Query about intrinsic verus orbital angular momentum
Jay R. Yablon
http://farside.ph.utexas.edu/teaching/qm/lectures/node40.html
it is stated as follows in about the middle of the page:
"Spin angular momentum clearly has many properties in common with
orbital angular momentum. However, there is one vitally important
difference. Spin angular momentum operators cannot be expressed in terms
of position and momentum operators, like in Eqs. (297)-(299), since this
identification depends on an analogy with classical mechanics, and the
concept of spin is purely quantum mechanical: i.e., it has no analogy in
classical physics. Consequently, the restriction that the quantum number
of the overall angular momentum must take integer values is lifted for
spin angular momentum, since this restriction (found in Sects. 5.3 and
5.4) depends on Eqs. (297)-(299). In other words, the quantum number is
allowed to take half-integer values."
Digging a little deeper, when it is stated that "the concept of spin is
purely quantum mechanical: i.e., it has no analogy in classical
physics," is this because the electron (or whatever fermion is under
consideration) is regarded to be a "point" particle with no "radius" for
the spin? In other words, does this restriction apply because of the
"point particle" notion or would this apply even if the electron had a
finite (albeit exceedingly tiny, e.g. Planck scale) spatial expanse
which permitted a "rotation"?
Related to this, how do we "know" that intrinsic spin does not in fact
involve rotation on an exceptionally small scale beyond the direct reach
of our experimentation? Is this just supposition?
Related to this, is the orbital angular momentum based on the l and m
quantum numbers for the electron considered to entail a small-scale
rotation of a sort that does not apply for the spin, or is this as
equally "quantum mechanical" as the intrinsic spin?
Finally, because each of orbital and spin angular momentum, together
with their conserved sum j = l + s, only have definite eigenvalues with
respect to a single chosen (z) axis of quantization, and the only other
good quantum number comes from the Casimir operator j(j+1), the angular
momenta (orbital and spin) around other than the z axis are uncertain.
Does this not place l and s on the same footing quantum mechanically?
Any clarification is appreciated.
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
Jordi Burguet Castell
> Digging a little deeper, when it is stated that "the concept of spin is
> purely quantum mechanical: i.e., it has no analogy in classical
> physics," is this because the electron (or whatever fermion is under
> consideration) is regarded to be a "point" particle with no "radius" for
> the spin? In other words, does this restriction apply because of the
> "point particle" notion or would this apply even if the electron had a
> finite (albeit exceedingly tiny, e.g. Planck scale) spatial expanse
> which permitted a "rotation"?
It has a lot to do with the electron beeing a "point particle".
Experimentally, the electron radius is known to be r < 10^-16 m (the
limits coming from electron-neutrino scattering, I think to remember,
but I would be glad if anyone points to the correct number and
experiment).
Now, if the electron spin was due to the mass m of "electron-material"
rotating around its axis (in the same way as the Earth's spin is due
to the addition of all the mass chunks rotating around), then its
moment of inertia would be
I ~ m r^2
(with a factor of 2/5 or so to have an equality, depending on the
distribution of the electron mass)
and the spin
S = I w ~ m r^2 w
for an angular velocity w = v/r where v is the velocity of a point in
the surface of the electron. But this velocity has a surprise:
v = w r ~ S/(m r^2) r = S/(m r)
If we now put the values we know for the electron spin (S=1/2 hbar),
mass (m=9e-31 kg) and experimentally maximal radius (r=1e-16 m), we
get v ~ 5e11 m/s... much faster than the speed of light!
I saw this argument in David Griffiths' very nice "Introduction to
Elementary Particles". My memory may have distorted it though :-)
> Related to this, how do we "know" that intrinsic spin does not in fact
> involve rotation on an exceptionally small scale beyond the direct reach
> of our experimentation? Is this just supposition?
It is just not consistent with the classical analysis above.
> Related to this, is the orbital angular momentum based on the l and m
> quantum numbers for the electron considered to entail a small-scale
> rotation of a sort that does not apply for the spin, or is this as
> equally "quantum mechanical" as the intrinsic spin?
Orbital angular momentum is related to rotation, even if it is not
small. It is just that quantum-mechanically it turns out to be
discrete. Spin also, and it contributes to the total angular momentum,
but it doesn't mean it is due to a rotation.
