Free Energy?
Ron Maimon
|> However, just because a model makes no guess at what these parameters
|> really are doesn't mean they aren't there. Electrons *ARE* physical
|> beasts. When we describe "spin" for example it isn't the electron's
|> rotation about an axis but it is something.
This is not a flame, but why do people consider the spin of an electron as
"not a spin about an axis" but the spin of a ping-pong ball as a spin
about an axis
I mean, they are described in the exact same way quantum mechanically, one
with small angular momentum, the other with large angular momentum.
Or is this just something that's supposed to make people stop thinking
classically?
Ron Maimon
Carl J Lydick
=This is not a flame, but why do people consider the spin of an electron as
="not a spin about an axis" but the spin of a ping-pong ball as a spin
=about an axis
=
=I mean, they are described in the exact same way quantum mechanically, one
=with small angular momentum, the other with large angular momentum.
No, they're NOT described "in the exact same way quantum mechanically." The
spin of a ping-pong ball is the motion of the components of the ping-pong ball.
Yes, the angular momentum of the aggregate is a useful quantity. But that is
NOT the same as the spin of an electron.
=Or is this just something that's supposed to make people stop thinking
=classically?
It's something that's supposed to make people understand that the "spin" of an
electron is NOT motion.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CA...@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL
Disclaimer: Hey, I understand VAXen and VMS. That's what I get paid for. My
understanding of astronomy is purely at the amateur level (or below). So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it. If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
Ron Maimon
|> In article <2b3ds5$d...@scunix2.harvard.edu>, rma...@husc9.Harvard.EDU (Ron Maimon) writes:
|> =This is not a flame, but why do people consider the spin of an electron as
|> ="not a spin about an axis" but the spin of a ping-pong ball as a spin
|> =about an axis
|> =
|> =I mean, they are described in the exact same way quantum mechanically, one
|> =with small angular momentum, the other with large angular momentum.
|>
|> No, they're NOT described "in the exact same way quantum mechanically." The
|> spin of a ping-pong ball is the motion of the components of the ping-pong ball.
|> Yes, the angular momentum of the aggregate is a useful quantity. But that is
|> NOT the same as the spin of an electron.
|>
|> =Or is this just something that's supposed to make people stop thinking
|> =classically?
|>
|> It's something that's supposed to make people understand that the "spin" of an
|> electron is NOT motion.
|>
Well, it may not be motion of the electron's constituent parts, but it most
definitely _is_ angular momentum. If I have enough electrons spinning up, I can
take an intially motionless beach ball, and start it spinning, and in the process
invert the spins of the electrons.
So in a sense this spin is a motion of the electron, not of any constituent parts,
of course, but of the electron as a whole. There is nothing wrong with calling
it spin.
Ron Maimon
Michael Weiss
This is not a flame, but why do people consider the spin of an electron as
"not a spin about an axis" but the spin of a ping-pong ball as a spin
about an axis
I mean, they are described in the exact same way quantum mechanically, one
with small angular momentum, the other with large angular momentum.
Or is this just something that's supposed to make people stop thinking
classically?
And in another message he writes:
So in a sense this spin is a motion of the electron, not of any constituent parts,
of course, but of the electron as a whole. There is nothing wrong with calling
it spin.
People DO refer to the spin of an electron, but inveigh against
PICTURING the electron as a little spinning ping-pong ball. So the real
question, how justified are these dire warnings? I agree with Ron that
they are usually overblown. Warnings, yes, but make them "non-dire".
Mathematically, the issue is representations of the group of rotations
(SO(3)). The Lie algebras of SO(3) and SU(2) are isomorphic, so in this
sense electron is spin is indeed "just like spin about an axis". But SU(2) and
SO(3) are not isomorphic, so treating the picture literally is bound to get
you in trouble if carried far enough.
Compare with all those "rubber plane" pictorial analogies in discussions of
GR. Yes, the analogy lends itself to abuse, but if you know your definite
from your indefinite metrics, mentally stretching rubber won't hurt you and
probably helps. (I can't imagine getting a real "feel" for GR without
going through the exercise at least once.)
So why do QM texts take a harder line? I think history is to blame. The
rubber sheet analogy never actually caused any confusion in the development
of GR, because the correct mathematical framework came first.
Electron spin, by contrast, bedevilled the quantum pioneers throughout the
history of the "old quantum theory". The first hint that "electron spin
doesn't inhabit R^3", so to speak, was the anomalous Zeeman effect.
Heisenberg and Land\'e, independently, introduced half-integer quantum
numbers to explain the spectral results. Heisenberg was a student in
Sommerfeld's spectral seminar at the time (so was Pauli). Sommerfeld and
Bohr were both shocked; Sommerfeld reportedly said, "If we know one thing
about quantum numbers, we know they are integers!" Pauli warned that today
we allow half-integers, tomorrow it's quarter-integers, then eighths,
sixteenths, and before you know it the quantum conditions have eroded away.
Nowadays we would say that Heisenberg ascribed spin-1/2 to the electron.
Heisenberg didn't put it that way. Still, he associated the half-integer
quantum number with an angular momentum of some kind, and the mathematics
of the time was quite up to showing that (in modern parlance) all orbital
angular momenta must be integral. Heisenberg had to wave his hands pretty
fast. He did have the spectra on his side: the splitting of certain
spectral lines into an EVEN number of finer lines showed clearly the need
for half-integer quantum numbers. (Lorentz once wrote that the splitting
of the two sodium D-lines into four and six components was "scarcely
credible".)
I think by the time Heisenberg, Pauli, Dirac, and the rest cleared up all
the confusion, the distinction between orbital angular momenta and
(half-integer) spin angular momenta seemed paramount. The just-corrected
mental error always looks like the key point. (That's why Einstein's
analysis of simultaneity deserves such credit, even though mathematically
it's just another term in the Lorentz transformation.) When Dirac and
others sat down to write textbooks, they couldn't write "Picture the
electron as a little spinning ball" without LOUD ALARM BELLS going off in
their heads. So they included dire warnings, and we've inherited this
punctilious attitude ever since.
Ron Maimon
thank you.
This makes perfect sense.
Ron Maimon
Pertti Lounesto
bilinear covariants (=observables)
\sigma = \psi\dagger\gamma_0 \psi
J_\mu = \psi\dagger\gamma_0 \gamma_\mu \psi
S_{\mu\nu} = \psi\dagger\gamma_0 i\gamma_{\mu\nu} \psi
K_\mu = \psi\dagger\gamma_0 i\gamma_{0123}\gamma_\mu \psi
\omega = -\psi\dagger\gamma_0 \gamma_{0123} \psi
where \gamma_{0123} = \gamma_0\gamma_1\gamma_2\gamma_3.
The vector J = J_\mu\gamma^\mu is the Dirac current,
and v = J/\rho is the velocity for non-zero \rho
= sqrt(\sigma^2+\omega^2).
