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
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm
Robert J. Kolker
> Quick question:
>
> For two commuting three-vectors A and B, the cross product
What is a "commuting" 3 vector? A three vector is an ordered triplet of
3 real numbers. The property of commutation refers to the -operator-,
not the operands.
>
> AxB = -BxA
>
> But what if A and B are non-commuting. Say they are the spin oprator S
> and the angular momentum operator L.
Say what? Look at the definition of X in terms of the vector components
of the vectors being multiplied by the operator X.
AxB = -BxA follows directly from the definition.
Bob Kolker
Oh No
>Jay R. Yablon wrote:
>> Quick question:
>> For two commuting three-vectors A and B, the cross product
>
>What is a "commuting" 3 vector? A three vector is an ordered triplet
>of 3 real numbers. The property of commutation refers to the -operator-
>, not the operands
For an example consider the Dirac gamma matrices, which are often
considered loosely as a 4-vector. More strictly the dot product of the
gamma matrices with a vector is invariant under coordinate
transformation.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Ken S. Tucker
> 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.
When A and B are parallel, AxB = BxA = 0.
Ken
Ken S. Tucker
Hi Jay, have a glance at this.
I'll use I,J,K as unit Cartesian Vectors.
I x J x J x J x J = I
(......K...-I.._K....I )
where the values in (..) are the results.
Allow me to write,
I = I x J^4 == I x J^exp(x4).
I prefer the latter, so the vector product is clearly
notated within the exponent.
Using "f" as a function.
A view of d^4 f / d^4 of (sin f) = sin f,
with each d/df (sin f) == the operation " x J ",
rotates 90.
Recall we can use fractional derivatives,
http://en.wikipedia.org/wiki/Fractional_calculus
to enable, when N=4,
df^N (sin f) /df^N = sin f,
to depart from N being a whole number.
Using the above, employs 1,2,3,4 in the rotations
to produce a 360 degree in place of the 2 pi rads,
to place the rotations onto a Calculus Field.
IOW's I use N=4 in place of 2 pi radians.
Regards
Ken S. Tucker
Ken S. Tucker
I may have inadvertently introduced a notation
that is unfamiliar to mainstream mathematicians,
specifically dealing with repeated application of
"xJ" with an exponent assignment.
I researched the web and have so far found no
specific document using that notation, therefore,
for now, it will be inventive.
Below I'll do a few minor corrections, and then
add an application.
On Apr 19, 2:34 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
...
Hi Jay, have a glance at this.
I'll use I,J,K as unit Cartesian Vectors.
I x J x J x J x J = I
(......K...-I...-K....I )
where the values in (..) are the running results
provided for clarity.
Allow me to write,
I = I x J^4 == I x J^(x4).
I prefer the latter, so the vector product is clearly
notated within the exponent.
Using "f" as a function.
A view of d^4 (sin f) / df^4 = sin f,
with each d/df (sin f) == the operation " x J ",
rotates 90.
Recall we can use fractional derivatives,
http://en.wikipedia.org/wiki/Fractional_calculus
to enable, when N=4,
df^N (sin f) /df^N = sin f,
to depart from N being a whole number, with
the result that a variation from N=4 will not
yeild sin f. For example N=2 yeilds,
df^N (sin f) /df^N = -sin f ...etc.
> Using the above, employs 1,2,3,4 in the rotations
> to produce a 360 degree in place of the 2 pi rads,
> to place the rotations onto a Calculus Field.
> IOW's I use N=4 in place of 2 pi radians.
An application addendum:
You may recall I used an aircraft (a/c) to describe spin.
Anyway I Yawed once (360, around the z-axis) and
Rolled the a/c twice (720, around the y-axis), while
doing so to return the a/c back to it's original orientation.
But I specified the Pitch (around the x-axis) would be
zero.
Let me define one cycle as
I = I x J^(x4),
to use J^(x4) to spin "I" 360.
Then using frequency as a cycle per time,
w2 = J^(x4/t) = I cycle / time.
equivalent of 1 cycle of Roll using the a/c.
The Yaw rate has be designated as 1/2 the
Roll rate,
w3 = K^(x2/t) = 1/2 cycle / time,
or Yawed in the opposite direction,
w3 = K^(-x2/t) = -1/2 cycle / time.
Applying the above to unit vector "I" gives,
Pitch rate = w1 = I x J^(x4/t) x K^(-x2/t) = I^(x2/t) = -I /t
= -1/2 cycle/time around x,
(A complete cycle would have returned "I" back to
"I", so a 1/2 cycle produces the "-I" unit X vector),
by using the exponential rule within cross products,
x4/t - x2/t = x2/t while using that in the J x K above.
I intend to produce concrete conclusions, if and
when that notation makes reportable sense.
Regards
Ken S. Tucker
Jay R. Yablon
may be of interest here too. Best, Jay.
"Aage Andersen (REMOVE)" <a...@email.dk> wrote in message
news:480d2434$0$2086$edfa...@dtext02.news.tele.dk...
>
> "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.
> > I think I answered my own question, and the answer is:
>>
>
>> 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
>
>
>
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.