# Intrinsic Spin

1. Apr 10, 2008

### Dahaka14

It is true that electrons don't actually "spin" on their own axes according to their intrinsic spin, but this trait is simply a mathematical entity, correct?

2. Apr 10, 2008

### tiny-tim

Hi Dahaka14!

Either Feynman or Aitchison & Hey (I can't find the actual quote) point out that if you fire a beam of electrons with a particular spin at a disc that is free to spin, then the disc will spin … thereby proving that electrons really do have an angular momentum, and not just a "theoretical spin"!

3. Apr 10, 2008

### genneth

4. Apr 10, 2008

### pam

Electron spin is angular momentum without rotation of the electron.
This angular momentum can be imparted to other objects that spin.
The notion that objects have to rotate to have angular momentum is classical,
but not in relativistic QM.

5. Apr 10, 2008

### Hans de Vries

The electron's wave function is considered to be a continuous distribution of charge
and spin. (Pauli-Weisskopf) Each point of the wave function is assigned a charge
density, a current density, but also an axial current density. (the spin)

So, each point of the wave function has a spin-density giving rise to a magnetic
moment density. The volume integral of the magnetic moment density gives the
total magnetic moment $\mu_e$.

Spin does represent an effective current cycling around the wave-function as
a whole. (for example, see Sakurai, Advanced QM, section 3-5, Gordon decomposition)

This is due to Stokes law and equivalent to what happens in a magnetic material:
The "little circular currents" inside the material cancel each other but at the edge
they do not and sum up to an effective electric current around the material as
a whole.

This effective current around the wave function is given by the curl of the spin density.
When this effective current is used to calculate the magnetic moment, then again
the integral over space leads to the magnetic moment $\mu_e$.

Regards, Hans

6. Apr 10, 2008

### peter0302

However, don't forget that the electron (in current physics) is a point particle, so it has no axis around which to spin. It's simply an inherent property that shows up in interactions with other quanta.

7. Apr 10, 2008

### Dahaka14

I did check the Einstein-de Hass effect, and it seemed convincing, but I found resources that continue to state that electron spin is not to be taken literally as with the classical analogue:
"The property called electron spin must be considered to be a quantum concept without detailed classical analogy...The term "electron spin" is not to be taken literally in the classical sense as a description of the origin of the magnetic moment described above. To be sure, a spinning sphere of charge can produce a magnetic moment, but the magnitude of the magnetic moment obtained above cannot be reasonably modeled by considering the electron as a spinning sphere."
I got this from hyperphysics, but I can't have the URL because I haven't made 15 posts yet....

8. Apr 10, 2008

### peter0302

9. Apr 10, 2008

### Hans de Vries

The magnetic moment can not be derived simply by "rotating" the charge density.
You would end up with speeds higher as the speed of light.

This doesn't mean there is a conflict with for instance Maxwell's laws. The charge we
measure is not the bare charge. It is screened by vacuum polarization effects where
the vacuum can contains equal amounts of negatively and positively charged virtual
particles. We don't know what the bare charge of the electron is, neither do we know
what exactly causes the magnetic moment, neutral particles can have a magnetic
moment as well. For instance a particle with a counter rotating antiparticle produces
a magnetic moment while being electrically neutral.

Regards, Hans

Last edited: Apr 10, 2008
10. Apr 10, 2008

### Hans de Vries

A "point particle" is more a way of saying that we do not observe a composite
structure for the electron in scattering experiments.

Strictly speaking, a point particle can not have an electric charge either since it would
lead to infinite energies and it can't have mass since a radius smaller as the Schwartz-
schild radius would result in a black hole.

For practical purposes, as for instance molecular modeling, one uses the distributed
charge and spin densities of the wave function and the electric and magnetic fields
they produce. The EM fields seen by other particles are then determined by integrating
over the entire wave function.

Regards, Hans

Last edited: Apr 10, 2008
11. Apr 14, 2008

### Lojzek

I remember a post on this forum, about detection of rotation of a "perfect sphere": an object with perfect spherical symetry. We could not assign any direction to such object, so we could not define angular velocity as a time derivative of the angle between that direction and coordinate system.
But if the sphere would have mass and charge, then it would have both angular momentum and magnetic moment: just like electron. I think we could imagine electron as a very small charged ball with perfect spherical symmetry.

12. Apr 14, 2008

### Hans de Vries

For some problems which arise from such a simple model, see for instance:

The Feynman lectures on physics, volume II, chapter 28.

Regards, Hans