The bivector S = 1/2 S_{\mu\nu}\gamma^{\mu\nu} is
called in some literature the SPIN bivector and in
some other literature the electromagnetic moment bivector.
The vector s = K/\rho is called the SPIN vector, when
K = K_\mu\gamma^\mu and \rho is non-zero.
These bilinear covariants satisfy certain quadratic equations
called the Fierz identities (discovered by Pauli and Kofink)
J^2 = \sigma^2+\omega^2 K^2 = -J^2
J.K = 0 J\wedge K = -(\omega+\gamma_{0123}\sigma)S
which enable us to recover the original spinor \psi up to a
phase, in case of a non-zero \rho.
A nice description of all this can be found in the paper by
J. Crawford: On the algebra of Dirac bispinor densities:
Factorization and inversion theorems. J. Math. Phys. 26
(1985), 1439-1441.
However, everything goes quite differently in case \rho = 0.
This concerns the case of Weyl spinors, where S = 0, and
Majorana spinors, where K = 0. In this null case there also
exists new kind of spinors which not have not been discussed
before, namely the case when both S and K are non-zero.
An analysis of the role of SPIN in case of Weyl, Majorana and
the NEW spinors can be found in the paper by P. Lounesto:
Clifford algebras and Hestenes spinors, Found. Phys. 23 (1993),
1203-1237.
Pertti Lounesto
Pertti....@hut.fi Institute of Mathematics
voice: +358-0-4513020 Helsinki University of Technology
fax: +358-0-4513016 SF-02150 Espoo, Finland
Lawrence R. Mead
: In article <1993Oct29....@novell.com> Dan Seeman <dse...@novell.com> writes:
: >In article <1993Oct29.1...@ke4zv.atl.ga.us> Gary Coffman,
: >ga...@ke4zv.atl.ga.us writes:
: >>There really aren't little
: >>spinning balls we call electrons. That's a metaphor. An electron is
: >>best described as a quantum mechanical wave function that can exhibit...
: >
: >(Excuse me, I am jumping into the middle of this conversation --work
: >avoidance. Please disregard this if it is too "out from left field.")
: >
[much that is correct omitted]
: QM established the idea of indivisible quanta of radiation and the related
: idea of forbidden zones in electron orbitals where an electron cannot be.
There is nowhere in an orbital where the electron cannot be. This statement
is utter nonsense. There is a finite probability for the postion of the
electron to be found at any location and none are "forbidden".
: So the only way for an electron to lose energy is in discrete jumps that
: corrgainespond to the allowed orbitals in an atom. Since an electron is
:Once again i have to explain that electrons do NOT make discrete jumps
from one "orbit" to another. The wavefunction evolves continuously from
one stationary state to another -- unless a measurement of the energy is
made at an intermediate time causing wave-function collapse. If left alone
under the influence of a (say) time-dependent field, the evolution of the
initial wavefunction in time is smooth, continuous and at no time is any
postion is space forbidden by any principle.
[more correct statements omitted]
: Gary
: --
: Gary Coffman KE4ZV |"If 10% is good enough | gatech!wa4mei!ke4zv!gary
: Destructive Testing Systems | for Jesus, it's good | uunet!rsiatl!ke4zv!gary
: 534 Shannon Way | enough for Uncle Sam."| emory!kd4nc!ke4zv!gary
: Lawrenceville, GA 30244 | -Ray Stevens |
--
Lawrence R. Mead (lrm...@whale.st.usm.edu) | ESCHEW OBFUSCATION !
Associate Professor of Physics
Lawrence R. Mead
: This is not a flame, but why do people consider the spin of an electron as
: "not a spin about an axis" but the spin of a ping-pong ball as a spin
: about an axis
: I mean, they are described in the exact same way quantum mechanically, one
: with small angular momentum, the other with large angular momentum.
: Ron Maimon
Spin is an internal degree of freedom, not associated with ordinary
xyz space in which we live: that is the operators are not functions
of x,y, or z. "Spin" is an angular momentum, which is Defined as any
triplet of operators satisfying
[Sx,Sy] = i Sz , and thus the spin degree of freedom is an angular
momentum just as much as L = r x p is, but it just does not "live"
in ordinary 3-space. Does this clarify matters any?
Dan S.
>|> People DO refer to the spin of an electron, but inveigh against
>|> PICTURING the electron as a little spinning ping-pong ball. So the real
>|> question, how justified are these dire warnings? I agree with Ron that
>|> they are usually overblown. Warnings, yes, but make them "non-dire".
[stuff deleted]
Just a few questions from someone who really should know.
1) Isn't it true that quantized 'spin-about-an-axis' will be integral?
So, if I have my little spinning top, and I decrease it's mass so that I
have to describe it quantum-mechanically, will I not find that it's spin
will be n(hbar)? and not 1/2(2n+1)(hbar)?
2) Aren't half-integral angular momenta exactly those that 'overcover' the
SO(3) group?
By this I mean that there is a homomorphism (you used iso twice, but you
meant homo one of those times) between SO(3) and SU(2). In fact, it seems
that SU(2) is 'twice as big' as SO(3). That is where the half integral
quantum numbers come from, and they cannot exist in SO(3). Since spinning
tops exist in SO(3), they cannot have half-integral angular momenta.
3) Are the above questions what you mean when you say 'warnings should
exist, just non-dire ones'?
Electron spin is not a spin like a top has spin; however, the mathematics
of the electron spin is very similar to that of to a 'real' spin, so we can
think of it as a little spinning top.
Maybe this is all trivial? I hope not.
Dan S.
Ron Maimon
Ok....
why does this answer why electron spin is different from ping-pong ball spin?
In principle there is a "ping-pong ball field" whose quanta are indistinguishable
ping pong balls, and this field has a huge number of internal states, most of
which are highly nondegenerate, and represent a ping pong ball that falls apart.
But, it is a representation of the Lorentz group with some huge spin, so the
whole quantum formalism works for it.
Ron Maimon
SCOTT I CHASE
>
>There is nowhere in an orbital where the electron cannot be. This statement
>is utter nonsense. There is a finite probability for the postion of the
>electron to be found at any location and none are "forbidden".
>If left alone
>under the influence of a (say) time-dependent field, the evolution of the
>initial wavefunction in time is smooth, continuous and at no time is any
>postion is space forbidden by any principle.
To pick a couple of nits, which I know you know but which you did not
say:
(1) There is zero probability for the electron to be found at any
specific location. There is finite probability for the electron to
be found in any region.
(2) In multielectron atoms, there is zero amplitude for finding an
electron exactly in the same position as any other electron. In
this sense there are forbidden locations, and the principle which
produces them is the Pauli exclusion principle which follows from
the rule that wavefunctions must be completely antisymmetric
under exchange of identical fermions.
