Spin +1 vs -1: Does Antimatter Spin Differently?

[lufc88]

does spin +1 mean a particle is 'spinning' in a different direction to a -1 particle or is it a magnitude of spin?
does antimatter spin in the opposite direction to normal matter?

 

[edguy99]

does spin +1 mean a particle is 'spinning' in a different direction to a -1 particle or is it a magnitude of spin?
does antimatter spin in the opposite direction to normal matter?

Hi there, Antimatter spin does indeed spin in the opposite direction to normal matter in the sense that right hand spin stands for magnetic up, or right hand spin stands for magnetic down. If the magnetic direction of the particle and antiparticle were the same, the direction of spin would be opposite. If the spin of the particle and antiparticle were the same, the magnetic direction (north/south) would be opposite.

Spin 1 means the particle has 3 different "states" of spin. Visualize a proton or a neuton alone, they are both spin 1/2 particles that have 2 states - up or down. Now visualize a proton and neutron combined in hydrogen2 (commonly known as deuterium). This is a spin 1 particle as both the proton and neutron can be "up" for +1, the proton and neutron can have opposite spins for 0, or the proton and neutron can both be spin down for the third state of -1. The +1 hydrogen2 particle would go "north", the -1 hydrogen2 would go "south" and the 0 hydrogen2 spin particle would not be affected by a magnetic field.

Hope this is of some assistance.

 

[Bill_K]

Antimatter spin does indeed spin in the opposite direction to normal matter in the sense that right hand spin stands for magnetic up, or right hand spin stands for magnetic down. If the magnetic direction of the particle and antiparticle were the same, the direction of spin would be opposite. If the spin of the particle and antiparticle were the same, the magnetic direction (north/south) would be opposite.

I believe what edguy99 is trying to say is that since an electron is negatively charged its magnetic dipole moment is antiparallel to its spin, while for a positron the magnetic moment will be parallel to the spin since its charge is positive.

 

[lufc88]

where would the 1/2 particle go half north south and thank you for your explanations but why 'If the magnetic direction of the particle and antiparticle were the same, the direction of spin would be opposite'

 

[andrewr]

where would the 1/2 particle go half north south and thank you for your explanations but why 'If the magnetic direction of the particle and antiparticle were the same, the direction of spin would be opposite'

? I'm not sure what you're asking.

Spin itself, the number, is a measure of energy. It is not energy seen in translational motion, so it is indeed some kind of circulating energy. (Standing wave.)

So, spin 1 -- in units of plank's constant (energy/rotation or Hz) has double the energy of a spin 1/2 particle. The 'location' of this energy is not easily obtainable -- just as the energy (power) in a poynting vector for coax cable power standing waves is intuitively wrong. There is a "problem" with locating the exact place that the energy resides.

There really isn't a spin 1/2 in isolation, though, as spin is has to be coupled no matter how weakly for it to exist (afaik). i(in QM all particles have spin, and only composite objects can have spin 0) For example, an electron in Hydrogen is coupled to the spin of the proton nucleus -- and the total magnetic effect is spin 1 or 0. (usually/eventually 0...) Thus experiments don't measure 1/2 bohr magneton -- they always measure in integer units as far as I know. And, don't forget (if you knew it before) that a magnetic field requires two dimensions to vary in order to exist -- (classically and QM) -- so there is no such thing as measuring "spin" in one direction. Most authors will cite a continuous distribution of mass (any distribution) rotating as the only classical model worth considering for "spin", but that is clearly an error -- for mass and momentum which are interacting do not have to rotate uniformly as the classical electron radii theory predicts. There are clearly ways to make mistakes in identifying the location of energy in a pair of interacting masses -- or even in two orthogonal waves. So, although visualizing an electron as spinning like the Earth is wrong in a certain sense -- that does not mean that a physical system can't be constructed that has rotational motion and mass but does not move in a continuous fashion with respect to that mass. It's jerky motion, though, only has to have the properties to allow the emergence of true rotational motion when it *interacts* with another object. Of course, you're free to be close minded and ignore the issue -- QM is perfectly workable without understanding the meaning of "spin".

