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
Timo Nieminen
> 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.
At this point, I would say that this is clearly wrong. Spin is a quantity
that exists very happily in classical field theory, and quantum mechanics
and QED and QFT inherit spin from classical field theory.
Still, it's a common statement, and it appears to be an overgeneralisation
from "spin doesn't exist in classical _mechanics_". But classical
mechanics is far from all of classical physics.
> 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?
AFAIK, the situation for electrons is analogous to photons. For
photons, the spin is a property of the classical EM field, and the
quantisation of spin is a result of the quantisation of the EM field.
However, one could argue that an electron is more "particle-like" than a
photon, the QED version of which has little in common with the classical
idea of "particle".
> 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?
I'd say so. I can suggest some references on photon spin/spin in classical
EM if you're interested, but I might not be able to come up with much on
electron spin, other than an old paper in Am J Phys by Romer (iirc).
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Peter
> On Mon, 7 Apr 2008, Jay R. Yablon wrote:
>
> > 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.
This is of the style 'eat or perish' and is a sequence of postulates rather
than arguments - poor students
> At this point, I would say that this is clearly wrong. Spin is a quantity
> that exists very happily in classical field theory, and quantum mechanics
> and QED and QFT inherit spin from classical field theory.
>
> Still, it's a common statement, and it appears to be an overgeneralisation
> from "spin doesn't exist in classical _mechanics_". But classical
> mechanics is far from all of classical physics.
This is not true in this generality
> > 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?
> AFAIK, the situation for electrons is analogous to photons. For
> photons, the spin is a property of the classical EM field, and the
> quantisation of spin is a result of the quantisation of the EM field.
Indeed
> However, one could argue that an electron is more "particle-like" than a
> photon, the QED version of which has little in common with the classical
> idea of "particle".
> > 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?
> I'd say so.
Me too, since all three, L, S and J=L+S, obey the same Lie algebra (iirc)
> I can suggest some references on photon spin/spin in classical
> EM if you're interested, but I might not be able to come up with much on
> electron spin, other than an old paper in Am J Phys by Romer (iirc).
I would feel indebted for that,
Thank you in advance,
Peter
Ken S. Tucker
Since we're dealing with spin I'll add my 3.1416 scents,
(my post might stink).
> Email: jyab...@nycap.rr.com
> co-moderator: sci.physics.foundations
> Weblog:http://jayryablon.wordpress.com/
> Web Site:http://home.nycap.rr.com/jry/FermionMass.htm
Let me start by using an airplane,
orientated from tail to nose by using Y axis,
with wings being wings being on the X axis ,
and Z upward.
The Pitch is around X, the Roll around Y and
the Yaw around Z.
Denote Pitch = A23, Roll=A31, Yaw=A12,
where for legalese,
Auv = x_u dx^v/dt - x_v dx^u/dt.
Presume our airplane has equal moments of
inertial in all 3 axes, to unitized the momentum.
I can maintain a Pitch=0, and Roll at twice the
rate of Yaw, meaning I can Roll 720 degs for a
Yaw of 360 degs.
It appears to me that Yaw/Roll =1/2 and I suggest
that is an intrinsic scalar.
To prove that, let's employ (Maxwell's 2nd set),
(a property of asymmetric tensors),
A23,1 + A13,2 + A21,3 = 0
(Pitch + Roll + Yaw , rates of change).
Let's use integers "n" for "rates of change",
n1 * A23 + n2 * A13 + n3 * A21 = 0,
so in the above example, n1=0, n2=2, n3=1,
with "intrinsic spin" being n3/n2 = 1/2.
A question arises as to why the assumption
of using "n" as an integer value is reasonable.
Suppose "n" is in units of action which can
be equated to units of angular momentum,
and spin, and the equation evolves to,
n1*h * A23 + n2*h * A13 + n3*h * A21 = 0,
in a clearer presentation, to permit,
h * (n1 + n2 + n3) = 0
by using the above assumption of equal moments
of inertial to allow unitizing A23 , A13 and A21.
Let n1=0 , n2 =2 , n3 = -1 then
n2/n2 + n3/n2 = 0 => 1 - 1/2 = 1/2
which is a scalar invariant.
