Spin questions
Dana Mackenzie
Hi all,
I'm thinking of writing a popular-style article about spin. I've seen a fair
amount of publicity about "spintronics." It seems to me that the writers
usually explain electron spin as if the electron were a little ball rotating
clockwise or counterclockwise. But that isn't right, is it? In my article
I'd like to take a stab at explaining what spin REALLY is. I might even try
to throw in words like "spinor" (although the editor is likely to recoil and
say that's jargon).
The reason I'm posting here is that I'm not a physicist; I'm a journalist
and a lapsed mathematician, and therefore I may have some really wrongheaded
notions about spin. To that end, I'd like to toss out a few questions and
I'd appreciate any and all thoughtful replies.
* Do you have a favorite way of explaining spin to non-mathematicians and
non-physicists?
* Does electron spin come from the distinction between the groups SO(3) and
SU(2)?
* Why is it that there are only two choices for the spin of an electron,
namely "up" or "down" (or "+1/2" and "-1/2")? My mathematical background
suggests to me that this comes from the fact that SU(2) is a double cover of
SO(3)--so that you can describe where you are in SU(2) by where your
projection is in SO(3) [would that correspond to your angular momentum?] and
a choice of one of the two preimages of that point [would the two preimages
correspond to "spin up" and "spin down"?]. Am I on the right track here, or
is this too naive?
* Is classical physics kind of like living on a Lie algebra (where so(3) and
su(2) are the same), while quantum physics is more like living on the actual
Lie groups?
* Do higher-dimensional spinors start entering the picture when you look at
particles other than the electron or forces other than electromagnetism? If
not, when do they enter the picture?
* Can someone explain to me why the particles with mass are the ones with
non-integral spin? Would it be overreaching to say that without spin, there
wouldn't be mass?
* Does anyone have some favorite examples of real-world manifestations of
spin? (I saw a couple mentioned in an article by Michael Weiss on John
Baez's Web page -- the Zeeman effect and phosphorescence.)
* Finally, I'd be interested in any reading suggestions that you have on
spin or spintronics, especially ones that explain the link between the
mathematics and the physics of spin.
Thanks for your help!
Dana Mackenzie
Freelance Mathematics and Science Writer
mack...@cruzio.com
Uncle Al
Dana Mackenzie wrote:
>
> Hi all,
>
> I'm thinking of writing a popular-style article about spin.
Good luck. Many are the models and few are the understandings.
Relativity and quantum mechanics are fundamentally incompatible. Spin
straddles both. The relativistically-corrected Dirac equation is
quite a tap dance - and still its full meaning is, ah, debated.
Relativistically-corrected Shroedinger is a whole "Riverdance."
> I've seen a fair
> amount of publicity about "spintronics." It seems to me that the writers
> usually explain electron spin as if the electron were a little ball rotating
> clockwise or counterclockwise. But that isn't right, is it? In my article
> I'd like to take a stab at explaining what spin REALLY is. I might even try
> to throw in words like "spinor" (although the editor is likely to recoil and
> say that's jargon).
If you don't want to calculate numbers, a "little spinning top" is as
good as any model for conceptualization (and vastly erroneous). 90%
of organic chemistry is done with the LCAO model, which is also mostly
a fairy dust heuristic. Anybody who tries to do synthesis (except for
the Woodward-Hoffmann rules) with a more sophisticated model will fall
on his face into a big pile of theoretical poo-poo. Minimizing the
structure of a decent molecule with MM+, which is OK but not sterling,
is a 15-minute PC wait. One can grow old and die chewing on minutia
while fast and sloppy gets the job done as fast as a pen can be
stroked across paper - and it works to spec in a roundbottom flask.
> The reason I'm posting here is that I'm not a physicist; I'm a journalist
> and a lapsed mathematician, and therefore I may have some really wrongheaded
> notions about spin. To that end, I'd like to toss out a few questions and
> I'd appreciate any and all thoughtful replies.
>
> * Do you have a favorite way of explaining spin to non-mathematicians and
> non-physicists?
"Little spinning top" "8^>)
> * Does electron spin come from the distinction between the groups SO(3) and
> SU(2)?
> * Why is it that there are only two choices for the spin of an electron,
> namely "up" or "down" (or "+1/2" and "-1/2")? My mathematical background
> suggests to me that this comes from the fact that SU(2) is a double cover of
> SO(3)--
[snip]
This looks like a job for John Baez. He will give you an answer
mathematically rigorous four ways orthogonal (at least!). Given that
the average product of American zero-goal education cannot make change
of a dollar without a calculator and a counselor, Id' say you have
quite an upstream swim.
Good luck.
(Remember that free space symmetries are broken by lattice
anisotropies when you start talking about real world devices - e.g.,
Fermi's golden rule).
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Ian Macmillan
direction that a simple entity spins depends only on from which side you
observe it. However if an entity has a counter rotating characteristic (like
two disks back to back) the direction can be unambiguously observed.
All the best
Ian Macmillan
[Moderator's note: Quoted text deleted. -TB]
Uncle Al
Axial vs. polar vectors.
Reflect an arrow in a mirror. If the arrow long axis is parallel to
the plane of the mirror, both the arrow and its reflection have the
same direction. If the arrow long axis is normal to the plane of the
mirror, the arrow and its reflection have opposite directions.
Reflect a screw in a mirror, as above. Does the comparison of the
direction of twist, screw vs. reflection, change with orientation?
If the spinning entity is relativistic, how do you get to the other
side to view it?
--
Uncle Al
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
puppe...@hotmail.com
[much text snipped]
> I'd like to take a stab at explaining what spin REALLY is.
Ok, I say it several times in this, but it needs emphasis:
there is a truckload of math I'm about to glibly skip over.
Spin is about how things behave under rotation.
Think about a set of four balls held in a square by rubber rods.
If the balls are all the same, and the rods are all the same, you
can do several rotations and reflections on this square and get
back and equivalent thing. There are four rotations, by 0, 90, 180,
and 270 degrees. And there are four reflections, in a vertical
axis, in a horizontal one, and in either of the two diagonal axis.
So those eight actions form a group. Page the mathematicians for
just a truckload of math.
Now think about the square again. Suppose we give it a little smack
so the balls start to wobble around on the rubber rods. Very soon
you can discover what are called "normal modes." For example, there
is a mode where all the balls are moving directly in or out from
the centre, all in phase. The square is being stretched or shrunk
symmetrically. This is often called the breathing mode. This vibration
is also invariant under the full group. If you rotate it or reflect
it, you get back the same thing.
Another mode is where balls 1 and 3 are moving 180 degrees out of
phase with balls 2 and 4. So 1 and 3 are moving inwards when 2 and
4 are moving out, and so on. Notice that this vibration is not
invariant under all the rotations and reflections. Under some of
them, it changes phase by 180 degrees. Under others, it is the same.
So what? Well, the rotation group is just that, rotations. And
the normal modes, when represented as displacement vectors, form
the basis set for a vector space. And then you can write down
the rotations acting on these as matrix operations. And suddenly,
you have got matrix representations of the rotations and reflections.
(HUGE amount of math getting glossed over here.)
So, that breathing mode is invariant. So rotations on it are just
multiplying it by 1. In other words, it gets a very simple representation
of rotations. And that rep is called the scalar rep. And every group
has a scalar rep.
Now the other vibration, the 1-3 in while 2-4 out, gets a more complicated
rep. It has either leaving the vibration the same, or multiplying it by -1.
So this rep is a set of eight integers, four of which are just 1, and
another four are just -1.
So you can do that, and build up a basis for the entire group. Lots
and lots of math is being glibly skipped over here.
