Any internal structure to the electron?
James Logajan
A while back, someone made an offhand remark in this newsgroup that
there is an internal structure to the electron. Does this mean that
the internal charge is smeared, sort of like in nuclei, or does it mean
something else?
Just what *is* known about the electron anyway? I know it has mass,
charge, spin, and magnetic moment, but not a lot else. At what point,
if any, can it no longer be treated as a point charge? Do electron-electron
collisions (or other experiments) show anything other than a 1/r potential
at very small r?
Sorry for the rather basic questions; I need to update my mental model
of the electron. :-)
Mark Hopkins
There is no known internal structure to the electron. Experiments go all
the way down to around 10^{-18} meters resolution for the elctron and it still
looks like a point.
john baez
In article <jameslE9...@netcom.com>,
James Logajan <jam...@netcom.com> wrote:
>A while back, someone made an offhand remark in this newsgroup that
>there is an internal structure to the electron. Does this mean that
>the internal charge is smeared, sort of like in nuclei, or does it mean
>something else?
I'm not sure what that person was referring to, but to the best of
our knowledge the electron is the quantum-field-theoretic version of
a point charge, unlike a proton, which is a kind of bag full of quarks
and gluons. The electron has no "constituents"; it is not a "bound
state".
Of course thanks to quantum mechanics you can't simultaneously
meaure the position and momentum of an electron. These effects
become very noticeable when you make measurements at about the length
scale of atoms, the Bohr radius (about 5 x 10^{-11} meters). This
changes ones idea of what a "point particle" is like, but people
who know quantum mechanics are used to this, and still think of the
electron as pointlike. You can measure its position more accurately as
long as you slack off on momentum.
More interestingly, if you try to measure the position of an electron
*very* precisely, you wind up using so much energy that there's enough
energy around to create an electron-positron pair. This effect cuts
in at about the Compton wavelength of an electron (about 4 x 10^{-13}
meters). This means that at this distance scale, it pays to visualize
the electron as surrounded by a swarm of virtual positron-electron pairs,
like moths attracted to a flame. As it turns out, this cloud "screens"
the charge of the electron, somewhat as moths would dim our view of the
bulb. (For a much more precise explanation, see below.) As you get
closer to the electron you see more of its charge, since you have
penetrated more of this screening. This means that the electromagnetic
field grows more quickly than the naive 1/r^2 force law predicts...
but only at quite short distance scales.
I should emphasize that this is not mere idle theorizing. These effects
have been carefully measured in particle accelerator experiments, and
serve as one of many importance pieces of evidence for quantum field
theory.
Now... how does screening actually work? In fact this effect is
important in other, less strange contexts than quantum field theory.
Here is a little explanation of screening taken from "week94":
Say you have an electron in some water. It'll make an electric field, but
this will push all the other negatively charged particles little bit *away*
from your electron and pull all the positively charged ones a little bit
*towards* your electron:
-
+
your electron: - +-
+
-
In other words, it will "polarize" all the neighboring water molecules. But
this will create a counteracting sort of electric field, since it means
that if you draw any sphere around your electron, there will be a bit
more *positively* charged other stuff in that sphere than negatively
charged other stuff. The bigger the sphere is, the more this effect
occurs - though there is a limit to how much it occurs. We say that the
further you go from your electron, the more its electric charge
is "screened", or hidden, behind the effect of the polarization.
This effect is very common in materials that don't conduct electricity,
like water or plastics or glass. They're called "dielectrics", and the
dielectric constant, epsilon, measures the strength of this screening
effect. Unlike in math, this epsilon is typically bigger than 1. If you
apply an electric field to a dielectric material, the electric field
inside the material is only 1/epsilon as big as you'd expect if
this polarization wasn't happening.
What's cool is that according to quantum field theory, screening occurs
even in the vacuum, thanks to "vacuum polarization". One can visualize it
rather vaguely as due to a constant buzz of virtual particle-antiparticle
pairs getting created and then annihilating - called "vacuum bubbles"
in the charming language of Feynman diagrams, because you can draw them
like this:
/\
e+/ \e-
/ \
\ /
\ /
\/
Here I've drawn a positron-electron pair getting created and then
annihilating as time passes - unfortunately, this bubble is square,
thanks to the wonders of ASCII art.
Greg Weeks
One can take alternative approaches to the question of whether electrons
are pointlike. Depending on how you define "pointlike", you can get an
interesting range of answers. For example, it is natural to view electron
spin not as some mystical property but simply as the angular momentum of a
nonpointlike object. From another angle, many people agree that it makes
no sense to imagine an electron as being localized to any greater precision
than its inverse mass. And then there is the following view, which I don't
understand:
Mark Hopkins (ma...@omnifest.uwm.edu) wrote:
: There is no known internal structure to the electron. Experiments go all
: the way down to around 10^{-18} meters resolution for the elctron and it
: still looks like a point.
How do experiments determine whether an electron "looks like a point"?
I presume the experiments are scattering experiments, and that they confirm
the predictions of QED. Are the electrons of QED pointlike by definition?
Or is there something about the particular (no pun intended) scattering
amplitudes of QED that means that electrons look like a point? Is so, what
properties do scattering amplitudes of point particles need to have?
In my youth, I thought the answer was "trivial form factors". (As I
vaguely recall, form factors are the scattering amplitudes after the
Coulomb scattering was factored out.) But surely electron form factors
aren't trivial, not with all those wild Feynman diagrams contributing.
So, I don't get it.
Greg
Jon Thaler
we...@orpheus.dtc.hp.com (Greg Weeks) writes:
> How do experiments determine whether an electron "looks like a point"?
> I presume the experiments are scattering experiments, and that they confirm
> the predictions of QED. Are the electrons of QED pointlike by definition?
This is an interesting question. An article by S. Weinberg in a
recent issue of Science, Nature, or Physics Today (I forget which) on
the 100th anniversary of the electron discusses it. His point is that
the answer to the question depends on the theory being used to analyze
the problem. The theory defines "pointlike particle" and tells you
how it would behave.
Jon
Jeremy Henty
Greg Weeks wrote:
> ... For example, it is natural to view electron
> spin not as some mystical property but simply as the angular momentum of a
> nonpointlike object.
Surely not. Rotational angular momentum is always integral (in units of
h/(2*pi)). The elecron's spin is 1/2 (in the same units). Ergo, it
cannot be rotational in origin (or else I'm missing something).
Jeremy Henty
Matt McIrvin
In article <5kjl9k$hdg$1...@agate.berkeley.edu>, we...@orpheus.dtc.hp.com
(Greg Weeks) wrote:
> How do experiments determine whether an electron "looks like a point"?
[...]
> In my youth, I thought the answer was "trivial form factors". (As I
> vaguely recall, form factors are the scattering amplitudes after the
> Coulomb scattering was factored out.) But surely electron form factors
> aren't trivial, not with all those wild Feynman diagrams contributing.
My definition of "pointlike electrons" is that the data may be described
by a field theory that
(a) has an electron field
and
(b) has local interactions (fields multiplied at the same space-time point)
down to the spatial resolution of the experiments.
By this definition, hadrons are not pointlike, because for sufficiently
short wavelengths, one has to go to a theory with quarks and gluons in it,
instead of hadrons, to describe scattering data. An effective field theory
of hadrons seems to have local interactions, but it also has an
ultraviolet cutoff scale which is within the realm of experiment.
Strings in string theories are not pointlike because the interactions are
not local.
I think that this adequately encapsulates what particle physicists mean
when they say something is treated as a point particle.