Regards
Ken S. Tucker
Jay R. Yablon
news:Pine.LNX.4.50.0804080821400.25834-100000@localhost...
> On Mon, 7 Apr 2008, Jay R. Yablon wrote:
>
>> 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.
>
> At this point, I would say that this is clearly wrong. Spin is a
> quantity
> that exists very happily in classical field theory, and quantum
> mechanics
> and QED and QFT inherit spin from classical field theory.
>
> Still, it's a common statement, and it appears to be an
> overgeneralisation
> from "spin doesn't exist in classical _mechanics_"
That is exactly my thought about this -- we are extrapolating physics
results from what are nothing more than sloppy linguistic statements
about classical mechanics, which is a great way to run into unintended
mischief. Do others perceive this as well?
Thanks,
Jay.
Ken S. Tucker
> "Timo Nieminen" <t...@physics.uq.edu.au> wrote in message
> news:Pine.LNX.4.50.0804080821400.25834-100000@localhost...
My 1st post to this thread was intended to support
Timo's conjecture. The *tricky* part (IMHO) is to
produce an "intrinsic" spin compatible with classical
mechanics.
In my understanding, we need to prove a classical
spin (aka angular momentum), is relativistically
invariant, IOW's there is no FoR (Frame of Ref) even
in GR to eliminate that so-called "intrinsic" spin.
Suppose you have 2 spinning disc's side by side.
You may attach a CS to either one, but, the relational
spin cannot be transformed away by any choice of CS.
It's the *difference* of the spins that's intrinisic,
that I posted on.
((An old canuck joke, "What's the difference
between a moose?")).
Ok Jay, I'll vote for Timo.
Cheers
Ken S. Tucker
Peter
...
> That is exactly my thought about this -- we are extrapolating physics
> results from what are nothing more than sloppy linguistic statements
> about classical mechanics, which is a great way to run into unintended
> mischief. Do others perceive this as well?
Yes
The extremely sharp thinking and aiming at consequent concluding using well-
defined notions that we have heritaged in classical mechanics is
underestimated in contemporary physics and its representations, respectively.
Sloppy thinking and sloppy language are quite close one to another.
Peter
Cl.Massé
>> 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.
The spin operator isn't expressed in terms of ordinary space, but of
internal space, and there is no a priori reason to metaphysically
distinguish them.
"Timo Nieminen" <ti...@physics.uq.edu.au> a écrit dans le message de
news:Pine.LNX.4.50.0804080821400.25834-100000@localhost...
> At this point, I would say that this is clearly wrong. Spin is a quantity
> that exists very happily in classical field theory, and quantum mechanics
> and QED and QFT inherit spin from classical field theory.
> Still, it's a common statement, and it appears to be an overgeneralisation
> from "spin doesn't exist in classical _mechanics_". But classical
> mechanics is far from all of classical physics.
The distinction classical/quantum isn't clean cut. Is the Dirac field
quantum mechanical, or rather it's second quantification? It is still
harder to say with the EM field, since historically it was classical. If
the Dirac field is classical, then the intrinsic spin is also classical, and
the orbital angular momentum has integer values. It may be considered the
quantization of a point particle, thus quantum mechanical. Either way, that
distinction is useless.
>> 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?
The "point" particle concept must be handled with care. It means "non
composite", and even a composite electron couldn't have a half integer spin
from the constituents motion, even if it is very small. Or it means
"obtained from the quantization of a point particle", and that is the exact
synonym of a field or wave.
Samely, "no classical analogy" is at best wordy, since we know Nature is
wholly quantum mechanical. It is but a historical term.
> AFAIK, the situation for electrons is analogous to photons. For
> photons, the spin is a property of the classical EM field, and the
> quantisation of spin is a result of the quantisation of the EM field.
No, it is the consequence of the internal symmetry.
> However, one could argue that an electron is more "particle-like" than a
> photon,
That would be true with the Schrödinger equation, but the Dirac equation
can't be written as the quantization of a classical particle. Today, both
are on the same footing, which isn't claimed too noisily for historical
reasons.
> the QED version of which has little in common with the classical idea of
> "particle".
