| Topic: |
Science > Physics |
| User: |
"andy everett" |
| Date: |
19 Dec 2006 07:29:17 AM |
| Object: |
A blast from the past: the thread Spin and SO(3) |
Spin and SO(3)
Has anyone formulated quantum mechanical spin and angular momentum in
SO(3) instead
of SU(2)? Why was SU(2) used? Did Pauli just think spinors are cool? ...
sci.physics - Mar 20 2002, 5:10 pm by George Jones - 7 messages - 4 authors
Part of that thread:
From: George Jones <george_llew_jo...@yahoo.com>
Newsgroups: sci.physics
Subject: Re: Spin and SO(3)
Date: Wed, 20 Mar 2002 18:11:37 -0400
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"Gregory L. Hansen" wrote:
Has anyone formulated quantum mechanical spin and angular momentum in
SO(3) instead of SU(2)? Why was SU(2) used? Did Pauli just think
spinors
are cool?
Short answer: SU(2) is the quantum rotation group because multiplication
of states by phase factors leaves transition probabilities unchanged.
Now the long-winded version.
Let |psi> and |phi> be any 2 (normalized) states in a (Hilbert) state
space H. Call |<phi|psi>|^2 a transition probability. Then if
|psi'> = c1 |psi>, |phi'> = c2 |phi>,
where c1 and c2 are phase factors (|c1| = |c2| = 1),
|<phi'|psi'>|^2 = |<phi|psi>|^2.
Consequently, multiplication of states by overall phase factors doesn't
affect transition probabilities.
Suppose observer 1 sees a physical system to be in (quantum) state |f>,
with |f> in state space H. Now suppose a observer 2 looks at the same
physical system. Suppose further that a rotation R in SO(3) takes the
spatial axes of 1 into the spatial axes of 2. Because observer 2's
perspective is different than 1's, the state |f'> of the physical system
as seen by 2 is different than the state |f> of the physical system as
seen by 1.
This is true for all possible states of the system, i.e., a rotation R
induces a map on states space H whose action is
|f> |-> |f'>
for every f in H. Call this map U_R (ASCII is ugly here), so
|f'> = U_R |f> .
The map is indexed by R because different maps induce different
operators (maps) on state space.
Consider a transition probability |<f|g>|^2. Looking at a system from
a different perspective doesn't change any physics, so transition
probabilities remain unchanged, i.e.,
|<f'|g'>|^2 = |<f|g>|^2 .
Now consider a rotation S that takes observer 2's axes into the axes of
observer 3. Since the rotation form a group, T = S R is also a rotation.
Physically, T is the rotation that takes 1's axes in to 3's. So in
physical space we can a 2-step to get from 1 to 3 by first rotating by
R and then rotating by S, or we can do a 1-step by just rotating by T.
It amounts to the same thing.
Therefore, it seems plausible that doing a 2-step or a 1-step is the
same thing in state space, i.e., that
U_S U_R = U_(SR) . (1)
However, because of the phase freedom involved in transition
probabilities, this isn't quite true. What is actually true is
U_S U_R = w(S,R) U_(SR), (2)
where w is a phase factor that depends on both R and S.
After much yucky, unenlightening math, it can be shown that the phases
w can always be chosen such that either w(S,R) = 1 or w(S,R) = -1, so
U_S U_R = +/- U_(SR) (3)
Now some terminology.
An association R |-> U_R such that (1) is true is called a
representation of the rotation group SO(3) on the the representation
space of states H. U_R is called a representative of R. In other words,
since R acts on 3-dimensional physical space, it can't act directly on
infinite-dimensional state space. A representative U_R that does the job
for R is needed.
------------------------------------------------------------------------
Now a mathematical digression about the relationship between SU(2) and
SO(3), then back to quantum theory.
SU(2) is an "unfolding" of SO(3). Think of SU(2) as a piece of paper,
and SO(3) as the identical paper folded tightly in half. ***** the
folded paper (SO(3)) with a pin, and hold it up to the light. Light is
visible through 1 hole, which corresponds to 1 element of SO(3). Unfold
the paper, thereby turning SO(3) into SU(2). Now light is visible
through 2 holes. Therefore every element in SO(3) corresponds to
precisely 2 elements of SU(2). In math terms, there is a 2-to-1 onto
mapping from SU(2) onto SO(3),
SU(2) -> SO(3).
This mapping has the exceedingly nice property that for any A in SU(2),
A and -A get taken to the same element of SO(3).
I can give explicit examples of the action this map if anyone is
interested.
-----------------------------------------------------------------------
Back to quantum stuff. The above exceedingly nice property of the
relationship between SU(2) and SO(3) means that if, in (3), SO(3)
elements are replaced by SU(2) elements, the +/- gets absorbed:
U_B U_A = U_(BA),
where +/- A in SU(2) correspond to R in SO(3), and +/- B in SU(2)
correspond to S in SO(3).
What all this means, is that rotations in SO(3) in physical space act on
state space via (representations) of SU(2), the "unfolding" of SO(3).
This is because of phase freedom.
Because of the 2-to-1 relationship between SU(2) and SO(3), there are
twice as many possible actions (representations) of SU(2) as their are
actions of SO(3). Not only that, the 2-to-1 relationship means that half
of the actions of SU(2) can be associated with actions of SO(3) in a
natural way.
Those actions of SU(2) that are also actions of SO(3) correspond to
integral angular momenta. Those actions of SU(2) that aren't also
actions of SO(3) correspond to half-integral angular momenta.
This is one reason why macroscopic behaviour is associated with bosons,
but not with fermions - fermions require the "quantum" rotation group
SU(2), not the "classical" rotation group.
Patrick, if you're reading, this is why (or at least a big chunk of
why) I said that quantum theory forces spinors upon us.
There have been nice experiments demonstrating this - something like the
following. Take a beam of spin up neutrons and split it into 2 spin up
beams. Rotate in physical space 1 of the beams by 2*pi. Combine the
beams and observe the interference; since a neutron has spin 1/2, in
state space the rotated beam just picks up a minus sign. If only SO(3)
acted on state space, and not SU(2), this wouldn't happen.
Regards,
George
.
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