# Spin and real space

1. Sep 17, 2015

### hokhani

If spin is a quantum degree of freedom represented out of real space, why the real coordinates (x,y,z) are attributed to it? In other words, how can one say $s_x, s_y, s_z$?

2. Sep 17, 2015

### ddd123

Spin is represented by eigenstates of the operator $\vec{L} = \vec{x} \times (- \textrm{i} \hbar \vec{\nabla})$, which is the quantum correspondent to the classical angular momentum. The eigenvalue equation for that operator yields semi-integer values (other than integer ones), which have no classical analogue and can be considered only for intrinsic angular momentum. If it's intrinsic (integer or semi-integer), the eigenfunctions don't have spatial arguments: there is no turning with respect to a center, like $\vec{r} \times \vec{p}$ would suggest, since in classical physics "spinning" makes sense only for extended objects; so the only acceptable conceptual definition for spin is the generator of rotations along an axis - it is abstract, the eigenfunctions have no spatial arguments, but the axis of rotation (which is a vector in real space) still defines it. Also, experimentally it is measured that way, along a direction, as it is involved in magnetic phenomena which also classically have to do with rotating charges; analogously, a charged particle with spin experiences a deflection in a magnetic field in a certain direction, but there are fundamental differences from the classical version.

3. Sep 17, 2015

### stevendaryl

Staff Emeritus
I would not say that. The total angular momentum operator is $\vec{L} + \vec{S} = \vec{x} \times (- \textrm{i} \hbar \vec{\nabla}) + \frac{\hbar}{2} \vec{\sigma}$, where the components of $\vec{\sigma}$ is the relativistic generalization of the Pauli spin matrices. $L$ usually is not considered to include spin, and it has integral eigenvalues, not half-integral.

4. Sep 17, 2015

### ddd123

If you do the algebra the half-integral eigenvalues do come out of $\vec{L}$. They're just ignored in the orbital treatment because they're inconsistent with it. The spin operator still has meaning as a generator of rotations around an axis, for which you can have an algebra through r x p.

5. Sep 17, 2015

### Lapidus

There is angular momentum and (intrinsic) spin of a particle. Elementary particles like a electron or a photon are point particles. So only angular momentum has a classical correspondence, intrinsic spin is a complete quantum concept. Nevertheless, it can be measured in real space (though, it is a quantized entity, i.e. just certain amounts of spin in real space are allowed). Also, angular momnentum and spin together are conserved. Of course, spin can not be visualised due to its quantum nature. But it can be measured in real space by macroscopic measurement devices.

6. Sep 17, 2015

### stevendaryl

Staff Emeritus
What you are saying is not my understanding. For a particle with spin, the full angular momentum operator, $\vec{J}$, is the generator of rotations, but $\vec{L} = \vec{r} \times \vec{p}$ is only the generator of rotations for spinless particles. In particular, a particle with spin can have momentum 0, but it still has angular momentum.

7. Sep 17, 2015

### stevendaryl

Staff Emeritus
For example, see this section in Wikipedia: https://en.wikipedia.org/wiki/Angul...Spin.2C_orbital.2C_and_total_angular_momentum

8. Sep 17, 2015

### HomogenousCow

Once you impose lorentz invariance on the system, you get a conserved current which corresponds to the total angular momentum of the system. When you take this to the non-relativistic limit, it separates into orbital and spin parts. In the same way that three orbital angular momentums come out corresponding to the three orthogonal rotations, three spin orbital angular momentums come out corresponding to the three orthogonal (is that the right word?) boosts.

9. Sep 17, 2015

### Mentz114

This reminds of an 'accidental' calculation I did when I got my vector field commutation software working. In rectangular coords $t,x,y,z$ one can write the boosts as vector fields $B_k=x^k\partial_t-t\partial_k$ and these commute correctly into rotations $[B_i,B_j]=R_k$. Imposing full Lorentz symmetry means transforming the spatial part of $B_k$ to spherical polar coordinates. This gives

$C_1= r\,\cos\left( \theta\right) \partial_t + \sin\left( \phi\right) \,t\,\sin\left( \theta\right) \partial_r + \frac{t\,\sin\left( \theta\right) }{r}\partial_\theta$
$C_2= \sin\left( \phi\right) \,r\,\sin\left( \theta\right) \partial_t + \sin\left( \phi\right) \,t\,\sin\left( \theta\right) \partial_r +\frac{sin\left( \phi\right) \,t\,cos\left( \theta\right) }{r}\partial_\theta + \frac{\cos\left( \phi\right) \,t\,E}{r\,\sin\left( \theta\right) }\partial_\phi$
$C_3=cos\left( \phi\right) \,r\,sin\left( \theta\right) \partial_t +cos\left( \phi\right) \,t\,sin\left( \theta\right) \partial_r +\frac{cos\left( \phi\right) \,t\,cos\left( \theta\right) }{r}\partial_\theta -\frac{sin\left( \phi\right) \,t}{r\,sin\left( \theta\right) }\partial_\phi$

By my calculation the $C_k$ are Killing vectors. Furthermore they commute like this
$[C1,C2]=\sin(\phi)\partial_\theta+\frac{\cos\left( \phi\right) \,\cos\left( \theta\right) }{\sin\left( \theta\right) }\partial_\phi$
$[C1,C3]=\cos(\phi)\partial_\theta-\frac{\sin\left( \phi\right) \,\cos\left( \theta\right) }{\sin\left( \theta\right) }\partial_\phi$
$[C2,C3]=-\partial_\phi$

I think this shows (rather crudely) that the boosts are mapped into Killing vector fields whose conserved currents coincide with the usual angular momentum.

Last edited: Sep 17, 2015
10. Sep 17, 2015

### HomogenousCow

Could you elaborate on this step? I don't quite understand. My knowledge of killing vectors and related stuff is sketchy at best.

11. Sep 17, 2015

### Mentz114

I'll PM you. I don't think that my post is very relevant to the OPs question.