Is Cross Product AxB always equal to -BXA? Even if A and B are non-commuting?
Jay R. Yablon
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: jya...@nycap.rr.com
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Jay R. Yablon
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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.
Aage Andersen
"Jay R. Yablon" >> Quick question:
>>
>> For two commuting three-vectors A and B, the cross product
>>
>> AxB = -BxA.
>>
>> But what if A and B are non-commuting.
>
> AxB = -BxA, always.
Then, how do you explain the formula for angular momentum?
M x M = i hbar M in quantum mechanics
M x M = 0 in classical mechanic
Aage
Jay R. Yablon
news:480d2434$0$2086$edfa...@dtext02.news.tele.dk...
1) I think you are right, and have swung back to my original view of
the problem. Let's start with
A X B
Now, with e_ijk being the Levi-Civita tensor and i-1,2,3, I believe we
may define:
(A X B)_i = e_ijk A_j B_k (1)
This means that if we just rename A <---> B, and then rename and
transpose indexes, we may write:
(B X A)_i = e_ijk B_j A_k = e_ikj B_k A_j = -e_ijk B_k A_j (2)
Then, we add (1) and (2) to obtain the commutator:
(A X B)_i + (B X A)_i = e_ijk [A_j, B_k] (3)
Whether A x B + B x A = 0 then depends on the properties of the
commutator, and this is *not necessarily* zero. One place it is not
zero, is for M x M = i hbar M, as you point out.
2) The specific problem I have been considering which got me started on
this thread, is as regards the canonical commutation relationship.
Here:
[x, p] = i hbar.
This is not a cross product per se. But, if one considers the cross
product:
L X S
where L is the angular momentum operator and S is the intrinsic spin
operator, and if:
L = x X p
where x and p are position and momentum operators (position x lowercase,
versus cross X uppercase), then:
L X S + S X L
has some intriguing properties because there is an implied commutation
of x and p embedded in this. When I first started the thread, I was
trying to ascertain what the canonical commutation relationship did to
this cross product, and whether it is or is not equal to zero.
What I have found after careful calculation is that:
L X S = (x X p) X S = i hbar S - x(p dot S) - p(x dot S) (4)
and that:
S X L = S X (x X P) = i hbar S +(S dot p)x + (S dot x)p (5)
Five points here:
a) The cross product picks up an extra i hbar S term that does not
appear in the usual expression for a triple cross product.
b) The i hbar S arises directly from [x, p] = i hbar, and so is solely
of quantum mechanical origin.
c) If one isolates i hbar S from all the other terms, then
i hbar S = L X S + x(p dot S) - p(x dot S) (6)
is, effectively, a "decomposition" of the spin operator into terms
involving x, p and itself.
d) Adding (4) and (5) as we did above with (1) and (2), it may look like
L X S + S X L <> 0, but in fact, L X S + S X L = 0.
To see this, add (4) and (5) to get:
L X S + S X L
= 2 i hbar S {-x(p dot S)+(S dot p)x} {-p(x dot S)+(S dot x)p}=0
Each of the terms in {} brackets turn out, via [x, p] = i hbar, to be
equal to
-x(p dot S)+(S dot p)x = -i hbar S
-p(x dot S)+(S dot x)p = -i hbar S
So at least in this case, L X S = - S X L, *despite the extra i hbar S
term* which was what I was specifically grappling with when I started
the thread.
e) The magnetic moment operator is just
-(e/m) gamma^0 i (hbar/2) S
see Ohanian's article at
http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf
which I have discussed in some other threads and which is a very
fundamental and very under-recognized article which clarifies some
difficult boundaries between classical and quantum theory and gives
classical theory more explanatory power over "non-classical, two-valued"
spin than it was previously understood to have. (Sorry Pauli ;-)
This means via (6), that one can also decompose the magnetic moment
operator into terms that involve x and p and the Heisenberg
relationships. I am at present studying Robertson-Schroedinger and
wondering if there is some way to use this to provide an alternate
explanation, particularly of the *anomalous* magnetic moment, based on
the Heisenberg *inequality* that would set 2 to be, not the gyromagnetic
ratio, but the *lower boundary* of this ratio, with the Schwinger
expansion telling us the degree to which delta x delta p exceeds hbar
/2, based on the Heisenberg inequality:
delta x delta p >= hbar / 2.
That should give an idea what I am wrestling with at the moment and how
this cross product question arose along the way.
Thanks,
Jay.
Rock Brentwood
> > For two commuting three-vectors A and B, the cross product
> > AxB = -BxA.
>
> > But what if A and B are non-commuting.
Then not.
> I think I answered my own question, and the answer is:
>
> AxB = -BxA, always.
In the covering algebra of a Lie group, where [u,v] = uv-vu, the
"homogeneous" field laws become (in 3-vector form):
div B + A.B - B.A = 0
curl E + AxE + ExA + dB/dt + B phi - phi B = 0
where ().() denotes scalar product; A, phi are respectively the vector
potential and scalar potential; E and B are respectively the Lorentz
forces for electricity and magnetism, given in terms of the potentials
by
B = curl A + A x A
E = -grad phi - dA/dt + phi A - A phi.
The components of AxE + ExA are the Lie brackets of the components.
For the z-component:
(AxE + ExA)_z = (Ax Ey - Ay Ex) + (Ex Ay - Ey Ax)
= (Ax Ey - Ey Ax) + (Ex Ay - Ay Ex)
= [Ax, Ey] + [Ex, Ay];
where Ex, Ey and Ax, Ay are, respectively, the x and y components of E
and A.