> Finally, because each of orbital and spin angular momentum, together
> with their conserved sum j = l + s, only have definite eigenvalues with
> respect to a single chosen (z) axis of quantization, and the only other
> good quantum number comes from the Casimir operator j(j+1), the angular
> momenta (orbital and spin) around other than the z axis are uncertain.
> Does this not place l and s on the same footing quantum mechanically?
They are both angular momenta, so in that sense they are related
(quantum mechanics enters specially in the fact that the electron spin
has no analogous in classical physics). The stuff about j=l+s and only
having a definite total and z projection angular momentum is
circumstantial (and not quite true, those quantities form a base, but
any state can be in a superposition of that kind of states and thus
have no definite total and/or z projection angular momentum).
I hope this clarifies things a little bit!
jordi
J. J. Lodder
> Referring to the link below:
>
> http://farside.ph.utexas.edu/teaching/qm/lectures/node40.html
>
> it is stated as follows in about the middle of the page:
>
> "Spin angular momentum clearly has many properties in common with
> orbital angular momentum. However, there is one vitally important
> difference. Spin angular momentum operators cannot be expressed in terms
> of position and momentum operators, like in Eqs. (297)-(299), since this
> identification depends on an analogy with classical mechanics, and the
> concept of spin is purely quantum mechanical: i.e., it has no analogy in
> classical physics. Consequently, the restriction that the quantum number
> of the overall angular momentum must take integer values is lifted for
> spin angular momentum, since this restriction (found in Sects. 5.3 and
> 5.4) depends on Eqs. (297)-(299). In other words, the quantum number is
> allowed to take half-integer values."
All this is just nonsense:
quantum numbers have no meaning in classical physics anyway.
> Digging a little deeper, when it is stated that "the concept of spin is
> purely quantum mechanical: i.e., it has no analogy in classical
> physics," is this because the electron (or whatever fermion is under
> consideration) is regarded to be a "point" particle with no "radius" for
> the spin? In other words, does this restriction apply because of the
> "point particle" notion or would this apply even if the electron had a
> finite (albeit exceedingly tiny, e.g. Planck scale) spatial expanse
> which permitted a "rotation"?
Clasically speaking point particles can have angular momentum.
It is just a matter of taking limits in the appropriate way.
> Related to this, how do we "know" that intrinsic spin does not in fact
> involve rotation on an exceptionally small scale beyond the direct reach
> of our experimentation? Is this just supposition?
On the contrary,
we do know that electron spin does involve rotation,
from angular momentum conservation.
For example the Einstein-De Haas effect:
inverting magnetization direction of a magnet
causes a macroscopic rotation of it.
Even more directly: firing a polarized electron beam
into a macroscopic target will cause it to rotate.
Best,
Jan
news.individual.net
"J. J. Lodder" <nos...@de-ster.demon.nl> wrote in message
news:1if6y3m.13j...@de-ster.xs4all.nl...
So, would it be off-based to believe that classically speaking, there
really is no such thing as "spin"? That is, to believe that spin is
just a composite of orbital angular momenta for individual infinitesimal
mass elements about an origin. I.e., one can only have a true "spin"
which is not orbital, if one has a true physical "point" that also
coincides with the rotational origin.
If one is applying classical analogies to quantum phenomena, this would
suggest that there can also really be no quantum mechanical spin, unless
the electron is truly a "point" particle. If it is merely very tiny,
such as on the order of the Planck length, then even its spin really has
to be orbital on the Planck scale.
Your thoughts?
Jay.
Gerard Westendorp
[..]
> So, would it be off-based to believe that classically speaking, there
> really is no such thing as "spin"? That is, to believe that spin is
> just a composite of orbital angular momenta for individual infinitesimal
> mass elements about an origin. I.e., one can only have a true "spin"
> which is not orbital, if one has a true physical "point" that also
> coincides with the rotational origin.
There are some interesting old threads on this (2001, "spin, what is it?")
A few quotes:
> What I see in my head is a circulating flow of energy and momentum density
> around the borders of the particle's quantum-mechanical wave function.