-Scott
-------------------- Physics is not a religion. If
Scott I. Chase it were, we'd have a much easier
SIC...@CSA2.LBL.GOV time raising money. -Leon Lederman
Ron Maimon
|> Ron Maimon (rma...@husc9.Harvard.EDU) wrote:
|>
|> : This is not a flame, but why do people consider the spin of an electron as
|> : "not a spin about an axis" but the spin of a ping-pong ball as a spin
|> : about an axis
|>
|> : I mean, they are described in the exact same way quantum mechanically, one
|> : with small angular momentum, the other with large angular momentum.
|>
|> : Ron Maimon
|>
|> Spin is an internal degree of freedom, not associated with ordinary
|> xyz space in which we live: that is the operators are not functions
|> of x,y, or z. "Spin" is an angular momentum, which is Defined as any
|> triplet of operators satisfying
|> [Sx,Sy] = i Sz , and thus the spin degree of freedom is an angular
|> momentum just as much as L = r x p is, but it just does not "live"
|> in ordinary 3-space. Does this clarify matters any?
|>
This is true for a ping pong ball as well, it has a bunch of internal
states and these are partially labelled by their angular momentum.
There was a good explanation a while ago concerning the fact that the
spin of the electron is 1/2 and this caused a lot of confusion at the
foundation stage of quantum mechanics, because all orbital angular momentum
must be integral spin.
This was a good enough explanation of the difference for me.
Ron Maimon
Michael Moroney
>In article <2b8isj$1t...@whale.st.usm.edu>, lrm...@whale.st.usm.edu (Lawrence R. Mead) writes...
>>
>>There is nowhere in an orbital where the electron cannot be. This statement
>>is utter nonsense. There is a finite probability for the postion of the
>>electron to be found at any location and none are "forbidden".
>To pick a couple of nits, which I know you know but which you did not
>say:
In addition to what Scott had to say, for electron orbitals more complex
than 1s, there are nodal planes or surfaces where the electron has zero
probability of existing (for example a p orbital pointing in the z direction
there is zero probability of the electron being exactly in the xy plane)
-Mike
Lawrence R. Mead
: In article <2b8j7o$1t...@whale.st.usm.edu>, lrm...@whale.st.usm.edu (Lawrence R. Mead) writes:
: |> Ron Maimon (rma...@husc9.Harvard.EDU) wrote:
: |>
: |> : This is not a flame, but why do people consider the spin of an electron as
: |>
: |> xyz space in which we live: that is the operators are not functions
: |> of x,y, or z. "Spin" is an angular momentum, which is Defined as any
: |> triplet of operators satisfying
: |> [Sx,Sy] = i Sz , and thus the spin degree of freedom is an angular
: |> momentum just as much as L = r x p is, but it just does not "live"
: |> in ordinary 3-space. Does this clarify matters any?
: |>
: This is true for a ping pong ball as well, it has a bunch of internal
: states and these are partially labelled by their angular momentum.
: There was a good explanation a while ago concerning the fact that the
: spin of the electron is 1/2 and this caused a lot of confusion at the
: foundation stage of quantum mechanics, because all orbital angular momentum
: must be integral spin.
I would add that the Theory of orbital angular momentum does *not*
forbid half-integer orbital angular momentum. Whether L =1/2 orbital
angular momentum occurs in nature was settled by experiment. Grab a
good quantum text and follow the argument.
Lastly, the ping-pong ball "spin" lives in ordinary 3-space; the spin
of the electron does not. Does this distintion not catch on with you?
: Ron Maimon
Lawrence R. Mead
: In article <2b8isj$1t...@whale.st.usm.edu>, lrm...@whale.st.usm.edu (Lawrence R. Mead) writes...
Consider them picked ;-)
Michael Weiss
rma...@husc9.Harvard.EDU (Ron Maimon) writes:
Wrong. Ron didn't write any of the stuff you quoted. I wrote it, replying
to a post of Ron's.
Dan S. also writes:
Just a few questions from someone who really should know.
1) Isn't it true that quantized 'spin-about-an-axis' will be integral?
"Spin-about-an-axis" isn't a technical term in QM, so far as I know. If
you mean "orbital angular momentum", then yes, this must be integral.
2) Aren't half-integral angular momenta exactly those that 'overcover'
the SO(3) group? [...] By this I mean that there is a homomorphism
(you used iso twice, but you meant homo one of those times) between
SO(3) and SU(2).
Did I say SU(2) was iso to SO(3)? If so, shame on me! I hope I said that
the Lie algebras of SU(2) and SO(3) are isomorphic. Anyway that's what I
meant to say.
Since spinning tops exist in SO(3), they cannot have half-integral
angular momenta.
Yes, that's what I meant when I said that any sort of spin that "inhabits
R^3" must be integral. We're both being vague, and counting on sympathetic
readers--- after all, real wooden tops contain gobs of electrons and other
fermions, which have spin one-half!
3) Are the above questions what you mean when you say 'warnings should
exist, just non-dire ones'?
Electron spin is not a spin like a top has spin; however, the mathematics
of the electron spin is very similar to that of to a 'real' spin, so we can
think of it as a little spinning top.
Yes. And we can be fairly precise about the sense in which the mathematics
is similar. The Lie algebras of SU(2) and SO(3) are isomorphic, but the
Lie groups are not (as you noted). This statement has a direct, intuitive
significance! Going back to Lie's original approach, one can think of
elements of the Lie algebra as infinitesimal rotations; thus angular
velocities and related concepts (like angular momentum) depend only on the
Lie algebra. But finite rotations live in the Lie group.
Much of the classical theory depends only on the Lie algebra. This
explains (partly) why the semi-classical Bohr-Sommerfeld theory lasted as
long as it did. For example, consider this formula from the theory of the
Zeeman effect:
mag J(J+1) - L(L+1) + S(S+1)
E = constant BM ( 1 + -------------------------- )
J 2J(J+1)
J, L, and S are respectively the total, orbital, and spin (or "intrinsic")
angular momentum quantum numbers, and M_J is the quantum number for the
z-component of total angular momentum.
You can give a thoroughly "pictorial" derivation of this formula, with
vectors precessing about other vectors. In fact, I believe Pauli gave such
a derivation some time before real QM arrived on the scene. And hindsight
justifies the argument, because if you look at the equations and not the
pictures, the whole derivation "lives in the Lie algebra". This is the
Heisenberg-Dirac reformulation of Bohr's Correspondence Principle: if the
equations look the same, you'll get the same results.
I don't want to exaggerate too far the trustworthiness of Newtonian
pictures. The double-covering of SO(3) is actually rather a fine point,
compared with the basic fact of the quantization of the z-component of spin
(called "space quantization" in the Bohr-Sommerfeld theory). I don't know
of any way to fit this into a truly Newtonian picture. But of course
spin-1 and spin-(1/2) are on the same footing here.
Maybe this is all trivial? I hope not.
I don't think so, seeing how it took a constellation of some of the most
brilliant physicists of this century about 15 years to get it right.