 

[lufc88]

thanks for your reply but my second question was in reply to

The +1 hydrogen2 particle would go "north", the -1 hydrogen2 would go "south" and the 0 hydrogen2 spin particle would not be affected by a magnetic field.

that's what i was asking about the half spin particle for

 

[andrewr]

thanks for your reply but my second question was in reply to that's what i was asking about the half spin particle for

I see. You're welcome. I hope your question is sufficiently answered, then.

As a side note:
I'm not entirely certain that a spin zero particle would be unaffected. A magnetic field can be demonstrated to have an effect on even a spin zero hydrogen atom. cf; fine and hyperfine / spin orbit coupling. Spin orbit coupling in hydrogen being *like* all other spin orbit coupling of all other atoms... and an integer spin as a whole; Suggests that one might be able to treat two or more hydrogens as two bosons when *far* apart so the antisymmetry requirement of the constituent fermions (proton and electron) are not dominant.

I am reasoning from the fact that protons have constituent sub-atomic particleS with spin and yet the proton, in its typical existence, is considered to have only a net spin equivalent to that of the electron which has "no internal structure".

It may not be gravity ... but there are going to be locations of lowest energy corresponding to the characteristic shape of the box (universe/local and/or local gas bottle) the hydrogens are in -- and they are BOTH/all going to be attracted to it/them. If there is spin based repulsion without considering the coulomb potential, I don't see anything stopping spin based attraction. (hall effect, too...)

 

[lufc88]

thank you and woah please elaborate on spin based repulsion and attraction

 

[andrewr]

thank you and woah please elaborate on spin based repulsion and attraction

I am not sure I will be very accessible to you.
The basics of QM can be found on many sites and doing some random Google searches will often bring up Russian, Spanish, French and other languages if you are more comfortable in one of those languages.

There are different approaches that can be taken, but the central idea in QM that one has to accept is associated with Heisenburg's uncertainty principle. There are philosophical and physical issues with what he proposed, and although it is certainly part of the problem and solution -- try to carefully study what other's say before jumping to conclusions. Science is rarely, if ever, advanced peaceably and seldom (if ever) have those who are popular been completely correct -- and I mean all throughout history. Much of history is written by those who are left or perceived more creditable -- for the experimenters themselves got caught in the experiment or politics. eg: Madame Curie, Heisenberg, and even Isaac Newton (see his rings and critiques about his actions)

Heisenberg's principle -- as all principles -- is one of exclusion. Something must be, because something else "CANT" be. It's both a powerful and useful way of arguing -- and it is also always subject to boundary conditions that can be quite slippery.

A mathematician's puzzle of the man going for execution who is granted a wish, is a good example of exclusion and inductive reasoning. A man is condemned to be executed before the anniversary of his crime -- but he is given a request. So, the man, thinking himself wise asks that he "not" know the day he will die. The judge assents. So the man argues with himself -- if I am not dead by the day before the anniversary -- then I will know that I must die the next day; for that's the law. Therefore, they can't come for me the day before. And then he reasons, so -- if I were to be alive the day before that -- and they are forbidden to kill me on the next day, they must kill me on this one. But that is not allowed --because I will know that I must die; therefore I am safe on that day. And so, he reasons back one more day, and comes to the same result. It is mathematically a solid argument -- and yet, obviously -- he really didn't know which day they would kill him on -- so he died.

Heisenburg argues that one "can't" know certain things without changing them. He is at least partially right -- but that doesn't mean one ought to be smug about the uncertainty; you never know.

The celebrated principle is put into mathematical form in what are called commutation operators. They try to put boundaries on what can and can't be known -- and if that isn't disturbing, it should be.
But it is useful as a "guess". The popular way that agrees with experiments in "one" dimension is called commutation operators; and the word "tensor" will come up -- which is a headache, that doesn't really mean much to me. Sometimes calculus operators such as derivatives can be treated as if they were algebraic variables; so instead of treating d/dx as a command to solve something, it is treated more like an unknown 'x' and one mixes and rearranges things until it looks easier (at least it gives you time to think while mixing, and mixing, and... well...). And problems that are really messy to figure out in one's head start to look more like basic algebra problems (or, more to the irritating problem, like matrix algebra if you are familiar with the idea of "non-commutation") Some things you just CAN'T do.

