electron spin
Andrew A Patricio
Apologies if this is a rather trivial question. I was reading about
how heisenberg's uncertainty principle means that the electron around
the atom is delocalized. If this is true, what exactly is meant by
electron spin? Is this another physics euphemism for some weird
physical attribute or is it actually mechanical spin? And if so,
how does one measure the spin of a probability cloud?
Also, can anyone recommend a good textbook on general relativity?
i have read some popular qualitative explanations and would like to
see some math. Thanks very much.
--
Andrew Patricio "Son, you got a panty on yur head."
West Samatta University "Just drive, fast."
patr...@uiuc.edu
http://www.cen.uiuc.edu/~patricio/home.html
Greg Weeks
: If this is true, what exactly is meant by
: electron spin? Is this another physics euphemism for some weird
: physical attribute or is it actually mechanical spin?
It is actual mechanical spin. However, this is not apparent until you
examine the energy-momentum density in quantum field theory. (In other
contexts, spin is best viewed as some weird physical attribute.)
: how does one measure the spin of a probability cloud?
You might similarly ask how do you measure the position or momentum or
angular momentum or energy of a probability cloud. The answer has less to
do with spin than with the unfathomable but necessary basics of quantum
mechanics.
Greg Weeks
Greg Weeks
: It is actual mechanical spin. However, this is not apparent until you
: examine the energy-momentum density in quantum field theory. (In other
: contexts, spin is best viewed as some weird physical attribute.)
This was a bit hasty. If you consider the simplest systems that are both
relativistic and quantum mechanical (ie, single particle systems), spin is
manifested as the angular momentum of particles at rest. I'm not sure if
I'd prefer to call this "actual mechanical spin" or "some weird physical
attribute", but I lean toward the former. In any case, once you embed the
particle states in a quantum field theory, the interpretation of spin
becomes straightforward.
Greg Week
Ben Bullock
As far as we can measure, the electron is a point particle.
"Mechanical spin" of a point-particle is a meaningless notion.
Therefore, the only answer to the original question which is justified
by present day knowledge is that spin is a "weird physical attribute".
--
Ben Bullock @ KEK (National Laboratory for High Energy Physics) / address:
1-1 Oho, Tsukuba, Ibaraki 305, JAPAN / TEL: 0298-64-5403 / FAX: 0298-64-7831 /
in japanese: ベン ブロック ＠ 高エネルギー研究所 茨城県つくば市大穂1ー1
Matt McIrvin
Ben Bullock <b...@theory1.kek.jp> wrote:
>Therefore, the only answer to the original question which is justified
>by present day knowledge is that spin is a "weird physical attribute".
It would be misleading, though, to imply that the identification of
spin with angular momentum is merely analogical, like isospin, or that
the word "spin" is as fanciful a metaphor as quark "color"; that's the
impression that many people seem to get from this kind of description.
Spin *is* a form of angular momentum--it can be converted into other
kinds of angular momentum, so to treat it as a completely separate
entity one would need to abandon angular momentum conservation.
It's true that the electron is not a classical spinning ball of
charge; there's no way to obtain its magnetic moment that way, for one
thing, and it would have to spin faster than light. However, an
examination of how the angular momentum arises from the electron
*field* makes it look much more "mechanical," a point Greg Weeks likes
to emphasize.
--
Matt 01234567 <-- Indent-o-Meter
McIrvin ^ Harnessing tab damage for peaceful ends!
Douglas A. Singleton
>> This was a bit hasty. If you consider the simplest systems that are both
>> relativistic and quantum mechanical (ie, single particle systems), spin is
>> manifested as the angular momentum of particles at rest. I'm not sure if
>> I'd prefer to call this "actual mechanical spin" or "some weird physical
>> attribute", but I lean toward the former. In any case, once you embed the
>> particle states in a quantum field theory, the interpretation of spin
>> becomes straightforward.
>
>As far as we can measure, the electron is a point particle.
>"Mechanical spin" of a point-particle is a meaningless notion.
