Quantum Mechanics: Spin

[JayMark]

I just wanted to gain basic informations about what we call "spin" in quantum mechanics.

So far, when I look at fundamental properties of elementary particles I understand pretty much what they (basically) mean but as for spin, it seems a little more abstract.

It is often refered to as beeing an "intristic degree of freedom" but what does that basically mean? How does that affect their behaviors, interactions etc? How is that observed and determined?

Please, I would like if you could explain that as simply as possible. I have checked wikipedia but it gets a little confusing for now.

Thanks a bunch.

Peace.

[Mr Right Wing]

I just wanted to gain basic informations about what we call "spin" in quantum mechanics.

So far, when I look at fundamental properties of elementary particles I understand pretty much what they (basically) mean but as for spin, it seems a little more abstract.

It is often refered to as beeing an "intristic degree of freedom" but what does that basically mean? How does that affect their behaviors, interactions etc? How is that observed and determined?

Please, I would like if you could explain that as simply as possible. I have checked wikipedia but it gets a little confusing for now.

Thanks a bunch.

Peace.

Atomic spin is different from the way planets spin.

A planet can rotate 360 degrees over time. At one point it might be 45 degrees into the rotation, another point it might be 187 degrees etc.

Particle rotation is quantised into fixed values and doesnt change. As an example the spin for an electron is always 1/2. Its a waste of time trying to picture it because they dont work like planets. The electron with a spin value of 1/2 requires 720 degrees rotation to end up back where it started.

Edited April 4, 2012 by Mr Right Wing

[Pyridium]

Let's start with the basic splitting of 2 hydrogen atoms, by direct collision of the nuclei. This results in 6 quarks to be "ripped" into smaller clumps of matter that are unstable. A clump, call it what you want, of negative charge has a specific twist and was a part of the stable quark from which it came from. The visual display is not a single photon being fired off from the collision, it is in fact thousands of fundamental matter breaking off 1 by 1. This fragment of the quark will continue to fire off single particles until if finds a stable state. Once stable it just floats around waiting to be picked up by other stable clumps. This was a negative fragment and we could see all the light show. It curved constantly in a counter clockwise circle, then it just dissappeared.

That is spin. An atom or fragment of an atom spins one way if negative and spins the opposite direction when positively charged. We can see the decay photons during a collision. A clump of matter with a neutral charge always fly, (decay), apart in a straight line.

Since there are thousands of types of clumps of matter smaller then the quark, we see them rarely in the LHC and CERN. Take a look at about 100 imagies of collisions and see just how many different results you get. Some particles spin in wider circles, some barely spin at all, others spin so fast it makes circles on the screen, look for the spins that are in the opposite direction. Watch all the different lengths of the straight lines. No 2 clumps are identical in their decay.

[StarMountainKid]

Spin is one of the four quantum numbers that describe a particle's quantum state. I think in quantum mechanics, you just have to primarily accept the mathematical description, then try to use your imagination to visualize what the mathematics describes.

This doesn't mean that one's visualization has any relationship to the real quantum world.

[JayMark]

Let's start with the basic splitting of 2 hydrogen atoms, by direct collision of the nuclei. This results in 6 quarks to be "ripped" into smaller clumps of matter that are unstable. A clump, call it what you want, of negative charge has a specific twist and was a part of the stable quark from which it came from. The visual display is not a single photon being fired off from the collision, it is in fact thousands of fundamental matter breaking off 1 by 1. This fragment of the quark will continue to fire off single particles until if finds a stable state. Once stable it just floats around waiting to be picked up by other stable clumps. This was a negative fragment and we could see all the light show. It curved constantly in a counter clockwise circle, then it just dissappeared.

That is spin. An atom or fragment of an atom spins one way if negative and spins the opposite direction when positively charged. We can see the decay photons during a collision. A clump of matter with a neutral charge always fly, (decay), apart in a straight line.

