# The spin of a particle

1. Dec 21, 2005

### ukmicky

can anyone help please

can anyone help,i'm trying to understand particle spin, is there an easy to understand explanation

2. Dec 21, 2005

### CarlB

Spin is an attribute of elementary particles. It's something they have.

A similar kind of thing that light has is called "polarization". If you take two pieces of polaroid filter, or two polaroid sunglasses lens, you can move them around and see that the relative angle between them effects how much light goes through them.

Now the light polarizations that sunglasses use are vertical or horizontal polarizations. The spin attribute of particles is more similar to the "circular" polarizations of light than to the horizontal and vertical polarizations.

In circular polarization, the electric and magnetic fields rotate around instead of being vertical and horizontal. If you think of spins as being similar to this you won't be that far off, either for the mathematics used to describe it, or the physical object itself.

You can also describe spin as the quantization of angular momentum, but that's tougher to understand. Other posters will undoubtedly comment on this after my post.

Carl

3. Dec 22, 2005

### marlon

Here's my attempt...

An electron has a spin (intrinsic angular momentum), but it does NOT actually rotate around its axis, guys. The 'rotational nature' of spin comes from the behavior of the Dirac wavefunction (this is a matrix that represents a physical state and arises when solving the Dirac equation. This equation describes a fermion : a particle with non-integer spin) under coordinate-transformations (which are called the rotations). Basically, what we want is our physical laws (expressed by formula's and probabilities of wavefunctions) to be INVARIANT under rotations. In order to obey this, the wavefunction must behave in a certain way when we apply these rotations onto it.

With behavior i mean : how does the physics change if we interchange the components of this Dirac spinor, if we change the parity, if we apply coordinate transformations to the wavefunction and so on....For example, if we rotate the wavefunction 360°, do we still get the same physical laws...You see the pattern ???

It is this specific behavior that yields the name SPINOR because if you rotate it 360°, you get the opposite value. Now, changing coordinates (represented by rotations) and looking how the physics changes or not, is NOT THE SAME as actually rotating. So, spin arises thanks to symmetries involved but there is no actual rotation of the particle around some axis.

regards
marlon

Last edited: Dec 22, 2005
4. Dec 22, 2005

### humanino

Dear Marlon,

you must concede that spin is some sort of angular momentum. What is (semi classical and) badly wrong is to picture spin as an orbital momentum. In particular, I believe electrons are point-like, so there is nothing turning around anything, yet thete is still some intrisic spin.

Spin is a very difficult thing to understand if you want to go technical... Many things have been posted here already, for instance many explanations have been provided by Marlon. If you understand "SU(2) is the universal covering of SO(3)" then it becomes suddenly much easier !

http://en.wikipedia.org/wiki/Table_of_Lie_groups

5. Dec 23, 2005

### marlon

Well, i have no problem admitting that. the point i am trying to make is that "angular momentum J" has nothing to do with rotational motion of objects. It is a direct consequence of the invariance of, for example, probabilities being invariant under rotations. In order to do so, the fundamental building blocks of QM, ie the wavefunctions, have to transform conform "a certain" way under these rotations of SO(3). This "certain way" is...: SPINORS. This means that if you apply a rotation of 360° onto the wavefunction, you must get the opposite function. If you again, apply the same rotation onto the opposite value you get the original function. So there is invariance of the wavefunction under rotations of 720°

That is it.

That is my point

regards
marlon

6. Dec 23, 2005

### Hans de Vries

Still, one can transfer the spin angular momentum from electrons
to macroscopic objects.

A magnetic material which becomes polarized, (The spins of the outher
electrons become aligned) will start rotating in the opposite direction
with the correct angular momentum. (Einstein, de Haas effect)

Nevertheless one should indeed be very careful with the classical picture
of a spinning ball. First of all, Quantum Mechanically things never spin
around an axis but always around a point. That's why the total angular
momentum is always larger as any arbitrary component. ($s = \sqrt{s_z(s_z+1}$)

"Spinor" is maybe a somewhat misleading name. It was Paul Ehrenfest
I think who proposed this name after he stimulated van der Waerden
to write his mathematical treatment.

Every 3-vector or even 4-vector can be rewritten as a spinor.
This by itself does not yet make these vectors "spinning objects"

What the spinors really do is that they tell you how to interpret
for instance "imaginary momentum" That is, The square of the radial
momentum in Dirac solutions $p_r^2$ is generally negative. (The same is
true for Schroedinger's equation)

This compensates for the relativistic mass of electrons close to
the nucleus so that the total energy becomes (just a little bit) less
than the rest-mass energy and the same every ware, for instance:

$$E^2= p_r^2 c^2 + p_\theta^2 c^2 + p_\phi^2 c^2 + m_0^2 c^4\ = (\ m_0 c^2 - 13.6eV)^2$$

(Simplifying out the all potential energy terms here except the 13,6 eV )

It's the radial term

$$\frac{l(l+1)}{r^2}\hbar^2 c^2$$

Which enters the radial equation of the Laplacian which accounts exactly
for the "missing" relativistic mass and it's the $p_r^2$ component which is negative
with the right function in r to compensate for this.