>Therefore, the only answer to the original question which is justified
>by present day knowledge is that spin is a "weird physical attribute".
>
>--
>Ben Bullock
Actually there is a way of seeing the electron's spin as being
ordinary mechanical spin. Starting with the Dirac equation calculate the
energy-momentum tensor, T^{uv}. Then one can calculate the angular
momentum density carried by the Dirac field (e^{ijk} x_j T_{0k},
where e^{ijk} is the usual antisymmetric tensor, x_j is just
position and T_{0k} is the momentum flow fo the Dirac field).
Integrating this density over all space one finds that the angular
momentum comes out to be hbar/2. You can do the same thing for
the photon and get that it must have spin 1. Loosely one can
think of the electron as a propagating wave which rotates around
its direction of propagation.
DOug
.
Greg Weeks
: Actually there is a way of seeing the electron's spin as being
: ordinary mechanical spin. Starting with the Dirac equation calculate the
: energy-momentum tensor, T^{uv}. Then one can calculate the angular
: momentum density carried by the Dirac field (e^{ijk} x_j T_{0k},
: where e^{ijk} is the usual antisymmetric tensor, x_j is just
: position and T_{0k} is the momentum flow fo the Dirac field).
: Integrating this density over all space one finds that the angular
: momentum comes out to be hbar/2. You can do the same thing for
: the photon and get that it must have spin 1. Loosely one can
: think of the electron as a propagating wave which rotates around
: its direction of propagation.
Right! By the way, you need to use the observable energy-momentum tensor
(symmetric and gauge-invariant), which might not be what you get directly
from Noether's theorem.
Greg Weeks
Ben Bullock
(... I wrote ...)
> >As far as we can measure, the electron is a point particle.
> >"Mechanical spin" of a point-particle is a meaningless notion.
> >Therefore, the only answer to the original question which is justified
> >by present day knowledge is that spin is a "weird physical attribute".
> Actually there is a way of seeing the electron's spin as being
> ordinary mechanical spin.
This is slightly nitpicking, but ....
It's not possible, ever, to see fermion spin as being ordinary
mechanical spin. Fermion "spin" transforms under rotations as if it
were in SU(2) space, which can have no analogy in the classical
physics of rotating bodies.
[[ By the way, only because all observables of fermions come about from
combining two wavefunctions of the particles is it possible for them
to have this property. ]]
> Starting with the Dirac equation calculate the energy-momentum
> tensor, T^{uv}. Then one can calculate the angular momentum density
> carried by the Dirac field (e^{ijk} x_j T_{0k}, where e^{ijk} is the
> usual antisymmetric tensor, x_j is just position and T_{0k} is the
> momentum flow fo the Dirac field).
My question here is "position relative to what ?" Where are you
measuring the origin from? Since you're using a momentum-space
representation of the quantum field, it's clear that the position is
arbitrary. Therefore x_j can be taken from any point. My crude
guessing is that if you move the origin of where you are integrating
from you will get a different answer.
This is actually the same as my initial objection to Greg Week's
original post - saying the electron is a point particle is exactly the
same as saying that in the momentum space representation we can't
ascribe it a particular position.
> Integrating this density over all space one finds that the angular
> momentum comes out to be hbar/2. You can do the same thing for the
> photon and get that it must have spin 1. Loosely one can think of
> the electron as a propagating wave which rotates around its
> direction of propagation.
I can't think of this yet very well and I would like some more
details. Do you have a reference for this?
Thanks very much for any suggestions.
(unfortunately I am new to this group basically having dropped in here
to see if I could find an address for Fermilab preprints, and saw this
discussion & decided to join in - excuse me if this is all
well-covered before).
Greg Weeks
: spin. Fermion "spin" transforms under rotations as if it were in SU(2)
: space, which can have no analogy in the classical physics of rotating
: bodies.
The Dirac Psi field does indeed transform under SU(2) "rotations" rather
than under O(3) rotations. Similarly, if you rotate an electron state by
2*pi, you pick up a factor of (-1). Nevertheless, the spin of an electron
(approximately) at rest is (approximately) the electron's angular momentum.