But the point is, if you start with a set of symmetries, and they act
on some object that can vibrate, you can then:
- construct the normal modes,
- collect the modes as a basis set, and
- build matrix represenations of the symmetries.
Just as an aside: Such normal modes are sometimes very useful in
- the study of molecules,
- the study of large symmetric boilers,
- the study of nuclear reactors,
- the study of turbines, etc.
Well, what about fundamental particles and spin? They turn out to
have normal modes as well, and these sit in basis sets, and can
be used to construct matrix representations of the symmetries.
There is a scalar rep, where a thing would be completely invariant
under rotations. This would be a spin 0 particle. So far, we have
not seen any of them.
There is a 3 by 3 matrix rep of rotations. This is the common one
everybody learns in high school. It is the usual one where 3 dimensional
vectors are rotated around. For example, to rotate around the z
axis by an angle theta, you would use this matrix.
cos(theta) sin(theta) 0
- sin(theta) cos(theta) 0
0 0 1
This would be the spin 1 rep, also called a vector representation
of the rotations. This is the photon and the gluon.
Now, in quantum mechanics, a wonderful thing happens to rotations.
They are not SO(3), but instead are SU(2). That means, we really
don't have just the possible rotations in regular 3-D space. We
have a doubly connected version. And when we look for representations
of those, we find a rep between the scalar and the vector, in terms
of two by two matrices. One common form of these is the Pauli spin
matrices. The important thing here is, you have to go all the way around
twice, a full 720 degrees, to get back where you started. And these
are the spin 1/2 particles such as electrons and quarks. And the rep
is called a spinor rep. (This gets doubled up to a 4 by 4 rep in the Dirac
equation, basically to account for anti-particles.)
Gravity turns out to sit in a tensor rep, and have spin 2. At least,
the wave-like solutions to general relativity do. And, for
reasons I'm going to glibly skip over, there are expected to be no
sensible particle theories for fundamental objects with more than
spin of 2.
So, the spin of a particle is related to how it transforms under
rotations, and how transformations are represented in terms of
the theory we use to describe the particle. Spin 1 particles in
a vector rep, for example photons and gluons, and spin 1/2 in a
spinor rep, for example electrons and quarks.
Socks
Mike Mowbray
> > I'm thinking of writing a popular-style article about spin. [...]
> > I'd like to take a stab at explaining what spin REALLY is. I might
> > even try to throw in words like "spinor" [...]
> >
> > * Do you have a favorite way of explaining spin to
> > non-mathematicians and non-physicists?
For a popular-style article, try challenging your readers by saying
that they don't really understand the 3D space in which they live. The
old "Dirac scissors (or belt)" experiment should show them this (or at
least puzzle them into thinking hard about it). Lay a belt on a table.
Put something heavy like a book on the non-buckle end to stop it moving.
Then rotate the buckle 360deg (axis of rotation corresponding to the
line of the belt). The challenge is then: without rotating the buckle,
try to straighten out the belt. You can move the buckle around in air,
up, over and around the belt, provided you don't rotate it. Can't be
done. But start over with a flat belt and rotate the belt by 720deg.
This time, by moving the buckle over and under the belt, you *can*
straighten out the belt without rotating the buckle.
I found this quite astonishing the first time I read about it, and
tried it at home. The point is that rotating by 720deg is not always
the same as rotating by 360deg, - which is very counter-intuitive.
I.e: our usual understanding of the 3D space in which we live is
at best incomplete.
The next fact is that there are also subatomic particles for which
rotation by 360deg and 720 deg are different, i.e: electrons and other
particles with intrinsic half-integral spin. A simple 2-component
spinor is a way of describing them. Given several electrons without
knowing their origin, we don't know which of them have been rotated
360deg relative to each other. So there are two possibilities,
expressed by a spinor with 2 components.
- MikeM.
pat mc cormack
Might be worth taking a look at
http://www.ucl.ac.uk/~ucesjph/reality/ga/intro.html
Not my speciality but the physical intuitions of spin etc seem clearer when
couched in more general geometric (Clifford algebra) algebra terms. David
Hestenes' "Clifford algebras and the interpretation of quantum
mechanics.pdf" http://modelingnts.la.asu.edu/pdf/caiqm.pdf is a nice
start.
Pat McCormack
"Dana Mackenzie" <mack...@nospam.cruzio.com> wrote in message
news:amvml...@enews1.newsguy.com...
>
> Hi all,
>
> I'm thinking of writing a popular-style article about spin. I've seen a
fair
> amount of publicity about "spintronics." It seems to me that the writers
> usually explain electron spin as if the electron were a little ball
rotating
> clockwise or counterclockwise. But that isn't right, is it? In my article
> I'd like to take a stab at explaining what spin REALLY is. I might even
try
> to throw in words like "spinor" (although the editor is likely to recoil
and
> say that's jargon).
[Moderator's note: Much unnecessary quoted text deleted. -MM]
Ian Macmillan
"Uncle Al" <Uncl...@hate.spam.net> wrote in message
news:anfhkf$ku6$1...@lfa222122.richmond.edu...
> Ian Macmillan wrote:
> Uncle Al wrote:
> Axial vs. polar vectors.
>
> Reflect an arrow in a mirror. If the arrow long axis is parallel to
> the plane of the mirror, both the arrow and its reflection have the
> same direction. If the arrow long axis is normal to the plane of the
> mirror, the arrow and its reflection have opposite directions.
>
> Reflect a screw in a mirror, as above. Does the comparison of the
> direction of twist, screw vs. reflection, change with orientation?
>
> If the spinning entity is relativistic, how do you get to the other
> side to view it?
>
> --
> Uncle Al
> http://www.mazepath.com/uncleal/eotvos.htm
> (Do something naughty to physics)
I must confess, I do not really grasp the relevance of Uncle Al's remarks
about mirrored rotations and translations to my thought about counter-
rotation, but I am far from being in his league, and usually get a 'now-
wait-a-minute' feeling when mirrors come into play!
To me, the essence of mirrors is that they do not reverse anything, but
reflect every point from where it is on the original.
A mirrored view can only be sensibly compared with a reality if the
observer stands in between, looking first at one and then the other.
The "reversal" in a reflection then arises from a rotation of the point
of view. The observer standing between mirror and object first looks at
the object and sees a red side (say) on his left. He turns 180 degrees
about a vertical axis and sees that in the reflection the red side is on
his right. The same result obtains from rotating a normal image 180
degrees about a vertical axis, or if you prefer, swapping left to right.
If you repeat this process about either the vertical Y or the horizontal
X axis, the image is restored to 'normal', except that in the latter
case, the image is inverted, but can be restored to normal by rotating it
180 degrees about the Z axis.
With this in mind, all rotations and screw threads are reversed in a
mirror, and do not change with orientation.
However, if the axis of a wheel is parallel to the mirror care is needed
because the direction of rotation depends on which side of the wheel you
look at. However if the sides are painted red and green, rather than
being named left and right, the reflected red side turns in the opposite
direction to the real red side.
I do not think that the direction of rotation of an entity can be decided
by looking at reflections of the "other side", because who knows which
side is THE side? However, if the rotating entity is a counter rotating
composite, THE side is the outside. Of course this leaves aside how to
measure, without comparison, which way is clockwise and which way is
widdershins, a much more difficult question.
In my fevered imagination, a kind of standing matter wave with say two
wavelengths around a cylinder could have a counter-rotating spin aspect,
with opposite handedness on orthagonal axes. Maybe aspirin will help!
As far as getting to the other side of a spinning relativistic entity to
view it is concerned, why, you would overtake it and have a look. That
is, of course, if I have properly understood the question.
I suppose the core of my question is, what does spin up and down mean?