Now, one can quibble. The interactions could look local in the Dirac
representation but *not* in the Newton-Wigner representation, or vice
versa. The former is actually the case with QED: the Dirac representation,
not the Newton-Wigner representation, is the one in which the electrons
appear "pointlike," in that the fundamental interactions in the correct
Lagrangian multiply fields at the same point (and are in fact given by the
minimal coupling prescription). In the Newton-Wigner representation, the
electron seems to have a charge radius on the order of the Compton
wavelength. As far as I know, there is no compelling reason why it
*couldn't* have been the other way around. So this definition of
"pointlike" is subject to some ambiguity.
--
Font-o-Meter! Proportional Monospaced
^
http://world.std.com/~mmcirvin/
Matt McIrvin
Rotational angular momentum is always an integer in nonrelativistic
quantum mechanics. It turns out, though, that finite wave-packet solutions
of the (relativistic) Dirac equation have a momentum density flowing
around their borders, which yields the spin angular momentum. So in the
Dirac theory, spin is not quite as mysterious as it is commonly made out
to be.*
This is not commonly known because one usually works in terms of
plane-wave solutions. However, it was apparently discovered by Belinfante
in the 1930s.
If you think about it, this stands to reason. I mean, the angular momentum
of a circularly polarized wave is supposed to correspond to photon spin in
QED, and *that* can be broken down into a circulating momentum density in
the electromagnetic field. The Dirac situation is similar, except that the
result is not an integer.
* The unusual nature of quantum spin is given a lot of stress in
introductory QM classes. I think that sometimes it is given *too much*
emphasis, because it sometimes engenders the belief that spin is not
really angular momentum, but is just something that bears a passing
mathematical similarity to it (specifically, the same commutation
relations). This is wrong; angular momentum is not conserved unless you
include spin, since it can be converted into other kinds of angular
momentum. There *are* things that bear this particular passing
mathematical resemblance to angular momentum, yet are *not* really
angular momentum. But spin is not one of them!
Brian J Flanagan
On 6 May 1997, Matt McIrvin wrote:
> In article <5kjl9k$hdg$1...@agate.berkeley.edu>, we...@orpheus.dtc.hp.com
> (Greg Weeks) wrote:
>
> > How do experiments determine whether an electron "looks like a point"?
(snip)
>
> Strings in string theories are not pointlike because the interactions are
> not local.
>
>
> Now, one can quibble.
BJ: Excuse me, but do you mean "not local" up to a certain point? Or are
you saying that string interactions are "nonlocal" in its EPR sense?
Jeremy Henty
Matt McIrvin wrote:
> In article <336EF5...@orl.co.uk>, Jeremy Henty <j...@orl.co.uk> wrote:
>
> > Greg Weeks wrote:
> >
> > > ... For example, it is natural to view electron
> > > spin not as some mystical property but simply as the angular
> > > momentum of a nonpointlike object.
> >
> > Surely not. Rotational angular momentum is always integral (in units of
> > h/(2*pi)). The elecron's spin is 1/2 (in the same units). Ergo, it
> > cannot be rotational in origin (or else I'm missing something).
>
> Rotational angular momentum is always an integer in nonrelativistic
> quantum mechanics. It turns out, though, that finite wave-packet solutions
> of the (relativistic) Dirac equation have a momentum density flowing
> around their borders, which yields the spin angular momentum. So in the
> Dirac theory, spin is not quite as mysterious as it is commonly made out
> to be.*
This is fascinating stuff I was not aware of, but I don't see how it can
entirely answer my objection. If I go to a frame in which the packet is
at rest, I can calculate (a component of) the angular momentum by
rotating it around my z-axis and seeing how it transforms. This is the
*same* in both non-relativistic and relativistic quantum mechanics.
After all, the generator of the rotation is L_z in both cases. And
surely the packet, being at rest, transforms the same way under L_z
whether or not I take account of relativity. So how can relativity make
a difference?
Of course I will get contributions from the spatial variation of the
wave function. But such contributions cannot by themselves give a
non-integral answer. Functions on spacetime *always* sit in the
integral spin representations of the rotation group. It is surely
necessary that the internal degrees of freedom of the electron
contribute the 1/2. At most I need to modify my statement by saying
that the electron's spin cannot be *solely* rotational in origin. The
angular momentum operator must be acting on the internal degrees of
freedom as well.
If it is possible to write the spin angular momentum of a wave packet as
the torque of a momentum density, then that momentum density must
(sticking my neck out here) contain contributions from the electron's
internal degrees of freedom. I'd expect that to be true in both
non-relativistic and relativistic quantum mechanics.
If you have any more details or references to the wave packet results
you mention, I'd be glad to have them.
> --
> Font-o-Meter! Proportional Monospaced
> ^
> http://world.std.com/~mmcirvin/
Very neat!
--
Jeremy Henty
john baez
Matt McIrvin wrote:
>> Rotational angular momentum is always an integer in nonrelativistic
>> quantum mechanics. It turns out, though, that finite wave-packet solutions
>> of the (relativistic) Dirac equation have a momentum density flowing
>> around their borders, which yields the spin angular momentum. So in the
>> Dirac theory, spin is not quite as mysterious as it is commonly made out
>> to be.
I am a bit confused here. How exactly are you transforming your solution
of the Dirac equation when you rotate it? It is a spinor-valued function,
right? It seems to me that the only sensible way to transform it involves
letting the rotation act on the spinor in the usual way, in addition to
dragging the function around. I.e. if you rotate the spinor-valued function
psi(x)
on space, it seems you should get
g(psi(g^{-1}x))
where the outside g acts on the spinor psi(g^{-1}x) via the usual
representation of the (double cover of) the rotation group on spinors.
Anyway, if the rotation is acting on the spinor-valued function psi(x)
in a way that involves the spin-1/2 representation, why should we
be surprised that we're going to get noninteger angular momenta here?
Also, what does this have to do with relativistic quantum mechanics?
Everything I said would apply to nonrelativistic quantum mechanics, too:
again we'd describe an electron by a spinor-valued function psi(x),
which would transform under rotations like
g(psi(g^{-1}x)).
Matt McIrvin
In article <5kpua0$il2$1...@agate.berkeley.edu>, Brian J Flanagan
<bfla...@sleepy.giant.net> wrote:
> On 6 May 1997, Matt McIrvin wrote:
> > Strings in string theories are not pointlike because the interactions are
> > not local.
[...]
>
> BJ: Excuse me, but do you mean "not local" up to a certain point? Or are
> you saying that string interactions are "nonlocal" in its EPR sense?
I probably should have been more specific.
I wasn't talking about the EPR sense of nonlocality, which applies even
to "local" quantum field theories. I mean that in string theory, the
interaction term in the Lagrangian isn't a product of fields multiplied
together at the same space-time point. The strings can interact with
each other even if they are in slightly different positions. So we
say that they are not points, that they have some nonzero size.
Greg Weeks
Jeremy Henty (j...@orl.co.uk) wrote:
: Greg Weeks wrote:
: > ... For example, it is natural to view electron
: > spin not as some mystical property but simply as the angular momentum of a
: > nonpointlike object.
: Surely not. Rotational angular momentum is always integral (in units of
: h/(2*pi)). The elecron's spin is 1/2 (in the same units). Ergo, it
: cannot be rotational in origin (or else I'm missing something).
You are using a representation with an X-coordinate plus remaining degrees
of freedom that you presume to be X-independent. This presumption is
unwarranted. A basketball has both a position and a spin -- and the spin
is associated with the motion of rubber (or whatever) in X-space.
Similarly, an electron has both position and spin degrees of freedom -- and
the spin is due to a whirling in the electron's energy-momentum tensor
(which has nonzero values in an extended region of X-space).
