# Possible measurements of z-component of angular momentum

• I
• Fosheimdet
In summary: 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.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$$#### 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$? Fosheimdet said: 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. 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##.