And it turns out (happily, IMO) that the angular momentum is the spatial
integral of the classical angular momentum density (formed from the
energy-momentum density).
I don't see a contradiction here (and I don't believe there is one),
although it feels like there should be one.
: My question here is "position relative to what ?" Where are you measuring
: the origin from? Since you're using a momentum-space representation of the
: quantum field, it's clear that the position is arbitrary. Therefore x_j
: can be taken from any point. My crude guessing is that if you move the
: origin of where you are integrating from you will get a different answer.
If the expectation value of the 3-momentum is strictly zero, it doesn't
matter where the origin is. Otherwise, it doesn't matter where the origin
is in the limit as the electron becomes more and more "at rest". (This
presumes the absence of singular behavior in the electron's momentum space
wave-function, a reasonable assumption).
I don't know if this is relevant, but it may be helpful to point out that
for an electron nearly at rest, the expectation value of the
energy-momentum is very small and very spread out. The nonzero
contribution to the integrated angular momentum comes from regions far away
from the origin. Indeed, if you "interchange the limits" by making the
electron strictly at rest before doing the spatial integral of the angular
momentum density, then you get the wrong answer (zero).
Greg Weeks
Greg Weeks
: It's true that the electron is not a classical spinning ball of
: charge; there's no way to obtain its magnetic moment that way, for one
: thing, and it would have to spin faster than light.
With a little trepidation, I have to disagree. You DO obtain the magnetic
moment of an electron (approximately at rest) by taking the expectation
value of the classical expression for the magnetic moment.
(Now, using the expectation value is a bit of a cheat; it is conceivable
that the expectation value is not finding an eigenvalue. But this doesn't
seem likely.)
Greg
Matt McIrvin
>Matt McIrvin (mci...@scws4.harvard.edu) wrote:
>
>: It's true that the electron is not a classical spinning ball of
>: charge; there's no way to obtain its magnetic moment that way, for one
>: thing, and it would have to spin faster than light.
>
>With a little trepidation, I have to disagree. You DO obtain the magnetic
>moment of an electron (approximately at rest) by taking the expectation
>value of the classical expression for the magnetic moment.
I was thinking about a classical calculation, not in the sense of
Dirac field theory, but in the naive sense of "electron as BB," in
which you treat the electron as a charge and mass distribution with
some finite size and the observed electron's total charge and mass,
spin it around like a rigid ball with an angular momentum of hbar/2,
and try to reproduce the observed gyromagnetic ratio by fiddling with
the radial distributions. I had been taught that this, at least, was
not possible for an electron rotating more slowly than c.
Am I wrong? I'm confused now.
Greg Weeks
: I was thinking about a classical calculation, not in the sense of
: Dirac field theory, but in the naive sense of "electron as BB," in
: which you treat the electron as a charge and mass distribution with
: some finite size and the observed electron's total charge and mass,
: spin it around like a rigid ball with an angular momentum of hbar/2,
: and try to reproduce the observed gyromagnetic ratio by fiddling with
: the radial distributions.
Ah, but in QED, the energy-momentum flow T and the current J are not
slavishly tied together as they are for classical matter fields. It turns
out that only half the terms in T contribute to the spin; and those that do
contribute in a fashion similar to the way J contributes to the magnetic
moment. The result (in free field theory) is g=2.
Cool?
Greg Weeks
we...@cdc.hp.com
Douglas A. Singleton
Ben Bullock <b...@theory1.kek.jp> wrote:
>Douglas A. Singleton (da...@fermi.clas.Virginia.EDU) wrote:
>
>(... I wrote ...)
>
>> >As far as we can measure, the electron is a point particle.
>> >"Mechanical spin" of a point-particle is a meaningless notion.
>> >Therefore, the only answer to the original question which is justified
>> >by present day knowledge is that spin is a "weird physical attribute".
>
>> Actually there is a way of seeing the electron's spin as being
>> ordinary mechanical spin.
>
>This is slightly nitpicking, but ....