> Belinfante showed back in the 1930s that spin in the Dirac theory could be
> interpreted this way, and H. C. Ohanian wrote a famous article about it in
> the American Journal of Physics:
>
> Hans C. Ohanian, "What is spin?" AJP 54 (6), 500-505 (1986)
>
So Ohanian shows that you *do* actually get spin half if you integrate
angular momentum density over a Dirac wave packet!
One way you might think spin 1/2 cannot arise from a wave function, is
de Broglie's formula,
momentum = h/ wavelength
So if you have a circular wave at distance R , the angluar momentum (L)
is
L = h R / (2 pi R) = h_bar
In other words, the quantization comes from the fact that circular waves
must have integer wave numbers in the tangential direction.
But for a Dirac spinor field, de Broglie's formula is a bit more
complex. Also, the solution of the Dirac equation in a central potential
is more complex: The spinor field is not locally aligned with the total
spin.
The puzzling thing is, outof quite complex calculation, you "magically"
get the spin half from the integral of angular momentum density. It
would be nice if we could see this in a more intuitive way
Gerard
to...@cc.usu.edu
> Related to this, how do we "know" that intrinsic spin does not in fact
> involve rotation on an exceptionally small scale beyond the direct reach
>
Ultimately, one should address these issues from the
point of view of quantum field theory.
But, I think a decisive argument - strictly in the context
of non-relativistic quantum mechanics - is simply that one
can show angular momentum of the r x p type (associated
with rotation) can have only integer quantum numbers. The
electron, of course has half-integer quantum numbers; its
spin angular momentum must therefore be interpreted in some
other way.
charlie
sr
> So, would it be off-based to believe that classically speaking, there
> really is no such thing as "spin"? That is, to believe that spin is
> just a composite of orbital angular momenta for individual infinitesimal
> mass elements about an origin. I.e., one can only have a true "spin"
> which is not orbital, if one has a true physical "point" that also
> coincides with the rotational origin.
Have a careful read of Box 5.6 in Misner, Thorne & Wheeler's
"Gravitation", which defines angular momentum,
and then explains its decomposition into orbital and
intrinsic parts. This is classical, of course, but it's an
essential piece of information needed when trying to
understand this stuff.
Separately, regarding another posting about spin-1/2 emerging
from Dirac wave packets, that's less surprising if one
recalls that the integral and half-integral values for
total spin are a direct consequence of angular momentum
commutation relations, represented unitarily, i.e., in a Hilbert
space.
Jay R. Yablon
"Gerard Westendorp" <wes...@xs4all.nl> wrote in message
news:47ff1ba4$0$14351$e4fe...@news.xs4all.nl...
Hi Gerard,
First, I'll note that I first learned general relativity from Dr.
Ohanian at RPI in the early 1980s. Thanks for pointing out this work,
which Dr. Thomas Love at USC had earlier brought to my attention and
which I cited in a draft paper on this topic at
http://jayryablon.files.wordpress.com/2008/04/intrinsic-spin-30.pdf
(comments always welcome). Hans DeVries also commented on your post
here, over at sci.physics.foundatons, see
http://groups.google.com/group/sci.physics.foundations/browse_frm/thread/c6fce6b9b281dd3c/21da7a5375d8c5cc#21da7a5375d8c5cc.
Now, the above L = h R / (2 pi R) = h_bar is based on tangential
integer wave numbers over a 2pi rotation. But what if we go to
orientation / entanglement -- i.e., "version," as laid out in Misner /
Wheeler / Thorne in Gravitation, section 41.5. Then we obtain a double
covering, and your equation above turns into:
L = h R / (4 pi R) = (1/2) h_bar
> But for a Dirac spinor field, de Broglie's formula is a bit more
> complex. Also, the solution of the Dirac equation in a central
> potential is more complex: The spinor field is not locally aligned
> with the total spin.
>
> The puzzling thing is, outof quite complex calculation, you
> "magically" get the spin half from the integral of angular momentum
> density. It would be nice if we could see this in a more intuitive way
Might the above be the "more intuitive" way to which you refer? Cross
reference to eqs. (5.3) and (5.4) in the file linked above.
Thanks,
Jay.
