Spinor spreading

[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! :smile:

A spinor doesn't spread. :wink:

[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]

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! :smile:

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. :smile:
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? :wink:

She gets around, doesn't she?