I'm looking through an old exam, and don't quite understand the solution given for one of the problems.(adsbygoogle = window.adsbygoogle || []).push({});

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

and are asked what possible measurements can be made of the z-component of the angular momentum.

My instinct is to use the operator [itex] \hat{L}_z = \frac{\hbar}{i} \frac{\partial}{\partial \phi} [/itex] on the wavefunction, which shows that g is not an eigenfunction of the operator.

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

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 [itex] \hat{L}_z [/itex] to the entire wavefunction? Has this got something to do with your choice of [itex] \phi [/itex]?

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# I Possible measurements of z-component of angular momentum

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