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I'm trying to get some intuition behind the spherical harmonics being the angular momentum eigenstates in quantum mechanics.

Firstly, am I correct in saying that we can imagine the angular momentum ($|L|^2$ and $L_z$ values) at every point in this angular momentum eigenstate wavefunction as being identical (and equal to the respective $|L|^2$ and $L_z$ eigenvalues)? Or instead are we meant to interpret the net 'sum' of the angular momenta at each point over the whole wavefunction as being equal to the eigenvalues? I'm assuming its the former since, with momentum eigenstate position wavefunctions for instance, we assume you'd measure that value of momentum at any position. I know we don't measure momentum at a single position but if the wavelength is constant everywhere I'd assume that means we can measure it at any position and get the same value.

Another reason why I assume it is the former is that I did some calculations using probability current and derived for the $Y_1^1$ spherical harmonic (assuming probability current $J=\rho v$ where $\rho = |Y_1^2|^2$), that the effective particle velocity was $\vec{v} = \frac{\hbar}{mR\sin\theta} \vec{\phi}$. When combined with the classical formula for the z-component of angular momentum (for a particle mass m on a shell of radius R so a distance $R \sin\theta$ from the -axis) you get:

$L_z = mR\sin\theta v = mR\sin\theta \times \frac{\hbar}{mR\sin\theta} = \hbar$ which implies that the effective $L_z$ is equal to $\hbar$ everywhere in the spherical harmonic. It seems like the effective particle velocity at points further from the z-axis reduces proportionally to keep the same $L_z$ everywhere.

(I did similar calcualtions for $Y_2^1$ and $Y_2^2$ and got $\hbar$ and $2 \hbar$ again so I think there is some merit to this method.) However I was unable to use this method so far to show that $|\vec{L}|^2$ was equal to $2\hbar^2$ everywhere. This makes me partially doubt my thinking that these values should be equal to the eigenvalues everywhere as opposed to when integrated over.

Am I wrong in my assumption? Also could anyone use this method to show the results for $|\vec{L}|^2$ too?

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Firstly, am I correct in saying that we can imagine the angular momentum ($|L|^2$ and $L_z$ values) at every point in this angular momentum eigenstate wavefunction as being identical (and equal to the respective $|L|^2$ and $L_z$ eigenvalues)? Or instead are we meant to interpret the net 'sum' of the angular momentua at each point over the whole wavefunction as being equal to the eigenvalues?

Neither. Angular momentum is not a property of a wavefunction at a point; it is a property of a wavefunction as a whole.

When you apply $L^2$ to an angular momentum eigenstate $\psi_l$, then you find $L^2 \psi_l = [l(l+1)\hbar^2] \psi_l$. That is, $l(l+1)\hbar^2$ is the value of $L^2$ which is associated to the eigenstate $\psi_l$. There's no sense in which it is associated to some region of space and then summed.

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  • $\begingroup$ Thank you very much for this clarification. What do you think of the probability current result then, essentially a coincidence? Or would I need to provide more information? $\endgroup$ – Alex Gower Sep 18 '20 at 23:19
  • $\begingroup$ Also would you say all other properties like momentum are also properties of the wavefunction as a whole too? $\endgroup$ – Alex Gower Sep 18 '20 at 23:25

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