From: kramsay@aol.commangled (Keith Ramsay)
Subject: Re: Spin of the Electron
Date: 28 Jan 2000 05:35:26 GMT
Newsgroups: sci.math
Summary: [missing]

In article <86ng6t$b5s$1@wanadoo.fr>, "denis.feldmann"
<denis.feldmann@wanadoo.fr> writes:
|I thought you, Pertti, of all people, would know a lot about those things,
|like, for instance, spin is not "real" angular momentum,

In what sense is it not? There was a thread some time ago in
sci.physics whose upshot that it is. Feynman had an example to
illustrate it-- a situation where you can exert a torque on an object
if you also flip the spins of enough particles to compensate. It's
an angular momentum in the sense in which angular momentum is
conserved.

|and a lot of
|quantic "conservation laws" have not the same exact meaning as their
|classical counterparts ; for instance, "tunnel effects" seem to violate the
|conservation of energy, which would anyway be approximative, as dE x dt >h
|is one of the Heisenberg relations, etc. etc.

One can't take conservation laws as being *exactly* the same as the
conservation laws in classical physics, but I recommend caution also
in supposing that they are too different either. A particle which
tunnels through a barrier isn't at a well-defined position, according
to quantum mechanics, and thus you shouldn't picture quantum mechanics
as describing it passing briefly through one of the places where it
needs temporarily to borrow a bit of energy to be there. There are
popular accounts which describe it that way, but I think it's a bit
misleading.

We can imagine a particle enclosed within a box, where there's a
potential field V(x,y,z) varying from point to point. It's possible
for the particle to be at a single well-defined energy level E, and
one doesn't speak of its switching briefly into other energy levels
at all; the state its in is like a standing wave which doesn't change
in shape. Portions of the wave will in general extend to places where
V(x,y,z)>E. That's unlike the analogous classical model, where the
particle would remain always in regions where V(x,y,z)<=E.

What this means is that there's a certain positive probability that if
you suddenly observe where the particle is, you will find it within a
barrier (i.e., where V(x,y,z)>E). This energy is not "borrowed".
Suppose you go back to keeping the system isolated, and later measure
the energy of the system to within a high enough accuracy. You will
find that the system has gained energy which it keeps. It has at least
the potential energy of the particle's location when you observed it
within the barrier, plus kinetic energy on top of that. The extra
energy doesn't disappear. What has happened, of course, is that you've
just *given* the system that extra energy by making the precise
measurement of the particle's position.

Tunnelling doesn't require giving the system that extra energy, though.
In principle you can put a particle within a high potential barrier,
measure the energy of the particle to reasonably high accuracy
(which is limited by the length of time required for the tunnelling,
according to the dE*dt uncertainty relation). You can come back much
later and find that the particle has "slowly" tunnelled out. The model
doesn't describe it as having had extra energy at any point in the
process. Instead of being like a chihuahua which on average is low to
the ground, but capable of occasional bursts of energy to jump fences,
it's more like a long snake which is able to slide over the barrier
while maintaining a low center of gravity, because most of it remains
low to the ground even though a piece of it is high. It is a delicate
kind of snake, though, because its state is disturbed by so much as
your observing part of it lying on one side of the fence or the other.

I think a good way to understand the dE*dt uncertainty relation is to
think about the dynamics of a system which is at a highly precise
energy level (dE bounded above). Looking at Schroedinger's equation,
you find that the quantum state of the system changes slowly (it has a
wavefunction changing in phase at a rate of E/h-bar, but two systems
which differ only in phase are in equivalent states). So dt is bounded
below because a certain elapsed time is required before the system
gets to a state which can be reliably distinguished from its original
state. If you leave it for only a time short compared to h-bar/E, the
probabilities of making any given measurement you might make will not
have changed by very much. In order to have rapid dynamics, a spread
of possible energies is required. You don't have to think of anything
as happening particularly fast in quantum tunnelling; the system just
gradually shifts from being in a state almost entirely on one side of
the barrier, to being in a state mostly on the other side of it.

|To try to understand "spin" (or at least get a good intuitive feeling for
|it ) as some rotation of the electron on its axis is a good way to a nervous
|breakdown :-)

Someone wrote a paper in which they pointed out that if you think of
the electron as a wave, extended in space, the angular momentum can be
understood in terms of the different portions of the electron
contributing to the momentum in different directions. Not of course
quite like a spinning classical ball, but not so completely alien to
the ordinary notion of angular momentum either.

I don't really understand how people could be content with knowing
just the part of the story which is accessible to them without knowing
much of the mathematics, though. Well, okay, I guess I can understand,
but it seems it should be one of those things which you should be
thinking would be very nice to experience first-hand if only you had
the time, like the way I'd like to go to Paris if I were given extra
vacation time and so on just for the purpose, but don't go there since
I have so many other places to go first.

A handy "toy" example of a quantum system, a particle with spin 1/2,
can be thought of as having a state in a two-dimensional space. The
space has complex coordinates, which makes it more interesting. The
state of the system is represented by a line through the origin. You
are allowed to observe the state by picking a pair of orthogonal lines
through the origin, and "ask" the system which it chooses to be in.
It will then switch to being in one or the other of those states, with
a probability of cos^2 of the angle between its former state and the
new state you are offering to it.

