Natural Science Forum / Physics / Research / April 2008

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.