Even without little spinning balls, if there is angular momentum, I
presume an axis and chirality are implied.
Oz
>There is a 3 by 3 matrix rep of rotations. This is the common one
>everybody learns in high school.
Unfortunately not so 30 years ago. However my daughter taught me some ..
>It is the usual one where 3 dimensional
>vectors are rotated around. For example, to rotate around the z
>axis by an angle theta, you would use this matrix.
>
> cos(theta) sin(theta) 0
>- sin(theta) cos(theta) 0
> 0 0 1
OK.
>This would be the spin 1 rep, also called a vector representation
>of the rotations. This is the photon and the gluon.
OK, so am I to assume that spin 1 can be represented as a vector in a
simple 3-D world, as if it were a spinning top?
But ISTR others pointing out that this is an artefact of a 3-D world and
in higher dimensions spin is in a plane.
>Now, in quantum mechanics, a wonderful thing happens to rotations.
>They are not SO(3), but instead are SU(2). That means, we really
>don't have just the possible rotations in regular 3-D space. We
>have a doubly connected version. And when we look for representations
>of those, we find a rep between the scalar and the vector, in terms
>of two by two matrices.
Since you seem to have the gift of the common tongue, I wonder if you
wouldn't mind going into this in a little more detail with some simple
examples? Just the pauli spin matrices would suffice, I think.
>One common form of these is the Pauli spin
>matrices. The important thing here is, you have to go all the way around
>twice, a full 720 degrees, to get back where you started. And these
>are the spin 1/2 particles such as electrons and quarks. And the rep
>is called a spinor rep. (This gets doubled up to a 4 by 4 rep in the Dirac
>equation, basically to account for anti-particles.)
OK, so crudely I am imagining something that is trying to 'catch up'
with me. As I turn it refuses to stay 'still' in its reference frame
but appears to turn theta for each 2 theta I turn. So when I have
rotated 360 degrees it has rotated 180deg in what I think ought to be
its rest frame. Hmmm, don't relativistic discs do funny things too?
>Gravity turns out to sit in a tensor rep, and have spin 2. At least,
>the wave-like solutions to general relativity do. And, for
>reasons I'm going to glibly skip over, there are expected to be no
>sensible particle theories for fundamental objects with more than
>spin of 2.
I think I'll leave gravity, for now anyway.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Note: soon (maybe already) only posts via despammed.com will be accepted.
Uncle Al
Ralph E. Frost
Uncle Al <Uncl...@hate.spam.net> wrote in message
news:anfhkf$ku6$1...@lfa222122.richmond.edu...
Go through the middle.
puppe...@hotmail.com
Oz <aco...@btopenworld.com> wrote in message news:<e0DIVvAa...@btopenworld.com>...
[snip]
> OK, so am I to assume that spin 1 can be represented as a vector in a
> simple 3-D world, as if it were a spinning top?
In somewhat over simplified terms, yes. It's the three-vector part
of the vector potential of the electro-magnetic field. And as needs
emphasizing repeatedly when trying to just talk about this stuff,
there is just tons of math being glibly ignored here.
> But ISTR others pointing out that this is an artefact of a 3-D world and
> in higher dimensions spin is in a plane.
Well, I would not know about that. Spin is angular momentum, so
I don't really know what it means for it to be "in a plane" as
opposed to any other condition it could have. And I really don't
know higher dimensions at all.
But, for example, if you are in a space-time in which the
rotation group is not SO(3) any more, then the reps are not
going to be SO(3)'s reps any more either. They may line up
fairly well, but I can't tell you what the reps will be, and
so can't tell you what spins will exist. (It was hard enough
for me to work out the reps for SO(3) lo these many years
ago when I was in grad school.)
[talking about me saying you get SU(2) in QM instead of SO(3)]
> Since you seem to have the gift of the common tongue, I wonder if you
> wouldn't mind going into this in a little more detail with some simple
> examples? Just the pauli spin matrices would suffice, I think.
Well, there aren't really any simple examples. You'd need to have
the full machinery of the P. spin matrices, talk about commutators
and anti-commutators, and see how these things fit into something
like a Dirac equation or some such. Otherwise I'd just be telling
you things like the multiplication properties of matrices that
I told you had something to do with electrons.
But basically, it's related to electrons having anti-symmetric
wave functions, and to the fact that the phase of a wave function
is a complex (that is, real and imaginary part) number. So,
rotating by 360 degrees gets you -1 instead of 1.
Think about odd and even functions. You can project out the
odd part of a function f(x) by building fo(x) =1/2 ( f(x) - f(-x)).
And the even part is just fe(x) = 1/2( f(x) + f(-x)). So fo(-x)
is -fo(x) and so on. Those are projection operators working
on f(x). You can build the same things for any set of symmetry
operations that form a group. And so you can divide all functions
up into the odd and the even ones. The even ones are like bosons
(in a highly superficial, glibbly-skipping-the-math way) and
the odd ones are like fermions. It's just that the fermions are
odd under a 360 degree rotation, not just a mirror of the x-axis.
I learned all this from Bjorken and Drell _Relativistic Quantum
Mechanics_, Tinkam _Group Theory and Quantum Mechanics_ and
a few others on group theory and quantum field theory, plus a
lot of reading and discussing with proffs and grad students.
The simplest set of mathematical efforts you'd need to go through
to have more than a "wordy" understanding of this is probably to
understand how to construct the normal modes of a symmetric object,
how to build the projection operators onto those normal modes,
and how to build the representations of the symmetry group that
goes with the object.
> OK, so crudely I am imagining something that is trying to 'catch up'
> with me. As I turn it refuses to stay 'still' in its reference frame
> but appears to turn theta for each 2 theta I turn. So when I have
> rotated 360 degrees it has rotated 180deg in what I think ought to be
> its rest frame. Hmmm, don't relativistic discs do funny things too?
No. When you rotate an electron 360 degrees, it has rotated 360 degrees.
It's just that doing so gives you that -1 I mentioned, not +1 like you
might expect.
Somebody else mentioned the "belt trick." There's also a cool thing
in Misner, Thorne, and Wheeler _Gravitation_ about a particle in a
box with a lot of string tied to it and the box. If you rotate the
particle by 720 degrees, then you can un-tangle the string without
moving the box or the particle. But if you only rotate 360 degrees,
you can't do it. I've done this with a rubber ball, a desk drawer,
some string and tape, and it's a lot of fun. There's an understandable
picture of this in MTW.
Another crude way to imagine this is the groups SU(2) and SO(3).
Both have the same tangent space, that is, the same Lie algebra.
But SU(2) consists of two disconencted parts. You could think
of the single SO(3) region as the rotations as they apply to
bosons, and the two disconnected parts of SU(2) as so: The first
part is the regular rotations 0 to 360. The second part is the
set from 360 to 720 as applied to fermions after they have gone
around 360 already.
Socks
Oz
>No. When you rotate an electron 360 degrees, it has rotated 360 degrees.
>It's just that doing so gives you that -1 I mentioned, not +1 like you
>might expect.
Well, I'm not absolutely sure how I would know if an electron had
rotated other than by looking at it's spin. Since I never studied spin
this is a problem but there doesn't seem to be a way in other than
studying some heavyweight stuff in depth. Strangely actual experimental
results are rarely discussed.
I imagine the following experiment.
A beam of polarised photons is sent to a detector that can be rotated.
It detects the number of spin up and spin down electrons it receives.
Initially it detects 100% +1.
It is slowly rotated (I presume around the axis of the beam).
The number of +1 it detects decreases and -1 increases.
When it has rotated 360deg they are all -1.