Gary Coffman
>Gary Coffman (ga...@ke4zv.atl.ga.us) wrote:
>: In article <1993Oct29....@novell.com> Dan Seeman <dse...@novell.com> writes:
>: >In article <1993Oct29.1...@ke4zv.atl.ga.us> Gary Coffman,
>: >ga...@ke4zv.atl.ga.us writes:
>: >>There really aren't little
>: >>spinning balls we call electrons. That's a metaphor. An electron is
>: >>best described as a quantum mechanical wave function that can exhibit...
>: >
>: >(Excuse me, I am jumping into the middle of this conversation --work
>: >avoidance. Please disregard this if it is too "out from left field.")
>: >
>[much that is correct omitted]
>
>: QM established the idea of indivisible quanta of radiation and the related
>: idea of forbidden zones in electron orbitals where an electron cannot be.
>
>There is nowhere in an orbital where the electron cannot be. This statement
>is utter nonsense. There is a finite probability for the postion of the
>electron to be found at any location and none are "forbidden".
Ah, I meant to say that there are zones *between* orbitals where an
electron cannot remain because it would represent a fractional energy
state. Of course the electron can be anywhere *in* an orbital since
that represents a constant energy state. Exactly where it is in the
orbital is a probability function that in essense *is* the wave function
of the electron.
>: So the only way for an electron to lose energy is in discrete jumps that
>: correspond to the allowed orbitals in an atom. Since an electron is
>Once again i have to explain that electrons do NOT make discrete jumps
>from one "orbit" to another. The wavefunction evolves continuously from
>one stationary state to another -- unless a measurement of the energy is
>made at an intermediate time causing wave-function collapse. If left alone
>under the influence of a (say) time-dependent field, the evolution of the
>initial wavefunction in time is smooth, continuous and at no time is any
>postion is space forbidden by any principle.
Now this is a bit more subtle. If an electron can only transition from
one orbital to another by absorbing or releasing discrete quanta, as
theory and experiment say, then it would seem intuitive that the transition
must be a discrete step. Whether we call it a jump or a collapse of the
wave function seems a matter of semantics. Now a *free* electron can
tunnel due to the probabilistic nature of the wave function, but I didn't
know electrons bound to an atom could tunnel from orbital to orbital. That
would seem to indicate that the electromagnetic catastrophy is possible.
Since quantum theory was developed to explain why the EMC doesn't happen,
that appears to be a contradiction.
john baez
>Ah, I meant to say that there are zones *between* orbitals where an
>electron cannot remain because it would represent a fractional energy
>state. Of course the electron can be anywhere *in* an orbital since
>that represents a constant energy state. Exactly where it is in the
>orbital is a probability function that in essense *is* the wave function
>of the electron.
This is pretty darn vague. An "orbital" is simply an energy eigenstate.
Electrons will not be found to have energies that aren't eigenvalues of
the energy operator. However, when you speak of "zones" it sounds like
your talking about regions in SPACE where the electron can't be found.
This would be wrong; when an electron is in an energy eigenstate, its
wavefunction never vanishes in any region of space (any region of
nonzero volume, to be precise). I.e., it has a nonzero probability of being
found ANYWHERE in space.
john baez
>I would add that the Theory of orbital angular momentum does *not*
>forbid half-integer orbital angular momentum.
I suppose it all depends on what you mean by "orbital" angular momentum.
I would tend to say the theory *does* forbig half-integer orbital
angular momentum, but I mean by this something very precise; it's a
mathematical fact that the spin-(n + 1/2) representations of SU(2) does not
occur in the decomposition of L^2(R^3) into irreducible reps of SU(2),
simply because the (usual) action of SU(2) on L^2 comes from an action
of the SO(3). I.e., a scalar wavefunction has angular momentum, but
never half-integer angular momentum.
SCOTT I CHASE
>
>Ah, I meant to say that there are zones *between* orbitals where an
>electron cannot remain because it would represent a fractional energy
>state. Of course the electron can be anywhere *in* an orbital since
>that represents a constant energy state. Exactly where it is in the
>orbital is a probability function that in essense *is* the wave function
>of the electron.
This is still not true. There are no forbidden zones between orbitals.
Each orbital has a different probability density distribution, but none
has regions of zero probability. So, for example, a hydrogen 1s electron
can be anywhere. It just has small probability of being far from the region
which you imagine to define the orbital.
Ron Maimon
|> In addition to what Scott had to say, for electron orbitals more complex
|> than 1s, there are nodal planes or surfaces where the electron has zero
|> probability of existing (for example a p orbital pointing in the z direction
|> there is zero probability of the electron being exactly in the xy plane)
|>
I hate to be anal and stupid, but there is a zero probability of the electron
being in the x-y plane in any orbital and any potential, and for all realistic
potentials there are no large regions where the wavefunction vanishes. It only
vanishes on sets whose volume is zero. I think this is what Scott meant in his
post, that the electron has a finite probability of being anywhere.
Ron Maimon
Ron Maimon
|>
|> I would add that the Theory of orbital angular momentum does *not*
|> forbid half-integer orbital angular momentum. Whether L =1/2 orbital
|> angular momentum occurs in nature was settled by experiment. Grab a
|> good quantum text and follow the argument.
I thought I did....
The idea is that every wavefunction can be expanded in terms of the angular
momentum wavefunctions with integer angular momentum only, so that a spinless
wavefunction will decompose into integer angular momentum eigenstates only.
Or, to be more physical, the wavefunction is single valued, since the phase
difference of the wavefunction between two distant points is observable, and
therefore must return to itself after a 2pi rotation.
I can see that total angular momentum and spin angular momentum can be half
integer, but I can't see how an orbital angular momentum can be half integer,
inasmuch as decomposing the angular momentum into an orbital part and a spin
part makes sense.
Ron Maimon
Ron Maimon
|>
|> Now this is a bit more subtle. If an electron can only transition from
|> one orbital to another by absorbing or releasing discrete quanta, as
|> theory and experiment say, then it would seem intuitive that the transition
|> must be a discrete step.
well, intuition is misleading here.
If you start off in a state |excited> and want to evolve to the state |ground>
amplitude slowly leaks from the state excited to the state ground.
In the meantime there is a corellated change in the electomagnetic field from
a state with no quanta to a state with one quanta.
If you are looking at the atom, you will be "splitting" into a collection of
"you"s each of which sees a quantum at a different time.
but the evolution of the system is continuous.
Ron Maimon
Gary Coffman
Well my understanding of orbitals stems from a 25 year old chemistry
education, but I was taught that orbitals do occupy volumes of space
and have definite shapes. That's why the bond angles of compounds are
as they are, and why elements with different numbers of electrons in
different electron shell arrangements occupy different volumes. Now if
I understand what you're saying, orbitals don't have definite volumes
or shapes, that an electron in a p orbital could be *anywhere* in the
Universe and still be forming a bond with a particular angle.
Or perhaps you are saying that an electron's wavefunction never has a
boundary in space, but that the probabilities become exceedingly small
for positions that are extremely far from the expected volume. Somewhat
like the idea that there is a finite probability that all the air molecules
in a room could move in the same direction at once and gather in a corner,
incidentially suffocating everyone in the rest of the room, but that the
odds are so slight that it doesn't matter in practice.