Since there are thousands of types of clumps of matter smaller then the quark, we see them rarely in the LHC and CERN. Take a look at about 100 imagies of collisions and see just how many different results you get. Some particles spin in wider circles, some barely spin at all, others spin so fast it makes circles on the screen, look for the spins that are in the opposite direction. Watch all the different lengths of the straight lines. No 2 clumps are identical in their decay.

Thanks a bunch. I'm strating to see the picture there. I think about those pictures showing the reslut of a collision (all the lines).

Now how is that so important? Is there anything "conservative" about spin numbers? If I look at a proton (uud) all quarks have 1/2 spin and so does the proton. But if I look at the omega- (sss) it has a 3/2 spin. Why is it diffrent for both baryons if all quarks have a half-integer spin? Does the electric charge has anything to do with this? Or mass? I guess so.

Now more specifically, I can see (so far in a basic model) that spin numbers are either 1/2, 3/2 or 1. Is the "curvature" more about the electric charge, only that a higher spin will result in a more "wider" curve simply because a higher spin is like a "higher degree of freedom"? Would it be fair to say that neutral particles will go off in a more straight line only that a higher spin will also result in a "longer line"?

I'll wait for your comeback bofore going further into it.

Peace.

[Pyridium]

A clump of subatomic particle that is stable, will act as a whole and all mass will share the charge as each fundamental particle rubs, vibrates, against its bonding partner in the chain, or string. When you have a stable string broken apart, there must be a decay, shedding off, pieces until the string is stable again. This break could produce 2, 3, 4 or more pieces of strings that must decay individually and this is what we see in the collision photos. This is what I call the sparkler affect. You are searching for hard core answers, I am just painting a picture.

[sepulchrave]

[This is a very qualitative explanation]

Mathematically, ``degrees of freedom'' are free parameters that can be chosen arbitrarily (in pure math) or determined experimentally (in real-world physics).

We have a world that is full of particles and forces. We have laws, like gravity, that describe how particles may move. For a particle to be affected by gravity, it must have some quantitative value that describes how it couples to the gravitational field. This value is ``mass''. But the theory doesn't predict what value the mass is... we need to do an experiment.

It is the same for ``charge'', ``spin'', and for some particles; ``colour'', and ``flavour'', and in some situations; ``angular momentum'', etc.

Spin-statistics theory predicts two classes of particles, called ``fermions'' and ``bosons'', which are separated by their symmetries. Bosons are symmetric, meaning that you can exchange two bosons without changing the system dynamics. Fermions are antisymmetric, meaning that if you exchange two particles the over-all description will be the negative of the original description.

If you ask the question ``what happens if I have two identical particles in the same place at the same time?'' you can see that this is allowed for Bosons (for particles a and b, symmetry requires ab = ba), but not allowed for Fermions (for particles a and b, antisymmetry requires ab = -ba, but since they are in the same place and identical we know that ab = ba, the only possibility that satisfies both ab = -ba and ab = ba is ab = 0).

From that it is easy to guess that Fermions represent matter (you can't have two objects in the same place at the same time), and Bosons represent forces or fields (you can have two forces in the same place at the same time).

Thanks a bunch. I'm strating to see the picture there. I think about those pictures showing the reslut of a collision (all the lines).Now how is that so important? Is there anything "conservative" about spin numbers?

Spin is conserved but it is directional.

If I look at a proton (uud) all quarks have 1/2 spin and so does the proton. But if I look at the omega- (sss) it has a 3/2 spin. Why is it diffrent for both baryons if all quarks have a half-integer spin?

The magnitude of a quark's spin is 1/2, but a quark can have a spin of -1/2 or +1/2.

