Commutators of vector operators

1. Jan 19, 2012

thecommexokid

I've been trying to work out some expressions involving commutators of vector operators, and I am hoping some of y'all might know some identities that might make my job a little easier.

Specifically, what is $\left[\mathbf{\hat A}\cdot\mathbf{\hat B}, \mathbf{\hat C}\right]$? Are there any useful identities to express this in terms of simpler commutators?

Any help is appreciated.

2. Jan 22, 2012

brydustin

Are you sure you mean a "vector operator", typically we talk about matrix operators when discussing the commutator relationships (or group elements in a more general setting).
Vector operator: http://en.wikipedia.org/wiki/Vector_operator

[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
to be an identity.
Does that answer your question.... you could have looked anywhere on the internet to get this.... so I'm guessing this isn't what you want.

3. Jan 22, 2012

brydustin

4. Jan 22, 2012

thecommexokid

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.

Last edited: Jan 22, 2012
5. Jan 23, 2012

brydustin

Yeah, sorry I don't know. I thought you were intending for matrix operators. Good luck.