Or is there some yet more subtle factor I'm missing?
SCOTT I CHASE
>
>Well my understanding of orbitals stems from a 25 year old chemistry
>education, but I was taught that orbitals do occupy volumes of space
>and have definite shapes. That's why the bond angles of compounds are
>as they are, and why elements with different numbers of electrons in
>different electron shell arrangements occupy different volumes. Now if
>I understand what you're saying, orbitals don't have definite volumes
>or shapes, that an electron in a p orbital could be *anywhere* in the
>Universe and still be forming a bond with a particular angle.
They do have definite shapes, probably exactly the ones you remember.
The subtle point which you are forgetting is that these distributions
have diffuse surfaces which trail off toward infinity. Any electron
has a nonzero probability density in all regions. But the distribution
is strongly peaked near the shape which you remember. So it both has a
definite shape *and* occupies all space, in the sense that there is
no finite volume of space over which you could integrate the probability
density and get a total of 1. There will always be some probability
of finding it outside your finite volume.
-Scott
-------------------- i hate you, you hate me
Scott I. Chase let's all go and kill barney
SIC...@CSA2.LBL.GOV and a shot rang out and barney hit the floor,
no more purple dinosaur.
Michael Moroney
A plane is an infinite length, infinite width and zero thickness box. It
would have an undefined volume, not necessarily zero. If the volume was
not zero it could have a finite probability of containing the electron,
but in the case I mentioned it still wouldn't contain it. This is because
the wave function is zero for the xy plane (for pz orbital), so psi squared
(probability density of electron) is zero there.
-Mike
Dan Seeman
sic...@csa5.lbl.gov writes:
>density and get a total of 1. There will always be some probability
>of finding it outside your finite volume.
As I remember, yes there is _some_ probability of finding the electron
outside the finite volume of a given orbital, but that probability is low
enough for many applied praticioners to assume the electron never
occupies volumes between orbitals (for an observable length of time). It
is a gross assumption and simplification, but useful. Stricktly speaking
though, the probability is non-zero.
I can understand why Gary said what he did. It is a very common way to
talk about simple atomic models.
dks.
Douglas A. Singleton
There is an excellent article by Ohanian (Am. J. Phys. 54{6} 1986 pg. 500)
that gives a more physical picture of spin than the usual "it's an
internal quantum number obeying an SU(2) algebra. That's all you need
to know; now go work problems". Ohanian first reviews the classical
derivation of a photon having spin 1. By calculating the momentum
density of an EM wave (G ~~ E X B) one can show that there's a circulating
energy flow which leads to an "internal" angular momentum for the
photon. Then he does the same thing for the Dirac field. he finds the
momentum density in the Dirac field and shows that here too there is a
circulating energy flow. Working through the math (basically take the
Dirac lagranian, find the symmetrized energy-momentum tensor, look at
the 0-i components) he shows that this circulating energy flow gives
the electron an internal angular momentum of 1/2 (h-bar = 1). Thus one
can think of the electron as a wave which "rotates" as it moves. Since
it obeys Dirac's eqn. rather than Maxwells eqn. this rotation is 1/2
rather than 1. Actually there is nothing new in this development, but
it does give a more physical picture of spin than what one ordinarily
gets, and shows that there is something (energy) circulating inside
the electron.
Doug
.
Ron Maimon
|> rma...@husc9.Harvard.EDU (Ron Maimon) writes:
|>
|> A plane is an infinite length, infinite width and zero thickness box. It
|> would have an undefined volume, not necessarily zero. If the volume was
|> not zero it could have a finite probability of containing the electron,
|> but in the case I mentioned it still wouldn't contain it. This is because
|> the wave function is zero for the xy plane (for pz orbital), so psi squared
|> (probability density of electron) is zero there.
|>
|> -Mike
while some of this is true, I meant that if you asked "what is the probabilty
that the electron is on the x-y plane?"
then for any continuous normalized wavefunction the answer is "zero"
this is true no matter what "volume" you assign to a plane, since the probability
of finding the electron in the x-y plane is the integral of the wavefunction along
this plane
(integral x from -inf to inf, y from -inf to inf, z from 0 to 0) psi(x)psi*(x)
and this is zero whenever psi is continuous and normalized.
BTW, there is a sense in which you can assign a zero volume to the plane. Just
notice that you can cover it with an infinite sequence of boxes such that the
sum of their volumes converges to a number which can be as small as you want.
This is a "lebesque measure" way of saying the volume of the plane is zero.
Ot you can say that the volume of the plane is the limit of the volume of the
section of the plane bounded by a large sphere, as the radius goes to infinity,
which again makes the volume of the plane zero. This is what you do in calculus
when you define the integral from infinite bounds- you define it as the limit of
the areas as the bounds become bigger and bigger.
Ron Maimon
Aephraim M. Steinberg
Dan Seeman <dse...@novell.com> wrote:
>In article <4NOV1993...@csa5.lbl.gov> SCOTT I CHASE,
>sic...@csa5.lbl.gov writes:
>>density and get a total of 1. There will always be some probability
>>of finding it outside your finite volume.
>
>As I remember, yes there is _some_ probability of finding the electron
>outside the finite volume of a given orbital, but that probability is low
>enough for many applied praticioners to assume the electron never
>occupies volumes between orbitals (for an observable length of time). It
Other responses have explained this bit about shapes of orbitals versus
their infinite extent, so I won't repeat that, except to say that the
contour plots you've seen of orbitals are related to the actual orbital
wave functions the same way a 2 cm line segment is to a Gaussian which
happens to have 1cm rms width (or something like that). I don't like
the use of the phrase "volume of an orbital," except as an approximate
description.
More importantly, the different orbitals _overlap_ (even in their high-
probability regions). There is no region which is between two orbitals
and not in any other orbital. Furthermore, orbitals are just a useful
way of breaking down the state of an electron, like Fourier analysis for
describing an arbitrary function. The electron need not be in a given
orbital; in fact, in the absence of a magnetic field, it is most likely
not to be. It can be in the superposition of any number of orbitals.
In fact, it can be localized arbitrarily well at any arbitrary point in
space, but this would correspond to a superposition of many different
orbitals, with a very uncertain energy. Since we are used to observing
things in terms of energy, we generally "force" things to pick one
energy state or another (and even at a given energy level, there's enough
symmetry that several states can mix).
>is a gross assumption and simplification, but useful. Stricktly speaking
>though, the probability is non-zero.
--
Aephraim M. Steinberg | "If the human brain were simple
UCB Physics | enough for us to understand, we
aeph...@physics.berkeley.edu | would be too simple to understand
| it." -- anonymous
Keith Ramsay
|A plane is an infinite length, infinite width and zero thickness box. It
|would have an undefined volume, not necessarily zero.
