an old exam, and don't quite understand the solution given for one of the problems.

We have given a wavefunction $g(\phi,\theta) = \sqrt{\frac{3}{8\pi}}(-cos(\theta) + isin(\theta)sin(\phi))$

and are asked what possible measurements can be made of the z-component of the angular momentum.
My instinct is to use the operator $\hat{L}_z = \frac{\hbar}{i} \frac{\partial}{\partial \phi}$ on the wavefunction, which shows that g is not an eigenfunction of the operator.

In the solution however, they rewrote g as $g(\phi,\theta) = \sqrt{\frac{3}{8\pi}}(-cos(\theta) + \frac{1}{2}sin(\theta)(e^{i\phi}-e^{-i\phi}))$ and said that the first term is an eigenfunction of $\hat{L}_z$ with the eigenvalue m=0, the second term has the eigenvalue m=1 and the last m=-1. Therefore the possible measurements are $m=0,\pm{1}$.

My question is why this is valid. Why does one of the components of the original wavefunction give you one of the measurement values? Don't you have to apply $\hat{L}_z$ to the entire wavefunction? Has this got something to do with your choice of $\phi$?

Nugatory
Mentor
My question is why this is valid. Why does one of the components of the original wavefunction give you one of the measurement values? Don't you have to apply $\hat{L}_z$ to the entire wavefunction? Has this got something to do with your choice of $\phi$?

They've just rewritten the wavefunction as a superposition of three eigenfunctions.
If you apply ##L_z## to the entire wave function you will get the expectation value of ##L_z##. The possible results and their probabilities are given by the Born rule: The result will be an eigenvalue, and the probability of getting a particular eigenvalue is the square of the coefficient of that eigenfunction.

vanhees71
$$\langle 1,m |g \rangle=\int_{\Omega} \mathrm{d}^2 \Omega [Y_1^m(\vartheta,\varphi)]^* g(\vartheta,\varphi) \quad \text{for} \quad m \in \{-1,1,0 \}.$$