How can you add together three of these spins? You have four options:

• -1/2 - 1/2 - 1/2 = -3/2
  
• -1/2 - 1/2 + 1/2 = -1/2
  
• -1/2 + 1/2 + 1/2 = +1/2
  
• +1/2 + 1/2 + 1/2 = +3/2
  

You can clearly see that the possible combination of spins is split into two classes¹, the ``low spin'' class with total spin +/- 1/2 and the ``high spin'' class with total spin of +/- 3/2. Since the difference between a spin of +1/2 and -1/2 is more or less just a property of the direction of the observer (just as if you look at a clock from the front, the hands go clockwise, but if you looked at it from the back the hands would go counter-clockwise), we expect only two different classes of particles made with 3 quarks:

• ``low spin'' particles with total spin magnitude of 1/2 (protons = uud, neutrons = udd)
  
• ``high spin'' particles with total spin magnitude of 3/2 (delta⁺ = uud, delta⁰ = udd, see the wiki)
  

Of course the higher the spin the higher the energy (note that a delta⁺ is made of the same quarks as a proton, but has a mass ~20% larger); so it is possible for a high spin 3/2 particle to collapse into low spin 1/2 particles, so we expect delta particles to be unstable (and they are).

Does the electric charge has anything to do with this? Or mass? I guess so.

Charge and mass have an effect on stability (for the same reasons, particles with a large charge magnitude and a large mass magnitude can probably decay into low charge, low mass particles) but not the total spin.

Now more specifically, I can see (so far in a basic model) that spin numbers are either 1/2, 3/2 or 1. Is the "curvature" more about the electric charge, only that a higher spin will result in a more "wider" curve simply because a higher spin is like a "higher degree of freedom"? Would it be fair to say that neutral particles will go off in a more straight line only that a higher spin will also result in a "longer line"?

I assume you are talking about the curve in the sense of trajectories of charged particles in magnetic fields after collisions? (Like in a bubble chamber image?)

Spin has nothing to do with the curve of a particle in a bubble chamber. However, high spin particles also tend to be high mass, and high mass makes these particles curve less sharply than low mass particles. A high spin will have no effect on the length of the line that a neutral particle will travel (but again, high spin tend to be high mass and tend to be unstable, the length of the line is related to the stability of the particle since it will eventually decay and therefore a single line will branch out into many lines).

CERN has a step-by-step article on how to interpret bubble chamber images, see here.

**************

[1] Actual spin-statistics is somewhat more complicated, for example 3 quark have a ``spin space'' of {1/2} x {1/2} x {1/2} which decomposes into the fundamental {1/2} + {1/2} + {3/2} spaces, but without getting into the Lie algebra of SU(2) the above explanation will suffice.

[JayMark]

[This is a very qualitative explanation]

Mathematically, ``degrees of freedom'' are free parameters that can be chosen arbitrarily (in pure math) or determined experimentally (in real-world physics).

We have a world that is full of particles and forces. We have laws, like gravity, that describe how particles may move. For a particle to be affected by gravity, it must have some quantitative value that describes how it couples to the gravitational field. This value is ``mass''. But the theory doesn't predict what value the mass is... we need to do an experiment.

It is the same for ``charge'', ``spin'', and for some particles; ``colour'', and ``flavour'', and in some situations; ``angular momentum'', etc.

Spin-statistics theory predicts two classes of particles, called ``fermions'' and ``bosons'', which are separated by their symmetries. Bosons are symmetric, meaning that you can exchange two bosons without changing the system dynamics. Fermions are antisymmetric, meaning that if you exchange two particles the over-all description will be the negative of the original description.

If you ask the question ``what happens if I have two identical particles in the same place at the same time?'' you can see that this is allowed for Bosons (for particles a and b, symmetry requires ab = ba), but not allowed for Fermions (for particles a and b, antisymmetry requires ab = -ba, but since they are in the same place and identical we know that ab = ba, the only possibility that satisfies both ab = -ba and ab = ba is ab = 0).

From that it is easy to guess that Fermions represent matter (you can't have two objects in the same place at the same time), and Bosons represent forces or fields (you can have two forces in the same place at the same time).

Spin is conserved but it is directional.

