This might sound pretty fundamental etc, but one thing i have yet to understand to any proper degree is the SPIN of particles.
Could anyone please explain what a particle' spin really is as a property?
I understand what a particle's spin means for that particle, but not what the spin itself is.
:(

Quantum mechanically, spin is just another quantum number (or set of numbers) which the particle possesses. The reason it is called "spin" is because it shares many of the characteristics of classical angular momentum.

Questions such as how a "point" particle such as an electron can spin are not really answered by quantum theory.

spin is an artifact of the fact that particles, even though we desccribe them as points in spacetime, do actually have some sort of structure. for example, the field that describes photons is actually a vector field, so it has several degrees of freedom at each point in spacetime. spin, then, describes how those internal degrees of freedom behave under rotations, just as angular momentum describes how the external degrees of freedom (i.e. the freedom to propagate through spacetime) behave under rotation.

Thank you.
*bow*

Originally posted by lethe
spin is an artifact of the fact that particles, even though we desccribe them as points in spacetime, do actually have some sort of structure. for example, the field that describes photons is actually a vector field, so it has several degrees of freedom at each point in spacetime. spin, then, describes how those internal degrees of freedom behave under rotations, just as angular momentum describes how the external degrees of freedom (i.e. the freedom to propagate through spacetime) behave under rotation.

what are some equatins involving spin? And what values can it take?

"Spin" is a possibility... such that say the circulation of say Earth around the Sun can go one of two directions in the ecliptic.

Spin is like a superconducting field circulation that is set up when charge is allowed to move in a magnetic/electric field. In some situations, such as an artificial satellite, spin is isotrophic, .... that is it can be non-inertially set, however inertial precessional forces will bring the spin in line with the inertial circulation given a suitable length of time.

IMO

:)

Zarkov,

You probably have angular momentum in mind. The spin (as a quantum mechanical variable) of the earth is unknown, and it will be for quite some time... not to say never.

Whatever relation you can make between superconduction and the earths orbit should require a little more explanation, and perhaps also a little more thought.

Bye!

Crisp

Originally posted by cephas1012
what are some equatins involving spin?


A couple of the more important ones are:

1.) Allowed Values of Spin Eigenvalues
This depends on the magnitude of the spin of the particle. When we say “spin-1/2” or “spin-1” particles, we mean that one of the components of the spin has a maximum magnitude of (1/2)(h/2p) J^.s and (h/2p) J^.s, respectively. In general, particles are “spin s”. The allowed values of the components are then:

S_i=m_s(h/2p),

where –s<=m_s<=s.

The magnitude of the total spin vector is:

|S|=(s(s+1))^1/2(h/2p).


2.) Commutation Relations of Spin Operators
Spin operators (S_i, S²) satisfy the same Lie algebra as orbital angular momentum operators (L_i, L²), namely:

[S_i,S_j]=ie_ijk(h/2p)S_k
[S_i,S²]=0

The first relation implies that all 3 components of spin cannot be known simultaneously (except of course when they are all zero). Thus, there is an uncertainty relation among the components of spin, as well as those of orbital angular momentum by virtue of the same Lie algebra.

3.) Properties of Spin States Under Rotation
The operator representing a rotation of q radians about an axis defined by a unit vector n is:

D(q;n)=exp(-iS^.nq/(h/2p))

One of the most bizarre predictions of quantum mechanics involves this operator on spin-1/2 states for rotations of 2p radians. The most general such state is:

|a>=c₊|+>+c_-|->

For simplicity, let the axis of rotation be the z-axis (since I am using the z-diagonal basis of the spin states). Let the operator D(2p;k) act on the state |a>:

D(2p;k)|a>= c₊exp(-iS_z(2p)/(h/2p))|+>+c_-exp(-iS_z(2p)/(h/2p))|->
D(2p;k)|a>=c₊exp(-ip)|+>+c_-exp(ip)|->
D(2p;k)|a>=c₊(-1)|+>+c_-(-1)|->
D(2p;k)|a>=-|a>

What this means is that if you rotate a spin-1/2 particle (such as an electron) about any axis by 2p radians, you do not get the same state back. This has no classical analog. Imagine if a friend of yours was standing facing you, and then he were to spin around one complete revolution. You would expect him to be facing you again afterwards. But this behavior of spin-1/2 particles is such that it is not “facing you” after one revolution. It has to go through two complete revolutions to get back to its starting position.

edit: fixed a few font brackets

Tom2,
thanks for the nice description.