Indeed.
>> 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?
The operators associated with l and s are both generators of a
representation of space rotation. But s need a very special representation
using the Pauli matrices as the basis of the representation space, on which
it acts on both sides at the same time, which mathematically amount to 2x2
complex matrices acting on a complex vector called a spinor. All the
details can easily be found in the literature.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
Cl.Massé
news:7df5f080-0e66-4d5e...@q27g2000prf.googlegroups.com...
> My 1st post to this thread was intended to support
> Timo's conjecture. The *tricky* part (IMHO) is to
> produce an "intrinsic" spin compatible with classical
> mechanics.
In that respect, topology shows that a half integer spin can emerge from
scalar fields only. A famous example is the skyrmion, on which there is an
extended literature.
Ken S. Tucker
> "Ken S. Tucker" <dynam...@vianet.on.ca> a écrit dans le message denews:7df5f080-0e66-4d5e...@q27g2000prf.googlegroups.com...
>
> > My 1st post to this thread was intended to support
> > Timo's conjecture. The *tricky* part (IMHO) is to
> > produce an "intrinsic" spin compatible with classical
> > mechanics.
>
> In that respect, topology shows that a half integer spin can emerge from
> scalar fields only. A famous example is the skyrmion, on which there is an
> extended literature.
Thank for the direction Mr. Masse.
A nice brief on "skyrmions" is here, (Author Dr. Wong),
http://arxiv.org/abs/hep-ph/0202250
Eqs.(7,8,9) are straightforward, (Eq.(9), (being what I tried to
clumsily post), Wong fixes the rotation around the z-axis to 0.
I my example (repeated below) I set the rotation about x to 0.
GR permits using rotating CS's as a reference, so by choice
in my airplane example, the Pitch was zeroed. Of course I
could have zeroed any rotation, leaving the other 2 rotations
to be relatively rotating. It's that difference in rotations that
I find to be "intrinsic".
I began by using this straighforward Eq.
A_bc,a + A_ac,b + A_ba,c = 0
that is true in 4D. I think Wong's attempt at generalization
in his Eq.(19) is awkward, (see his footnote on it).
+++++++++++++++++++
Let me start by using an airplane,
orientated from tail to nose by using Y axis,
with wings being wings being on the X axis ,
and Z upward.
The Pitch is around X, the Roll around Y and
the Yaw around Z.
Denote Pitch = A23, Roll=A31, Yaw=A12,
where for legalese,
Auv = x_u dx^v/dt - x_v dx^u/dt.
Presume our airplane has equal moments of
inertial in all 3 axes, to unitized the momentum.
I can maintain a Pitch=0, and Roll at twice the
rate of Yaw, meaning I can Roll 720 degs for a
Yaw of 360 degs.
It appears to me that Yaw/Roll =1/2 and I suggest
that is an intrinsic scalar.
To prove that, let's employ (Maxwell's 2nd set),
(a property of asymmetric tensors),
A23,1 + A13,2 + A21,3 = 0
(Pitch + Roll + Yaw , rates of change).
Let's use integers "n" for "rates of change",
n1 * A23 + n2 * A13 + n3 * A21 = 0,
so in the above example, n1=0, n2=2, n3=1,
with "intrinsic spin" being n3/n2 = 1/2.
-----------------------------------------------------------------
-
> ~~~~ clmasse on free F-country
> Liberty, Equality, Profitability.
Cheers
Ken S. Tucker
David Rutherford
Jay R. Yablon wrote:
> 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.
In my theory, "New Transformation Equations and the Electric Field
Four-vector" at,
http://www.softcom.net/users/der555
which is an extention of classical physics, my angular momentum density
four-vector L is
L = XP
where X is the position four-vector, and P is the momentum density
four-vector. The four-vector product XP can be written
XP = X.P + XxP + X:P
where X.P is the four-dimensional dot product, which I think represents
some kind of (unkown to me) oscillation or angular momentum density, XxP
is the classical orbital angular momentum density, and X:P is the spin
angular momentum density. In rectangular coordinates, these can be written
X.P = (X_1 P_1 + X_2 P_2 + X_3 P_3 + X_4 P_4)e_4
XxP = (X_2 P_3 - X_3 P_2)e_1
+ (X_3 P_1 - X_1 P_3)e_2
+ (X_1 P_2 - X_2 P_1)e_3
X:P = (X_1 P_4 - X_4 P_1)e_1
+ (X_2 P_4 - X_4 P_2)e_2
+ (X_3 P_4 - X_4 P_3)e_3
where the e_\mu are orthonormal basis vectors and \mu=1,2,3 are spatial
indices, \mu=4 is a temporal index. So my spin angular momentum density
X:P is described by classical quantities, but not by existing classical
equations (as far as I know).
