Monday, 16 November 2015

quantum mechanics - Commutation of abstract $O(3)$ generators and vectors



I've been given the following problem, and I'm quite lost with it.



Let $L_1$, $L_2$, and $L_3$ denote the abstract $o(3)$ algebras. You are given that $\vec{A} = (A_1, A_2, A_3)$ and $\vec{B} = (B_1, B_2, B_3)$ transform as vector operators of $o(3)$.


Show that $[L_j, \vec{A} \cdot \vec{B}] = 0$



I know that $L_j = \varepsilon_{jlm} q_l p_m$, and I can obviously determine the dot product, but I'm not sure where to go from there.



I do, however, know that $\vec{A} = \frac{1}{Ze^{2}\mu}(\vec{L} \times \vec{p}) + (\frac{1}{r})\vec{r}$, but I'm not sure how to integrate that into this problem.



Answer



A collection $\{V_1, V_2, V_3\}$ of operators on a vector space $V$ is called an $\mathfrak{o}(3)$ vector operator with respect to a representation $\rho$ of $\mathfrak{o}(3)$ acting on $V$ provided \begin{align} [V_i, L_j] = i\epsilon_{ijk}V_k \end{align} where $L_1, L_2, L_3$ are the representatives of the standard basis on $\mathfrak{o}(3)$. This means, in your case, that merely knowing that $A_i$ and $B_i$ are the components of vector operators tells us that \begin{align} [A_i, L_j]=i\epsilon_{ijk}A_k, \qquad [B_i, L_j]=i\epsilon_{ijk}B_k \end{align} These commutation relations alone are sufficient to demonstrate the desired result using commutator identities.


As a tangential note, you will probably find the Wikipedia page on tensor operators to be generally, conceptually helpful for understanding this stuff. Also, a while back I asked the following question on physics.SE dealing with how to generalize and formalize the notion of tensor operators in a less basis-dependent way than how I defined them at the beginning of this answer. In case you're interested in math, here is that question:


Tensor Operators


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...