Then the -1's decrease and the +1's increase until at 720deg it detects
only +1 again.
Is this what we are talking about?
[Moderator's note: No. When you rotate a spin-up electron
180 degrees around the x or y axis, you get a spin-down electron.
When you rotate it 360 degrees it's spin-up again. The subtlety
lies in the *phase*, and that will require more explanation. - jb]
Danny Ross Lunsford
> * Do you have a favorite way of explaining spin to non-mathematicians and
> non-physicists?
Yes - the same one Dirac made, namely:
A relativisitic theory requires that space and time enter on an equal
footing. So, if the space derivatives are 1st order, then so must the time
derivatives be. Combined with the relativistic formula for rest mass,
momentum, and energy E^2=p^2 + m^2, this forces the introduction of
multi-component wave functions, that is, spinors.
> * Does electron spin come from the distinction between the groups SO(3)
and
> SU(2)?
No.
> * Why is it that there are only two choices for the spin of an electron,
> namely "up" or "down" (or "+1/2" and "-1/2")?
Because in general, spin comes in half-integral multiples of Planck's
(reduced) constant that differ by an integer, so one has the possibilities
for physical particles
0
-1/2 +1/2
-1 0 +1
-3/2 -1/2 +1/2 +3/2 etc.
One must then match up what is observed with this list. Asking "why" is not
really appropriate, it's a question of "what" and "how".
> My mathematical background
> suggests to me that this comes from the fact that SU(2) is a double cover
of
> SO(3)--so that you can describe where you are in SU(2) by where your
> projection is in SO(3) [would that correspond to your angular momentum?]
and
> a choice of one of the two preimages of that point [would the two
preimages
> correspond to "spin up" and "spin down"?]. Am I on the right track here,
or
> is this too naive?
No, this is not the right way to think about it. One can combine irreducible
representations of the Lorentz group according to definite rules, so spin 1
can be thought of as a certain aspect of two combined spin 1/2s.
> * Is classical physics kind of like living on a Lie algebra (where so(3)
and
> su(2) are the same), while quantum physics is more like living on the
actual
> Lie groups?
I don't think so, but I may be misunderstanding the question.
> * Do higher-dimensional spinors start entering the picture when you look
at
> particles other than the electron or forces other than electromagnetism?
If
> not, when do they enter the picture?
One can make hypotheses about the existence of higher spin particles and
then work out the equations they have to satisfy, in analogy with the
equation for the electron, then look for their experimental traces. All the
fundamental particles in physics seem to be either spin 0, spin 1/2, or spin
1. (The hypothetical graviton is not included here.)
> * Can someone explain to me why the particles with mass are the ones with
> non-integral spin? Would it be overreaching to say that without spin,
there
> wouldn't be mass?
There are massive spin 0 and spin 1 particles.
> * Does anyone have some favorite examples of real-world manifestations of
> spin? (I saw a couple mentioned in an article by Michael Weiss on John
> Baez's Web page -- the Zeeman effect and phosphorescence.)
The Stern-Gerlach experiment.
> * Finally, I'd be interested in any reading suggestions that you have on
> spin or spintronics, especially ones that explain the link between the
> mathematics and the physics of spin.
>
> Thanks for your help!
Mr. Baez?
Dana Mackenzie
e-mail from Urs Schreiber, who referred me to an earlier thread called
"Dirac for Dummies." At the very end of that thread (on 2002-01-16), John
Baez gave a very interesting explanation of what the four components of a
Dirac wave function signify. He said that if you take the Fourier transform
of the wave function, then the four components are amplitudes (or
probability densities) for the particle to be a spin-up particle, a
spin-down particle, spin-up antiparticle, or spin-down antiparticle.
New question: Okay, but how does this relate to the classical concept of
angular momentum? This is the flip side of the complaint in my earlier post.
Popular accounts make spin and angular momentum seem like basically the same
thing, which I feel, to quote "Uncle Al," is "vastly erroneous." John Baez's
explanation goes too far the other direction: It leaves me groping to see
any connection at all between spin and angular momentum. To be fully
satisfied, I'd like to see an explanation that lets me see both the
connection and the difference at the same time. Anyone want to take a shot
at this?
Kizhevur Kandaswamy Diravidan
Hello Oz,
> >No. When you rotate an electron 360 degrees, it has rotated 360 degrees.
> >It's just that doing so gives you that -1 I mentioned, not +1 like you
> >might expect.
> Well, I'm not absolutely sure how I would know if an electron had
> rotated other than by looking at its spin. Since I never studied spin
> this is a problem but there doesn't seem to be a way in other than
> studying some heavyweight stuff in depth. Strangely actual experimental
> results are rarely discussed.
Let us first consider the rotation of the spin vector. Consider a rotation
by an angle phi about the z axis. The rotation operator is
R = exp(-i.Sz.phi/hbar) , Sz being the z component of the spin (operator).
Let us act this on a general spin state,
|alpha> = |-> <-|alpha> + |+> <+|alpha> ,
|+> and |-> representing the eigen kets of Sz.
(i.e, Sz|-> = -(hbar/2)|-> , Sz|+> = +(hbar/2)|+> )
So, R|alpha> = exp(i.phi/2)|-><-|alpha> + exp(-i.phi/2)|+><+|alpha>
Putting phi=2.pi, u see that R|alpha> = -|alpha> !! and u will also see
that u need phi to be 4.pi to get back |alpha> after rotation! This is a
very quantum mechanical result, and if you notice it occured because of
the hbar/2 eigenvalues of the electron (fermion) spin operator.
This prediction has been tested too! The way they test it is to rotate
the spin vector (the way u rotate it is by passing it through a
magnetic field.. the spin vector precesses in a magnetic field. The
amount of rotation depends on the time the spins encounter the
field). If you can run an interference experiment where only one path
is affected by the field, there will be interference pattern beacuse
of the phase difference induced by the magnetic field. The intensity
in the interference region can be shown to have a sinusoidal
dependence on the applied magnetic field. It has been shown from such
interference experiments that succesive maxima occur for a change in
magnetic field corresponding to a phase shift (rotation) of 4.pi ! I
have rushed through the experimental part. (also the spin precession
part) U might want to look it up in a text book, or I'd be willing to
help out with the details if u so require.
Cheers,
kandy
Mikael Johansson
Let's see if this one makes it all the way this time (I guess the domain
in
the From:-field needs to exist?)
Dana Mackenzie wrote:
> I'm thinking of writing a popular-style article about spin. I've seen a fair
> * Does anyone have some favorite examples of real-world manifestations of
> spin? (I saw a couple mentioned in an article by Michael Weiss on John
> * Finally, I'd be interested in any reading suggestions that you have on
> spin or spintronics, especially ones that explain the link between the
> mathematics and the physics of spin.
I think the spin-"forbidden" transitions of the northern/southern
lights are very impressive.
Sin-itiro Tomonaga's "The story of spin" might be of interest.
Unfortunately, I have only quickly browsed through it as of yet (no fun
until my Master's degree is done...) At least from a historic viewpoint
(without forgetting the mathematics) it seems quite an enjoyable book.
A very interesting and well written (spoken) lecture by Goudsmit on the
discovery of (electron) spin is found at:
http://www.lorentz.leidenuniv.nl/history/spin/goudsmit.html
I quote one encouraging part from the end (replace "physics" with your
favourite brand of science :-)
"""
First: you need not be a genius to make an important contribution to
physics because, I do admit, the electron spin is an important
contribution. That I know now, then we did not know, but now I do. They
all told me so.
"""
Have a nice day,
Mikael J.
http://www.helsinki.fi/~mpjohans/
John Devers
> 180 degrees around the x or y axis, you get a spin-down electron.
