What is the definition of the spin of a particle?

1. Jun 21, 2012

AbsoluteZer0

Hi,

I am scratching the surface of information regarding particle physics. I have a basic understanding of standard model. What I am not quite understanding is what 'spin' is. I know that all fermions have a spin of 1/2, but what exactly is spin?

Thanks

2. Jun 21, 2012

Staff: Mentor

Particles have intrinsic angular momentum whose magnitude is constant and the same for all particles of the same type. It's one of the defining properties of a particular type of particle (electron, muon, pion, etc.).

The magnitude of the intrinsic angular momentum, S, is determined by a quantum number, s, which can be integer (including zero) or half-integer: $S = \sqrt{s(s+1)} \hbar$. When we talk about the "spin" of a particle, we usually mean the quantum number s. For example, an electron has s = 1/2, so $S = (\sqrt{3} / 2) \hbar$. S has the usual units for angular momentum, e.g. kg-m2/s.

Last edited: Jun 21, 2012
3. Jun 22, 2012

sigma.alpha

http://arstechnica.com/science/2012/04/electrons-like-gaul-come-in-three-parts/

4. Jun 25, 2012

cygnet1

Doesn't another definition pertain to how fast the quantum state of a particle rotates in phase, compared to how fast the observation reference frame rotates? So for a fermion, (s=1/2), if your reference frame rotates by 180 degrees, the state function (and hence its "probability waves") will be multiplied by i. For a boson, (whose s=1), the state function will be multiplied by -1. For a spin-0 particle, the state function won't change with measurement reference frame.

Disclaimer: I'm no expert in this field, so anything I say could be in error. A confirmation from someone who is an expert would be appreciated.

Last edited: Jun 25, 2012
5. Jun 29, 2012

JeremyEbert

Is this a correct view?

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]"

spin S = ((n+1)/2) - 0.5

for a set of complex numbers Z

z=(((n+1)/2)-k) + i sqrt(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

visual:
http://dl.dropbox.com/u/13155084/particle%20spin%205-2.png [Broken]

This is related to Julian Voss-Andreae's "Father and Mother"

"Father and Mother of the series Spin Family (2009) by physicist-turned-sculptor Julian Voss-Andreae"

Last edited by a moderator: May 6, 2017
6. Jun 29, 2012

tensor33

I'm not an expert in this subject, but I thought "spin" referred to the rotational symmetry of a particle. For example, If you looked at a fermion of spin 1/2 then rotated it 180 degrees, it would look the same in both cases.
Someone correct me if I'm wrong.

7. Jul 6, 2012

JeremyEbert

I finally found reference material that makes some sence to me about spin, I thought I would share it.

"square root spin operator":
http://www.easyspin.org/documentation/spinoperators.html

spin:
http://www.columbia.edu/itc/chemistry/photochem/spin/01.pdf
through
http://www.columbia.edu/itc/chemistry/photochem/spin/18.pdf

Last edited by a moderator: May 6, 2017
8. Jul 6, 2012

unchained1978

I've seen a "new" interpretation of spin within the context of QM as the circulating flow of energy, and subsequently the magnetic moments are the circulating flow of charge within a wave field. This is essentially to say that spin and these other peculiar results from QM may be just a mathematical consequence of wave mechanics.

http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf

It may be more enlightening to hunt down Belinfante's original paper from 1939 cited in this paper for a more detailed explanation.

9. Jul 7, 2012

scijeebus

This spin isn't exactly physical spin, if it was physical spin the particles would just radiate their energy away, and these are wave-like objects we're talking about, how would a wave "spin"? It's more about orientation of oscillations, similar to which direction a particle oscillates in. You can put a ripple in the water, and if you focus on one part off the wave, you'll notice it's only traveling on one direction, it's only traveling outward from a specific point, as apposed to other waves in the water which might be traveling towards it, but physical descriptions cannot completely describe properties of quantum particles.

10. Jul 7, 2012

unchained1978

I suggest you read the paper first. The author remarks that this intrinsic angular momentum of an electron, for example, can be regarded as a circulating flow of energy in space. There are some fairly extensive calculations done in the paper to demonstrate this using only classical E&M, but I don't care to parrot all of the information entailed here. It's an interesting perspective and one that I think deserves a glance.

11. Jul 9, 2012

JeremyEbert

updated view from the z-axis:
https://dl.dropbox.com/u/13155084/SPIN/particle%20spin%205-2-detail.png [Broken]
-90 DEG Rotation
https://dl.dropbox.com/u/13155084/SPIN/particle%20spin%205-2.png [Broken]

Last edited by a moderator: May 6, 2017
12. Jul 10, 2012

JeremyEbert

This is a very interesting read and I very much like the notion of spin states being defined by wave mechanics.