So, in any event -- like the prisoner -- many people today are starting to argue a "multi-verse" instead of a "uni-verse", etc. And, well -- they can die in as many dimensions as they'd like in their imagination and I have no mind control or tension knob to help them with -- so I hope they make their money fast and with me in some other cell...)

In any event, I'm possibly going to be off topic -- I'm not certain -- so perhaps I'm really safe --perhaps NOT. It's sort of like spelling on this site, try to spell Heisenburg's name with the phonetic "U" like Germans do ... It won't work. It must be Heisenberg when anyone here sees it -- ΝΟ θνψερταιντυ.
And before you correct me --- really ---

Let's say I live in "Scappoose", and my sister's GERMAN (and wealthy, and don't you forget it!) mom in-law, malice, malise,?? (no joke) don't remember the spelling -- well -- she writes to us in Skappose -- and the letter STILL gets here. It's that kind of thing; so do I put her in her place -- or do I let her think she can put me and my sister anyplace?

So, to get to the point -- now that I have made it vague enough -- If you are working with "bosons" the commutation operators work one way -- and if you are working with "fermions" they work the OTHER way. The argument, half way, is that these things "can't" be in the same place; and when you go all the way -- it's there together. When one asks "why", well it's because they can't or we would know something we can't know. That's the only QM reason there *is* for the PRESSURE when FERMIONS get put together -- but Boson's can occupy each other's space, happily, gaily (in a good way); those are the whole spin things. Only the ones who spin half way are called ODD spin. Now the odd ones cause pressure and the even ones like to be in the same place -- so even though it would seem there is a simple consensus since one turns into the other when combined, and the even ones go together -- that everything would eventually be just EVEN. But that doesn't work. You see? It's a natural dilemma.

But Heisenburg CAN'T be wrong -- we're told. (Of course, those aren't Heisenburg's words -- any more than this site is a US military site trying to deny that any of this exists by filling it with garbage and pretending to be a foreign country which is the most effective way to hide the truth. Or perhaps they just want to be able to say the truth without being executed.? ) Well I can't help it... or I'll be shot. So, just be gentle -- is all I ask.

Now, the formula and argument goes like this:
Spin odd, 1/2, 3/2, 5/2, so forth... those things must occupy DIFFERENT states. They will fight otherwise. Therefore, since they must all be in different States -- there has to be "ROOM" for them all. The closer they are together the more likely one of them is going to violate the different "space" thing -- final frontier is ex-election. So, do the math -- and one gets a change in energy for a change in space -- and that's it. dE over a change in surface (fermi - level) equals pressure.

If you don't believe me -- study -- study -- study... I don't have to prove it -- it will be beaten into you by the WHOLE world. Now, at least I didn't make you do the math the hard way -- and although you may not thank me now; it doesn't matter. Peace to you, I'll stay in my space thanks.

 

[lufc88]

thank you for your time

 

[SpectraCat]

? I'm not sure what you're asking.

There really isn't a spin 1/2 in isolation, though, as spin is has to be coupled no matter how weakly for it to exist (afaik). i(in QM all particles have spin, and only composite objects can have spin 0) For example, an electron in Hydrogen is coupled to the spin of the proton nucleus -- and the total magnetic effect is spin 1 or 0. (usually/eventually 0...)

No .. I don't think that is really correct. The response of a particle is based on it's magnetic moment, not its total spin. The Bohr magneton (which is the unit for magnetic moments of electrons) is about 2000 times larger than the nuclear magneton (which is the unit for measuring magnetic moments of nuclear particles and whole nuclei). Thus, when you subject a free H atom to a magnetic field, it behaves *almost exactly* as would be expected for a particle with an electronic spin of 1/2. There is a small correction based on the non-zero nuclear magnetic moment, but it is orders of magnitude smaller than the response due to the electronic magnetic moment. See http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nspin.html for more info.