> When you rotate it 360 degrees it's spin-up again. The subtlety
> lies in the *phase*, and that will require more explanation. - jb]
Do photons also have the property of angular momentum?
Is this property related to the polarization or is it independant?
How can one go about testing whether a photon is spin up or spin down?
http://math.ucr.edu/home/baez/physics/Quantum/bells_inequality.html
John Baez
>Well, I'm not absolutely sure how I would know if an electron had
>rotated other than by looking at its spin.
Good: that's the main way!
And for the most part, it works a lot more simply than you seem
to think. If I have an electron whose spin angular momentum points
in some direction, and I rotate the electron, the spin angular
momentum just gets rotated in the obvious way! No funny business.
For example, suppose I have an electron whose spin angular momentum
is pointing straight up: the famous "spin-up state". And suppose
I rotate the electron 180 degrees around the x axis. Then its
spin angular momentum points straight down! It doesn't do anything
weird. It doesn't, like, rotate half as much as you'd expect.
The weirdness comes in here: besides having an angular momentum,
a spin-1/2 particle also has a "phase". And this acts a bit funny.
In particular, if you rotate a spin-1/2 particle 360 degrees
around any axis, its phase gets multiplied by -1. Its angular
momentum comes right back to where it was, but the phase is
changed! This is the sort of thing that can be detected by
interference experiments.
To understand this "phase" versus "angular momentum" stuff in
more detail, a little math comes in handy. Spinors - they're
just pairs of complex numbers. Pauli matrices - they're just
2 x 2 complex matrices. A decent place to learn this is the
very beginning of volume 3 of the Feynman lectures. He builds
it all up starting from the Stern-Gerlach experiment.
Philip Charlton
<snip>
>The weirdness comes in here: besides having an angular momentum,
>a spin-1/2 particle also has a "phase". And this acts a bit funny.
>In particular, if you rotate a spin-1/2 particle 360 degrees
>around any axis, its phase gets multiplied by -1. Its angular
>momentum comes right back to where it was, but the phase is
>changed! This is the sort of thing that can be detected by
>interference experiments.
>
I thought the phase change wasn't detectable phyically? It's not an
observable.
Philip
[Moderator's note: The phase of an individual particle is not
observable, but phase *differences* are. You can in principle take a
beam of particles, send it through a beam splitter, rotate one half
360 degrees and leave the other half alone, and recombine. The way
the two beams interfere will be different than if you hadn't done the
rotation. -TB]
Ralph E. Frost
John Baez <ba...@galaxy.ucr.edu> wrote in message
news:ao07r2$euc$1...@glue.ucr.edu...
Besides these approaches [which, I must admit I am not familiar with at
all], I'm wondering if there are also any mathematical structures, that,
when "rotated" or manipulated as you describe, fold inside out or otherwise
morph so as to exhibit features analogous to what you say the experimental
evidence is for angular momentum and spin for the point electron?
I'm wondering if electron was not a point particle but had substructure,
say, that of a spiraling torus, are there any handy math structures that
match up these spin issues better than ?
Mind you, I'm asking not for some fight over whether electron has point
structure or not, but more like if that question were COMPLETELY open, yet
folks were given the electron behaviors, what other mathematical structures
would also map to those behaviors?
--
--
Ralph Frost
Looking for a desktop model to help you ponder this topic?
http://flep.refrost.com -- now with secure online ordering
Use more robust symbols
Seek a thought worthy of speech.
puppe...@hotmail.com
johnd...@froggy.com.au (John Devers) wrote in message news:<2978f9d5.02100...@posting.google.com>...
> Do photons also have the property of angular momentum?
Yes. Photons are spin 1.
> Is this property related to the polarization or is it independant?
Yes. Photons can be polarized to have their angular momentum
either parallel or anti-parallel to their direction of motion.
Gaugue symmetry kills off the third possibility of having a zero
projection onto the direction-of-motion axis.
> How can one go about testing whether a photon is spin up or spin down?
By intereacting them with electrons.
Socks
Oz
>The weirdness comes in here: besides having an angular momentum,
>a spin-1/2 particle also has a "phase". And this acts a bit funny.
>In particular, if you rotate a spin-1/2 particle 360 degrees
>around any axis, its phase gets multiplied by -1. Its angular
>momentum comes right back to where it was, but the phase is
>changed! This is the sort of thing that can be detected by
>interference experiments.
<Oz smacks himself on the head>
Right. That's really made things very clear indeed.
Brilliant. Nothing that inscrutable at all.
A property of the particle, nothing to do with the particle rotating
(necessarily) at all?
=============
A check though, if I may.
An electron is spin(1/2).
It also has some angular momentum, which presumably (but now I must
check) ties in with it's magnetic field?
So in a helium atom we could align the electron magnetic axes by
applying a suitably strong magnetic field and this would align the
angular momentum of both electrons.
But each electron can still happily exist in the same orbital because
one will be spin(1/2) and the other spin(-1/2)?
Correct?
=============
Follow up:
Is there a way of measuring spin 'absolutely' to say 'this electron is
spin up', or is it 'this electron has opposite spin to that electron'?
Is there any way of producing beams of electrons (or other fermions)
with an enriched number in a particular spin state?
>To understand this "phase" versus "angular momentum" stuff in
>more detail, a little math comes in handy. Spinors - they're
>just pairs of complex numbers. Pauli matrices - they're just
>2 x 2 complex matrices.
No chance of a brief introduction to one of these is there?
>A decent place to learn this is the
>very beginning of volume 3 of the Feynman lectures.
Can't.
My copy has been filched by my son on the pretext he might actually do
some learning. I live in hope. He reports going to 'quite a lot of
lectures' which doesn't bode well.
>He builds
>it all up starting from the Stern-Gerlach experiment.
Sounds very feynmanesque.
John Baez
>John Baez <ba...@galaxy.ucr.edu> writes
>>The weirdness comes in here: besides having an angular momentum,
>>a spin-1/2 particle also has a "phase". And this acts a bit funny.
>>In particular, if you rotate a spin-1/2 particle 360 degrees
>>around any axis, its phase gets multiplied by -1. Its angular
>>momentum comes right back to where it was, but the phase is
>>changed! This is the sort of thing that can be detected by
>>interference experiments.
><Oz smacks himself on the head>
>
>Right. That's really made things very clear indeed.
Great!
>Brilliant. Nothing that inscrutable at all.
Right: this part is not tricky. The tricky part is just how
the phase gets affected by rotation. That's where people need
to mumble about spinors and Pauli matrices and the coffee cup
trick and SU(2) and so on.
>A property of the particle, nothing to do with the particle rotating
>(necessarily) at all?
Well... a particle must have nonzero spin for you to be able
to tell if you've rotated it. The spin defines a vector, the
angular momentum vector. Rotations rotate that vector. For a
spin-0 particle like a pion that vector is zero! So the particle
doesn't change at all when you rotate it.
>A check though, if I may.
Okay.
>An electron is spin(1/2).
Yes.
>It also has some angular momentum,
That's not "also"; that's practically the same thing.
Being a spin-1/2 particle means that it has a certain
definite amount of angular momentum even when it's not moving.
Even when it's not moving around, it's spinning!
This angular momentum is called "spin angular momentum".
>which presumably (but now I must check) ties in with its magnetic field?
A charged spin-1/2 particle acts like a little bar magnet,
whose poles point along the axis of spin. This is the main
way you can tell which way the axis of spin is pointing.
>So in a helium atom we could align the electron magnetic axes by
>applying a suitably strong magnetic field and this would align the
>angular momentum of both electrons.
