Natural Science Forum / Physics / Research / April 2008

neurop...@yahoo.com.au wrote:

Dear Neuropulp,
I agree with you, there does not appear to be any connection between a
spinor and Dirac's spanner illustrated by figure 41.6 on page 1148 of
MTW. Dirac's spanner illustrates the fact that the rotation group
SO(3) is not simply connected. The rotations in MTW's figure are
examples of the defining representation of SO(3) which is the spin 1
rep. A spinor rep such as spin 1/2 is not carried by ordinary 3-d
Euclidean space (for example).

Stephen Blake
http://www.stebla.pwp.blueyonder.co.uk

On Apr 19, 3:52 am, neurop...@yahoo.com.au wrote:

I haven't got MTW here, but I do have a bit of a take on looking at
spinors (for spin 1/2 systems) - which might not be perfectly rigorous
(be warned!).  First though, it's important to keep in mind that
spinors are what are acted on by SO(3) (rotations), they are not the
rotations themselves.

You can think of spinors as a pair of objects:  a unit direction in
space, and a +1 or -1 sign.  So, the spinor can be represented as a
pair (n, +/-), where n.n=1.

What then is important is to ask is how the rotation group acts on a
spinor.  For that, you need a picture of the rotation group, and how
it is doubly connected.  Intuitively, one would think *at first* that
a good way of representing rotations is as a ball in 3-d space,
centred at the origin, and of radius pi.  A point a in the ball then
represents a right-handed rotation about the a-direction, (i.e., in
the direction a/|a|), by an amount equal to the distance of a from the
origin (i.e., by an angle equal to |a|).  Since a rotation by pi about
some direction a  is equivalent to a rotation by -pi about the
direction -a, one has to identify opposite points on the surface of
this ball.  The effect of a rotation R_a, represented by point a of
the ball, would then change a spinor (n,+/-) to the spinor (R_a n,
+/-).

However, this picture obviously doesn't quite work, as nothing happens
to the +/- sign of the spinor.  The doubly connected nature of SO(3)
has not shown up.  To fix this, one needs a *second* 3d ball of radius
pi.  One can sit this ball in a different space to the first, but it
is mentally convenient to think of the second ball as sitting close
by.  However, the second ball is inverted with respect to the first.
In particular, a point b in the second ball corresponds to a *left-
handed* rotation about the direction b/|b|, by an angle |b|.  Now,
instead of identifying antipodal points of the first ball with each
other, one instead simply identifies the surface of the first ball
with the surface of the second ball (for essentially the same reason:
a right-handed rotation of pi about a given direction is equivalent to
a left-handed rotation of pi about the negative direction).  This
gives us what we need.

A rotation is now characterised by a point  a in *one* of the two
balls.  If it is a point in the first ball, the effect on the spinor
(n,+/-) is to map it to the spinor (R_a n, +/-) as before.  However,
if it is in the second ball, then the effect on the spinor is to map
it to the spinor (R'_a n, -/+), i.e., the second component of the
spinor is reversed in sign.  Here, R'_a denotes a left-handed rotation
by angle |a|.

To see the 4 pi effect, consider the action on a spinor as one moves
through rotation space, starting at the centre of the first ball
(i.e., no rotation).  First, move up along the z-axis to the north
pole.  This is the same as the north pole is of the second ball, so
move over to the second ball's north pole (you can imagine a "virtual
trajectory" through the intervening 3d space if you like).  Second,
move down the z-axis of the second ball to its centre.   What has
happened to a spinor (n,+) ?  Well, you ended up in the centre of the
second ball, so R'_0 is the identity matrix, and n is mapped to n - no
change there.  But, you are in the second ball, so the second
component of n changes sign.  Hence, (n,+) is mapped to (n,-).

Physically, what has happened?  Well, first you rotated to the right
by pi, corresponding to moving up to the north pole of the first
ball.  Then you hopped over to the second north pole, as you had
equivalently rotated to the left by pi.  But then you undid the
rotation to the left by pi, by moving down to the centre of the second
ball, which is the same as rotating to the right by pi again.  Hence,
overall, you rotated by 2 pi, but the spinor flipped its sign.

To get back to the original spinor, (n,+), one can imagine keeping
moving down to the south pole of the 2nd ball, then over to the south
pole of the 1st ball (draw another virtual trajectory between the two
balls), and up to the origin of the first ball.  This corresponds
rotating by another 2 pi angle, i.e., 4 pi altogether, and the spinor
flips back from (n,-) to (n,+) again.

Hope this picture helps.  The important thing is remember that the
balls aren't spinors, they are ways of visualising the action of the
rotation group on spinors.

The above picture also helps seeing how the rotation group  SO(3) is
connected.  Drawing a path from one ball over to the other, as in the
first 2 pi rotation above (with "virtual" portions), one clearly can't
shrink this path to a point.  However,extending such a path to go back
to the original ball allows a loop to be formed, which can clearly be
continuously shrunk to a point.