Originally posted by Tom2
3.) Properties of Spin States Under Rotation
The operator representing a rotation...

...
What this means is that if you rotate a spin-1/2 particle (such as an electron) about any axis by 2p radians, you do not get the same state back. This has no classical analog. Imagine if a friend of yours was standing facing you, and then he were to spin around one complete revolution. You would expect him to be facing you again afterwards. But this behavior of spin-1/2 particles is such that it is not “facing you” after one revolution. It has to go through two complete revolutions to get back to its starting position.

specifically:...This has no classical analog.and then:Imagine if a friend of yours was standing facing you, and then he were to spin around one complete revolution. You would expect him to be facing you again afterwards. But this behavior of spin-1/2 particles is such that it is not “facing you” after one revolution. It has to go through two complete revolutions to get back to its starting position.Inconsistent? What do you mean by this analogy? What is the meaning of "facing you?"

Originally posted by errandir
Inconsistent?


What exactly is your question?


What do you mean by this analogy?


Macroscopic objects are in the same state after being rotated by 2p radians.

Spin-1/2 particles are not.


What is the meaning of "facing you?"


What do you think? :rolleyes:

Tom 2 has given a good mathematical description of spin, now I would like to give a good verbal one. Spin is associated with the phase of a quantum mechanical amplitude, it is the relation between the phase and the angle of rotation in the plane that spin is defined in. A spin of one simply represents one radian of quantum mechanical phase per radian of plane angle.


So it is strictly a quantum concept and has no classical analog.

Tom,
Sorry; I just wasn't getting the point of your analogy, but I think I see it now. (I have a hard time with analogies.)

<font color="#FF0000"><quote><i>What this means is that if you rotate a spin-1/2 particle (such as an electron) about any axis by 2pi radians, you do not get the same state back. This has no classical analog. Imagine if a friend of yours was standing facing you, and then he were to spin around one complete revolution. You would expect him to be facing you again afterwards. But this behavior of spin-1/2 particles is such that it is not “facing you” after one revolution. It has to go through <b>two complete revolutions</b> to get back to its starting position.</i></quote></font>


<font color="#008000"><b>Exactly;

Well, then there are particles (mostly theoretical) that have a spin of about 1/4 to many others as well. This a property lately associated with the holomorphic nature of the wave functions of the particles. Another consequence of this morphing property is that there are particles with spin 2 or 4 which can be turned by just about pi/2 or pi/4 radians to get the * same state back. *
This is as difficult, for some, as to imagine time dilation or length contraction. But, then there is this book by Mir Publishers ( I don’t rem. the name/authors but it’s like a debate book and expels a lot of knowledge. Sorry! ) ,which has nice diagrammatic explanation of all this.

It’s like Tom explains it all, and with maths too! (before anyone has a chance !). Man, that was great !! ;) </b></font>

"Well, then there are particles (mostly theoretical) that have a spin of about 1/4 to many others as well."

I'll have to ask a reference on that.

Bye!

Crisp

Well, then there are particles (mostly theoretical) that have a spin of about 1/4 to many others as well

Nature allows only particles whose states are wholly symmetric or antisymmetric, mixed states are forbidden (mixed states in 2-D have been observed in simulations and predicted theoretically, but we are not 2 dimensional are we). Spin-statistics theorem tells us that particles with half integer spin obey Fermi-Dirac statistics, and particles with integer spin obey Bose-Einstein statistics. At high temperatures these 2 converge to the classical Maxwell boltzmann distribution.

I cannot see how the particles with spin other than these 2 cases can exist, and if they do what stats they obey because from this side of the fence, all possibilities seem to be exhausted.