Greg
Thomas N. Lockyer
In article <5kskp1$qol$1...@agate.berkeley.edu> we...@orpheus.dtc.hp.com (Greg Weeks) writes:
>From: we...@orpheus.dtc.hp.com (Greg Weeks)
>Jeremy Henty (j...@orl.co.uk) wrote:
>: Greg Weeks wrote:
>: > ... For example, it is natural to view electron
>: > spin not as some mystical property but simply as the angular momentum of a
>: > nonpointlike object.
>: Surely not. Rotational angular momentum is always integral (in units of
>: h/(2*pi)). The elecron's spin is 1/2 (in the same units). Ergo, it
>: cannot be rotational in origin (or else I'm missing something).
> Stuff deleted:
>Similarly, an electron has both position and spin degrees of freedom -- and
>the spin is due to a whirling in the electron's energy-momentum tensor
>(which has nonzero values in an extended region of X-space).
I want numbers, please calculate the spin angular momentum from your
postulated energy momentum tensor.
Spin angular momentum is equal to 1/2 h bar. If you can get that number:
0.527 x 10^-34 Joule seconds, then I will believe you.
Regards: Tom: http://www.best.com/~lockyer/home3.htm
Greg Weeks
Thomas N. Lockyer (loc...@best.com) wrote:
: I want numbers, please calculate the spin angular momentum from your
: postulated energy momentum tensor.
: Spin angular momentum is equal to 1/2 h bar. If you can get that number:
: 0.527 x 10^-34 Joule seconds, then I will believe you.
Well, it just happens that I've kept the calculation lying around from
a posting 3 years ago:
--------------------------------------------------------------------------
In classical particle mechanics, angular momentum is obtained from momentum
(and position):
L(i) = epsilon(i,j,k) x(j) p(k)
In classical field theory, angular momentum density is obtained from the
energy-momentum density:
M(m,a,b) = x(a)T(m,b) - x(b)T(m,a)
In quantum particle mechanics, angular momentum is NOT obtained solely
from momentum (and position):
J(i) = epsilon(i,j,k) X(j) P(k) + S(i)
There is both an "orbital" and a "spin" contribution.
What about quantum field theory? It is usually suggested that the angular
momentum density has both an orbital and a spin part (eg, equation 3-155 of
Itzykson and Zuber; see also Ryder, Bjorken and Drell, etc):
M(m,a,b) = x(a)T(m,b) - x(b)T(m,a) + S(m,a,b)
THIS IS WRONG. In fact, just as in classical field theory, angular
momentum density is obtained solely from the energy-momentum density:
M(m,a,b) = x(a)T(m,b) - x(b)T(m,a)
Isn't that COOL? Or don't you believe me? The derivation is
straightforward, and I'm surprised I haven't seen it. It follows below.
For definiteness, the argument uses Itzykson and Zuber's "Quantum Field
Theory" text. The calculation part of the argument is no more than a
sketch, due to the limitations of ascii. (I'd be happy to help anyone
interested over the trouble spots.)
Itzykson and Zuber note that "the energy-momentum density is in principle
measurable and, moreover, is coupled to the gravitational field." It
follows that T(m,n) should be symmetric and gauge-invariant. Itzykson and
Zuber note that T(m,n) obtained directly from Noether's theorem may be
unsymmetric and gauge-variant. It is desirable to remedy this by adding a
term that doesn't affect the spatial integral of T(0,n).
Strangely enough, I&Z don't take their own advice when dealing with the
free Dirac field. Here, gauge-invariance is not an issue, but
symmetricness is. The result of Noether's theorem is [with D(n) denoting
partial differention]:
T'(m,n) = (i/2)[Psi_bar gamma(m) D(n)Psi - D(n)Psi_bar gamma(m) Psi]
The "'" in T' is to remind us that this result is not symmetric in (m,n).
Consequently, T' is not the observable energy-momentum density. For some
reason, I&Z don't take the next step and symmetrize T' (which is what they
DID do with the Maxwell field). An obvious candidate for the symmetric
energy-momentum density is:
T(m,n) = (i/4)[Psi_bar gamma(m) D(n)Psi - D(n)Psi_bar gamma(m) Psi
+
Psi_bar gamma(n) D(m)Psi - D(m)Psi_bar gamma(n) Psi]
A direct computation shows that the spatial integral of T(0,n) equals the
spatial integral of T'(0,n). So we take T to be our energy-momentum
density. [I've since encountered a better derivation of T in Weinberg's
QFT textbook.]
What about angular momentum? For now, lets assume:
M(m,a,b) = x(a)T(m,b) - x(b)T(m,a)
To test this, consider a state |s> consisting of an electron approximately
at rest with spin up (along the z-direction). What is value of the spatial
integral
x_integral[ <s| M(0,1,2) |s> ] ?
If M doesn't require an additional spin part, the spatial integral of
M(0,1,2) should be J(3), and the expectation value in state |s> ought to be
1/2. Is this the case? Lets see. Here is a SKETCH of the calculation.
For starters, consider the first term:
x_integral[ <s| x(1)T(0,2) |s> ]
Naively, the result is zero, since <s|T(0,2)|s> is zero for a particle at
rest. However, the electron can't be strictly at rest. Instead, the
momentum-space wave function is sharply peaked at zero. In this case,
<s|T(0,2)|s> is still very small, and most of the contributions to the
spatial integral vanish. But one survives! Consider the piece:
x_integral[ x(1) <s| PsiBar gamma(2) D(0)Psi |s>
(At this point we start to get sloppy about trivial multiplicative
factors.) Recalling that
Psi(x) = p_integral[u(p)b(p)exp(-ipx) + v(p)d*(p)exp(ipx)] ,
one of the contributions is:
p_integral q_integral
[ x_integral [x(1) exp(i(p-q)x)] f(p,q) u_bar(p) gamma(2) u(q) ]
where q is another 3-momentum and f(p,q) is sharply peaked around p=q=0.
The x_integral yields:
D(1) delta(p-q)
Integrating by parts (applying the D(1) to the remaining factors in the
integrand) yields various terms, most of which are negligible due to the
sharp peak in f(p,q). But one term survives. Applying the Gordon identity
(I&Z A-38) to u_bar(p) gamma(2) u(q) yields a term:
u_bar(p) sigma(3) u(q) (p(1) - q(1))
The (p(1) - q(1)) factor is removed by the D(1), and the remaining factors
do NOT vanish due to the sharp peak in f(p,q). u_bar(p) sigma(3) u(q) is
just 1. Collecting omitted factors -- in particular, the suppressed i/4
and a factor of 2 from including the second term of M -- yields
x_integral[ <s| M(0,1,2) |s> ] = 1/2
as desired.
To summarize, the spin of an electron is obtained from the angular momentum
density obtained from the _observable_ energy-momentum density in the
_classical_ manner.
Greg
PS: A better derivation would demonstrate that the spatial integral of
M(0,a,b) above equals the spatial integral of the angular momentum density
in I&Z. I haven't done the calculation, at least not in the last ten
years.
---------------------------------------------------------------------------
Not perfect. (It uses expectation values instead of eigenvalues.) But
good enough to make the point, I hope.
Greg
Jeremy Henty
In article <5kskp1$qol$1...@agate.berkeley.edu>, we...@orpheus.dtc.hp.com (Greg Weeks) writes:
|> Jeremy Henty (j...@orl.co.uk) wrote:
|> : ... The elecron's spin is 1/2 (in the same units). Ergo, it
|> : cannot be rotational in origin (or else I'm missing something).
|>
|> of freedom that you presume to be X-independent. This presumption is
|> unwarranted. A basketball has both a position and a spin -- and the spin
|> is associated with the motion of rubber (or whatever) in X-space.
|> the spin is due to a whirling in the electron's energy-momentum tensor
|> (which has nonzero values in an extended region of X-space).