So, it really doesn't make sense to talk about an H-atom with a spin of 0 (or 1) in the context of measurements with magnetic fields ... they behave essentially as spin-1/2 particles. Consider also that they are able to use magnetic fields to trap Rb-87 atoms (or certain other alkali-metal isotopes) to make BEC's ... even though the particles are spin-0 bosons for the purposes of the condensation, the trapping still occurs through the large magnetic moment arising from the unpaired electron.

 

[andrewr]

thank you for your time

Very GOOD! You're welcome.
And now that you have shown mastery of the principle, be careful about the next time you reply -- lest you make a mistake, and everyone think this answer you gave was just dumb luck!

 

[andrewr]

No .. I don't think that is really correct. The response of a particle is based on it's magnetic moment, not its total spin.

The Bohr magneton (which is the unit for magnetic moments of electrons) is about 2000 times larger than the nuclear magneton (which is the unit for measuring magnetic moments of nuclear particles and whole nuclei). Thus, when you subject a free H atom to a magnetic field, it behaves *almost exactly* as would be expected for a particle with an electronic spin of 1/2.

Generally silver, or sodium are used for the actual experiments. I haven't seen one done using Hydrogen -- I'd love to if you have a link to a real experiment.

There is a small correction based on the non-zero nuclear magnetic moment, but it is orders of magnitude smaller than the response due to the electronic magnetic moment. See http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nspin.html for more info.

So, it really doesn't make sense to talk about an H-atom with a spin of 0 (or 1) in the context of measurements with magnetic fields ... they behave essentially as spin-1/2 particles.

I think you missed my point. I said I don't believe it exists in isolation; In order to do a Stern Gerlach experiment a magnet has to be used -- what is it that generated the magnetic field? In a permanent magnet (non superconductor based, non-magnetic coil based) the simple answer is that unpaired spin in all the atoms in the magnet cause the magnetic field. Now, if there were no magnet -- the particle in question would not be deflected at all. It would not INTERACT with a pure electric field -- but only a magnetic field. Spin 1/2 has no meaning without another object which also has spin.

In the Hydrogen atom, real experiments, the same magnitude of magnetic moment exists in a spin 1/2 electron which is NOT orbiting, as is found in the first electron in an orbital with angular spin of 1 added. The Bohr magneton is the same in both cases. Let me quote an old text (White's college physics, Harve E. White -- there is also a branch in London.) P. 462:

"The magnetic moment for a P electron orbit is one Bohn magneton, fo d-electron orbit is two bohr magnetons, etc. An s electron orbit with l=0 has no mechanical moment and no magnetic moment. "

Now the idea of spin hasn't changed in these years, so this isn't "out of date" -- though it is quite old.
The author also says in another place: "Since this is the same as equation. (52E) we see that while a spinning electron has a mechanical moment of 1/2h(bar), it has a magnetic moment of one Bohr magneton."

Thus, pointing back to what I said -- even though it is supposedly spin 1/2 -- it has the same magnetic field as an integer spin effect. (0,1,2) That is, the S orbital and the P orbital have the same magnetic strength multiplier minus some small deviation due to relativistic effects. Your questioning of my calling the particle as a whole "spin" 1 is fair -- I probably need to consider how to state what I meant in clearer terms.

All I need to show the attraction (which is clearly going to be weak) is that there is a coupling of "odd" spin with "odd" spin to make an even spin. eg: if it happens with two electrons of different spin -- it must also happen with the nucleus. The objects will align or anti-align.

The proton is assigned a "spin" of 1/2h(bar) identical to that of the electron. So, I can't agree that they are different in the sense of the energy stored in them/HERTZ -- but I CAN agree that they are different in terms of mass and how quickly they will react to a magnetic field -- which appear to be what you are referring to. It's a puzzle to me as well -- so don't think I am proclaiming an defined answer.

If you know of a free electron (1 electron) experiment that measures it's spin, then you would have my attention and I would happily reverse my remark. It may be possible, but I have never seen a true 1/2 spin experiment -- and that's why I don't believe with CERTAINTY that such a thing is possible.