--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555
Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
======================================= MODERATOR'S COMMENT:
JRY: Approved for post following review. There are moderator concerns
about the physics, but we will leave it to group participants to offer
their critiques publicly.
--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555
Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
Hans de Vries
See my comment on a similar thread her:
http://www.physicsforums.com/showthread.php?t=227852
This is handled and illustrated in chapter 1 of my graduate textbook
on Quantum Field theory:
http://physics-quest.org/Book_Chapter_EM_basic.pdf
(sections 1.8 and 1.9)
Also take note of the post of Gerard Westendorp on your thread
at sci.physics.research.
http://groups.google.com/group/sci.physics.research/msg/850dc44679f1fcb4
Regards, Hans
Regards, Hans
Jay R. Yablon
news:f329ab02-8f60-4b19...@p25g2000pri.googlegroups.com...
Good to hear from you. Doing US taxes this week (ugh -- the national
plague of early April), so not much time for physics. But, I do
appreciate the references.
I note and very much agree with your comment at the physicsforums thread
that:
"The magnetic moment can not be derived simply by "rotating" the charge
density.
You would end up with speeds higher as the speed of light.
This doesn't mean there is a conflict with for instance Maxwell's laws.
The charge we
measure is not the bare charge. It is screened by vacuum polarization
effects where
the vacuum can contains equal amounts of negatively and positively
charged virtual
particles. We don't know what the bare charge of the electron is,
neither do we know
what exactly causes the magnetic moment, neutral particles can have a
magnetic
moment as well. For instance a particle with a counter rotating
antiparticle produces
a magnetic moment while being electrically neutral. "
You will see that I have developed this point at length in sections 9
and 10 of my draft paper at:
http://jayryablon.wordpress.com/files/2008/04/intrinsic-spin-30.pdf
If you get some time to take a look, I would very much appreciate your
feedback.
Best regards,
Jay.
David Rutherford
David Rutherford wrote:
>
> In my theory, "New Transformation Equations and the Electric Field
> Four-vector" at,
>
> http://www.softcom.net/users/der555
>
> which is an extention of classical physics, my angular momentum density
> four-vector L is
>
> L = XP
>
> where X is the position four-vector, and P is the momentum density
> four-vector. The four-vector product XP can be written
>
> XP = X.P + XxP + X:P
>
> where X.P is the four-dimensional dot product, which I think represents
> some kind of (unkown to me) oscillation or angular momentum density, XxP
> is the classical orbital angular momentum density, and X:P is the spin
> angular momentum density. In rectangular coordinates, these can be written
>
> X.P = (X_1 P_1 + X_2 P_2 + X_3 P_3 + X_4 P_4)e_4
>
> XxP = (X_2 P_3 - X_3 P_2)e_1
> + (X_3 P_1 - X_1 P_3)e_2
> + (X_1 P_2 - X_2 P_1)e_3
>
> X:P = (X_1 P_4 - X_4 P_1)e_1
> + (X_2 P_4 - X_4 P_2)e_2
> + (X_3 P_4 - X_4 P_3)e_3
I'd like to show that my spin angular momentum density has a more
physical interpretation (when analyzed using quantum mechanical methods)
than does the conventional quantum mechanical one (aside from the fact
that it's a part of the same four-vector in my theory).