Sorta kinda, yeah - but notice, now we're getting into more complex
territory because both *orbital* angular momentum and *spin*
angular momentum matter here. In addition to spinning, the
electrons are orbiting around the nucleus. Applying a magnetic
field will affect both of these things. I don't think we want
to get into such a complicated problem just yet. Textbooks usually
break things down into bite-sized pieces, and I think that's what
we should do too.
>But each electron can still happily exist in the same orbital because
>one will be spin(1/2) and the other spin(-1/2)?
The Pauli exclusion principle prevents two electrons
from being the same in *all* ways. But being the same in
*all but one way* is okay. So:
Two electrons in a helium can be in what's otherwise the same
state - same orbital angular momentum, same energy level -
as long as their spin angular momentum is different. In fact
this is what helium in its ground state is like!
Note I'm no longer talking about your problem of helium in
an external magnetic field; I'd rather talk about simple things
first.
>Is there a way of measuring spin 'absolutely' to say 'this electron is
>spin up' [...]
Sure! It's a little magnet; you see if the north pole is pointing
up, and you deduce that the spin angular momentum is pointing up.
There could be a sign error here because some idiot named Ben
Franklin made the electron *negative*, but let's not worry about
that: the point is, the spin angular momentum can be measured using
the fact that the electron is a little magnet.
>Is there any way of producing beams of electrons (or other fermions)
>with an enriched number in a particular spin state?
Sure: a "polarized electron beam". But I'm a theorist; I don't
actually make such things here in my office.
>>To understand this "phase" versus "angular momentum" stuff in
>>more detail, a little math comes in handy. Spinors - they're
>>just pairs of complex numbers. Pauli matrices - they're just
>>2 x 2 complex matrices.
>No chance of a brief introduction to one of these is there?
Not until you stop making outlandish claims about photon
wavefunctions and let Michael Weiss teach you about them.
Arkadiusz Jadczyk
On Fri, 11 Oct 2002 01:33:12 +0000 (UTC), puppe...@hotmail.com wrote:
>
>johnd...@froggy.com.au (John Devers) wrote in message news:<2978f9d5.02100...@posting.google.com>...
>> Do photons also have the property of angular momentum?
>
>Yes. Photons are spin 1.
>
More precisely, because photons are of rest mass zero, the concept of
"spin" does not excatly applies (there is no "rest frame"). Instead we
have "helicity".
In terms of generators of the Poincare group we have the following
formulas:
For particles with mass m>0:
Spin S:
S_j=(1/m)(wj - (w_0Pj/(m+P0))
For massless particles:
Helicity Lambda
W_mu = Lambda P_mu
where W is Pauli-Lubansky vector:
http://www.lns.cornell.edu/spr/2000-06/msg0026104.html
Seee also
http://spin.riken.bnl.gov/rsc/write-up/Boer/SPINdisc_112499bweb.ps.gz
for some useful formulas.
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Michael Weiss
The Wiz:
: >To understand this "phase" versus "angular momentum" stuff in
: >more detail, a little math comes in handy. Spinors - they're
: >just pairs of complex numbers. Pauli matrices - they're just
: >2 x 2 complex matrices.
Oz:
: No chance of a brief introduction to one of these is there?
Sure! Have a look at
http://math.ucr.edu/home/baez/lie/lie.html
The Wiz:
: >A decent place to learn this is the
: >very beginning of volume 3 of the Feynman lectures.
Oz:
: Can't.
: My copy has been filched by my son on the pretext he might actually do
: some learning. I live in hope. He reports going to 'quite a lot of
: lectures' which doesn't bode well.
OK, have a look at:
http://math.ucr.edu/home/baez/spin/spin.html
In particular, you need to learn about the difference between j and j_z,
which is explained in the very first section, "The Facts".
Radu Grigore
Oz wrote:
> So in a helium atom we could align the electron magnetic axes by
> applying a suitably strong magnetic field and this would align the
> angular momentum of both electrons.
>
> But each electron can still happily exist in the same orbital because
> one will be spin(1/2) and the other spin(-1/2)?
>
> Correct?
AFAIK this happens if the magnetic field is not TOO strong.
The state with spin + will have a lower energy than the state
with spin -. The difference is proportional to the intensity
of the magnetic field. Therefore the tendency to 'jump' to state
spin + is higher if the magnetic field is stronger. At a certain
moment this will overwhelm the Columb atraction and the two electrons
will both be in state spin +, on different orbitals.
But you need a really big magnetic field to do this. A problem
in Feynman is to estimate it.
bye,
radu
Oz
Radu Grigore <radug...@ieee.org> writes
>The state with spin + will have a lower energy than the state
>with spin -. The difference is proportional to the intensity
>of the magnetic field. Therefore the tendency to 'jump' to state
>spin + is higher if the magnetic field is stronger. At a certain
>moment this will overwhelm the Columb atraction and the two electrons
>will both be in state spin +, on different orbitals.
<sigh>
OK, so spin is in fact closely related to angular momentum.
OK, glad to have that sorted, then.
Richard Saam
Radu Grigore wrote:
>
> AFAIK this happens if the magnetic field is not TOO strong.
> The state with spin + will have a lower energy than the state
> with spin -. The difference is proportional to the intensity
> of the magnetic field. Therefore the tendency to 'jump' to state
> spin + is higher if the magnetic field is stronger. At a certain
> moment this will overwhelm the Columb atraction and the two electrons
> will both be in state spin +, on different orbitals.
>
> But you need a really big magnetic field to do this. A problem
> in Feynman is to estimate it.
>
> bye,
> radu
2 Helium Energy 24.58 ev or 3.94E-11 erg
Bohr Magneton 9.27401E-21 erg/gauss
2 Helium Energy/Bohr Magneton = 4.25E+09 gauss
Yes, a really big magnetic field to do this.
Richard Saam
Oz
John Baez <ba...@galaxy.ucr.edu> writes
>In article <ao7ilt$1gc$1...@panther.uwo.ca>, Oz <ozac...@despammed.com> wrote:
>
>>John Baez <ba...@galaxy.ucr.edu> writes
>
>
>Okay.
>
>>An electron is spin(1/2).
>
>Yes.
>
>>It also has some angular momentum,
>
>That's not "also"; that's practically the same thing.
>Being a spin-1/2 particle means that it has a certain
>definite amount of angular momentum even when it's not moving.
>Even when it's not moving around, it's spinning!
>
>This angular momentum is called "spin angular momentum".
>
>>which presumably (but now I must check) ties in with its magnetic field?
>
>A charged spin-1/2 particle acts like a little bar magnet,
>whose poles point along the axis of spin. This is the main
>way you can tell which way the axis of spin is pointing.
Now you seem to be saying spin and angular momentum are 'the same'.
Lets try some more clarification.
Let's describe the direction of the angular momentum as a vector
perpendicular to the plane of rotation using (say) a right hand rule.
I know (or at least I have been told) that whenever we measure the
angular momentum of a particle, we measure a single quantised quantity
which can be either 'positive' or 'negative'.
Let's call these am+ and am-, it's angular momentum here.
Now, the first and most basic thing is to ask precisely what we mean by
a direction of angular momentum.
It seems to me to be two possibilities that a measurement can make.
am pointing along the line of travel.
am pointing perpendicular to the line of travel.
I imagine only one of these is measured, for aesthetic reasons I would
like it to be the latter, but I doubt it is.
where +am and -am refer to opposite directions in each case.
Now we discuss 'spin'.
I am unclear what the relationship between spin and angular momentum is.
It would appear from earlier posts that it's related, but not 'the same
as'. I could well be under a misunderstanding here.
Let's try another little example.
I get a nice beam of electrons and force their angular momentum to point
in one direction, let's call it +am.