>
>It's not possible, ever, to see fermion spin as being ordinary
>mechanical spin. Fermion "spin" transforms under rotations as if it
>were in SU(2) space, which can have no analogy in the classical
>physics of rotating bodies.
Well this is the normal thinking, and in chosing this approach
you're in good company (namely Pauli), but this is simply not
correct. Ohanian has an excellent article in the American Journal
of Physics (it's in 1985 - or maybe 1984 or 1986) called "What
is Spin" where he gives the details of the rough outline I've
given below. Or you could try looking for Greg Week's original
postings on this stuff. In any case the Ohanian article gives all
the details, and the math is very straightforward so it's hard
to see how one could avoid the conclusions that he draws (which
aren't earth shattering, but it gives you a nice physical way
to think about spin). Also there have been some attempts to link
the spin of the electron with its zitterbewegung, which are also
very interesting. Two articles on this are K. Huang AJP on page
479 of 1952 or 1953, and A. Barut (and someone I forgot) in
Phys. Rev. Letts. of 1984.
>> Starting with the Dirac equation calculate the energy-momentum
>> tensor, T^{uv}. Then one can calculate the angular momentum density
>> carried by the Dirac field (e^{ijk} x_j T_{0k}, where e^{ijk} is the
>> usual antisymmetric tensor, x_j is just position and T_{0k} is the
>> momentum flow fo the Dirac field).
>
>My question here is "position relative to what ?" Where are you
>measuring the origin from? Since you're using a momentum-space
>representation of the quantum field, it's clear that the position is
>arbitrary. Therefore x_j can be taken from any point. My crude
>guessing is that if you move the origin of where you are integrating
>from you will get a different answer.
You're answer won't depend on where you chose to set up your
origin. I'm going to wimp out and just refer you to the details
in the Ohanian article.
>I can't think of this yet very well and I would like some more
>details. Do you have a reference for this?
>
>Thanks very much for any suggestions.
>
>(unfortunately I am new to this group basically having dropped in here
>to see if I could find an address for Fermilab preprints, and saw this
>discussion & decided to join in - excuse me if this is all
>well-covered before).
>
Doug
.
Matt McIrvin
>Ah, but in QED, the energy-momentum flow T and the current J are not
>slavishly tied together as they are for classical matter fields. It turns
>out that only half the terms in T contribute to the spin; and those that do
>contribute in a fashion similar to the way J contributes to the magnetic
>moment. The result (in free field theory) is g=2.
Right-- thinking about my last post, I realized that if the current
and energy-momentum distributions aren't necessarily tied to each
other, you could have any gyromagnetic ratio you like. I think that
the "mechanical" treatment I saw as an undergrad treated the electron
as made of some sort of charged matter with a fixed relation between
the mass and charge densities in its rest frame.
Ben Bullock
> >It's not possible, ever, to see fermion spin as being ordinary
> >mechanical spin. Fermion "spin" transforms under rotations as if it
> >were in SU(2) space, which can have no analogy in the classical
> >physics of rotating bodies.
>
> Well this is the normal thinking, and in chosing this approach
> you're in good company (namely Pauli), but this is simply not
> correct.
The above statements are definitely correct, so I don't see what you
mean - you can't make a rotating system in classical mechanics which
obeys SU(2) rather than O(3) - but anyway I guess what you mean is
that "this argument doesn't count against Greg Weeks / Ohanian's idea"
than "it is wrong".
> Ohanian has an excellent article in the American Journal
> of Physics (it's in 1985 - or maybe 1984 or 1986) called "What
> is Spin" where he gives the details of the rough outline I've
> given below.
I'll have a look for that then. If anyone does have a complete
reference please let me know so that I don't have to search through
all the indices etc.
> Or you could try looking for Greg Week's original postings on this
> stuff.
I don't have a clue where to find this though! As I said I just
jumped into this group - I long since gave up reading unmoderated
sci.physics groups because of the huge number of articles with very
little content in them.