But, as I argued elsewhere in this thread, if the "whirling" is solely
due to a rotation in space (ie. without affecting any internal degrees
of freedom) then it must result in integral spin. So I am *not*
presuming that the only position-dependent degree of freedom is the
X-coordinate. The electron's spin must have a contribution from
internal degrees of freedom, it cannot be *purely* rotational.
Jeremy Henty
Michael Weiss
Rutherford knew the nucleus had stuff in it because of occasional
large angle-of-deflection scatterings of the alpha particles he shot
at the Au foil. [Reference to proton scattering experiments snipped.]
Yup. (Actually, Geiger and Marsden shot the particles, and a year or
two later Rutherford came up with the nuclear model. Geiger and
Marsden were students of Rutherford at the time.)
The basic idea is that if a particle has internal constituents, then
some collisions will cause them to rattle around.
Is this supposed to apply to the Geiger-Marsden experiment, or just to
the proton scattering experiments?
The Geiger-Marsden collisions were elastic, as far as I know. The
point of the large-scale deflections was that they suggested the atom
had to have a compact core. Otherwise, the alpha particles could pass
right through, suffering minimal deflection.
Greg Weeks
Jeremy Henty (j...@orl.co.uk) wrote:
: But, as I argued elsewhere in this thread, if the "whirling" is solely
: due to a rotation in space (ie. without affecting any internal degrees
: of freedom) then it must result in integral spin.
It might be worth noting that:
The "whirling" does not correspond to the motion of any material substance
in space. It is exactly analogous to the energy-momentum of the classic
electromagnetic field. There is energy-momentum flow, but there is no
underlying stuff that is flowing. In the case of the electron, if there
WERE some underlying stuff that was flowing, then the momentum flow T[0,.]
would be proportional to the electric current J, and the gyromagnetic ratio
g would be 1.
Greg
Jeremy Henty
|> The basic idea is that if a particle has internal constituents, then
|> some collisions will cause them to rattle around. This rattling
|> energy and momentum will apparently be lost as far the incident and
|> scattered particles are concerned.
No it won't. The energy of motion of the internal constituents will
manifest itself in the mass of the collision products. Only if some
of the products go undetected will energy appear to be lost. This has
nothing to do with whether the particles involved have internal
constituents.
--
Jeremy Henty
Alison Chaiken
j...@orl.co.uk (Jeremy Henty) writes
>Only if some of the products go undetected will energy appear to be
>lost.
I agree. I was thinking of an experiment where the internal degrees
of freedom of a composite particle have been excited. Eventually the
internal degrees of freedom will relax to their ground state,
presumably by the emission of radiation or of a particle. I'm talking
about the case where the final relaxation is not observed as part of
the scattering experiment. This was true, for example, in
Rutherford's experiment; the gamma ray associated with de-excitation
of the nuclear vibration was not observed as part of his experiment,
which detected only the scattered alpha particles. Yet the fact that
the nucleus in the Au foil had internal constituents that were excited
was clearly implied by his experimental results.
>The energy of motion of the internal constituents will manifest
>itself in the mass of the collision products.
True, but Rutherford didn't weigh the nucleus. Neither, so far as I
know, did Hofstadter weigh all the constituents when he discovered the
proton. I was trying to point out that an *apparent* non-conservation
of energy and momentum in a scattering is a sign of structure in a
particle. I wasn't trying to imply that energy and momentum actually
aren't conserved.
--
Alison Chaiken ali...@wsrcc.com
(510) 422-7129 [daytime] http://www.wsrcc.com/alison/
Everywhere I hear, "You're just an engineer," and life goes on without me.
Jeremy Henty
Apologies for the delay in replying: real life (i.e. work) keeps
interfering.
Greg Weeks wrote:
> ... Here is all that I
> really wanted to say: In QED, electron spin is the at-rest
> angular-momentum, and the angular momentum is obtained from the
> energy-momentum tensor field exactly as in classical field theory.
>
> If I'd said only that in the first place, would there have been a
> disagreement?
No, the formal relationship between the stress-energy tensor and the
angular momentum is the same in each case. But it seems to me you
have retreated a little from your original position, which was IIRC
that the spin of an electron can be regarded as purely rotational in
origin. (Perhaps I jumped the gun in criticising you before it was
sufficiently clear what you were saying.) The point I made previously
about the angular momentum of an electron also applies to its
stress-energy: it must explicitly involve internal degrees of freedom
in a way that is not true of a scalar object (whether or not that
object is pointlike or extended). So there *is* a vital difference
between the two cases, although it does not appear in the relationship
between stress-energy and angular momentum. And there *must* be such
a difference, because the electron has half-integer spin.
An excellent point. But doesn't it argue *against* the idea that the
electron can be treated as an extended rotating object?
Regards,
--
Jeremy Henty
Jeremy Henty
Alison Chaiken <ali...@wsrcc.com> wrote:
> > <snip> in
> Rutherford's experiment; the gamma ray associated with de-excitation
> of the nuclear vibration was not observed as part of his experiment,
> which detected only the scattered alpha particles. Yet the fact that
> the nucleus in the Au foil had internal constituents that were excited
> was clearly implied by his experimental results.
True. But that did *not* follow from conservation of energy. It
followed because the momentum transfer Rutherford observed was so
large.
> >The energy of motion of the internal constituents will manifest
> >itself in the mass of the collision products.
>
> True, but Rutherford didn't weigh the nucleus. Neither, so far as I
> know, did Hofstadter weigh all the constituents when he discovered the
> proton. I was trying to point out that an *apparent* non-conservation
> of energy and momentum in a scattering is a sign of structure in a
> particle.
But there cannot be an apparent violation of violation of conservation
of energy if there constituents are not weighed! That's like an
accountant saying "the books don't appear to balance" without having
calculated the income.
Furthermore it's not true that apparent non-conservation of energy
says anything about structure. The loss of energy during beta decay
was not a sign of structure, it was a sign of a hitherto unobserved
particle: the neutrino. Rutherford did not observe non-conservation
(apparent or otherwise), he observed momentum transfer, which is a
different thing.
> Everywhere I hear, "You're just an engineer,"
You won't hear it from me! I admire engineers!
--
Jeremy Henty
Jim Carr
Alison Chaiken <ali...@wsrcc.com> wrote:
}
} <snip> in
} Rutherford's experiment; the gamma ray associated with de-excitation
} of the nuclear vibration was not observed as part of his experiment,
} which detected only the scattered alpha particles. Yet the fact that
} the nucleus in the Au foil had internal constituents that were excited
} was clearly implied by his experimental results.
j...@cam-orl.co.uk (Jeremy Henty) writes:
>
>True.
Not true. The classic "Rutherford" experiment was elastic
scattering of alpha particles. It did not excite the nucleus.
In addition, the range of momentum transfer explored was too
narrow to say much more than that the nucleus is consistent
with scattering from a point charge. A radius came later.
Another experiment, this one done by Rutherford as I recall,
was an (alpha,p) transfer reaction that definitely showed there
were constituents in the nucleus. Of course, that was not
controversial given the observation of alpha decay.
>But that did *not* follow from conservation of energy. It
>followed because the momentum transfer Rutherford observed was so
>large.
Not quite. It was because the cross section at that momentum
transfer was so large. In modern language, that says that the
transition density extends out in q-space so it must be small
in r-space. Rutherford did not go far enough out to see a node
or slope of the q-space distribution to be able to deduce a radius.
>Furthermore it's not true that apparent non-conservation of energy
>says anything about structure. The loss of energy during beta decay
>was not a sign of structure, it was a sign of a hitherto unobserved
>particle: the neutrino.
Beta decay, which takes you from one ground state to another,
does not necessitate thinking about internal structure like
seeing excited states does, although it is obvious in retrospect,
just as was the case in particle physics.
