View Single Post
P: 38
Commutators of vector operators

Thank you for the replyreplies.

 Quote by brydustin [S,T] = ST - TS (by definition) Start with [AB,C] = ABC - CAB (+ ACB - ACB ) = ABC - ACB + ACB - CAB = A(BC - CB) + (AC - CA)B = A[B,C] + [A,C]B Therefore we conclude [AB,C] = A[B,C] + [A,C]B
It seems to me that you're being pretty cavalier about vector multiplication, what with the way you're just putting vectors in a row next to each other without any dots or parentheses. For instance, what do you mean when you write “ABC”, when A, B and C are vector operators?

I would think that you should define $[{\bf{\hat S}},{\bf{\hat T}}] = {\bf{\hat S}} \cdot {\bf{\hat T}} - {\bf{\hat T}} \cdot {\bf{\hat S}}$, and therefore start your derivation with
$$[{\bf{\hat A}} \cdot {\bf{\hat B}},{\bf{\hat C}}] = ({\bf{\hat A}} \cdot {\bf{\hat B}}){\bf{\hat C}} - {\bf{\hat C}}({\bf{\hat A}} \cdot {\bf{\hat B}}).$$
But from there, I'm not sure how you can safely proceed, if you're being rigorous with your dots and parens. For instance — and correct me if I'm wrong on this — but I don't think $({\bf{\hat A}} \cdot {\bf{\hat C}}){\bf{\hat B}}$ is equal to ${\bf{\hat A}}({\bf{\hat C}} \cdot {\bf{\hat B}})$, so your next step seems iffy.