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thecommexokid
#4
Jan22-12, 09:52 PM
P: 32
Commutators of vector operators

Thank you for the replyreplies.

Quote Quote by brydustin View Post
[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 [itex][{\bf{\hat S}},{\bf{\hat T}}] = {\bf{\hat S}} \cdot {\bf{\hat T}} - {\bf{\hat T}} \cdot {\bf{\hat S}}[/itex], and therefore start your derivation with
[tex][{\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}}).[/tex]
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 [itex]({\bf{\hat A}} \cdot {\bf{\hat C}}){\bf{\hat B}}[/itex] is equal to [itex]{\bf{\hat A}}({\bf{\hat C}} \cdot {\bf{\hat B}})[/itex], so your next step seems iffy.