--
James A. Carr <j...@scri.fsu.edu> | "What are those, Daddy?"
http://www.scri.fsu.edu/~jac/ | Young girl at Smithsonian
Supercomputer Computations Res. Inst. | exhibit of objects and letters
Florida State, Tallahassee FL 32306 | left at the Vietnam Memorial.
Douglas A. Singleton
In article <5llrb3$1dn$1...@agate.berkeley.edu>,
Jeremy Henty <j...@orl.co.uk> wrote:
>Apologies for the delay in replying: real life (i.e. work) keeps
>interfering.
>
>Greg Weeks wrote:
>
>> ... Here is all that I
>> really wanted to say: In QED, electron spin is the at-rest
>> angular-momentum, and the angular momentum is obtained from the
>> energy-momentum tensor field exactly as in classical field theory.
>>
>> If I'd said only that in the first place, would there have been a
>> disagreement?
>
>No, the formal relationship between the stress-energy tensor and the
>angular momentum is the same in each case. But it seems to me you
>have retreated a little from your original position, which was IIRC
>that the spin of an electron can be regarded as purely rotational in
>origin. (Perhaps I jumped the gun in criticising you before it was
>sufficiently clear what you were saying.) The point I made previously
>about the angular momentum of an electron also applies to its
>stress-energy: it must explicitly involve internal degrees of freedom
>in a way that is not true of a scalar object (whether or not that
>object is pointlike or extended). So there *is* a vital difference
>between the two cases, although it does not appear in the relationship
>between stress-energy and angular momentum. And there *must* be such
>a difference, because the electron has half-integer spin.
>
You may want to take a look at the article "What is Spin ?" by
Hans Ohanian (Am. J. Phys. Vol. 54 pg. 500 (1986)) which is basically
in accord with what Greg Weeks is saying. Since I just got back
from vacation and am still a bit lazy let me just quote from the
abstract instead of doing the proper thing and outlining the idea
(besides I bet Greg Weeks already gave a sketch of this).
"According to the prevailing belief, the spin of the electron or of
some other particle is a mysterious internal angular momentum for which
no concrete physical picture is available, and for which there is no
classical analog. However, on the basis of an old calculation by
Belinfante, it can be shown that the spin may be regarded as an angular
momentum generated by a circulating flow of energy in the wave field
of the electron ..."
In the conclusion Ohanian specifically says that spin is not an
internal feature of the electron, but that it is an intrinsic
feature of the wave equation which the electron is thought to
obey. In other words, pick a different wave equation and you'll
get a different answer for the internal spin. Actually just take a
look at the Ohanian article he explains this much better than I just
did.
>> The "whirling" does not correspond to the motion of any material substance
>> in space. It is exactly analogous to the energy-momentum of the classic
>> electromagnetic field. There is energy-momentum flow, but there is no
>> underlying stuff that is flowing. In the case of the electron, if there
>> WERE some underlying stuff that was flowing, then the momentum flow T[0,.]
>> would be proportional to the electric current J, and the gyromagnetic ratio
>> g would be 1.
Hmm. I would regard the wave field as "underlying stuff" and in the
last section of Ohaian's article he shows "Likewise, the magnetic
moment may be regarded as generated by a circulating flow of
charge in the wave field" (again being lazy and quoting from the
abstract). In other words one also gets the g=2 correct in this way.
BTW is the Dirac equation the only equation which gives g=2 ?
(I know the answer to this and surprisingly - well surprising to
me anyway - the answer is no).
>An excellent point. But doesn't it argue *against* the idea that the
>electron can be treated as an extended rotating object?
>
Doug
Greg Weeks
Jeremy Henty (j...@orl.co.uk) wrote [in e-mail, actually]:
: In QFT the electron is described in exactly the same way as a scalar
: particle, except that there are extra internal degrees of freedom. The
: electron's spin derives from the fact that the angular momentum operator
: contains a *new* term that acts on those internal degrees of freedom.
You're viewing the angular momentum operator as it acts on single-particle
states in the "position"+spin representation. This represention does not
precisely mean what you think it does. Here's why:
In nonrelativistic quantum (particle) mechanics, the electron is viewed as
a point particle with a well-defined concept of position. In this case
there is no question that the electron is a point. The theory is
_formulated_ in terms of pointlike particles.
If you move to the simplest examples of relativisic quantum mechanics, you
obtain the irreducible representations of the Poincare group. These have
natural momentum and spin operators. But there is no particularly natural
position operator. It may seem that the position representation may be
obtained simply by Fourier transforming the momentum-space wave functions.
But at this point in the discussion there are no momentum-space wave
functions! The momentum-space wave function of a state |S> is defined as
<p|S>. But the momentum operator P determines the <p| vectors only up to a
phase. How can we fix these phases?
One answer is the following: Consider the finite-dimensional "little
Hilbert space" of (pseudo)states with p=0. This space is preserved by the
rotation group, which acts irreducibly on the space (since we assumed an
irreducible representation in the first place). Choose a natural
represention of the little Hilbert space, eg:
|P=0 Sz=1/2> and |P=0 Sz=-1/2>
(Here, S is the angular momentum operator on the little Hilbert space, Sz
it its z-component, and the particle has spin 1/2).
Now, for a given mass and any momentum p, there is a "simplest" element
A(p) in the Lorentz group that boosts a particle at rest to momentum p. We
may then define momentum space wave functions using the basis DEFINED by:
|P=p Sz=sz> = A(p) |P=0 Sz=sz> [LHS defined as RHS]
Now that we have a basis, we can DEFINE "position" be requiring
"position"-space wave-functions to be Fourier transforms of momentum-space
wave-functions.
The "position" operator defined in this way is known as the Newton-Wigner
position. If you insist on having a position operator, this is the most
natural choice. But is there any reason to insist a priori that an
electron has a position? In nonrelativisic quantum mechanics, you had no
choice. But perhaps in real life the concept of electron position is only
approximately defined.
This is in fact a natural point of view, for a handful of reasons:
1. Intuitively, an electron at position x will have all its charge at
position x. But this is not the case with the Newton-Wigner position
operator (nor, I would wager, with any other reasonable candidate for
"position").
2. From Haag's "Local Quantum Physics": Newton-Wigner localization is
dependent on the Lorentz frame. A state strictly localized at t=0 at the
point x=0 is not strictly localized at any time viewed in a different
Lorentz frame. However, for a massive particle, the ambiguity in defining
the localization is small, namely of the order of the inverse electron
mass.
3. The nonrelativistic QFT of macroscopic materials also has "particles"
known as phonons. No one (I hope) would say that phonons are point
particles, even though there is no doubt an approximate notion of phonon
localization.
4. A priori assumptions are to be distrusted.
So, in the absence of presuppositions, it is natural to view electrons as
being localizable down to the precision of the inverse electron mass, but
no further. (In the nonrelativistic limit, this precision becomes
infinite, hence the pointlike particles in quantum (particle) mechanics.)
So, the "position" in the "position"+spin representation only approximately
corresponds to what we think of as the position of the electron. The
notion of electron position makes sense only as an approximation.
Therefore the distinction between external degrees of freedom (quantities
observed in space-time, eg: position, momentum, and their cross product)
and internal degrees of freedom (quantities that are not observed in
space-time, eg, spin) is not a fundamental one. Fundamentally, all the
observable attributes of an electron (eg, its "position", momentum, and
spin) may be obtained from observable quantum fields defined on space-time
(specifically, from the energy-momentum density). In this sense, all
degrees of freedom of the electron are external.
Of course, people still refer to spin as an "internal" degree of
freedom, which makes sense in the nonrelativistic limit. However, the
nonrelativistic limit does not determine whether or not an electron is
pointlike.