We could for example make a basis of the states of the system
consisting of "spin pointing up" (corresponding to the x-axis) and
"spin pointing down" (corresponding to the y-axis). You can ask the
particle what its spin on the vertical axis is, and it will tell you
either +1/2 or -1/2 with certain probabilities depending on its
previous state. But there are many other possible observations you
could make, one for each direction through the origin. If we ask the
particle what it's spin is in the east-west axis, it will now tell us
either 1/2 to the east or 1/2 to the west. Because of the way we
prepared the particle by the previous question, the probability of its
answering either way is 1/2. (In the process of being observed this
way, it transfers an indeterminate amount of angular momentum to the
observer, in a way which preserves conservation of angular momentum.
If by a miracle of experimental science we managed to get an observer
isolated well and in a state where the total angular momentum with the
particle along the z-axis is zero, which is possible in principle,
when the particle was observed to be spin-up, the observer and rest of
the apparatus would measure as having 1/2 spin down if it were measured
the same way).

This east-west measurement corresponds to asking the system whether
its state in abstract state space lies on the line x=y (1/2 spin to
the east) or whether it lies on the line x=-y (1/2 spin to the west).
These are both at a 45 degree angle to both of the previous lines, and
the probabilities all work out to cos^2 of half a right angle, or 1/2.

Now it may seem odd that we have a 3-dimensional space in which we're
measuring spins, while the abstract state space is only 2-dimensional.
That may make it seem like we're not going to have enough different
ways of measuring the abstract state to correspond to our real
physical ways of measuring the spin. But the abstract state space has
complex coordinates. So we have also, for example, the question we
can ask the particle, whether its state lies on the line y=ix or on
the line y=-ix (which is at right angles to it), which will correspond
to measuring the spin of the particle on the north-south axis (either
angular momentum of 1/2 to the south, or north respectively). The
angles in state space are nicely symmetric, so if we limit ourselves
to measuring along one axis at a time, the particle will appear as
though it is making random coin flips to decide its spin each time we
switch to measuring it along a different axis from the previous one.

The angular momentum on an axis is related to the process of rotating
the particle along that axis. The wave-function of a particle which
has a well-defined angular momentum of b on an axis changes in phase
by b*theta if it is rotated by an angle of theta. This does not really
change the state of the particle, but has its meaning in terms of the
*relative* change of phase of various components of the state of a
particle whose angular momentum along that axis is not determinate.

(If you succeeded in measuring the angular momentum of any one of us
to such precision, we too would be left in an axially symmetric state
unchanged by rotation. This seems unlikely to occur in real life. Of
course it wouldn't necessarily *kill* you to be in such a state. Your
sister who went to that IV league school probably gets into such states
all the time, while you just hang around asymmetrically... but I
digress. :-) It would be a superposition of various more easily
recognized states, which would be easily decohered from each other by
any normal observation of, e.g., where one of your arms is, roughly.)

For example, suppose we take a particle with angular momentum 1/2 to
the east. It corresponds to the line x=y through our abstract state
space. If we now rotate the particle by 180 degrees around the z-axis,
the x component changes in phase by half as much, 90 degrees, which is
the same as multiplication by i, while the y component changes in phase
by -90 degrees, multiplication by -i. Applying that transformation to
the line x=y (the point (1,1) on it, for example) gives us the line
x=-y (the point (i,-i) lying on it), which corresponds to the state of
a particle with angular momentum of 1/2 to the *west*.

If instead of rotating by 180 degrees we rotate by 90 degrees, that
corresponds to multiplying the x component by e^(pi*i/4)=(1+i)/sqrt(2)
and the y component by (1-i)/sqrt(2). That carries our original line
x=y to the line y=-ix (it has the point ((1+i)/sqrt(2), (1-i)/sqrt(2))
on it, for example). That line corresponds to a particle with angular
momentum of 1/2 to the north. That is what you would expect if you
rotated a particle with angular momentum to the east by 90 degrees
around a vertical axis.

Similarly, if we apply a rotation of 90 degrees around the east-west
axis to a particle whose spin is 1/2 in the up direction, we get (1,0)
=(1/2,1/2)+(1/2,-1/2) transformed to [(1+i)/sqrt(2)]*(1/2,1/2)
+[(1-i)/sqrt(2)]*(1/2,-1/2)=(1/sqrt(2),i/sqrt(2)) which lies on the
line y=ix. (What I'm really doing is doing a cross check against the
possibility of my having misremembered the relationships.) And the
line y=ix corresponds to a state of spin 1/2 to the south.

It's not obvious at first glance, but consistency considerations force
the change in phase as a state is rotated through a full 360 degrees
to be either nothing or a change in sign (180 degree change in phase).
To begin with, we only know that a full rotation has to give us back
an equivalent state (different only in phase). But we have various
other consistency conditions: the superposition principle says that
it's possible to superpose states, to add them, and that when we make
our transformation it has to respect the superposition. That implies
that each transformation can individually be represented as a linear
transformation of the abstract quantum state space. Theory also
requires that the angles between lines remain the same. We also have
rotations around all axes, not just a single one, and they have to be
related as they actually are.

There's quite a bit of general admiration for both the fact that we
are left with angular momentum along an axis always coming in discrete
multiples of 1/2, as well as the fact that one actually gets particles
whose angular momenta are half odd integers.

Keith Ramsay