First I'll define the components of (my) spin angular momentum density
S_1, S_2, and S_3 as
S_1 = X_1 P_4 - X_4 P_1
S_2 = X_2 P_4 - X_4 P_2
S_3 = X_3 P_4 - X_4 P_3
and the conventional components of orbital angular momentum density
(same as mine) as
L_1 = X_2 P_3 - X_3 P_2
L_2 = X_3 P_1 - X_1 P_3
L_3 = X_1 P_2 - X_2 P_1
Please excuse my using L for both my four-vector and the conventional
three-vector. Now replace the P_\mu with the quantum mechanical operators
P_\mu -> - i \hbar \partial_\mu, \mu=1,2,3,4
so that the spin angular momentum density operators (please excuse my
using the same notation for operator components as for non-operator
components) become
S_1 = - i \hbar (X_1 \partial_4 - X_4 \partial_1)
S_2 = - i \hbar (X_2 \partial_4 - X_4 \partial_2)
S_3 = - i \hbar (X_3 \partial_4 - X_4 \partial_3)
and the orbital angular momentum density operators become
L_1 = - i \hbar (X_1 \partial_4 - X_4 \partial_1)
L_2 = - i \hbar (X_2 \partial_4 - X_4 \partial_2)
L_3 = - i \hbar (X_3 \partial_4 - X_4 \partial_3)
The components of (my) spin angular momentum density satisfy the
commutation relations (after messy calculations)
[S_1, S_2] = i \hbar L_3
[S_2, S_3] = i \hbar L_1
[S_3, S_1] = i \hbar L_2
The components of orbital angular momentum density satisfy the
commutation relations
[L_1, L_2] = i \hbar L_3
[L_2, L_3] = i \hbar L_1
[L_3, L_1] = i \hbar L_2
So
[S_1, S_2] = [L_1, L_2]
[S_2, S_3] = [L_2, L_3]
[S_3, S_1] = [L_3, L_1]
This shows that my spin angular momentum density (analyzed using the
methods of quantum mechanics) shares an intimate _physical_ relationship
with orbital angular momentum density.
Cl.Massé
>> 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.
See also the Zitterbewegung [Shaking motion].
--
David Rutherford
Sorry, this should be
L_1 = - i \hbar (X_2 \partial_3 - X_3 \partial_2)
L_2 = - i \hbar (X_3 \partial_1 - X_1 \partial_3)
L_3 = - i \hbar (X_1 \partial_2 - X_2 \partial_1)
Cl.Massé
news:bf3701bb-6d34-47f3...@24g2000hsh.googlegroups.com...
(from spr)
> 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?
Remembering that p = -i@x, along a circle of radius r it reads p = -i/r
@theta. As the wave function is single valued, when integrated over the
circle, L can yield only a integer multiple of 2pi. The pi value is only
possible if the wave function is double valued, like the Dirac spinor. In
fact SU(2) is the double covering of SO(3).
In topology, the integer spin corresponds to the first topological group of
S^1 (one dimensional sphere) while the half-interger spin correspond to the
one of CP^1 (complex projective space.)
Intuitively, when going along the circle and returning to the starting
point, the phase of the wave function must have changed by a multiple of
2pi. At each point of the circle ~ S^1 is associated a phase ~ S^1. The
orbital angular momentum is thus a winding number.
Jay R. Yablon
"Cl.Massé" <sp...@spam.com> wrote in message
news:48037784$0$11485$426a...@news.free.fr...
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 thoughts?
Jay.
Hans de Vries
> "Cl.Massé" <s...@spam.com> wrote in message
Hi Jay,
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 my book is straightly
from Sakurai's treatment of the Dirac equation in his book:
Advanced QM, chapter 3.5 "Gordon decomposition"
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
Jay R. Yablon
"Cl.Massé" <sp...@spam.com> wrote in message
news:48037784$0$11485$426a...@news.free.fr...
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.
Jay.
Hans de Vries
> 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 ashttp://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.
>
> Jay.
Thanks for posting that Jay! Isn't it remarkable that this is so
little known? The same is true for the spin-density of the
EM field. Ohanian discusses the EM spin in section II where
you'll find it in eq(4). in the term (E X A).
The total covariant expression is Si = D X A + HV, So = H.V/c
I devoted an entire chapter of my book to the EM spin:
http://physics-quest.org/Book_Chapter_EM2_ChernSimonsSpin.pdf
Regards, Hans
Peter
...