Assuming they will stay in this state for some reasonable time, I split
the beam, rotate the *angular momentum vector* of one beam by some angle
theta and obtain a diffraction pattern. Is the below what I see?
theta phase seen in diff patt
0 0
pi 50%=0, 50%=pi
2pi pi
4pi 0
If not, what do I find?
That's probably an excellent place to stop.
Let's sort out the very basics first.
>>>To understand this "phase" versus "angular momentum" stuff in
>>>more detail, a little math comes in handy. Spinors - they're
>>>just pairs of complex numbers. Pauli matrices - they're just
>>>2 x 2 complex matrices.
>
>>No chance of a brief introduction to one of these is there?
>
>Not until you stop making outlandish claims about photon
>wavefunctions and let Michael Weiss teach you about them.
You miss the point.
The point is to show where my model breaks down.
I would have thought that would be easy if it's wrong.
Once destroyed, then I must learn where it needs modifying or even
rebuilding from scratch.
Ted did this with 'energy of the universe', but it seems to have
permanently exhausted him. I am eternally grateful to him though.
I would have thought one simple experimental evidence would be enough.
Somehow nobody can put one together. This worries me.
John Baez
Oz <ozac...@despammed.com> wrote:
>>Being a spin-1/2 particle means that it has a certain
>>definite amount of angular momentum even when it's not moving.
>>Even when it's not moving around, it's spinning!
>>
>>This angular momentum is called "spin angular momentum".
>Now you seem to be saying spin and angular momentum are 'the same'.
Not quite: in the quantum universe we inhabit there are two forms of
angular momentum: spin angular momentum and orbital angular momentum.
Orbital angular momentum is familiar from classical mechanics: it's
the angular momentum due to the motion of a particle through space.
More precisely, it's q x p where q is the particle's position and p is
its momentum. But watch out: in quantum position and momentum don't
commute! One consequence is that the different components of orbital
angular momentum cannot be simultaneously measured with complete precision.
Another consequences is that orbital angular momentum is quantized!
If you measure any component of the orbital angular momentum, you'll
get an integer times hbar.
Spin angular momentum is the angular momentum intrinsic to a particle.
Most people are unfamiliar with spinning point particles in the context
of classical mechanics, so they think of spin as a purely quantum business.
This isn't quite right: there's a perfectly nice notion of a spinning
point particle in classical mechanics. However, nobody thought of it
until after quantum mechanics came along.
As with orbital angular momentum, the different components of spin
angular momentum cannot be simultaneously measured with complete precision.
Also, spin angular momentum is quantized... but in a different way from
than orbital angular momentum. If you measure any component of the spin
angular momentum, you'll get an integer or HALF-integer times hbar.
>Lets try some more clarification.
>Let's describe the direction of the angular momentum as a vector
>perpendicular to the plane of rotation using (say) a right hand rule.
>
>I know (or at least I have been told) that whenever we measure the
>angular momentum of a particle, we measure a single quantised quantity
>which can be either 'positive' or 'negative'.
This is the case when
1) we have a spin-1/2 particle
and
2) we measure a specific component of the angular momentum.
>Let's call these am+ and am-, it's angular momentum here.
Okay.
>Now, the first and most basic thing is to ask precisely what we mean by
>a direction of angular momentum.
>
>It seems to me to be two possibilities that a measurement can make.
>
>am pointing along the line of travel.
>am pointing perpendicular to the line of travel.
I don't know what the "line of travel" is. Nothing is travelling:
I'm imagining a particle just sitting there, so there's no orbital angular
momentum, only spin angular momentum.
If you want to imagine the particle as spinning, you can think of it as a
tiny little spinning ball, and then the angular momentum points along the
axis of spin in the usual way. But beware: this is just a visualization
for the sake of getting some intuition about how spinning point particles
in quantum mechanics are related to spinning balls in classical mechanics.
>Now we discuss 'spin'.
>I am unclear what the relationship between spin and angular momentum is.
I hope I made that a bit clearer above. Spin angular momentum is one sort
of angular momentum. When you have a spin-j particle, and you measure
any component of its angular momentum, you get one of these numbers:
-j hbar, (-j + 1) hbar, ....... (j-1) hbar, j hbar
In general you need to add that to its orbital angular momentum to get
its total angular momentum. But if we're trying to master spin, it may
be good to ignore orbital angular momentum for a little while before
trying to throw that back into the picture.
>>>>Spinors - they're just pairs of complex numbers. Pauli
>>>>matrices - they're just 2 x 2 complex matrices.
>>>No chance of a brief introduction to one of these is there?
>>Not until you stop making outlandish claims about photon
>>wavefunctions and let Michael Weiss teach you about them.
>You miss the point.
>The point is to show where my model breaks down.
I hope someone works with you on that, but it won't be me.
Norm Dresner
> In article <aonhuq$hup$1...@clyde.its.caltech.edu>,
> Oz <ozac...@despammed.com> wrote:
>
> >>Being a spin-1/2 particle means that it has a certain
> >>definite amount of angular momentum even when it's not moving.
> >>Even when it's not moving around, it's spinning!
> >>
> >>This angular momentum is called "spin angular momentum".
>
> >Now you seem to be saying spin and angular momentum are 'the same'.
>
> Not quite: in the quantum universe we inhabit there are two forms of
> angular momentum: spin angular momentum and orbital angular momentum.
>
Aren't they separately conserved? Wouldn't that imply that they are
fundamentally different things?
Norm
Oz
>This prediction has been tested too! The way they test it is to rotate
>the spin vector (the way u rotate it is by passing it through a
>magnetic field.. the spin vector precesses in a magnetic field. The
>amount of rotation depends on the time the spins encounter the
>field). If you can run an interference experiment where only one path
>is affected by the field, there will be interference pattern beacuse
>of the phase difference induced by the magnetic field. The intensity
>in the interference region can be shown to have a sinusoidal
>dependence on the applied magnetic field. It has been shown from such
>interference experiments that succesive maxima occur for a change in
>magnetic field corresponding to a phase shift (rotation) of 4.pi ! I
>have rushed through the experimental part. (also the spin precession
>part) U might want to look it up in a text book, or I'd be willing to
>help out with the details if u so require.
OK. Let me see if I have this right.
We take electrons and line them up by using a magnetic field.
We assume (given a strong enough magnetic field) that a rotating charged
electron produces a consequential magnetic dipole. So those with spin
+1/2 point up (say) and those with spin -1/2 point down. Hmmm. With some
suitable jiggery-pokery I'm sure you should be able to get a near-pure
beam of spin +1/2, probably if you space them out a bit.
We now split the beam and send half through a helical magnetic field
that rotates them by some amount theta.
We then join the beams and observe a diffraction pattern.
Now, assuming that the path lengths are such that we start with a
maximum at some handy point in the diffraction pattern, where (I guess)
the path lengths differ by half a wavelength so the +1/2 spin are in
phase. Sounds a tricky setup to me, damned smart these experimentalists.
We now rotate one of the beams.
If we rotate one beam by 2pi then we find we have a minimum.
That is the electrons arrive precisely out of phase.
Just a simple question, are the rotated electrons now in a -1/2 state or
not?
Whilst you might imagine that this would happen after a rotation of only
pi.
So my question is a simple one. What do you consider you were doing when
you rotated the beam by an angle? What was it you rotated? Was it the
direction of the magnetic dipole you rotated? So that in effect you can
have an electron with spin +1/2, but with it's magnetic dipole pointing
either up, or down?
I think I'll stop there.
Oz
>In article <aonhuq$hup$1...@clyde.its.caltech.edu>,
>Oz <ozac...@despammed.com> wrote:
>
>>>Being a spin-1/2 particle means that it has a certain
>>>definite amount of angular momentum even when it's not moving.