Greg
Matt McIrvin
Douglas A. Singleton <da...@erwin.phys.virginia.edu> wrote:
> BTW is the Dirac equation the only equation which gives g=2 ?
> (I know the answer to this and surprisingly - well surprising to
> me anyway - the answer is no).
In J. J. Sakurai's _Advanced Quantum Mechanics_, he gets g=2 by a
nonrelativistic trick apparently introduced by Feynman, in which you
first interpret the Laplacian in the Schrodinger equation as the square
of sigma dot grad, then invoke minimal coupling. (sigma dot grad)^2
is the same thing as grad^2 as far as the free wave equation is
concerned, but they lead to different interactions when the vector
potential is introduced by the minimal coupling recipe.
If this seems arbitrary, really it is no more arbitrary than what we do
in *relativistic* QED. The choice of the Dirac representation over the
Newton-Wigner representation as what we use when invoking minimal
coupling in QED is exactly the same kind of choice. The Dirac rep, or in
the nonrelativistic limit the choice of (sigma dot grad)^2, is simply
the choice that happens to give the right answer for the electromagnetic
coupling from the minimal-coupling prescription. Do a Foldy-Wouthuysen
transformation on QED and what you get no longer looks "minimally
coupled."
g=2 is often described as a consequence of the application of relativity
to QM, but really it is more a consequence of a particular choice of
wave equation, plus minimal coupling. Treatments of gauge theory often
describe QED and similar theories as falling out of the free wave
equation with a certain inevitability, given the assumption of gauge
symmetry, but this is an additional choice that is often swept under the
rug; minimal coupling is not really uniquely defined.
Greg Weeks
Greg Weeks (we...@orpheus.dtc.hp.com) wrote:
: In nonrelativistic quantum (particle) mechanics, the electron is viewed as
: a point particle with a well-defined concept of position. In this case
: _formulated_ in terms of pointlike particles.
I perhaps should have mentioned that the same is true of the RELATIVISTIC
quantum (particle) mechanics that you get when you (mis)interpret the Dirac
equation as a wave-function rather than as a field. There are indeed at
least 4 quantum-theoretical views of the electron floating around:
A) nonrelativistic quantum mechanics
B) the Dirac equation viewed as a wave-function
C) irreducible representations of the Poincare group
D) quantum field theory
Only C and D are thought to be correct, and those are the views in which an
electron is only approximately pointlike. (Actually, in view C, there is
no indication of electron localization at all as far as I can see.)
Greg
James Logajan
Greg Weeks (we...@orpheus.dtc.hp.com) wrote:
: There are indeed at
: least 4 quantum-theoretical views of the electron floating around:
: A) nonrelativistic quantum mechanics
: B) the Dirac equation viewed as a wave-function
: C) irreducible representations of the Poincare group
: D) quantum field theory
: Only C and D are thought to be correct, and those are the views in which an
: electron is only approximately pointlike. (Actually, in view C, there is
: no indication of electron localization at all as far as I can see.)
As the person who posed the original question, I appreciate the answers and
discussion, even if I don't understand some of what is posted (but I'm
trying!)
Of the four quantum-theoretical views listed, I can only claim some knowledge
of 'A'. It isn't clear to me if the answer to the following question is
the same for all four views:
If two electrons, P and Q, collide at some collision energy E, is there
any value of E for which the recoil is not determined by assuming a 1/r
potential for the two electrons?
The answer to this for 'C' and 'D' (assuming the question is well
posed for those quantum-theoretical views) eludes me. On the other hand,
I have been informed that there are experimental hints that suggest the
answer to the question is yes.
Greg Weeks
James Logajan (jam...@netcom.com) wrote:
: If two electrons, P and Q, collide at some collision energy E, is there
: any value of E for which the recoil is not determined by assuming a 1/r
: potential for the two electrons?
At any energy above the hard photon generation threshold, I don't think a
1/r potential alone can be used to derive the differential cross sections
for the inelastic collisions.
I'd be curious to know the answer for the elastic collisions. For example,
there is a contributing Feynman diaagram that looks like electron2 collides
not with electron1 but with the electron in a virtual electron-positron
pair emitted and absorbed by electron1. (A two-photon exchange is required
to circumvent Furry's theorem.) This diagram indicates that the electron
has all sorts of structure, but that the probability of probing it
elastically is tiny. (Or so it seems to me.)
Greg
Matt McIrvin
Greg Weeks <we...@orpheus.dtc.hp.com> wrote:
> Greg Weeks (we...@orpheus.dtc.hp.com) wrote:
> : In nonrelativistic quantum (particle) mechanics, the electron is viewed as
> : a point particle with a well-defined concept of position. In this case
> : there is no question that the electron is a point. The theory is
> : _formulated_ in terms of pointlike particles.
>
> I perhaps should have mentioned that the same is true of the RELATIVISTIC
> quantum (particle) mechanics that you get when you (mis)interpret the Dirac
> equation as a wave-function rather than as a field.
Actually, things start to get complicated as soon as you do that. It's
impossible to make an arbitrarily localized wave-function configuration
unless you include the negative-energy components, which opens a huge
can of worms: have you *really* got an electron? It is, of course, the
foot in the door of the idea that the wave-function interpretation
doesn't really work, that QED can really only be handled with a
multiple-particle theory. Electron localization is almost as screwed up
in the Dirac wave theory as it is in real live quantum field theory.
> There are indeed at
> least 4 quantum-theoretical views of the electron floating around:
>
> A) nonrelativistic quantum mechanics
> B) the Dirac equation viewed as a wave-function
> C) irreducible representations of the Poincare group
> D) quantum field theory
>
> Only C and D are thought to be correct, and those are the views in which an
> electron is only approximately pointlike. (Actually, in view C, there is
> no indication of electron localization at all as far as I can see.)
There are different notions of point-like-ness. You seem to be
identifying it with the existence of an arbitrarily precise position
operator, which is one way to think about it, but not the way that
particle physicists (especially experimentalists) usually talk about it.
The electron is precisely pointlike in QED in the following sense: that
QED is based on an electron field that has local interactions, that is,
the electron field (*if* written down in the Dirac way) is multiplied by
other fields at a single space-time point. Of course, this induces all
sorts of mathematical difficulties, because the operation is not really
a legitimate one to do with operator-valued distributions, which
introduces the whole headache of ultraviolet divergences. So one
introduces a short-distance regularization cutoff, and takes limits
after renormalizing. The theory could be thought of as a theory of
finite-size electrons, in the limit where the size goes to zero.
The particle physicist's notion of "pointlike" objects within some
theory is that, in the domain covered by the theory, the particle is the
quantum of a field with this kind of local interaction in the
Lagrangian. The Standard Model, in this sense, treats electrons as
pointlike but not protons. One can write down an effective theory, good
in some limit, that represents protons as pointlike, but at some scale
within the domain of the Standard Model, on the order of a fermi, it is
necessary to get rid of the proton field and replace it with quark and
gluon fields.
If string theory is correct, nothing is pointlike; string theory doesn't
have fields in the usual sense. Presumably it would have something along
the lines of what John Baez mentioned once, a sort of field defined on
the loop space of spacetime; the interactions of this thing would not be
quite local. The strings have some nonzero size and can touch even when
they're in slightly different spots.
Greg Weeks
Matt McIrvin (mmci...@world.std.com) wrote:
: The particle physicist's notion of "pointlike" objects within some
: theory is that, in the domain covered by the theory, the particle is the
: quantum of a field with this kind of local interaction in the
: Lagrangian.
Oooh, where do I begin to disagree? :-)
1. This notion of "pointlike" is independent of whether or not electrons
are pointlike in any geometric sense!