> > 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.
> >
> > Jay.
> Thanks for posting that Jay! Isn't it remarkable that this is so
> little known?
Yes. It's perhaps due to the fact that the usual representations stress the
differences between classical and quantum physics. On the contrary, I think
that it is both physically and pedagogically much better to stress their
commons. (For this, my book starts with the classical pendulum and just
removes the turning points.)
> The same is true for the spin-density of the
> EM field. Ohanian discusses the EM spin in section II where
> you'll find it in eq(4). in the term (E X A).
>
> The total covariant expression is Si = D X A + HV, So = H.V/c
> I devoted an entire chapter of my book to the EM spin:
> http://physics-quest.org/Book_Chapter_EM2_ChernSimonsSpin.pdf
Very interesting! I hope the chapters before provide sufficient introduction
to understand it fully ;-)
Looking forward,
Peter
Hans de Vries
Hi, Peter
The "intuitive picture" of the spin density is given in 6.4 and 6.5.
Chapter 6 basically checks two elementary cases:
1) The EM spin density field of the static electron. Interesting is
that
the electron needs both an electric charge and a magnetic moment
for its static EM fields to have a spin density. It relates the two.
You can find the derivation of the A and B fields corresponding to
magnetic moment in chapter 1.4 and 1.5
2) The EM spin density field of a photon emitted by an atomic
(spin 1) transition. This involves only the fields caused by the
charge since the fields related to the magnetic moment can be
neglected at sufficient distance.
Regards, Hans
http://www.physics-quest.org/
Jay R. Yablon
Hi Hans et al.:
If my former professor Ohanian in
http://jayryablon.wordpress.com/files/2008/04/ohanian-what-is-spin.pdf
can do a classical rxP analysis for intrinsic spin, then there has to be
an rxP operator for the associated group theory.
I took a stab at that problem in the file below. Straightforward
calculation, just over one page. Still a draft, but let me know what
you think.
http://jayryablon.files.wordpress.com/2008/04/ohanian-and-spin-operators.pdf
Best,
Jay.
Jay R. Yablon
the dot product as associative over multiplication, which it is not,
unless the term outside the parenthesis is a scalar, which it is not.
That is, A(B dot C) <> (A dot B)C
when all of A, B, C are vectors.
Jay.
Cl.Massé
>> 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?
Yes.
>> 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 thoughts?
Through entanglement, there should be a double-valued function since a 2pi
rotation doesn't give an identical wave function. That is similar to the
Dirac spinor, for which a 2pi rotation doesn't restore the initial value,
but it isn't in the rxp form.
SO(3) is a subgroup of SO(1,3), so its fundamental representation has the
same properties. Let's spell it out:
A 3-vector V can be written as a self-adjoint 2x2 matrix:
M(V) = x s_x + y s_y + z s_z
where s_x, s_y, s_z are the Pauli matrices.
A rotation of the vector is written as an S matrix, element of the
fundamental representation of SU(2):
M(V) -> S M(V) S+
but it is actually the fundamental representation of SO(3).
Now the transformation:
M(V) -> S M(V)
Is also a representation, but of SU(2) which is the double covering of
SO(3). It is called the Cartan, or spinorial, representation of SO(3) and
corresponds to the spin 1/2.
Notice that M can be reduced to its first column without loss of
information. It is then a spinor. The Dirac one contains two such spinors.
In the case of SO(3,1), a 4-vector A can be written as a matrix:
M(A) = t I + x s_x + y s_y + z s_z
But now a transformation need two matrices S and T:
M(A) -> S M(A) T+
If S = T, it is a space rotation, since S+ I S = I and thus t is unchanged.
But we have a fundamental representation of SO(3), not a spinorial one,
although written with a "SU(2) matrix".
Taking a similar transformation as above:
M(A) -> S M(A)
We have not a representation of SO(3), since t is involved.
The only possibility I think to have an rxp form is to use double valued
wave functions.
"Hans de Vries" <Hans.de....@gmail.com> a écrit dans le message de
news:e71a4470-7681-428d...@h1g2000prh.googlegroups.com...
> 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.