So I would really love to see that reference.:)

<b><font color="#008000">Dear ryans and Crisp.

I’ll be glad to be able to provide the ref. but at present I’m on vacation. So there...
However, I don’t want to jump this, so as soon as I do get back, I’ll visit my Lib. and put up the ref.
<u>I just want to remind you that it is quiet an old book of Russian publication, and I read it long time back.</u> So although I believe I’m correct, I don’t want to be dogmatic. If I am late on posting on this thread, I’ll manage to PM you the ref. ; do allow me. !
Though I was hoping to have a view by James as well, but I see he has been busy else where. </font></b>


<i><font color="#FF0000"><b>To ryans :</b> If you have some comprehensive links on that BE and FD stats. for the spin theorem and their convergance with high temp. - can you please post them here or PM me. I would love to update or maybe even correct myself..!! Will wait for the links if possible. </font></i>

<b><font color="#008000">Bye…</font></b>

No theory has spin-1/4 particles. The Lie algebra I presented for angular momentum has as its lowest dimensional representation spin-1/2.

<b><font color="#008000">To save myself from further embarresment, I decided to look to the net for something about spins.
And they do have quiet a lot on whole-integer and non-half-integer spins. But it would be tiring to put out all the links.. So look through these ones for now and I’ll try to look up that book I wrote of.
BTW tell me if what’s written on these links is valid, in your opinion, that is. </font></b>

http://www.innerx.net/personal/tsmith/See.html
http://www.innerx.net/personal/tsmith/PDS3.html#symspaces

http://www.innerx.net/personal/tsmith/SegalConf.html

http://www.ebtx.com/ntx/ntx25.htm


<b><font color="#008000">I figured that 1/2 , 3/2 , 5/2 were not the only spins possible. So I searched the net and fortunately or unfortunately enough, I found no reference to a spin qno. of -1/4. However, I did, find some content on new advances in the field of supersymmetry, that advocate the thought that spin states that are not allowed by Lie Algebra are possible for the hypothesized (s)particles. Even particles having counterparts with differing spins are now being theorized, e.g. Graviton with s-2 and s-3/2 (or something like that.!!) And you can find them, listed in Google if not in the links above. :D


Oh. And BTW, here is something quoted from: </font></b> http://www.math.unc.edu/Faculty/jds/moskva.ps. <b><font color="#008000">In txt format… </font></b>

<quote><font color="#FF0000"><i><b>As field theories, a Yang-Mills `particle' can be described by a field of spin 1, while the graviton can be described by a field of spin 2. Somehow (Mother Nature?) this is related to the strict Lie algebra structures just described. For higher spin particles, however, we have quite a different story, which first caught my attention in the work of Burgers, Behrends and van Dam [Bur85, BBvD84], though I have since learned there was quite a history before and after that and major questions still remain open. By higher spin particle Lagrangians, I mean that the fields are symmetric s-tensors (sections of the symmetric s-fold tensor product of the tangent bundle). If the power is s, the field is said to be of spin s and represents a particle of spin s. Burgers, Behrends and van Dam start with a free theory with abelian gauge symmetries and calculate all possible infinitesimal (cubic) interaction terms up to the appropriate equivalence (effectively calculating the appropriate homology group, as in formal deformation theory). They then sketch the problem of finding higher order terms for the Lagrangian, but do not carry out the full calculation. In fact, according to the folklore in the subject, a consistent theory for s * 3 will require additional fields of arbitrarily high spin s. For s = 3, the statement is that all higher integral spins are needed, while for s = 4, all higher even spins. </b></i></font></qoute>

<b><font color="#008000">Looks like my self saving efforts didn’t work much… but then, I'm still working on the book………
Cya’ :) </font></b>

Tyger,

So it is strictly a quantum concept and has no classical analog.

You may have answered my question here but I'll ask it anyway.

Something that has always been a point of confusion with me, even as an analogy is the issue of describing spin as being + or -.

