Spinor Spreading: Relation between \psi and \chi

[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?

 

[tiny-tim]

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?

Hi intervoxel!

A spinor doesn't spread.

 

[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?

 

[tiny-tim]

… on the road to Morocco …

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?

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.

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?

Is that the Dorothy Lamour precession?

She gets around, doesn't she?