I have a big confusion with group theory terminology. I know that orbital angular momentum (OAM) is $\mathrm{SO}(3)$-symmetric in 3D-space. Let's define QM orbital angular momentum (OAM) conventionally:

$$\pmb{L} = -i \pmb{r} \times \pmb{\nabla}$$

This definition can also be written using a set of $\mathrm{SO}(3)$ generators:

$$L^{\mu} = -i r_i \; S_{ij}^{\mu} \; \nabla_j$$

where $\mu = \{x,y,z\}$ for 3D space, and $S_{ij}^{\mu}$.

So... generators stand for the definition of a vector product in given space, essentially, definition of orthogonality? Or this is only in this case, I suppose, in which case why such a coincidence?

If I proceed with this:

$$\pmb{r} e^{-iS^{\mu} \phi} \pmb{p}= \pmb{r} \cdot \pmb{p} - i \delta \phi \; \pmb{r} S^{\mu} \pmb{p} + \cdots = \mathrm{const} \; e^{- i \pmb{r} \cdot \pmb{p}} + \delta \phi L^{\mu}$$

Matter wave in zeroth order and OAM in first? Does it have any interpretation?

  • $\begingroup$ A generic element $g$ of a continuous group would be $g=e^{i \mathbf{T}_i \theta_i}$ where $\mathbf{T}_i$ are infinitesimal generators of the group, and $\theta_i$ is a real parameter. $\endgroup$ – Oktay Doğangün Jul 19 '18 at 19:08
  • $\begingroup$ Where are the elements and where is the operation here? Admittedly, I'm pretty stupid, but it's unclear to me if you are referring to all possible orbital densities under (SO(3)) or the spherical ones alone? $\endgroup$ – user198207 Jul 19 '18 at 22:08
  • $\begingroup$ @StudyStudyStudy , honestly, I have no idea what you are talking about. I took Arfken&Weber and read the Group Theory Chapter, literally. I think I got that question though. Now, next question is why the action of group generators end up being identical to vector product operation, that I don't get. $\endgroup$ – MsTais Jul 19 '18 at 23:08
  • $\begingroup$ So, generators of group representation is/can be an operation? $\endgroup$ – MsTais Jul 19 '18 at 23:09

I don't know if I've understood your question well. The math may be too complicated for me, but the ideas are

  1. The usual definition of OAM satisfies the conmutation rule

$$\left[\frac{L_x}{\hbar}, \frac{L_y}{\hbar}\right]=i \frac{L_z}{\hbar}$$

  1. This means that they are infinitesimal generators of rotations in the space they live in.

    1. And that's correct: in fact $\exp(-i\theta L_n/\hbar)$ is actually the "rotation" operator of an angle $\theta$ about an axis $n$ in Hilbert's space.

You can prove this by stablishing

$$ \varphi_F (\vec{x}) = \varphi_0(R^{-1}\vec{x})$$

Which means "the updated wavefunction in any point is equal to the old wavefunction evaluated in the point before the rotation".

$R$ is the rotation matrix in $\mathbb{R}^3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.