Timeline for What are spherical tensors?
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| Jun 29, 2023 at 10:19 | comment | added | ZeroTheHero | @Solidification. Think of the (Cartesian) inertia tensor in class.mech, and remove the trace (which is invariant under rotation). What's left can be expressed in terms of components of a spherical tensor with $L=2$. If you rotate the coordinate system, the Cartesian components of the inertia tensors will rotate one into a linear combo of the others, as will the spherical components. The difference is that the combos for the spherical case are $D^2_{mm'}$ functions. | |
| Jun 29, 2023 at 8:20 | comment | added | Solidification | Thanks, your comments are very helpful in clearing some dust. I'll have to do some reading before I can come back to accept the answer :-) I have to convince myself that Sakurai's derivation of how spherical harmonics transform has nothing to do with quantum mechanics (though it uses "bra and ket" language to derive the transformation of spherical harmonics under rotation). | |
| Jun 29, 2023 at 7:32 | comment | added | Jagerber48 | But I guess I'll emphasize that the definition of $D^k_{q, q'}$ as the matrix elements for rotations of spherical harmonics need not have ANYTHING to do with quantum mechanics. Also, once we have spherical harmonics, we could calculate the elements of $D$ numerically if needed. The fact that analytic methods/tricks used to give an analytic expression for $D$ us notations and intuitions from quantum mechanics doesn't mean we NEED quantum mechanics to understand $D^k_{q,q'}$ or spherical tesnors. | |
| Jun 29, 2023 at 7:12 | comment | added | Jagerber48 | Using the formula I've given and orthogonal you can see that $D$ is directly related to the overlap integral between $Y_l^m$ and a rotated version of $Y_l^{m'}$. See e.g. physics.stackexchange.com/questions/383594/…. If this integral can be used to calculate the explicit formula on the linked Wikipedia page, I'm not sure. | |
| Jun 29, 2023 at 7:07 | comment | added | Jagerber48 | @Solidification so it's a well known fact or definition (en.wikipedia.org/wiki/…) that $Y_l^{'m} = Y_l^{m'} D^l_{m', m}$ But it does remain to be seen if we can use this relation to "straight-forwardly" derive an explicit formula for $D^l_{m', m}$. I imagine it should be possible, but I don't know how to do it and I don't have a reference off-hand. Or at least not one that moves into group theory, hilbert spaces, and representation theory (which I think is to be avoided per the spirit of the question..) | |
| Jun 29, 2023 at 7:02 | comment | added | Solidification | "...can be defined in terms of transformation properties of spherical harmonics" Is there a reference where a discussion can be found? | |
| Jun 29, 2023 at 7:01 | comment | added | Jagerber48 | $D(R)^k_{q, q'}$ is a just a $2k+1 \times 2k+1$ array, similar to how $R$ is a $3\times 3$ array | |
| Jun 29, 2023 at 7:00 | comment | added | Jagerber48 | @Solidification No, $D(R)^k_{q, q'}$ does not require quantum mechanics to define. It does not require $U(R)$. $D(R)^k_{q, q'}$ can be defined in terms of transformation properties of spherical harmonics which are the family of solutions on the sphere of a certain differential equation which can arise in contexts unrelated to quantum mechanics | |
| Jun 29, 2023 at 6:54 | comment | added | Solidification | Here is my problem. Can't we talk about spherical tensors without talking about quantum mechanics? If the definition is in terms of $D(R)$ (or $U(R)$, in your notation), suggests to me that we are using the knowledge of quantum mechanics to define a spherical tensor. This is weird! Certainly, for defining how Cartesian components of a tensor transform, we didn't need that. | |
| Jun 29, 2023 at 6:47 | history | answered | Jagerber48 | CC BY-SA 4.0 |