1. Nov 2, 2008

intervoxel

Since a full description of a particle is the product \psi \chi, what's the relation between the spreading of the spatial factor \psi and of its spinor \chi?

Last edited: Nov 2, 2008
2. Nov 3, 2008

tiny-tim

Hi intervoxel!

3. Nov 5, 2008

intervoxel

Ok, tiny-tim, a spinor doesn't spread. Thank you. :)

So, I suppose, we can visualize the spinor "cloud" as a conical surface (s=1/2, up state, say) made of up vectors with length hbar/2 centered around an axis in a certain direction in space so that its projection on that direction alone always returns the value hbar/2, while for others directions we might obtain any value, positive or negative. Is this picture correct?

In the case when we apply a magnetic field in that direction the Lamour precession means that the cone is denser (greater probability) around a vector precessing at the Lamour frequency. Is it?

Last edited: Nov 5, 2008
4. Nov 5, 2008

tiny-tim

… on the road to Morocco …

Hi intervoxel!

Any "cloud" that you see in diagrams of electron distributions has nothing to do with the spinor.

There is nothing conical about a spinor.

The spinor just defines the direction of spin …

if you want to visualise a spinor as a volume, then use a sphere, rotating about the axis defined by the spinor.
Is that the Dorothy Lamour precession?

She gets around, doesn't she?