With Spinors arising in the "Matrix Square Root" of the Klein Gordon equation:

$$\left( \begin{array}{cccc} E & 0 & -p_z & -(p_x-ip_y)\\ 0 & E & -(p_x+ip_y) & p_z \\ p_z & (p_x-ip_y) & -E & 0 \\ (p_x+ip_y) & -p_z & 0 & -E \end{array} \right) \left( \begin{array}{c} C_1 \\ C_2 \\ C_3 \\ C_4 \end{array} \right)\ \ =\ \ m_0c^2 \left( \begin{array}{c} C_1 \\ C_2 \\ C_3 \\ C_4 \end{array} \right)$$

After taking the Square at both sides this becomes:

$$\left( \begin{array}{cccc} E^2-p^2 & 0 & 0 & 0 \\ 0 & E^2-p^2 & 0 & 0 \\ 0 & 0 & E^2-p^2 & 0 \\ 0 & 0 & 0 & E^2-p^2 \end{array} \right)\left( \begin{array}{c} C_1 \\ C_2 \\ C_3 \\ C_4 \end{array} \right)\ \ =\ \ (m_0c^2)^2 \left( \begin{array}{c} C_1 \\ C_2 \\ C_3 \\ C_4 \end{array} \right)$$

The normal 3-vector representations become:

$$\left( \begin{array}{c} p_z \\ p_x+ip_y \end{array} \right) \quad \mbox{and} \quad \left( \begin{array}{c} p_x-ip_y \\ -p_z \end{array} \right) \quad \mbox{and} \quad$$

For the two opposite spin representations. That is: An imaginary
momentum $ip_x$ is interpreted via a 90 degrees rotation in the xy plane.

Regards, Hans

Last edited: Dec 23, 2005
7. Dec 23, 2005

### Hans de Vries

I might just as well give the proof for this here:

Let the angular momentum be:

$$\mathbf{p} \times \mathbf{r}\ =\ \sqrt{l(l+1)} \hbar$$

Entering the speed v via SR this becomes:

$$\frac{m_0 v}{\sqrt{1-v^2/c^2}}\ =\ \sqrt{l(l+1)} \frac{\hbar}{r}$$

We can write with rc as the Compton radius:

$$\frac{v/c}{\sqrt{1-v^2/c^2}}\ =\ \sqrt{l(l+1)} \frac{\hbar}{m_0 c r}\ =\ \sqrt{l(l+1)}\ \frac{r_c}{r}$$

$$\gamma \ =\ \frac{1}{\sqrt{1-v^2/c^2}}\ =\ \sqrt{1+l(l+1) r_c^2/r^2}$$

We can now assign a corresponding speed as a function of the
radius:

$$v/c \ =\ \sqrt{\frac{l(l+1)}{\frac{r^2}{r_c^2}+l(l+1))}}$$

The apparent increase in mass needed to produce the angular
momentum at this speed:

$$\gamma m_0\ =\ \frac{m_0}{\sqrt{1-v^2/c^2}}\ =\ m_0\sqrt{1+ l(l+1) \frac{r_0^2}{r^2}}$$

In general we have:

$$E^2= p_r^2 c^2 + p_\theta^2 c^2 + p_\phi^2 c^2 + m_0^2 c^4\$$

We now want a negative square radial momentum term which
compensates for the relativistic mass increase:

$$E^2 \rightarrow \ E_0^2 \ =\ m_0^2 c^4\$$

$$\mbox{So we need:}\ \ \ p_r^2 c^2 \ =\ - (p_\theta^2 c^2 + p_\phi^2 c^2)\$$

$$p_r^2 c^2 \ =\ m_0^2c^4 - E^2 \ =\ (1-\gamma^2)E_0^2$$

From the expression for gamma above:

$$(1-\gamma^2)\ =\ 1 - (\sqrt{1+l(l+1) r_c^2/r^2})^2 = -\frac{l(l+1)}{r^2}\ r^2_c$$

The last term is exactly the term which enters the radial equation
from the theta equation when solving the Spherical Laplacian in the
Schroedinger or Dirac equations:

$$-\left( \frac{l(l+1)}{r^2}r^2_c \right)\ E_0^2\ =\ -\frac{l(l+1)}{r^2} \hbar^2 c^2$$

This term you can find in any textbook which handles the derivation
of the Hydrogen solutions.

Regards, Hans

Last edited: Dec 23, 2005
8. Dec 25, 2005

### ukmicky

thankyou for your replies
and
HAPPY CHRISTMAS

9. Jan 4, 2006

### Neoliten

Interesting descriptions, but there is still some curious question:
Is it possible to imagine spin property of the particle or this is the same unimaginable thing as 6D space? If yes, then in simple words, what is spin? What is "rotating" and how?
Thank you and
Happy New Year!

10. Jan 6, 2006

### marlon

I am sorry, but have you read the previous posts ? All you are asking for here has been answered. Spin arises because in QM, the expectation values of observables need to be invariant under rotations. If there is anything rotating, it's the spatial coordinates of the wavefunctions, but not the particle or whatever...

marlon

11. Jan 6, 2006

### Neoliten

Thanks and sorry for stupid question, I have readed carefully this thread and realized that (but didn't delete the question).
Have anyone some visual information or sort of graphical representation of rotated wavefunction? I can imagine that, but I want to make sure.

12. Jan 6, 2006

### marlon

What do you mean by a rotated wavefunction ? Do you mean how the wavefunction would look like after a rotation has been performed onto it ? In that case, just plot a wavefunction after the coordinate transformation (ie the rotation). But in all honesty, i don't see why you would wanna "see" this ? You are not going to learn anything from it. It's the symmetry that counts, not the visualization.

marlon

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