>
> Gerard
>
Jay R. Yablon
news:bf3701bb-6d34-47f3...@24g2000hsh.googlegroups.com...
Thank you for the post. What you have done, effectively, is asserted as
an answer, the main query I am posing. You say "one can show angular
momentum of the r x p type (associated with rotation) can have only
integer quantum numbers." My question simply, is, HOW does one show
this? Especially, how does one *exclude* spin 1/2 from also being of
the rxp type?
As a counterpoint, take a look at Hans Devries' Figure 1.5 at
http://physics-quest.org/Book_Chapter_EM_basic.pdf. This seems to
suggest that one can also consider half-integer spins to also be of the
rxp type. You thoughts?
Thanks,
Jay
d...@frogfly.org
> You say "one can show angular
> momentum of the r x p type (associated with rotation) can have only
> integer quantum numbers." My question simply, is, HOW does one show
> this? Especially, how does one *exclude* spin 1/2 from also being of
> the rxp type?
Isn't it just as simple as finding eigenvalues of L = r x p operator?
eigenvalues of L^2 are l(l+1) h-bar^2, with l an integer. There are no
other solutions to the e-value problem. Thus half-integer spin can't
come from an r x p operator.
Is there a problem with this line of argument?
Dan
to...@cc.usu.edu
> momentum of the r x p type (associated with rotation) can have only
> integer quantum numbers." My question simply, is, HOW does one show
> this? Especially, how does one *exclude* spin 1/2 from also being of
> the rxp type?
>
>
This comes from trying to define position, momentum,
and orbital angular momentum as self-adjoint operators.
A nice discussion of some of the issues is in
E. Merzbacher
"Single-valuedness of Wave Functions"
American Journal of Physics, Vol. 30, pgs. 237-247 (1962).
charlie torre
Jay R. Yablon
<d...@frogfly.org> wrote in message
news:f662d120-f3fa-4cf4...@24g2000hsh.googlegroups.com...
Dan
[Yablon]
Not at all, Dan:
Finally, somebody has answered my question directly. Let me rephrase
what I think you mean:
We generate a rotation through angle psi about the z-axis by acting on a
wavefunction phi via with the z-axis rotation generator m, according to:
exp[i psi m] phi.
Because psi has a period of 2 pi, the eigenvalues of m must be an
integer, not a half integer. Am I correct so far?
If so, then my next question is this: Go to section 41.5 of Misner,
Wheeler, Thorne, where they discuss orientation / entanglement. There,
a rotation through 4pi is needed to restore not only orientation, but
the entanglement version. So, if psi needs a period of 4pi rather than
2pi to restore the entanglement relationship with the surrounding
environment, then the Eigenvalues of m can be half integers, and can be
generated -- not by the "rotation group" -- but by what one might call
the "entanglement / rotation" group. This would make it possible, I
believe, to create half-integer spin as well, from an rxp operator, with
the understanding that entanglement counts. This would appear to pave
the way to place the fundamental representation SO(1,3) of the Lorentz
group, which uses the SU(2) generators, onto an rxp operator footing as
well.
Your further thoughts.
Thanks,
Jay.
Hans de Vries
>
> Finally, somebody has answered my question directly. Let me rephrase
> what I think you mean:
>
> We generate a rotation through angle psi about the z-axis by acting on a
> wavefunction phi via with the z-axis rotation generator m, according to:
>
> exp[i psi m] phi.
>
> Because psi has a period of 2 pi, the eigenvalues of m must be an
> integer, not a half integer. Am I correct so far?
>
> If so, then my next question is this: Go to section 41.5 of Misner,
> Wheeler, Thorne, where they discuss orientation / entanglement. There,
> a rotation through 4pi is needed to restore not only orientation, but
> the entanglement version. So, if psi needs a period of 4pi rather than
> 2pi to restore the entanglement relationship with the surrounding
> environment, then the Eigenvalues of m can be half integers, and can be
> generated -- not by the "rotation group" -- but by what one might call
> the "entanglement / rotation" group. This would make it possible, I
> believe, to create half-integer spin as well, from an rxp operator, with
> the understanding that entanglement counts. This would appear to pave
> the way to place the fundamental representation SO(1,3) of the Lorentz
> group, which uses the SU(2) generators, onto an rxp operator footing as
> well.
