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Let $ |\psi\rangle$ be the normalised state vector of a spin 1/2 particle. Let the two function $\psi_\pm (r)$, where $\pm$ correspond to the values $S_z = \pm1/2$, be

$$\psi_+ = \langle r,+|\psi \rangle = R(r) \left[Y^0 _0 (\theta,\phi)+\frac{1}{\sqrt{3}} Y^0 _1 (\theta,\phi)\right]$$

$$\psi_- = \langle r,-|\psi \rangle = \frac{R(r)}{\sqrt{3}}\left[Y^1 _1 (\theta,\phi)-\frac{1}{\sqrt{3}} Y^0 _1 (\theta,\phi)\right]$$

the radial function R(r) is such that $|\psi\rangle$ is normalised.

Question : if we are asked to find the results of $L_z$ measured on the state and with the corresponding probabilities, do we need to take the expectation, or just find the resulting state by the action of the angular momentum operator? I am confused about the meaning of the measurement on the state? Do we need to consider the change to the present state by the measurement or just find the sum of squares of coefficients of different $L_z$ value?

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  • $\begingroup$ For a simple harmonic oscillator that has energy eigenfunctions $\psi_n$ with energy eigenvalues $E_n$, do you understand what energies you would measure in a state like $\sqrt\frac13\psi_2+\sqrt\frac23\psi_5$, and what the probabilities are of measuring these energies? If so, angular momentum works similarly. All observables have the same rules for quantum measurement. $\endgroup$
    – Ghoster
    Nov 28, 2022 at 20:11

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I've explained quantum measurements more formally in my other answer, but let me just reproduce what you need here.

There is a difference between applying some operator and measuring a state. A measurement of some operator $A$ on some system in the state $\left| \psi \right\rangle$ can be expressed if we know the eigenvalues of $A$ (which are also the possible outcomes upon measurement) and corresponding eigenvectors of $A$.

Supposing $A$ has $M$ unique eigenvalues labelled $a^{\,}_m$, I can write $A$ as $$ A \, = \, \sum\limits_{m=0}^{M-1} \, a^{\,}_m \, \mathbb{P}^{\,}_m \, ,$$ where the projectors $\mathbb{P}$ sum to the identity, square to themselves, and are orthogonal to each other.

If there is only one eigenvector $\left| \phi_m \right\rangle$ for the eigenvalue $a^{\,}_m$, then $\mathbb{P}^{\,}_m = \left| \phi_m \middle\rangle \hspace{-0.3mm} \middle\langle \phi_m \right|$. If there are multiple eigenvectors $\left| \phi^{k}_m \right\rangle$ for eigenvalue $a^{\,}_m$, then $\mathbb{P}^{\,}_m = \sum_k \left| \phi^k_m \middle\rangle \hspace{-0.3mm} \middle\langle \phi^k_m \right|$.

The possible measurement outcomes are the unique $a^{\,}_m$. The corresponding probability (for outcome $a^{\,}_m$) is just $p^{\,}_m = \left\langle \psi \middle| \mathbb{P}^{\,}_m \middle| \psi \right\rangle$. Technical explanation below:

Upon measuring $A$ on a system in the state $\left| \psi \right\rangle$, the new state is $$ \left| \psi \right\rangle \, \to \, \left| \psi' \right\rangle \, = \, \sum\limits_{m=0}^{M-1} \, \left( \mathbb{P}^{\,}_m \, \left| \psi \right\rangle \right)^{\,}_{\rm ph} \otimes \left| m \right\rangle^{\,}_{\rm meas} \, ,$$ where the state labelled "ph" is the post-measurement state of the physical system, and the state labelled "out" records the outcome—it is the state of the measurement apparatus if you like.

Note that $\mathbb{P}^{\,}_m \, \left| \psi \right\rangle$ is the unnormalized Copenhagen wavefunction following measurement, given that outcome $m$ obtained.

The probability of getting outcome $m$ is $$p^{\,}_m \, = \, \left\langle \psi' \middle| \mathbb{1} \otimes \widetilde{\mathbb{P}}^{\,}_{m} \middle| \psi' \right\rangle \, ,$$ where the tilde denotes an operator on the apparatus. The probability of outcome $m$ is just the expectation value that the detector is in the state $m$ after measurement.

Anyway, if you want to measure $L^{\,}_z$, you just need to work out the projectors onto its eigenstates in some basis. If you're already working in the $L^{\,}_z$ eigenbasis, then it's easier. But the possible measurement outcomes are always the eigenvalues, and the probability of an outcome is the square of the overlap of the pre-measurement state with the eigenstate corresponding to that outcome. If there are multiple such states, sum over each individual overlap squared. $$ p^{\,}_m = \left| \left\langle \phi^{\,}_m \middle| \psi \right\rangle \right|^2~~~\text{or}~~~p^{\,}_m = \sum_k \left| \left\langle \phi^{k}_m \middle| \psi \right\rangle \right|^2\,,$$ for nondegenerate $m$ and degenerate $m$, respectively.

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A measurement of an observable with corresponding operator $A$ on a state $|\psi \rangle$ doesn't have anything to do with the operation $A |\psi \rangle$. I would interpret a question asking for "the results of $L_z$ measured on $|\psi \rangle$" to be asking for the probabilities of getting $L_z=\hbar /2$ and $L_z = -\hbar /2$. I agree that if this was the wording of the question it's a bit vague and would have been more clear if the probabilities had been asked for directly.

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