# I Possible measurements of z-component of angular momentum

1. May 4, 2016

### Fosheimdet

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 $\phi$?

2. May 4, 2016

### Staff: Mentor

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.

3. May 5, 2016

### vanhees71

I think, it's just formulated in a somewhat strange way. I guess, what they wanted to know is, which possible values $L_z$ can take, if the system is prepared in the given state. Since the state belongs obviously to $l=1$, you can just calculate
$$\langle 1,m |g \rangle=\int_{\Omega} \mathrm{d}^2 \Omega [Y_1^m(\vartheta,\varphi)]^* g(\vartheta,\varphi) \quad \text{for} \quad m \in \{-1,1,0 \}.$$
Then all $m$ are possible outcomes of an $L_z$ measurement, for which this product is different from 0. Of course this is equivalent to the answer given in #1, because indeed you can immediately read off the decomposition of $g$ in terms of $Y_1^m$.