Matt McIrvin
Christophe Fraser <py...@swansea.ac.uk> wrote:
>
>The Cut-the-crap short version:
>
> Spin is called spin because it has the same
> mathematical algebraic properties as angular momentum.
More than that: angular momentum is not conserved in the observed real
world unless you include spin as a kind of angular momentum.
>More precisely, the Lorentz group can be decomposed into
>two copies of SU(2) (related by spatial inversion : parity),
>which have the same algebraic structure as the rotaion group
>SO(3). So we see that left-handedness and right-handeness
>are also automatically built into the picture.
>
>Further, if we examine the full Poincare group, we see that the
>representations are classified also by their mass.
>
>So contrarily to what is taught in many elementary quantum
>mechanics courses, spin, mass, left- or right-handedness
>are not put in by hand, but are in fact REQUIRED by special
>relativity.
But it is conceivable that, in the correct description of physics, all
*actual* fundamental particles could have belonged to the spin 0
representation, or that they could have been in massless
representations. I suppose it is possible that all interacting
spinless field theories have pathological properties of some sort; you
could also argue that composite bound states should be counted as
particles too. But it is not true that special relativity alone
requires the existence of particles with nonzero spin or mass. After
all, special relativity is consistent with a *free* theory of
massless, noninteracting, scalar bosons, though the physics of such a
situation is not very interesting.
Douglas A. Singleton
>
>Spin is not a 'mechanical' property of the electron.
>
>So where shall we start?
>
>The Cut-the-crap short version:
>
> Spin is called spin because it has the same
> mathematical algebraic properties as angular momentum.
> theory should obey special relativity. So
> though unintuitive, it is not just put there
> to make things work (this was however the historical
> reason).
Well not to put too fine a point on it, but actually spin can be
seen as "mechanical" angular momentum. See the 1985 or 1986 American
Journal of Physics article by H. Ohanian entitled "What is Spin?"
(I can e-mail you the exact ref. if you want). Just calculate the
energy-momentum tensor for the Dirac field, look at the T^{0i}
component and you'll see that there is a real angular momentum
in the field. You might object that this is angular momentum carried
by the field rather than a mechanical "particle", but from E&M we
know that fields must be able to carry angular momentum or that
conservation law will run into trouble in some situations.
Also there have been attempts to give the electron spin a physical
interpretation in terms of the electrons zitterbewegung. A. Barut
is big into this stuff, and his earliest work on it is in PRL 1984
around page 2000 or so. He's written a lot of subsequent stuff
on this which you can find in Phys. Letts. A after 1984.
Discuion of Lorentz group decomposition deleted
>So contrarily to what is taught in many elementary quantum
>mechanics courses, spin, mass, left- or right-handedness
>are not put in by hand, but are in fact REQUIRED by special
>relativity.
Spin isn't really required by SR. A Klein-Gordon particle has
no spin but is fine as far as SR is concerned. Same goes for
mass you can have massless and/or massive particles, but
SR will tell you nothing about what mass any particular particle
should have if any.
>So how do these concepts fit in with solitons, such as
>magnetic monopoles?
>Especially if they are to be dual to ordinary QFT particles,
>they should have similar properties. Or is the swapping
>of topological and Noether charges accompanied by an interchange
>of for example spin and angular momentum.
This is the old spin from isospin trick where you start out
with a theory with only gauge particles and scalar (spin-0)
particles and end up with a fermion (the refs. are Jackiw and
Rebbi PRL 36, 1116 (1976), Hasenfratz and 't Hooft PRL 36, 1119
(1976), A. Goldhaber PRL 36, 1122 (1976)). This comes about
because you bind a 't Hooft-Polyakov magnetic monopole to a
boson carrying isospin charge and the resulting composite system
has a "mechanical" angular momentum of hbar/2 that is basically
the non-Abelian version of the E X B angular momentum in an E&M
electric charge-magnetic charge system. This thing also obeys
Fermi-Dirac statistics so you get a fermion as a derived object
rather than it being fundemental.
>
>Cheers,
>
> Christophe Fraser
> Swansea
> (py...@swansea.ac.uk)
>
Doug
.