2. Are photons pointlike? I don't think Rudolph Haag would say so. Are
phonons pointlike? I wouldn't be surprised if no one would say so.
3. Would you really want to call a soliton particle in a fundamental
theory with local interactions a pointlike particle? (I'm assuming here
that a soliton can be created by the fundamental field. I might be wrong
about that.)
4. I believe there are theories with two species of particle that can be
formulated (with local interactions) in such a way that either of the two
particles may be viewed as fundamental with the other being a bound state.
Nevertheless, if there is general agreement that "pointlike" means "created
by a fundamental field in a local quantum field theory", then electrons are
"pointlike".
The question then remains: Given that electrons are "pointlike", are they
pointlike (in the geometric sense)? People often assume that they are,
viewing spin as essentially nongeometric and attributing renormalization to
the infinite Coulomb self-energy of a point charge. This is the view I've
been arguing against. And given that electrons are not pointlike, I would
also argue that it is confusing to define them to be "pointlike".
Greg
Jeremy Henty
In article <5mg0u5$1...@agate.berkeley.edu>, we...@orpheus.dtc.hp.com
(Greg Weeks) writes:
|> Jeremy Henty (j...@orl.co.uk) wrote [in e-mail, actually]:
|> : In QFT the electron is described in exactly the same way as a scalar
|> : particle, except that there are extra internal degrees of freedom. The
|> : electron's spin derives from the fact that the angular momentum operator
|> : contains a *new* term that acts on those internal degrees of freedom.
|>
|> You're viewing the angular momentum operator as it acts on single-particle
|> states in the "position"+spin representation. This represention does not
|> precisely mean what you think it does. Here's why:
|>
|> <snip fascinating discussion of the Newton Wigner position operator>
Greg, you're certainly right that there's more to the idea of
"position" in relativistic quantum theory than I had realised. But
it's not directly relevant to what I'm saying (though I'm still going
to learn it up). What I argued was that you can't explain away the
weirdness of an electron's spin by considering it as an extended
object. Well, I'm no longer so sure of that, as I see (thanks largely
to your efforts) that the subject is much more subtle than I realised.
But you seem to think that I argued that you *cannot* consider the
electron as an extended object. That's *not* my position. I feel (I
won't say I am certain until I've studied all this a bit more) that
even if you do think of the electron as extended, that will *not*
eliminate the weirdness.
Perhaps we should try and settle just what the "weirdness" of the
electron is supposed to be. For me there are two "weird" facts about
electrons: (1) rotating the electron through a full turn reverses the
sign of the electron's wave function, (2) the rotation operator not
only rotates the electron in space, it also mixes the internal degrees
of freedom of the electron as well. Both these facts are directly
related to the spin of the electron (almost by definition in case
(2)). Now I *think* that you can't explain this behaviour *just* by
thinking of the electron as extended. After all, all the comments
you've made about relativistic quantum theory apply equally well to
spin zero particles as to electrons, yet spin zero particles do not
exhibit (1) or (2). So the kind of extension you are discussing does
not seem to be sufficient. You have to add something else.
I'll also point out that both (1) and (2) have direct physical
consequences: (1) has been demonstrated for neutrons by the neutron
interferometer, (2) manifests itself in that there are two electron
states of every momentum (which is what I *mean* by the electron
having internal degrees of freedom, note that this does require me to
assert that the electron "is a point" in any sense) and in the angular
dependence of scattering amplitudes (minor handwave: I've not done the
calculations, but surely you cannot determine the spin of particles
from the results of scattering experiments unless this is so). So I
don't think you can dismiss (1) and (2) as merely features of whatever
representation I'm using.
Hopefully this clarifies things a bit.
Regards,
Jeremy
Greg Weeks
Jeremy Henty (j...@orl.co.uk) wrote:
: Perhaps we should try and settle just what the "weirdness" of the electron
: is supposed to be. For me there are two "weird" facts about electrons: (1)
: rotating the electron through a full turn reverses the sign of the
: electron's wave function, (2) the rotation operator not only rotates the
: electron in space, it also mixes the internal degrees of freedom of the
: electron as well.
I don't see the problem with (2). A basketball has both a position and a
spin. Applying a classical rotation to a basketball typically affects both
its position and its spin. I don't see either how this is weird or how
it differs significantly from an electron.
(1) is a deeper issue. Take a deep breath -- here we go:
Suppose that I have a "source" apparatus that can serve up ensembles of
electrons in any state I want. Suppose I have a "detector" apparatus that
can determine the quantum mechanical states of the electrons.
I can assign a grad student the task of using the detector to determine the
state of the electron and to represent the result as a vector in the
Hilbert space, eg, as a wave-function. If I assign the same task to a
different student, I might get a different wave-functions. But the two
wave functions will differ at most by an overall phase.
Now, at night when the grad student is asleep, I can rotate the detector.
(Note that source apparatus and the electron are undisturbed.) The grad
student will then come up with a different wave-function. According to a
theorem by Wigner, the mapping from the the original wave-functions to the
rotated wave-functions may be represented by a unitary operator, uniquely
determined up to a phase.
There is no natural way to fix these phases!
However, for infinitesimal rotations, the phases may be fixed by requiring
the effects of successive rotations to satisfy the same multiplication law
as the rotation group. (This result is also due to Wigner.) If I build up
a finite rotation from infinitesimal rotations, the phase ambiguity is
eliminates so long as I keep track of the path of rotations built up from
the successive application of the infinitesimal rotations.
So, the unitary operator associated with a rotation is well-defined only if
you allow it to depend on the path of intermediate rotations. Furthermore,
the results are the same for all paths that may be continuously deformed
onto each other. (Wigner again.) So, the unitary operators associated
with rotations are well-defined not on the rotation group SO(3), but on its
covering group SU(2). Furthermore, the unitary operator associated with
the nontrivial path from the identity to the identity equals -1.
As Haag phrased it, "the occurrence of the covering group results from the
fact that states correspond to rays rather than vectors in H".
Note again that the electron itself has not been rotated. Only the
detector was moved. If you rotate the detector only a little bit at a time
AND you fix the phase ambiguity (in a way that requires you to consider the
other two rotation axes!), then after 360 degrees your operator is -1
instead of one.
Please note:
The detector gives exactly the same results (statistically, of course)
in its original position as it does after it is rotated 360 degrees.
The -1 does not arise from the detector changing its behavior in any
way. It arises from keeping track of intermediate results and fixing
the intermediate phases in a particular way.
The effect achieved by rotating the detector apparatus can also be achieved
by rotating the source apparatus. Clearly, the electrons coming out of the
source apparatus are no different after the apparatus has been rotated
through 360 degrees!
Now things get more interesting. Evidentally there are dynamical processes
applicable to electrons (or is it neutrons?) that are represented by the
same unitary operators as are obtained by infinitesimally rotating the
detector (or source) apparatus and fixing phases in the way described
above. In other words, there are physically ways of "grabbing" an electron
and infinitesimally rotating it in a way that satifies the rotation group
multiplication law. Therefore the -1 above can be demonstrated physically
by some sort of beam-splitting + dynamical-rotation + beam-merging. Cool!
Does all this say something weird about the electron?
To me it says something interesting about the rotation group. It is also
strange that the distinction between SO(3) and SU(2) -- ie, between
rotations and _paths_ of rotations -- manifests itself for some kinds of
particles but not others. But I'm not convinced -- one way or the other --
that there is anything strange about the fact that, for some particles, the
distinction between SO(3) and SU(2) _does_ manifiest itsself.
I've had quite a time thinking about this, and I'm far from settled in my
thoughts. I appreciate the stimulation.