Just as an aside, by the standard mathematical definition of volume,
the xy-plane in R^3 definitely has zero volume. It is possible for
every epsilon>0 to enclose the plane within a union of a sequence of
balls, whose total volume is <epsilon.
Keith Ramsay
ram...@math.ubc.ca
Ron Maimon
Actually, it's not so obvious that you can have a consistent definition of the
notion of volume if you allow infinite collections of covering balls. The
standard definition of "volume" only allows a finite number of balls.
When you allow an infinite number of balls to define volume, you are defining
the lebesgue measure, really, and it is not such a trivial task to prove that
this is a consistent idea.
Ron Maimon
Keith Ramsay
||> Just as an aside, by the standard mathematical definition of volume,
||> the xy-plane in R^3 definitely has zero volume. It is possible for
||> every epsilon>0 to enclose the plane within a union of a sequence of
||> balls, whose total volume is <epsilon.
|
|Actually, it's not so obvious that you can have a consistent definition of the
|notion of volume if you allow infinite collections of covering balls.
|The standard definition of "volume" only allows a finite number of balls.
As far as I'm concerned, Lebesgue measure *is* the standard definition
of volume. What definition would you propose to regard as "standard"
in mathematics, if not Lebesgue measure?
|When you allow an infinite number of balls to define volume, you are
|defining the lebesgue measure, really, and it is not such a trivial
|task to prove that this is a consistent idea.
I don't suppose that it is. I think one can reasonably inform people
about the definition being used by mathematicians without assuming
that everyone will need to check to see that what I am saying is
correct according to an obviously consistent definition of volume. If
they feel the need, they can no doubt find a good textbook which will
give them the details.
If you like, take what I have said as assuring the readers that there
*is* a good way to define the volume, and that the volume once defined
is definitely zero. If you're going to talk about such things as "the
probability that the electron is on the xy-plane" as the previous
poster did, it doesn't make much sense then to shy away from concepts
such as "measure zero". You don't have to dig very deeply into real
analysis, by the way, to talk meaningfully about "sets of measure
zero", such as the xy-plane in R^3.
Keith Ramsay
ram...@unixg.ubc.ca
Ron Maimon
|> |notion of volume if you allow infinite collections of covering balls.
|> |The standard definition of "volume" only allows a finite number of balls.
|>
|> As far as I'm concerned, Lebesgue measure *is* the standard definition
|> of volume. What definition would you propose to regard as "standard"
|> in mathematics, if not Lebesgue measure?
|>
I just assume that the standard definition of volume is the one that allows
a covering by finite number of boxes only. So you define the "lower volume"
of a set as the supremum of the volume of all the finite sets of boxes it
contains, and the "upper volume" as the infinimum of all the finite sets of
boxes that contain it. If these are both equal, then the set has volume equal
to their common value.
In this definition, the set of all rational numbers has an undefined measure,
which is what we expect intuitively, and the area under a function is equal
to it's Riemann, not Lebesgue, integral, which is again what you would expect,
intuitively.
I guess the lebesgue integral is more standard in higher mathematics, but I
think the intuitive definition of area is closer to the Riemann approach.
Ron Maimon
Silver Omega
rma...@husc9.Harvard.EDU (Ron Maimon) writes:
> In article <ramsay.7...@unixg.ubc.ca>,
> ram...@unixg.ubc.ca (Keith Ramsay) writes:
> |> As far as I'm concerned, Lebesgue measure *is* the standard definition
> |> of volume. What definition would you propose to regard as "standard"
> |> in mathematics, if not Lebesgue measure?
>
> I just assume that the standard definition of volume is the one that allows
> a covering by finite number of boxes only. So you define the "lower volume"
> of a set as the supremum of the volume of all the finite sets of boxes it
> contains, and the "upper volume" as the infinimum of all the finite sets of
> boxes that contain it. If these are both equal, then the set has volume equal
> to their common value.
I think you can extend this to include finite areas with infinite dimensions
without going to the Lebesgue measure, in the same way that the Riemann
integral can be defined with infinite limits. It probably isn't trivial to
prove that this is a well-defined idea, but I don't think it's too
counterintuitive.
In this case, you'd allow an infinite sequence of boxes provided that the sum
of their volume converged. I don't think it's difficult to show that this
extension defines the volume of the plane as zero.
Harry.
--
"A foolish consistency is the hobgoblin of little minds" - Emerson
Harry Johnston, uda...@bay.cc.kcl.ac.uk
Ron Maimon
|> In article <2bi651$7...@scunix2.harvard.edu>,
|> rma...@husc9.Harvard.EDU (Ron Maimon) writes:
|>
|> > In article <ramsay.7...@unixg.ubc.ca>,
|> > ram...@unixg.ubc.ca (Keith Ramsay) writes:
|>
|> > |> As far as I'm concerned, Lebesgue measure *is* the standard definition
|> > |> of volume. What definition would you propose to regard as "standard"
|> > |> in mathematics, if not Lebesgue measure?
|> >
|> > I just assume that the standard definition of volume is the one that allows
|> > a covering by finite number of boxes only. So you define the "lower volume"
|> > of a set as the supremum of the volume of all the finite sets of boxes it
|> > contains, and the "upper volume" as the infinimum of all the finite sets of
|> > boxes that contain it. If these are both equal, then the set has volume equal
|> > to their common value.
|>
|> I think you can extend this to include finite areas with infinite dimensions
|> without going to the Lebesgue measure, in the same way that the Riemann
|> integral can be defined with infinite limits. It probably isn't trivial to
|> prove that this is a well-defined idea, but I don't think it's too
|> counterintuitive.
No it's not too counterintuitive. The way you would extend it is to take the
set whose volume you want, intersect in with a ball of radius R, determine
the volume of that section, and take the limit of the volume as R---> infinity.
This gives you a meaning of volume that is equivalent to the Riemann integral
with infinite limits allowed. It is still much less inclusive then the lebesgue
measure in terms of which sets have a defined volume.
|>
|> In this case, you'd allow an infinite sequence of boxes provided that the sum
|> of their volume converged. I don't think it's difficult to show that this
|> extension defines the volume of the plane as zero.
|>
|> Harry.
Actually, that extension _is_ the lebesgue definition of volume, and showing
that it is consistent involves one nontrivial proof- you must prove that if an
object can be covered with an infinite sequence of rectangles whose volume
can be made as small as you like, then it itself contains no rectangles. Or,
equivalently, that a rectangle cannot be covered by an infinite number of
rectangles whose total volume is less then it's own volume. This may seem
"obvious" but it isn't. Consider that all the points with rational coordinates
in a rectangle _can_ be covered with an infinite collection of rectangles whose
total volume can be made as small as you like, in particular, you can cover a
dense subset of the rectangle with a sequence of rectangles of volume e for e as
small as you like.
How do you do that? Well, the rationals are countable, so that you can order
all the points in the box with rational coordinates in a sequence. Then you
can cover the first one with a rectangle of volume e/2, the second with a
rectangle of volume e/4, the third with a rectangle of volume e/8, etc. so that
the total volume of all the rectangles is e. e can be as small as you like, so
you see that the points with rational ordinates in a rectangle, no matter how
large, have zero volume.