>>>Even when it's not moving around, it's spinning!
>>>
>>>This angular momentum is called "spin angular momentum".
>
>>Now you seem to be saying spin and angular momentum are 'the same'.
>
>Not quite: in the quantum universe we inhabit there are two forms of
>angular momentum: spin angular momentum and orbital angular momentum.
OK.
>Orbital angular momentum is familiar from classical mechanics: it's
>the angular momentum due to the motion of a particle through space.
>More precisely, it's q x p where q is the particle's position and p is
>its momentum. But watch out: in quantum position and momentum don't
>commute! One consequence is that the different components of orbital
>angular momentum cannot be simultaneously measured with complete precision.
>Another consequences is that orbital angular momentum is quantized!
>If you measure any component of the orbital angular momentum, you'll
>get an integer times hbar.
OK. Standard job for doing a hydrogen atom ignoring spin.
I did that once. Lots of solutions that couldn't be solved analytically
very nicely until someone said 'bessel functions'.
>Spin angular momentum is the angular momentum intrinsic to a particle.
>Most people are unfamiliar with spinning point particles in the context
>of classical mechanics, so they think of spin as a purely quantum business.
>This isn't quite right: there's a perfectly nice notion of a spinning
>point particle in classical mechanics. However, nobody thought of it
>until after quantum mechanics came along.
Being a trusting sould, I will believe you.
>As with orbital angular momentum, the different components of spin
>angular momentum cannot be simultaneously measured with complete precision.
OK. Here I guess you are talking x, y, z -type components.
>Also, spin angular momentum is quantized... but in a different way from
>than orbital angular momentum. If you measure any component of the spin
>angular momentum, you'll get an integer or HALF-integer times hbar.
OK. I can live with that.
But I most strongly suspect that redefining hbar as 1/2 hbar rather
misses the geometrical point of a spin 1/2 particle.
>I don't know what the "line of travel" is. Nothing is travelling:
>I'm imagining a particle just sitting there, so there's no orbital angular
>momentum, only spin angular momentum.
Ah.
Um.
Good point.
>If you want to imagine the particle as spinning, you can think of it as a
>tiny little spinning ball, and then the angular momentum points along the
>axis of spin in the usual way. But beware: this is just a visualization
>for the sake of getting some intuition about how spinning point particles
>in quantum mechanics are related to spinning balls in classical mechanics.
OK.
And whenever (somehow) we try and measure one of the components it
always comes out as hbar or 1/2hbar.
OK.
>>Now we discuss 'spin'.
>>I am unclear what the relationship between spin and angular momentum is.
>
>I hope I made that a bit clearer above. Spin angular momentum is one sort
>of angular momentum. When you have a spin-j particle, and you measure
>any component of its angular momentum, you get one of these numbers:
>
>-j hbar, (-j + 1) hbar, ....... (j-1) hbar, j hbar
OK. Sounds very reasonable.
Maxima at -+j and any integer differences in between.
>In general you need to add that to its orbital angular momentum to get
>its total angular momentum. But if we're trying to master spin, it may
>be good to ignore orbital angular momentum for a little while before
>trying to throw that back into the picture.
I approve of as much simplification as possible.
>>You miss the point.
>>The point is to show where my model breaks down.
>
>I hope someone works with you on that, but it won't be me.
Actually I have never studied spin.
So I don't have a model.
First I need to know how these spin 1/2 particles behave in 'real life'.
John Baez
Norm Dresner <nd...@worldnet.att.net> wrote:
No. I shudder to think where you got that impression.
Kevin A. Scaldeferri
Norm Dresner <nd...@worldnet.att.net> wrote:
No!
> Wouldn't that imply that they are
>fundamentally different things?
They are different, yet the same.
Study this mantra until you acheive enlightenment:
spin : angular momentum :: mass : energy
--
======================================================================
Kevin Scaldeferri Calif. Institute of Technology
The INTJ's Prayer:
Lord keep me open to others' ideas, WRONG though they may be.
Ralph Hartley
Lets call the initial state UP, and the state rotated clockwise by pi DOWN.
Oz wrote:
> We now split the beam and send half through a helical magnetic field that
> rotates them by some amount theta.
I think it is a little simpler than that. A magnetic field
perpendicular to the original spin direction. The electron will
precess like a little top, around the axis perpendicular to both the
original spin direction and the magnetic field.
> We then join the beams and observe a diffraction pattern.
>
> Now, assuming that the path lengths are such that we start with a maximum at
> some handy point in the diffraction pattern
> We now rotate one of the beams.
>
> If we rotate one beam by 2pi then we find we have a minimum. That is the
> electrons arrive precisely out of phase. Just a simple question, are the
> rotated electrons now in a -1/2 state or not?
"-1/2" is a bit ambiguous. The rotated electron is in state -UP, this
is absoutely NOT the same as DOWN. The state -UP is exactly the same
as UP, except for what happens to the diffraction pattern.
[Moderator's note: in other words, only the phase differs,
by a factor of -1. - jb]
> Whilst you might imagine that this would happen after a rotation of only pi.
A rotation of pi clockwise gives you DOWN, the same amount of rotation
widdershins gives you -DOWN. Again, both of those states have the
little magnet pointing down.
> So my question is a simple one. What do you consider you were doing when
> you rotated the beam by an angle? What was it you rotated? Was it the
> direction of the magnetic dipole you rotated?
Basically yes. But remember that there is also a phase. Changing the
phase of everthing never makes any difference. The only way to tell it
is there is by some sort of interference.
Ralph Hartley
Oz
Ralph Hartley <har...@aic.nrl.navy.mil> writes
>
>Lets call the initial state UP, and the state rotated clockwise by pi DOWN.
>
>Oz wrote:
>
>> We now split the beam and send half through a helical magnetic field that
>> rotates them by some amount theta.
>
>I think it is a little simpler than that. A magnetic field
>perpendicular to the original spin direction. The electron will
>precess like a little top, around the axis perpendicular to both the
>original spin direction and the magnetic field.
Ah. I suppose it would.
Those horribly messy gyroscope models.
Ugh!
Makes me dizzy just thinking about it.
The below assumes (I think) a starting state of 'UP'.
>> If we rotate one beam by 2pi then we find we have a minimum. That is the
>> electrons arrive precisely out of phase. Just a simple question, are the
>> rotated electrons now in a -1/2 state or not?
>
>"-1/2" is a bit ambiguous. The rotated electron is in state -UP, this
>is absoutely NOT the same as DOWN. The state -UP is exactly the same
>as UP, except for what happens to the diffraction pattern.
Right. OK this is what I was hoping someone would say.
>A rotation of pi clockwise gives you DOWN, the same amount of rotation
>widdershins gives you -DOWN. Again, both of those states have the
>little magnet pointing down.
OK. The other key thing I needed to know.
Now I can picture it better.
Many thanks for actually saying what happens.
So if we start with UP and rotate it we get
Theta Spin
-pi -DOWN
0 UP
pi DOWN
2pi -UP
3pi -DOWN
OK, glad that's sorted.
>> So my question is a simple one. What do you consider you were doing when
>> you rotated the beam by an angle? What was it you rotated? Was it the
>> direction of the magnetic dipole you rotated?
>
>Basically yes. But remember that there is also a phase. Changing the
>phase of everthing never makes any difference. The only way to tell it
>is there is by some sort of interference.
Hmmm.
I'm going to have to think about this.
Tricky.
Hmmmm.
It's almost as if you are rotating it in the spatial dimensions but not
doing so quite right in the time one.
I'll need to mull on this.