The magnitude of a quark's spin is 1/2, but a quark can have a spin of -1/2 or +1/2.

How can you add together three of these spins? You have four options:

• -1/2 - 1/2 - 1/2 = -3/2
  
• -1/2 - 1/2 + 1/2 = -1/2
  
• -1/2 + 1/2 + 1/2 = +1/2
  
• +1/2 + 1/2 + 1/2 = +3/2
  

You can clearly see that the possible combination of spins is split into two classes¹, the ``low spin'' class with total spin +/- 1/2 and the ``high spin'' class with total spin of +/- 3/2. Since the difference between a spin of +1/2 and -1/2 is more or less just a property of the direction of the observer (just as if you look at a clock from the front, the hands go clockwise, but if you looked at it from the back the hands would go counter-clockwise), we expect only two different classes of particles made with 3 quarks:

• ``low spin'' particles with total spin magnitude of 1/2 (protons = uud, neutrons = udd)
  
• ``high spin'' particles with total spin magnitude of 3/2 (delta⁺ = uud, delta⁰ = udd, see the wiki)
  

Of course the higher the spin the higher the energy (note that a delta⁺ is made of the same quarks as a proton, but has a mass ~20% larger); so it is possible for a high spin 3/2 particle to collapse into low spin 1/2 particles, so we expect delta particles to be unstable (and they are).

Charge and mass have an effect on stability (for the same reasons, particles with a large charge magnitude and a large mass magnitude can probably decay into low charge, low mass particles) but not the total spin.

I assume you are talking about the curve in the sense of trajectories of charged particles in magnetic fields after collisions? (Like in a bubble chamber image?)

Spin has nothing to do with the curve of a particle in a bubble chamber. However, high spin particles also tend to be high mass, and high mass makes these particles curve less sharply than low mass particles. A high spin will have no effect on the length of the line that a neutral particle will travel (but again, high spin tend to be high mass and tend to be unstable, the length of the line is related to the stability of the particle since it will eventually decay and therefore a single line will branch out into many lines).

CERN has a step-by-step article on how to interpret bubble chamber images, see here.

**************

[1] Actual spin-statistics is somewhat more complicated, for example 3 quark have a ``spin space'' of {1/2} x {1/2} x {1/2} which decomposes into the fundamental {1/2} + {1/2} + {3/2} spaces, but without getting into the Lie algebra of SU(2) the above explanation will suffice.

Woah! That helped! Thanks bro!

So baryons will have a spin number that would be the sum of the quark's spin numbers right? So a proton (1/2) would have to be composed of a -1/2, 1/2 and 1/2 quark right? Just like the omega^- will be composed of three 1/2 quarks? Fair enough.

But how is the "direction" (positive or negative spin number) determined within a baryon? How do we know which quark has a positive spin and which one a negative (if any like in the case let's say of a proton)?

So the fact that whether or not the spin number will be positive or negative for a same, let's say quark, depends on the position of the observer relative to the particle? Is that how it is? It would make sense because if I look at a standard model of elementary particles/forces, spin numbers are all positive so my guess is that they only are negative depending on their relative (to the observer) movement/direction. Would that mean that 2 of the same hadrons moving in the LHC with opposing directions have equivalent but opposed spin numbers?

Thanks for the help.

Peace.

[sepulchrave]

So baryons will have a spin number that would be the sum of the quark's spin numbers right? So a proton (1/2) would have to be composed of a -1/2, 1/2 and 1/2 quark right? Just like the omega^- will be composed of three 1/2 quarks? Fair enough.

It is better to say ``a proton is two parallel quarks and one antiparallel quark, while an omega^- is three parallel quarks''.

But how is the "direction" (positive or negative spin number) determined within a baryon? How do we know which quark has a positive spin and which one a negative (if any like in the case let's say of a proton)?

We don't, but remember direction of spin is just a question of which direction the observer is looking. In other words, we define what ``positive spin'' and ``negative spin'' are in the laboratory.