Greg
Greg Weeks
Jeremy Henty (j...@orl.co.uk) wrote:
: For me there are two "weird" facts about electrons: (1) rotating the
: electron through a full turn reverses the sign of the electron's wave
: interferometer
How indeed does the neutron receive this full turn?
As mentioned in a previous posting, Haag evidentally sees nothing profound
in the appearance of -1 from a 360 degree rotation: "The occurence of the
covering group results from the fact that states correspond to rays rather
than vectors of H." (Section I.3.1 of "Local Quantum Physics".) I would
be inclined to agree except for a curious fact: The generators of the
representation of the covering group SU(2) are physically significant in
their own right. Furthermore, under some circumstances, time evolution
acts to some degree like these SU(2) "rotations"; ie, the Hamiltonian apes
(a component of) the angular momentum. And further still, this SU(2)
"rotation" may be applied to one term of a vector superposition, making the
-1 observable.
Or so I've heard. But what are the details? How experimentally do you
grab (part of) a neutron and rotate it?
Greg
Greg Weeks
Jeremy Henty
Posted and mailed
I'm expecting this to be my last post on this thread. *Not* because I
think it's a waste of my time, but for the opposite reason: the
arguments have reached the limits of my present understanding and I
need to go away and learn and think a bit more before I can add
anything useful. I can, however, answer one of Greg's questions,
hence this post. Thanks to Greg and to everyone else who contributed
to a fascinating thread.
In article <5nnifm$b...@agate.berkeley.edu>, we...@orpheus.dtc.hp.com
(Greg Weeks) writes:
|> Jeremy Henty (j...@orl.co.uk) wrote:
|> : For me there are two "weird" facts about electrons: (1) rotating the
|> : electron through a full turn reverses the sign of the electron's wave
|> : function, ... (1) has been demonstrated for neutrons by the neutron
|> : interferometer
|>
|> How indeed does the neutron receive this full turn?
With a magnetic field. The experiment I'm thinking of used a neutron
interferometer. It's quite a while ago, so the details may be vague,
but here goes. The neutron interferometer is a perfect crystal of
silicon, about 10cm long by a few cm wide. It's cut into the shape of
a flat base with three ears rising vertically upwards, one at each end
of the base and one in the middle. The ears extend across the entire
width of the base. Because the crystal is perfect, the atomic planes
in any one ear are perfectly parallel to those in the others.
A beam of neutrons is fired at one end at an angle to long axis of the
base (which is also the normal to the face of the first ear). It
splits into two at the far face of the ear: one beam travelling
parallel to the direction of the incident beam, and the other relected
off the normal to the ear. This happens again at the next two ears.
I'd been hoping to avoid ascii art, but it looks like this (neutrons
enter at the top and leave at the bottom):
\
\
Ear 1 --------
/\
/ \
Ear 2 --------
/\ /\
/ \/ \
Ear 3 --------
/\
/ \
? ?
The interesting point is the direction of the neutrons emerging from
the centre of the third ear. Along which of the two paths marked '?'
do they emerge? It turns out, for this "vanilla" setup, interference
ensures they all emerge along only one of them (exercise for the
reader: which?, I forget). In other words, there is completely
constructive interference along one of the exit paths, and completely
destructive interference along the other.
Now we rotate one of the interfering neutron beams with respect to the
other by introducing a magnetic field along one of the two paths
leading to the centre of the third ear. This rotation alters the
interference, changing the proportions of neutrons appearing on the
two exit paths. When the phase of the neutron wave function is
*reversed*, we get destructive interference where we first had
constructive, and vice versa, so all the neutrons emerge on the
*other* exit path (relative to the results of vanilla setup). And, lo
and behold, when the neutrons are rotated through a *full* turn, this
is what happens! It takes *two* full turns to restore the original
interference pattern.
Hope this helps,
Jeremy
--
Jeremy Henty
Paul Arendt
Greg Weeks wrote:
>
>As mentioned in a previous posting, Haag evidentally sees nothing profound
>in the appearance of -1 from a 360 degree rotation: "The occurence of the
>covering group results from the fact that states correspond to rays rather
>than vectors of H." (Section I.3.1 of "Local Quantum Physics".)
Interesting quote... I think I would disagree with Haag, however, unless
I'm missing something:
In QM, we have probablities given by (complex) squares of wavefunctions;
this is what allows "wavelike" behavior, i.e. interference. Let's consider
a one-component (scalar) wavefunction: multiplication by exp(i t) (t real)
gives the same wavefunction. We can expand this to include multiplication
by ANY complex number, provided we "renormalize" whenever making a physical
measurement. So the fact that states correspond to rays rather than vectors
really says there are some freedoms we can exploit:
1) the U(1) rotation... which allows for electromagnetic behavior
2) Absolute scaling: the "renormalization group"
Let's now make our waveFUNCTION an actual function on some coordinate
space (manifold): R^3. We can consider the action of SO(3) on this
manifold: rotate our axes (or the space) about the origin and see how
our wavefunction transforms. Well, SO(3) is a compact Lie group, so
we can exploit a whole bunch of math to conclude that ANY function breaks
into the "irreducible representations" of the group: these are a set
of functions which only mix amongst themselves during rotations. The
wonderful thing is, the irreps. form a complete and orthonormal set,
just like Fourier transforms (which come from translations on the line).
From compactness, we also can make these finite-dimensional and unitary.
The number of functions is the "same" as the number of elements in the
group... a (3-parameter) continuum! So we go to the Lie algebra to
construct the irreps. of the Lie group. From then on, it's "easy"...
(Campbell-Baker-Hausdorff again!) and one gets representations of each
integer dimension:
dim = {1,2,3,4,5,...}
These are usually labeled by "spin j" instead, where dim = 2j+1.
However, a Lie algebra can be shared by more than one group, and in
fact we have found the irreps. of the "universal covering group", called
Spin(3) = SU(2). Locally, the groups are the SAME...
>(GW): ........ I would
>be inclined to agree except for a curious fact: The generators of the
>representation of the covering group SU(2) are physically significant in
>their own right. ~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~
Well, "same" is a bit strong; they're actually "locally isomorphic". But
that means that we can use the same group generators, anyway.
Enough digression. Globally, the groups are topologically different, and
only the odd-dimensional ones can be put single-valuedly on a space
acted upon by SO(3). Scalar fields don't intrinsically rotate, vector
and above fields do: draw a vector field in 3-space, and rotate your axes.
The vectors "move" to the rotated location, and they also turn... this
is what I mean by "intrinsic".
What about the even-dimensional irreps? Let's try to put a spin-1/2
function on R^3... just assign two numbers to every point, and require
that they transform correctly under SO(3)... you can't do it. But if
you assign two numbers _and their negatives_ to each point, you can
satisfy the requirements of the spin-1/2 representation. This is
called a "double-valued" representation, and it is that way for all
the even-dimensional irreps. So you can think of fermion fields
as _either_ single-valued reps. of SU(2) or double valued reps. of
SO(3). (Think of sqrt(z) mapping to +/- sqrt(z) simultaneously
to remain continuous under rotations in the complex plane, for what it
means to be "double-valued.")
Why does this work? Back to QM again: physical measurements _all_
have 2 "\psi" functions in them. When taking the product of two
even-dimensional irreps, we decompose into only odd-dim. irreps.,
which can be physical things in 3-space. (For example, a spin-1/2
times another spin-1/2 gives a vector plus a scalar.) And so...
Conclusion: even products of "double-valued" irreps. can be physical, and
we must allow for them to keep things completely general! QM can exploit
these half-integer spins in a way that classical mechanics could not.
I would therefore say it is the fact that wavefunctions are "square roots"
of probabilities that makes us consider SU(2) in good 'ol normal 3-space!