spin direction
ba...@rosencrantz.stcloudstate.edu
If I'm not mistaken then if the magnetic moment of an electron (but it is
not essential _what_ particle) is measured with respect to some axis the
result will be either -hbar/2 or +hbar/2. This is why the names "up" and
"down" are used".
But this implies that there is no way of determining the spin direction even
if it is a "vector" (magnetic moment). In other words the spin cannot be
measured simultaneously with respect to more than one axis. Is this true?
[Moderator's note: yes. - jb]
Radu Grigore
> if it is a "vector" (magnetic moment). In other words the spin cannot be
> measured simultaneously with respect to more than one axis. Is this true?
>
>
> [Moderator's note: yes. - jb]
One way to measure spin is to use a Stern-Gerlach apparatus which
spatialy separates particles with different spins (with respect to one
axis, let's say Z). A Stern-Gerlach apparatus is basicaly something
that creates a magnetic field described by:
B_x=0
B_y=0
dB_z/dz>0
(First question: Any mistake until now?)
What happens then if you put particles in:
B_x=0
dB_y/dy>0 (dB_y/dz=0)
dB_z/dz>0 (dB_z/dy=0)
Danny Ross Lunsford
news:5c77fd19.02022...@posting.google.com...
Hang on Willis, that B won't hunt - div B <> 0.
-drl
Pertti Lounesto
>"Radu Grigore" <radug...@ieee.org> wrote:
>>What happens then if you put particles in:
>>
>>B_x=0
>>dB_y/dy>0 (dB_y/dz=0)
>>dB_z/dz>0 (dB_z/dy=0)
>Hang on Willis, that B won't hunt - div B <> 0.
In Stern-Gerlach, the magnetic field must be inhomogeneous.
Inhomogeneity seems strange, because the interaction of spin
and a magnetic field B is given by the term -heB\psi, where
there is no derivative of B.
Question: Has anyone carried out an experiment to distinguish
the following two cases:
1) B = B_z*k and dB_z/dz > 0 but dB_z/dx = dB_z/dy = 0, and
2) B = B_z*k and dB_z/dz = 0 and dB_z/dx = 0, but dB_z/dy > 0.
Pertti Lounesto
> Danny Ross Lunsford wrote:
>> "Radu Grigore" <radug...@ieee.org> wrote:
Sorry, 1) is not possible, because B is sourceless, nabla.B = 0.
But, why B must be inhomogeneous in Stern-Gerlach?
Radu Grigore
> > B_x=0
> > dB_y/dy>0 (dB_y/dz=0)
> > dB_z/dz>0 (dB_z/dy=0)
>
> Hang on Willis, that B won't hunt - div B <> 0.
Yes indeed, I wasn't paying enough attention. My question is what is
observed if a there is a gradient in two directions. I think this
"works":
Danny Ross Lunsford
Well, let's start with Bz = 0 (I rotated your problem from x). Then
Bx,x + By,y = 0
and since
Lapl. B = 0
we also must have
Bx,y - By,x = 0
This means the allowable B's are essentially complex functions in the w=x+iy
plane,
B(w) = By + iBx
dB/dw = By,x + iBx,x = -iBy,y + Bx,y
You can pick any function you want now - it will have a certain number of
poles of various orders - they correspond to infinitely long, infinitely
thin solenoids parallel to the z-axis through the points where there are
poles. The order of the pole corresponds to the whatever multipole field the
solenoid generates - i.e. a simple pole is a dipole moment from a simple
solenoid, a pole of order 2 is a quadrupole moment from superposed simple
solenoids, etc. through the multipole moments.
All this assumes that there are no non-trivial charge distributions on a
local boundary.
-drl
Walter Pedersen
How did you discover this?? and why.
Mike Mowbray
> > [...] But, why B must be inhomogeneous in Stern-Gerlach?
Since no one else appears to have replied I'll have a go...
The force on a particle with magnetic moment u in a magnetic
field B is proportional to (u . grad(B)). For the beam of
electrons in a Stern-Gerlach experiment to be split on the
basis of each electron's spin we need grad(B) to be non-zero
so that electrons with different spins can experience
different forces.
- MikeM.
Pertti Lounesto
>The force on a particle with magnetic moment u in a magnetic
>field B is proportional to (u . grad(B)). For the beam of
>electrons in a Stern-Gerlach experiment to be split on the
>basis of each electron's spin we need grad(B) to be non-zero
>so that electrons with different spins can experience
>different forces.
What is grad(B)?
Andrew Scott
This statement is true only if we are restricted to projective measurements.
However more general types of measurements exist, each described by a
so-called positive operator-valued measure or POVM.
In theory one can measure the 3 components of spin simultaneously, but not
with arbitrary accuracy (since Sx, Sy, Sz do not commute). One can also
measure position and momentum simultaneously. However the accuracy of
such measurements will be restricted by the Heisenberg uncertainty
principle.
see quant-ph/9911021 and quant-ph/0110065 for more.
andrew.
A.J. Tolland
That should be parsed as (u dot grad)(B). u dot grad is the
covariant derivative in the u direction.
--A.J.