I've been trying to find the original paper but access is restricted. It looks like this has been of interest before:

another "classic view" regarding QM Spin:
http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec6_compressed.pdf

Last edited: Jul 10, 2012
13. Jul 12, 2012

JeremyEbert

unchained1978,

In Figure 1 of the article, it mentions the field amplitude as a function of radius. I cannot find any more information about this concept. Can you expound? I'm working on a visual that might be related:
http://dl.dropbox.com/u/13155084/SPIN/Conic%20Projection.png [Broken]

Last edited by a moderator: May 6, 2017
14. Jul 13, 2012

unchained1978

Jeremy,
The particular solution he's talking about is just a wave packet of the form $\vec S=\int c(k)e^{i\omega(t-z/c)}d^{3}k$ where the amplitudes $|c(k)|^{2}$ are given by $c(k)=\frac{1}{(2\pi)^{3}}\int \vec S(\vec r, 0)e^{-i\textbf{k}\cdot \textbf{r}}d^{3}x$. So depending on your initial conditions for $\vec S(\vec r, 0)$ you could have different waves, which could conceivably vary with radius. At least that's what I imagine. You could solve for S given the form of a circularly polarized wave as he gives which should give you something like that. I think it's only until you introduce the fourier coefficients that you actually get anything that depends on r and not just z.

15. Jul 13, 2012

Jim Kata

Spin is kind of like the square root of a vector. So for example 3D real vectors can be rotated ,and the group of 3D rotations is SO(3). Well for any group of the form SO(m,n) or O(m,n) there is an associated covering group Spin(m,n) and Pin(m,n) respectively. For example Spin(3) is SU(2) and as a 3D real vector is rotated by SO(3) there is a corresponding 2D complex spinor on which Su(2) acts. And when they say a particle has spin 1/2 they are referring to what mathematicians would call the highest weight of the representation.

16. Jul 13, 2012

JeremyEbert

Thank you unchained1978. I'm starting to get a better idea of whats being presented. This note helped as well:
http://rxiv.org/pdf/1204.0102v1.pdf

17. Jul 13, 2012

JeremyEbert

The relative intensities could fall in line with a helical standing wave packet of sorts.
Animation:
https://dl.dropbox.com/u/13155084/SPIN/2D/Fourier.html [Broken]

Last edited by a moderator: May 6, 2017
18. Jul 19, 2012

JeremyEbert

The triangular multiplication table is embedded in quantum mechanics Pauli matrix:
"In quantum mechanics, each Pauli matrix is related to an operator that corresponds to an observable describing the spin of a spin 1/2 particle, in each of the three spatial directions."
"It is possible to form generalizations of the Pauli matrices in order to describe higher spin systems in three spatial dimensions"
http://en.wikipedia.org/wiki/Pauli_matrices#Quantum_mechanics

spin = n/2
k*(n+1-k) = triangular multiplication table.
Spin = j = n/2
"For arbitrarily large j, the Pauli matrices can be calculated using the ladder operators"
Jy diagonals = sqrt(k*(n+1-k))

"In physics, an operator is a function acting on the space of physical states. As a result of its application on a physical state, another physical state is obtained, very often along with some extra relevant information."
http://en.wikipedia.org/wiki/Operator_(physics [Broken])

"In linear algebra (and its application to quantum mechanics), a raising or lowering operator (collectively known as ladder operators) is an operator that increases or decreases the eigenvalue of another operator. In quantum mechanics, the raising operator is sometimes called the creation operator, and the lowering operator the annihilation operator. Well-known applications of ladder operators in quantum mechanics are in the formalisms of the quantum harmonic oscillator and angular momentum."

visual:
http://dl.dropbox.com/u/13155084/SPIN/Conic%20Projection.png [Broken]
So I guess my equation is a part of a generalized Pauli matrix or a ladder operator or both?.
http://en.wikipedia.org/wiki/Generalizations_of_the_Pauli_matrices

Last edited by a moderator: May 6, 2017
19. Jul 29, 2012

JeremyEbert

This is a VERY good paper.
http://www.ejournal.unam.mx/rmf/no546/RMF005400609.pdf

Last edited by a moderator: May 6, 2017
20. Aug 9, 2012

JeremyEbert

Last edited by a moderator: May 6, 2017