Re: Is Cross Product AxB always equal to -BXA? Even if A and B are non-commuting?

"Jay R. Yablon" <jyablon@xxxxxxxxxxxx> wrote in message
news:66o1o3F2l7e3mU1@xxxxxxxxxxxxxxxxxxxxx
Quick question:

For two commuting three-vectors A and B, the cross product

AxB = -BxA.

But what if A and B are non-commuting. Say they are the spin oprator
S
and the angular momentum operator L.

Might it be that LxS <> -SxL in this case?

Or, would the related commutators:

[L,S]+[S,L]=0

cause us to have LxS = -SxL here as well?

Thanks.

Jay.
____________________________
Jay R. Yablon
Email: jyablon@xxxxxxxxxxxx
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

I think I answered my own question, and the answer is:

AxB = -BxA, always.

See
http://jayryablon.wordpress.com/files/2008/04/intrinsic-spin-decomposition.pdf,
which will explain the context in which I raised this question, because
of the canonical commutation relationship

[x,p]=i hbar Kronecker-delta

which does not permit commutation between co-aligned x, p.

Jay.

.