If we define a spin axis and measure a bunch of protons, we will find some with spin +1/2 (i.e. composed of +1/2, +1/2, -1/2 quarks) and some with spin -1/2 (i.e. composed of +1/2, -1/2, -1/2 quarks), relative to that axis. The same is true with the omega^- [1].

So we can convert a spin +1/2 proton into a spin -1/2 proton quite easily (put the proton through an appropriate magnetic field), but we can't convert a spin +1/2 proton into a +3/2 spin state (or -3/2 spin state) without changing the internal structure (i.e. converting the proton into a delta⁺ - which is possible, of course, but only in high-energy particle collisions).

So the fact that whether or not the spin number will be positive or negative for a same, let's say quark, depends on the position of the observer relative to the particle? Is that how it is? It would make sense because if I look at a standard model of elementary particles/forces, spin numbers are all positive so my guess is that they only are negative depending on their relative (to the observer) movement/direction.

Exactly, the magnitude of the spin is important. The actual sign (or technically the projection) of the spin is a consequence of the external environment.

Would that mean that 2 of the same hadrons moving in the LHC with opposing directions have equivalent but opposed spin numbers?

Maybe, maybe not. Spin is not coupled to the direction the particle is moving (in the non-relativistic sense, at least). If the particles were created from a single decay, then it depends on the spin of that particle.

For example, a gamma-ray photon (spin = 1) might separate into a positron and an electron. In this case both would have the same spin alignment (if the photon was +1, then both e^- and e⁺ would be +1/2, to add to +1, if the photon was -1, then both would be -1/2 to add to -1).

**********

[1] Things are actually more complicated for the omega^-. With a spin magnitude of 3/2, one could never-the-less measure spin projections of -3/2, -1/2, +1/2, or +3/2. Measuring a spin projection of -1/2 does not mean that the omega^- has a total spin of 1/2 though. Quantum mechanics is more complicated than just adding together fractions, after all.

[JayMark]

It is better to say ``a proton is two parallel quarks and one antiparallel quark, while an omega^- is three parallel quarks''.

We don't, but remember direction of spin is just a question of which direction the observer is looking. In other words, we define what ``positive spin'' and ``negative spin'' are in the laboratory.

If we define a spin axis and measure a bunch of protons, we will find some with spin +1/2 (i.e. composed of +1/2, +1/2, -1/2 quarks) and some with spin -1/2 (i.e. composed of +1/2, -1/2, -1/2 quarks), relative to that axis. The same is true with the omega^- [1].

So we can convert a spin +1/2 proton into a spin -1/2 proton quite easily (put the proton through an appropriate magnetic field), but we can't convert a spin +1/2 proton into a +3/2 spin state (or -3/2 spin state) without changing the internal structure (i.e. converting the proton into a delta⁺ - which is possible, of course, but only in high-energy particle collisions).

Exactly, the magnitude of the spin is important. The actual sign (or technically the projection) of the spin is a consequence of the external environment.

Maybe, maybe not. Spin is not coupled to the direction the particle is moving (in the non-relativistic sense, at least). If the particles were created from a single decay, then it depends on the spin of that particle.

For example, a gamma-ray photon (spin = 1) might separate into a positron and an electron. In this case both would have the same spin alignment (if the photon was +1, then both e^- and e⁺ would be +1/2, to add to +1, if the photon was -1, then both would be -1/2 to add to -1).

**********

[1] Things are actually more complicated for the omega^-. With a spin magnitude of 3/2, one could never-the-less measure spin projections of -3/2, -1/2, +1/2, or +3/2. Measuring a spin projection of -1/2 does not mean that the omega^- has a total spin of 1/2 though. Quantum mechanics is more complicated than just adding together fractions, after all.

Well thank you. That was very informative and quite easy to undertsand.

I'll post back if anything else comes to my mind.

Peace.