>
> Your further thoughts.
>
> Thanks,
>
A free Dirac electron has a phase which is equal throughout
its rest frame, so there is no such varying phase around the
wave-function.
Now, the effect which I describe in the first chapter in my
book is straightly from Sakurai's treatment of the Dirac
equation in his book: Advanced QM, chapter 3.5 "Gordon
decomposition"
http://physics-quest.org/Book_Chapter_EM_basic.pdf
section 1.9
For the analogy with magnetized media, Sakurai refers
to Jackson. The referred section has become section 5.8
in my 1999 copy, equation 5.79.
Regards, Hans
Matej Pavsic
Other relevant papers are:
C. Van Winter, Annals of Physics 47, 232 (1968).
S.S. Sanikov, Nucl. Phys. 87, 834 (1967).
For an old discussion see
W. Pauli, Helv. Phys. Acta 12, 147 (1939).
Van Winter provides a very detailed discussion. He chooses
a set of spherical harmonics with half integer values of l,
and shows that orbital angular momentum operator is not
self-adjoint with respect to such a set of functions.
In the case of l = 1/2, his set of functions YW_{lm}
(up to a normalization constant which I omit here) is:
YW_{1/2,1/2} = sin^{1/2} \theta e^{i phi/2}
YW_{1/2,-1/2} = sin^{1/2} \theta e^{-i phi/2} (1)
But there is a problem with the above set of functions,
because the second function does not come from the first
by the application of the "ladder" operator L_. Namely,
L_ YW_{1/2,1/2} is not equal to YW_{1/2,-1/2}.
Instead, we have that L_ YW_{1/2,1/2} is proportional to
cos \theta sin^{-1/2} \theta e^{-i phi/2}.
Other authors, including E. Merzbacher, also use the above
set of functions (and its generalization for higher l and m
values), and consequently they do not obtain self adjoint
orbital angular momentum operator.
In this respect, it is more natural to take functions
Y_{1/2,1/2} = (i/\pi) sin^{1/2} \theta e^{i phi/2}
Y_{1/2,-1/2} = -(i/\pi) cos \theta sin^{-1/2} \theta e^{-i phi/2}
for which it holds L_ Y_{1/2,1/2} = Y_{1/2,-1/2}.
We also have:
L+ Y_{1/2,1/2} = 0
L+ Y_{1/2,-1/2} = Y_{1/2,1/2}
But there is a catch: The action on L_ on the function
with lowest m value, i.e., m = -1/2 in our example,
we find
L_ Y_{1/2,-1/2} = Y_{1/2,-3/2}
The result is not zero, as it should be. And yet, if we
act on the latter state with the raising operator L+,
we obtain zero:
L+ Y_{1/2,-3/2} = 0.
The last result is crucial. It has for a consequence,
that the operators L_x, L_y, L_z are self adjoint,
provided that one employs a suitable renormalization
of the inner products between certain states. Then
everything appears to be OK, and the spherical harmonics
with half-integer l-values do form an irreducible
representation of the 3-dimensional rotation group.
All this has been discussed in papers
D. Pandres, J. Math. Phys. 6, 1098 (1965),
D. Pandres and D.A. Jacobson, J. Math. Phys. 9, 1401 (1968),
I followed Pandres' approach in a paper of mine
"Rigid particle and its spin revisited"
Published in Found.Phys.37:40-79,2007.
http://arxiv.org/abs/hep-th/0412324
in which I tried to clarify some other aspects as well,
and also used a complementary functions Z_{lm} in
a superposition with Y_{lm}, in order to obtain
"good behavior" under rotations, parity and time reversal.
Pandres and I do not claim that the ordinary
orbital angular momentum can have half integer l-values.
A reason of why to reject Y_{lm} and Z_{lm}
with half-integer l-values in the description of orbital angular
momentum was given correctly by Dirac in his textbook on QM.
In the free case, a complete set of solutions to the
Schr\" odinger equation consists of plane waves,
which are single valued. The latter property has to be preserved
when we use another representation, i.e., one with spherical
harmonics.