If lebesgue measure is to be a consistent notion, I had better have missed some
points in the rectangle when I covered all the rationals! Or else this is a proof
that all rectangles have volume zero- since they can be covered by a sequence of
rectangles of arbitrarily small volume.
The way to prove this is to use the fact that the rectangle has compact closure.
You then can see that the intersection of a bunch of closed sets in it is nonempty
if the intersection of any finite number of them is nonempty. Then you prove an
easy (but tedious) lemma that says that a rectangle cannot be fully covered by
a finite number of rectangles whose total volume is less then it's own.
But you have to go through a whole song-and-dance about how you can replace the
rectangles in the definition of area with open rectangles (so the proof will work)
without any loss of generality, and at the end you get a non-constructive proof.
Meaning that someone could one day give you a sequence of rectangles of
arbitrarily small volume that he claims covers all of a given rectangle, and you
could never construct a point that is not in his rectangle. As a matter of fact,
since the set of all constructible points in a rectangle is countable, it has
lebesgue measure zero! so it's pretty easy to "present" a covering of a rectangle
with a set of rectangles of near zero volume that misses only non-constructible
points.
So you see, Lebesgue measure is not so trivial an extension of Riemannian measure,
and I'm not so sure that its necessary to mention it at all for applied
math.
Ron Maimon
Lawrence R. Mead
: I thought I did....
: Ron Maimon
I meant something very simple here. There are two solutions possible to
the standard orbital angular momentum equations. One corresponds to
integer the other half-an-odd-integer angular momentum [ consider
yourself pondering the spinless hydrogen atom for the first time in
history ]. Both are mathematically consistent [ the easy way to show
this is too work directly with the raising and lowering ops, not the
differential equations]. It was spectrographic data which showed that
L = integer was correct.
A question in passing. It is only |Psi|^2 which need be single-valued
,it seems to me. Can you tell me why Psi itself must be single-valued
when it is not directly measurable?
Most books gloss over this point.
Michael Weiss
I would add that the Theory of orbital angular momentum does *not*
forbid half-integer orbital angular momentum. Whether L =1/2 orbital
angular momentum occurs in nature was settled by experiment. Grab a
good quantum text and follow the argument.
Lastly, the ping-pong ball "spin" lives in ordinary 3-space; the spin
of the electron does not. Does this distintion not catch on with you?
What's your definition of "orbital angular momentum"? (Obviously not one
which implies an SO(3) representation!)
Do you consider Dirac's "Foundations of QM" a good quantum text? As Dirac
defines orbital angular momentum, it must be integral--- this is not a
question of experiment, but of terminology plus mathematics.
john baez
>A question in passing. It is only |Psi|^2 which need be single-valued,
>it seems to me. Can you tell me why Psi itself must be single-valued
>when it is not directly measurable?
>Most books gloss over this point.
First one nitpick; Riemann invented Riemann surfaces to do away with the
horror of "non-single-valued functions" (i.e., functions that aren't);
more recently folks use the device of line bundles to make the
concept of a "non-single-valued wavefunction Psi" precise. These indeed
do come up, for example in the study of an electron/magnetic monopole
system (see my big fat article on braids available by ftp where all my other
stuff is). There are a lot of nice relationships between topology and
physics owing to wavefunctions Psi living in some line bundle.
But it is not true that only |Psi|^2 is directly measurable. When one
measures the momentum of a particle one obtains information about the
phase of Psi, for example. Multiplying Psi by a complex number of
magnitude 1 doesn't change any physics, but multiplying Psi(x) by a
complex FUNCTION of magnitude 1 certainly does. E.g. it can change a plane
wave of some momentum into that of some other momentum.
Ralf Muschall
>In article <2bb4io$13...@whale.st.usm.edu>, lrm...@whale.st.usm.edu (Lawrence R. Mead) writes:
>|>
>|> I would add that the Theory of orbital angular momentum does *not*
>|> forbid half-integer orbital angular momentum. Whether L =1/2 orbital
>|> angular momentum occurs in nature was settled by experiment. Grab a
>|> good quantum text and follow the argument.
>
[lines deleted]
>Or, to be more physical, the wavefunction is single valued, since the phase
>difference of the wavefunction between two distant points is observable, and
>therefore must return to itself after a 2pi rotation.
>
experiments - the method of Kaplan/Wu (cited after:
J. Formanek, Uvod do kvantove teorie (in czech)
(this is what I consider a "good quantum text", but I don't know
whether it has been translated :-( )):
Introduce the complex vector operator A=(p-i q)/sqrt(2)
and the complex scalar operator B=A_2+i A_1.
After some boring calculations one finds that
+ + +
J =B B - A A - A A is the z-part of the orbital momentum,
3 1 1 2 2
and the operators A , A and B/sqrt(2) have the same commutation
1 2
relations with their adjoint couterparts as the creation
and annihilation operators of the harmonic oscillator use to have.
Therefore, all three terms on the right-hand-side of the above
equation have only integer eigenvalues, so J is integer-valued.
3
This proof explicitly uses the fact that the orbital angular momentum
is expressed via p and q.
Ralf
--
--
to get PGP key: finger p...@hpux.rz.uni-jena.de (or from public key servers)
pgp public key fingerprint=A3 93 A6 0B E2 8F 04 25 5B 3C 00 F6 55 E5 64 20
Ron Maimon
|>
|> I meant something very simple here. There are two solutions possible to
|> the standard orbital angular momentum equations. One corresponds to
|> integer the other half-an-odd-integer angular momentum [ consider
|> yourself pondering the spinless hydrogen atom for the first time in
|> history ]. Both are mathematically consistent [ the easy way to show
|> this is too work directly with the raising and lowering ops, not the
|> differential equations]. It was spectrographic data which showed that
|> L = integer was correct.
The differential equation gives you a terrible solution for L half integer
which is both double valued and extremely singular. I will argue in a second
that this solution is physically as well as mathematically unappealing.
|>
|> A question in passing. It is only |Psi|^2 which need be single-valued
|> ,it seems to me. Can you tell me why Psi itself must be single-valued
|> when it is not directly measurable?
|> Most books gloss over this point.
|>
This is not true, but most authors don't really know why so they leave it
out of their books, or else they think it's trivial, and then they write
more advanced books.
this is because when you say "angular momentum" you really mean constructing
operator vector L such that if you do an infinitesimal rotation around the
axis e by the angle t, you get a new state R(e,t)|psi>, which is given
by |psi> + it e.L |psi> (e.L means e dot product L). This means that you
can know all the properties of angular momentum from the properties of
rotation of psi.
so say you have a |psi> whose wavefunction is psi(x). Then the rotated
wavefunction R|psi> is going to be psi(Rx) where R is the rotation matrix that
rotates the vector x, not to be confused with the operator R which has the
same name (you gotta love internet notation). Well, at least, it will be that up
to an arbitrary phase- this is what you pointed out.