If I assume a basket ball is a particle and spin it and look down on it from an upper view and its is spinning CW (assign that a + spin number), I can then take a position from below it and in that view the rotation is CCW or would be a - spin.

What reference is used to assign + or - spin. It can't be the same as for the basketball.

Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius.

Spin is a vector quantity. Put your right hand in the thumb-up position. If you curl the fingers of that hand in the direction the object is spinning, then your thumb points in the direction of the vector spin.

James R.,

Spin is a vector quantity. Put your right hand in the thumb-up position. If you curl the fingers of that hand in the direction the object is spinning, then your thumb points in the direction of the vector spin.

ANS: This is still unclear. It is simular to the right hand rule but in the above statment you specify "right hand" and then say curl your fingers in the direction of spin. My right hand fingers only curl one direction. If I also use my left hand to represent opposite spin or if I become a contortionist and curl my right hand fingers backwards, it doesn't alter the direction that the thumb points.

How are you getting + or - spin from this description?

If you mean point your thumb down to make the fingers curl in the direction of spin then I see the change; except that change appears arbitrary since the direction of rotation changes depending if I am standing upright or on my head when I view the particle. That makes every particle the same physically but different for different observers?


Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius.

That makes every particle the same physically but different for different observers?"

No, the quantum mechanical spin is totally not related to something "moving around" or "spinning" (dispite the naming).

Bye!

Crisp

MacM:

<i>If you mean point your thumb down to make the fingers curl in the direction of spin then I see the change; except that change appears arbitrary since the direction of rotation changes depending if I am standing upright or on my head when I view the particle.</i>

That's exactly what I mean, and no, the direction doesn't change if you stand on your head.

Sit a basketball on the floor and spin it anticlockwise as viewed from above. According to the right hand rule, the spin vector points upwards, towards the ceiling. Now stand on your head. The direction of the spin vector is unchanged, because your fingers only curl one way and you now have to turn your hand over to have them curl in the direction of spin. The thumb still points towards the ceiling when you're done.

James R.,

That's exactly what I mean, and no, the direction doesn't change if you stand on your head.


The rotation direction does change from CCW to CW but you have to point your thumb toward you toes when standing on your head. That makes the thumb still point the same direction.

So the spin VECTOR stays the same but the physical rotation appears reversed.

Thanks


Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius.

Originally posted by MacM
What reference is used to assign + or - spin. It can't be the same as for the basketball.Convention.

There is a Hilbert space (a portion of the Hilbert space) for describing spin. The x,y,z coordinate system is arbitrary, so, the +x,+y,+z directions are arbitrary (the + sign in this case would have no physical significance). If you have something centered at the origin in this cartesian system, then the relationship between + and - may (or may not) have physical significance, though. Now consider a spin-1/2 particle. It has two possibilities (a two dimensional Hilbert space): up or down. Up in what sense? Well, by convention, call one of its states "up" and then that leaves only one option for the other state, namely "down." (Usually this is referenced to the z-axis, which is, as has been mentioned, also arbitrary).

errandir,

Clear as mud. "Arbitrary" seems to to my hangup. I have trouble making use of arbitrary directions since they seem to change with perspective.

But thanks anyway. I'll work on it.

Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius.

Originally posted by James R
That's exactly what I mean, and no, the direction doesn't change if you stand on your head.


I think this is confusing. When you say "the direction doesn't change" you mean that the vector still points towards the ceiling, which is true. But when you stand on your head, the ceiling is in fact in a different direction!

Spin, like all vectors^*, do in fact change direction under rotations. This is both intuitively clear, and easily shown formally with rotation matrices.
__________

^*I say "vector" when I actually mean "pseudovector". A pseudovector transforms exactly like a vector under rotations, but not under parity (P).

Under parity, a vector transforms as:
x-->x'=Px=-x.

Because momentum is just p=dx/dt, it transforms in the same way:
p)-->p'=Pp=-p.

Now as for pseudovectors, a typical one is orbital angular momemtum which is L=xxp. It transforms as:
L-->L'=P(xxp)=(-x)x(-p)=+L.