Hence, such Schr\" odinger basis for spinor representation of the
3-dimensional rotation group cannot refer to the ordinary configuration
space of positions, but to an internal space associated with
every point of the ordinary space.
Double valued spherical harmonics are admissible,
if they do not refer to the ordinary configuration space in
which the usual quantum mechanical orbital angular momentum is defined,
but if they refer to an internal space in which a spin angular momentum
is defined. An example of such an internal space is the space of
velocities, associated with the so called {\it rigid particle}
whose action contains the square of the
extrinsic curvature of a particle's world line.
`Fermion' spherical harmonics are discussed also in papers:
G. Hunter, P. Ecimovic, I.M Walker, D. Beamish, S. Donev, M. Kowalski, A. Arslan
and S. Heck, J. Phys. A: Math. Gen. 32 795 (1999); G. Hunter and M. Emami-
Razavi, \Properties of Fermion Spherical Harmonics", [arXiv:quant-ph/0507006].
Jay R. Yablon
Dan
[Yablon]
What I replied with before is actually half of the answer. The other
half is as follows:
As to the issue of whether rxp can be applied to intrinsic spin, I call
your attention to Ohanian's article "What is spin?," linked as
http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf.
Go to equation (16) and the discussion following. With $ = integral,
and G=momentum density (see (10) through (13)), equation (16) can be
schematically rewritten as:
J = $ [(r x G)_orbital + (r x G)_spin] d^3 x
Then a few steps later after converting the triple cross product, the
second term turns directly into the expectation value of the sigma
operator, and by (19), we have spin 1/2.
Via the (r x G)_spin term in (16), it is clear that both spin and
orbital angular momentum *can* be treated with an rxp type operator,
contrary to what is often asserted. That is *exactly* what Ohanian does
here.
If one then applies entanglement so that one has to cycle through 4pi
rather than 2pi, the distinction between orbital and spin angular
momentum in terms of group theory (fundamental versus non-fundamental
Lorentz group representations of SO(1,3)) can, it appears, be
eliminated, and each can be treated via and rxp analog to classical
theory.
BTW, I mentioned earlier that Ohanian taught me general relativity while
he was an adjunct professor over at nearby RPI (and also taught at Union
College here in Schenectady where I first met him) in the 1980s. With
Wheeler's passing, I should also note if memory serves, that Ohanian is
a student of Wheeler's, and should be added to Wheeler's genealogies.
Jay.
Jay R. Yablon
news:48049E46...@ijs.si...
Dear Dr. Pavsic:
Looking at your article, you seem to be one of the few folks who is
seeking an rxp (rotation) view of intrinsic spin 1/2.
It seems that Ohanian's equation (16) in "What is spin?" (linked at
http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf)
provides direct support for using rxp for Dirac particles (see a pending
post at sci.physics.research which will hopefully be up soon).
If my former professor Ohanian in can do a classical rxP analysis for
intrinsic spin and end up with (1/2) h-bar, then there has to be a
quantum mechanical rxP operator for the group theory associated with
Ohanian's spin analysis.
I took a stab at that problem in the file below. Straightforward
calculation, just over one page. Still a draft, but please let me know
what you think.
http://jayryablon.files.wordpress.com/2008/04/ohanian-and-spin-operators.pdf
Best,
Jay.
Matej Pavsic
> Dear Dr. Pavsic:
>
> Looking at your article, you seem to be one of the few folks who is
> seeking an rxp (rotation) view of intrinsic spin 1/2.
I am building on an old papers [D. Pandres J. Mathem. Phys. 6 (1965)
1098] in which the Schroedinger basis for spinor representation
of the 3-dimensional rotation group is considered.
Here r and p are not the coordinates and momentum operators
in our usual 3-space, but in an 'internal space'.
>
> It seems that Ohanian's equation (16) in "What is spin?" (linked at
> http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf)
> provides direct support for using rxp for Dirac particles (see a pending
> post at sci.physics.research which will hopefully be up soon).
In that paper r and p are in usual space, so this is different
from what I consider.
>
> If my former professor Ohanian in can do a classical rxP analysis for
> intrinsic spin and end up with (1/2) h-bar, then there has to be a
> quantum mechanical rxP operator for the group theory associated with
> Ohanian's spin analysis.
