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What is the definition of the spin of a particle? |
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| Jun21-12, 03:05 AM | #1 |
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What is the definition of the spin of a particle?
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 |
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| Jun21-12, 07:48 AM | #2 |
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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. |
| Jun22-12, 12:22 PM | #3 |
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| Jun25-12, 11:53 AM | #4 |
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What is the definition of the spin of a particle?
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. |
| Jun29-12, 12:02 PM | #5 |
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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/par...spin%205-2.png 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" http://upload.wikimedia.org/wikipedi..._Family%22.jpg related thread: http://www.physicsforums.com/showthr...=601918&page=3 |
| Jun29-12, 10:58 PM | #6 |
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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. |
| Jul6-12, 04:41 PM | #7 |
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"square root spin operator": http://www.easyspin.org/documentatio...operators.html spin: http://www.columbia.edu/itc/chemistr...em/spin/01.pdf through http://www.columbia.edu/itc/chemistr...em/spin/18.pdf |
| Jul6-12, 07:08 PM | #8 |
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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.co...at-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. |
| Jul7-12, 12:10 AM | #9 |
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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.
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| Jul7-12, 12:50 AM | #10 |
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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.
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| Jul9-12, 08:40 PM | #11 |
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https://dl.dropbox.com/u/13155084/SP...5-2-detail.png -90 DEG Rotation https://dl.dropbox.com/u/13155084/SP...spin%205-2.png |
| Jul10-12, 09:35 AM | #12 |
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I've been trying to find the original paper but access is restricted. It looks like this has been of interest before: http://www.physicsforums.com/showthread.php?t=60845 another "classic view" regarding QM Spin: http://www.tcm.phy.cam.ac.uk/~bds10/...compressed.pdf |
| Jul12-12, 02:38 PM | #13 |
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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/SPI...Projection.png |
| Jul13-12, 12:05 AM | #14 |
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Jeremy,
The particular solution he's talking about is just a wave packet of the form [itex]\vec S=\int c(k)e^{i\omega(t-z/c)}d^{3}k[/itex] where the amplitudes [itex]|c(k)|^{2}[/itex] are given by [itex]c(k)=\frac{1}{(2\pi)^{3}}\int \vec S(\vec r, 0)e^{-i\textbf{k}\cdot \textbf{r}}d^{3}x[/itex]. So depending on your initial conditions for [itex]\vec S(\vec r, 0)[/itex] 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. |
| Jul13-12, 12:49 AM | #15 |
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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.
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| Jul13-12, 09:01 AM | #16 |
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http://rxiv.org/pdf/1204.0102v1.pdf |
| Jul13-12, 04:33 PM | #17 |
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Animation: https://dl.dropbox.com/u/13155084/SPIN/2D/Fourier.html |
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| elementary particles, fermions, spin, standard model |
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