0

I'm trying to solve this exercise.

Angular momentum $\mathbf{L}$ satisfies the relation $\mathbf{L\times L}=i\mathbf{L}$. Given two vectors $\mathbf{a}$ and $\mathbf{b}$ which satisfy $[\mathbf{a,b}]=[\mathbf{a,L}]=[\mathbf{b,L}]=0$, show that $$[\mathbf{a\cdot L,b\cdot L}]=i(\mathbf{a\times b})\cdot\mathbf{L}$$ How is it that a vector doesn't commute with itself?, and how does the vector product $(\mathbf{a\times b})$ appears? Here's my try: \begin{align*} [\mathbf{a\cdot L,b\cdot L}]&=\mathbf{a}\cdot[\mathbf{L},\mathbf{L}]\cdot\mathbf{b}\\ &=(\mathbf{a\times b})\cdot[\mathbf{L},\mathbf{L}]\\ &=(\mathbf{a\times b})\cdot i\mathbf{L} \end{align*} But I think there something wrong with my try.

David Barrero González
  • 1
  • I don't think there's enough context here to answer this question. Since you use a cross product I assume you are working on some copy of the Euclidean space $\mathbb{R}^3$ but what Lie bracket are you imagining on this space? What does $\cdot$ represent, the dot product or some kind of action? – Callum Apr 16 '22 at 22:00
  • There's no further information in the problem, but since a and b are common vectors, I think it's just a dot product. – David Barrero González Apr 17 '22 at 23:26
  • Then the Lie bracket $[a \cdot L, b \cdot L] = 0$ automatically since $a\cdot L$, etc. must be in $\mathbb{R}$ (or whatever field we are in) and the only possible Lie bracket on $\mathbb{R}$ is the trivial one $[x,y] \equiv 0$. – Callum Apr 18 '22 at 10:08
  • But L is not properly a vector, but an angular momentum operator which is being multiplied by a trivial vector on each side of the brackets, so that (I think so) [a⋅L,b⋅L]=(a⋅L)(b⋅L)-(b⋅L)(a⋅L), and I believe L is acting on b in the first term and on a in the second one, or is it acting on the dot product?. I forgot to say that the exercise asks for solving it using index notation. – David Barrero González Apr 19 '22 at 23:24
  • I think you may have better luck asking this question on the physics stack exchange. It seems to me there must be some standard meanings for some of this in physics which are not clear from a maths standpoint – Callum Apr 21 '22 at 13:29

0 Answers0