>
> I took a stab at that problem in the file below. Straightforward
> calculation, just over one page. Still a draft, but please let me know
> what you think.
>
> http://jayryablon.files.wordpress.com/2008/04/ohanian-and-spin-operators.pdf
Also in your pages above your r and p refer to our usual space.
Gerard Westendorp
[..]
> As to the issue of whether rxp can be applied to intrinsic spin, I call
> your attention to Ohanian's article "What is spin?," linked as
> http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf.
[..]
>
> Via the (r x G)_spin term in (16), it is clear that both spin and
> orbital angular momentum *can* be treated with an rxp type operator,
> contrary to what is often asserted.
[..]
> If one then applies entanglement so that one has to cycle through 4pi
> rather than 2pi,
I am not sure this is necesary.
[Thanks for putting the article on line, btw. I have it somewhere, but I
would have to tydy up the houde to find it]
Consider the 'p' part in 'rxp'.
For a *scalar* particle,
p_x = h_bar psi (d_x psi)
But if look at equation 10-13 of Ohanian,
Factors 1/2 and 1/4 now appear in the expression for the momentum
denstity. The 1/4 factor ends up in the end result of h_bar/2 spin.
So already, the half-valued spin is in the momentum itself.
A confusing thing is that this factor does not appear in plane waves,
you need to look at a *finite* wave packet to get the result, and the
circulating momentum density is on the fringes of the wave packet.
A similar thing is the case with electromagnetic waves, which I
understand a bit better. Consider a circularly polarized beam, entering
through your window. The electric field hitting the window will be of
the form
( 0 )
(Ex cos wt)
(Ey sin wt)
The arrow of E on your window is rotating. But most electrons in the
window are *not* made to rotate around the center of the beam. Rather,
the E of the beam is like a wiping sponge, whose friction force vector
rotates, while the sponge itself does circular translations, not rotate
about is own axis. Comparing electrons in the window pushed by E-fields
to little specs pushed about by the sponge, we see that specs in the
middle of the sponge move around in little circles, but on average, they
do not get very far. But now, consider a spec that is right on the edge
of the region that is wiped by the sponge, so that it is not always
wiped by the sponge, only when the sponge is in an extreme position.
Picture that for a while, and you can see that the spec on the outer
edge can actually be slowly pushed around in a large circle. (It sticks
more to the window than to the sponge). Going back to electrons, a
circular current is generated around the outside of the region
illuminated by the circularly polarized beam. In the middle, there is no
time averaged circular current, and hence no time averaged angular
momentum transfer.
The sponge wiping analogy is about the fact that rxp type angular
momentum is present on the *fringes* of the wave packet, not in the
center, or indeed in a plane wave, even if the the plane wave is
circularly polarised. That is why it often overlooked. The larger a
pseudo-plane wave, the larger the part that has no circulating momentum,
but the greater the 'r' part when you finally get to the fringes. So as
a beam diverges, it puts a lower and lower part of its energy in the
circular flow pattern on the fringes, but because of the increasing r,
rxP stays constant.
Still, the fact that you always get exactly h_bar/2 for any electron
wave packet, or orbital, is hard to visualize.
Gerard
Jay R. Yablon
news:66lf6vF...@mid.individual.net...
>
> Looking at your article, you seem to be one of the few folks who is
> seeking an rxp (rotation) view of intrinsic spin 1/2.
>
> http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf)
> provides direct support for using rxp for Dirac particles (see a
> pending
> post at sci.physics.research which will hopefully be up soon).
>
> intrinsic spin and end up with (1/2) h-bar, then there has to be a
> quantum mechanical rxP operator for the group theory associated with
> Ohanian's spin analysis.
>
> I took a stab at that problem in the file below. Straightforward
> calculation, just over one page. Still a draft, but please let me
> know
> what you think.
>
> http://jayryablon.files.wordpress.com/2008/04/ohanian-and-spin-operators.pdf
>
>
> Jay.
>
I took a better stab at this problem than the earlier one, please see:
http://jayryablon.wordpress.com/files/2008/04/intrinsic-spin-decomposition.pdf
Thanks,
Jay.