So I will include the phaze explicitly.
R|psi> has wavefunction Rpsi(x)= phaze(R) psi(Rx)
now I need to use the group property of R, and this says that if I do two
rotations in succession, I get a third rotation. Let A, B be rotations, and
let C=AB
then I find that
C|psi> = AB|psi> so that Cpsi(x) = phase(B)phaze(A) psi(ABx)
and in order to be consistent, I had better have phase(B)phase(A)=phaze(C).
I'll explain this more later, but this is the key.
the only way that can happen is if phase(A)=phase(B)=1. This says that the only
"number" representation of the rotation group is the scalar 1. If I had the
"phase" be a multi-component object, I could make it all work out, but I don't
so I can't.
this is confusing because you think- I can stick in any phase I want to into
the wavefunction! why am I being restricted here?
the reason is that there are two different types of phases you can "stick in".
There is a global phase, and a local phase.
A global phase means take the wavefunction of the entire universe, and multiply
it by an arbitrary phase, and it won't make a difference. This is true, and
obvious.
There is also the local phase of a part of the universe. You can multiply this
by an arbitrary phase so long as it doesn't interact with the rest of the
universe. So that if I had only one particle in the world, I could make it
change the sign of it's wavefunction every time I sneezed, it wouldn't matter.
But there are other particles in the world, and if I don't want the interaction
between them to change when I sneeze, I had better change their phases in the
exact same way as I changed the phaze of my first particle. So if I do a rotation
and I decide to stick in an "arbitrary phase" I have to stick it into all the
wavefunctions of the universe.
This arbitrary phase doesn't have to obey phase(A)phase(B)=phase(C), because
it really is arbitrary. It can be whatever. If the product of two rotations
isn't well defined down to a phase, who cares? You can choose it whichever way
you like.
the important phase, however, the local phase, or the phase difference between
the particle's wavefunction and another particle's wavefunction, this has to
change in a consistent way when you do rotations. This is what we say is the
"angular momentum" operator of a particle.
Finally to convince you that the sign of the wavefunction _is_ detectable if
it is a _local_ change, I will describe a scattering experiment.
Say I have an pi meson (a spin zero particle) in some state, and I scatter
another pi meson off it. I can in principle determine the state of the first
pi meson relative to the state of the second. For example, they might interfere
in a certain way, and cause the scattering to have peaks in certain directions.
If I rotate my apparatus and use an identical probing pi meson, I find that
it scatters differently. When I finish rotating, however, I had better find
that it scatters in the same way, because there is no consistent way to
insert a phase difference between the apparatus pi meson and the pi meson
I am probing with. The wavefunction doesn't have enough components.
This doesn't happen with electrons. I can set up a situation where when you
slowly turn a table with an apparatus on it 360 degrees, all the while
scattering electrons with a given wavefunction off the aparatus. the scattering
results change and change and change, and don't return to their original values
once I go 360 degrees around. It takes another turn of 360 degrees to restore
the original scattering problem. This has been done somewhere, too, and it
works.
Ron Maimon
Lawrence R. Mead
: then I find that
: Ron Maimon
Thanks Ron for the overview. I am not sure that my undergrads can handle
the rotation matrices, etc., but for a grad course it is fine [and falls
in with what is already covered]. The
full arguement you give above in fact should appear in a well-written text
as the phase difficulty is too often glossed over. Indeed , it is more
relevant than ever considering local phase changes as you point out above.
Lawrence R. Mead
Thanks for pointing out the relevance of nonlocal phase changes (gauge
transformations).
David Wayne Ring
>This doesn't happen with electrons. I can set up a situation where when you
>slowly turn a table with an apparatus on it 360 degrees, all the while
>scattering electrons with a given wavefunction off the aparatus. the
>the original scattering problem. This has been done somewhere, too, and it
>works.
The theory is obvious, but I am very surprised that it can actually be done.
Can you provide a reference?
Dave Ring
Cd...@phys.tamu.edu
Mark Dalton
study of physics/3 years of Chem - I am a biologist/genetics (^8 ).
Forgive me if you do not like the "Tao of Physics". It was the first thing
on physics that I have read besides my college physics text books. And I must
say that I agree that it handles things much more thoroughly. Also I tend
to be more or an Eastern thinker (at least more than my teachers).
My understanding of "electron orbitals" everyone has been partially
correct. The basic concept is that the are regions that are highly probable
and regions that are highly in probable. You can not state where the will
be at any particular moment, but you can predict the likelihood that it
will be at a point at any given moment. I think of a 'electron' as a
"bundle of energy". (feel free to straighten me out).
As far as, "free energy" there is no energy that is "free". Think
of it this way, if you set up solar panels over the entire earth, you will
get energy, but the earth will not. There is no way to escape the system
if you take energy, somewhere else there will be energy taken. Energy
is continually 'cycling', but I would not call it free.
These opinions are my own/in that, please hold no one else accountable
for these (^8.
Mark
___________
Mark Dalton AUG-GCU-AGA-AAG H
Cray Research, Inc. M A R K |
Eagan, MN 55121 CH3-S-CH2-CH2-C-COOH
Internet: m...@cray.com |
NH2
SCOTT I CHASE
>Okay, I will attempt to share what I know (with very little acedemic
>study of physics/3 years of Chem - I am a biologist/genetics (^8 ).
>
>Forgive me if you do not like the "Tao of Physics". It was the first thing
>on physics that I have read besides my college physics text books. And I must
>say that I agree that it handles things much more thoroughly. Also I tend
>to be more or an Eastern thinker (at least more than my teachers).
Everything that you subsequently explained about electron orbitals was
exactly right. But I can't pass up the chance to comment on your
introduction of _Tao of Physics_ into the discussion.
The situation is analogous to if I had come to you and said "Most of what
I know and believe about genetics I learned from a pamphlet from the
Institute for Creation Research. It was much more thorough than my
high school biology textbook, and besides, I tend to be more of a
Christian thinker."
I think that you should take the advice which you would probably give
me under the circumstances. Throw it away, and read a real physics
book. Although _Tao of Physics_ has a few true statements sprinkled
in among the garbage, it is basically misguided, misleading, and
hopelessly out of date. I'm not sure why you defiantly declare your
affinity for a book which you apparently know is completely
dismissed by real physicists as full of untruths. I doubt that you
would take such an antirational view of your own field of research.
-Scott
--------------------
Scott I. Chase "It is not a simple life to be a single cell,
SIC...@CSA2.LBL.GOV although I have no right to say so, having
been a single cell so long ago myself that I
have no memory at all of that stage of my
life." - Lewis Thomas
Ron Maimon
Sakurai in his book "Advanced Quantum Mechanics" describes an experiment of
this type. It wasn't exactly a scattering experiment, and it involved neutrons,
but it was checking the fact that the experimental results of a setup depend on
whether it's been rotated 360 degrees or not.
Ron Maimon