That is, there is no sign change under parity.


edit: fixed sizing brackets

Tom2,

But when you stand on your head, the ceiling is in fact in a different direction!

That was what it seemed like to me. that is to me the sign should change since from your new view the vector is opposite your sense of direction.

I'm glad to have some company here with the confusion.:D

Knowing to believe only half of
what you hear is a sign of
intelligence. Knowing which
half to believe will make you a
genius.

Originally posted by MacM
...arbitrary directions... ...seem to change with perspective.That's why one goes through the trouble of the coordinate system. Once you have that, then, with respect to the cartesian coordinate system, you may define a meaningful spin-up and spin-down. It is with respect to a coordinate system (and, as a matter of fact, with respect to a particular direction in a particular coordinate system) that spin-up is defined. This coordinate system may move with you (it may be your rest frame), or it may be "fixed" (the lab frame). The point is, whether or not the coordinate system is moving is beside the point. With respect to that system, you define the spin "directions."

Consider this:
How do you define "regular" up. Probably towards the ceiling/sky, or maybe, more rigorously, perpendicular to flat ground, or even more rigorously, the direction of the gradient of gravitational potential energy.

The point is that, this direction is ARBITRARY. Is it the same for the people in the US as it is for the people in Australia? No. As a matter of fact, it has unit probability for being different for ANY two observers. If I were to stand "right next to" you, even then, if we really wanted to probe the issue, we would find that the direction that you call "up" is slightly different than the direction that I call "up." This is what I mean by arbitrary. They may be a very rigorous way to define the term, but, like you said, it depends on the observer. There is nothing wrong with this, you just have to establish your coordinate system.

Originally posted by James R
Spin is a vector quantity. Put your right hand in the thumb-up position. If you curl the fingers of that hand in the direction the object is spinning, then your thumb points in the direction of the vector spin.

Spin is actually a pseudovector, or even more accurately a bivector. It changes direction upon rotation, but not upon reflection.

so, spin has no physical properties? I've been pretty confused about this and yet the fermion-boson thiny hit me even harder.... Why is it that particles have that fixed spin? Is that because of the quarks? If so, What are this quarks actually (they start to become something other than sub-atomic particles for me)?

Originally posted by curioucity
so, spin has no physical properties?
...
Why is it that particles have that fixed spin?
...
What are this quarks actually ...? Spin is physically, kind of like charge or mass or something. The total spin of a particle (even of an entire system) is fixed (as far as I know), but the individual components of spin can change. Nevertheless, it is definitely a strange concept (to me anyway). I don't know why particles have fixed spin any more than I know why they have fixed mass or charge. I think they just do (for now, but someday we may find the answer). I don't know anything more about the quarks escept that they compose the nucleons, and that they are fundamentally different than electrons, electrons being leptons and quarks being ... (I think that they are part of some major category of fermions).

Originally posted by cephas1012
And what values can it take?

an interesting fact: particles that live in 2 dimensions can have any spin they like. such particles are called anyons.

Originally posted by errandir
Spin is physically, kind of like charge or mass or something. The total spin of a particle (even of an entire system) is fixed (as far as I know), but the individual components of spin can change. Nevertheless, it is definitely a strange concept (to me anyway). I don't know why particles have fixed spin any more than I know why they have fixed mass or charge. I think they just do (for now, but someday we may find the answer). I don't know anything more about the quarks escept that they compose the nucleons, and that they are fundamentally different than electrons, electrons being leptons and quarks being ... (I think that they are part of some major category of fermions).

we name particles according to their transformation quantum numbers, which means mass, and spin for spacetime transformations, color, weak isospin, and hypercharge for internal symmetries. particles live in "irreducible representations" of these transformation laws, which means that if you have a state that changes charge under a particular transformation, then you can break it up into pieces, each of which does not change charge under the transformation.

by construction, we only call particles those things which live in the unchanging part of the representation, because this is convenient. so in a sense, you can say that spin, charge, mass, etc. are fixed for a given particle by definition of that particle, not by any physical principle.