The orbital angular momentum is then 0, but the reasoning is the same. That
would be so with a plane wave too, where the phase changes but the winding
number is 0.
Cl.Massé
news:7df5f080-0e66-4d5e...@q27g2000prf.googlegroups.com...
> In my understanding, we need to prove a classical
> spin (aka angular momentum), is relativistically
> invariant, IOW's there is no FoR (Frame of Ref) even
> in GR to eliminate that so-called "intrinsic" spin.
>
> Suppose you have 2 spinning disc's side by side.
> You may attach a CS to either one, but, the relational
> spin cannot be transformed away by any choice of CS.
>
> It's the *difference* of the spins that's intrinisic,
> that I posted on.
That is related to the would-be Dirac's "belt trick", which is rather a
sleight of hand, and that is intended to show with a classical system what
pertains to the quantum realm. There are two versions, both are fallacious:
You have a tray in your hand, and you rotate it in a horizontal plane. You
have to make 4pi to get the initial position again. But the axis of
rotation, although always about vertical, is not topologically the same in
each 2pi rotation. In the first one it passes behind the shoulder, and in
the second one before. If the same movement is made with the hand aligned
with the forearm, it is easy to see that the 2pi rotations are actually in
reverse direction, and not in the same as you are made to believe.
You take a belt (or better a tie) and fix one end at the ceiling. You fold
it and put both ends parallel and near to each other. Now, keeping them so,
you move the loose end around the other one so that it describes a full
horizontal circle. When you unfold the belt, it is twisted by 4pi. That is
supposed to show that a 4pi twist can be untwisted, but not a 2pi one. But
if the process is seen in a rotating frame where the centers of both ends
are at rest, you see that both ends are simply twisted by 2pi in the
opposite direction owing to the U-turn of the belt axis at the fold.
Explaining that usually leads to some aggressive behavior from physicists or
mathematicians that have been fooled because they think that all what is
printed or said by Dirac is necessarily true.
Returning to the two discs side by side, if the axis of one of them is
reverse, there is a frame where they both have a spin 0, but one spin has
had its sign changed before.
But here, let's not forget that it is angular velocity, and not angular
momentum. There is no frame where the intrinsic spin of an electron can be
canceled, since the angular velocity is "infinite", vaguely speaking.
Ken S. Tucker
On Apr 22, 12:22 pm, "Cl\.Massé" <s...@spam.com> wrote:
> "Ken S. Tucker" <dynam...@vianet.on.ca> a écrit dans le message denews:7df5f080-0e66-4d5e...@q27g2000prf.googlegroups.com...
Agreed 101% to this point, (more below*1).
> Returning to the two discs side by side, if the axis of one of them is
> reverse, there is a frame where they both have a spin 0, but one spin has
> had its sign changed before.
I'm not sure you can do that, (more below*2).
> But here, let's not forget that it is angular velocity, and not angular
> momentum. There is no frame where the intrinsic spin of an electron can be
> canceled, since the angular velocity is "infinite", vaguely speaking.
*1) I did a post on Mar.1, "An Electron Structure?"
in this group, that tries to cover exactly the objections
you mentioned, have a glance at it, (let me know if
you can't find it), it's fairly brief, anyway I think I can
simplify that.
Suppose we orbit two Satelites (S1 and S2) around the Sun.
For easy math the period of S1 will be 1 year and S2's will
be 2 years, and S1 is in the zy plane, and S2 is the zx plane.
The Sat's will be coincident after two revolutions of S1, (4pi)
and 1 one revolution of S2, (2pi) , cyclically.
I maintain there is NO relativitistic Frame of Reference that
will change the ratio of that coincidence, thus that ratio (1/2),
is intrinsic as determined by the rates of revolution.
I further maintain the Angular Momentum of that system,
defined by Sun, S1 and S2 is intrinsically non-zero,
though I cannot quite prove that off the top of my head #.
2*) That orbital variation (to make spin 0) would require
an absolute variation of either S1 or S2's angular momentum,
such as a rocket thrusting.
# please let me know if you have a means to do so, it
"looks" like common sense.
> ~~~~ clmasse on free F-country
> Liberty, Equality, Profitability.
Cheers
Ken S. Tucker