Consider also that they are able to use magnetic fields to trap Rb-87 atoms (or certain other alkali-metal isotopes) to make BEC's ... even though the particles are spin-0 bosons for the purposes of the condensation, the trapping still occurs through the large magnetic moment arising from the unpaired electron.

I need to be careful; Even though I have spoken of fermions creating pressure (which is calculated in models of the "free electron gas") the easiest boson particle I know of (light photon) will also create pressure -- but due primarily to it's propagation as far as I know. That is, light being a particle in motion -- hits the walls of the container and is reflected and transfers a very small amount of momentum in doing so. This pressure is TINY compared to the pressure of fermion spin exclusion. The entire pressure of photons hitting the Earth is pretty SMALL. But, I would look at that as potential evidence that the SPIN 1 boson -- light, has some relationship to spin 1/2 particles which isn't perfectly neutralized. eg: it could be a relativistic effect within the spin 1 particle itself.

The example you give of Rb-87 simply shows a far MORE shielded particle interacting with MANY more electrons being spin 1/2. That is, the single unpaired electron is not ISOLATED but attached quite nicely (if loosely?) to an atom. None of this is showing spin 1/2 in isolation which is all I was trying to say; you are certainly free to deal with the issues I am bringing up using a different vocabulary. I am simply referring to the standard conventions used in my QM introduction book with respect to adding the spin of Nucleus and Electron. Spins add. That the addition may not manifest itself the same as a non-composite particle of spin-1, I have no way to argue against. But I am still going to maintain my SUSPICION that if twoparticle spins add to '1' they may in fact be attracted to other particles with the same configuration -- although very weakly. Do you disagree? If so -- elucidate -- I'm listening.

 

[SpectraCat]

Generally silver, or sodium are used for the actual experiments. I haven't seen one done using Hydrogen -- I'd love to if you have a link to a real experiment.

Taylor and Phipps did the experiment in 1927. It was published in the Physical Review, volume 29, p. 309, (1927). It's hard to find layman's summaries of that experiment, however it showed similar behavior to the Stern-Gerlach experiment with silver atoms, namely, the beam of H-atoms split into two components, due to the interaction of the magnetic field gradient with the magnetic moment of the atoms arising from the unpaired electron (i.e. the Bohr magneton).

I think you missed my point. I said I don't believe it exists in isolation; In order to do a Stern Gerlach experiment a magnet has to be used -- what is it that generated the magnetic field? In a permanent magnet (non superconductor based, non-magnetic coil based) the simple answer is that unpaired spin in all the atoms in the magnet cause the magnetic field. Now, if there were no magnet -- the particle in question would not be deflected at all. It would not INTERACT with a pure electric field -- but only a magnetic field. Spin 1/2 has no meaning without another object which also has spin.

How then would you explain magnetic deflection experiments with neutrons? A beam of thermal neutrons is also deflected by magnetic fields. See for example, Phys. Rev. B, volume 7, p. 4142, (1973).

Also, magnetic fields can be generated by the flow of current in conductors that are non paramagnetic or ferromagnetic, so I don't understand your point about spin-1/2 particles always being the generators of magnetic fields.

In the Hydrogen atom, real experiments, the same magnitude of magnetic moment exists in a spin 1/2 electron which is NOT orbiting, as is found in the first electron in an orbital with angular spin of 1 added. The Bohr magneton is the same in both cases. Let me quote an old text (White's college physics, Harve E. White -- there is also a branch in London.) P. 462:

"The magnetic moment for a P electron orbit is one Bohn magneton, fo d-electron orbit is two bohr magnetons, etc. An s electron orbit with l=0 has no mechanical moment and no magnetic moment. "

You have misinterpreted that quotation. All he is saying is that the contribution of the orbital angular momentum to the magnetic moment depends on the l-quantum number. He also says that the contribution is the different based on l ... for l=0 (s-orbital) there is no contribution from the orbital angular momentum, for l=1 (p-orbital), the contribution is one Bohr magneton, for l=2 (d-orbital) the contribution is 2 Bohr magnetons, etc.

The orbital angular momentum (L) is then coupled to the spin angular momentum (S), using angular momentum coupling rules, not simple arithmetic, to obtain the total angular momentum J=L + S.

The overall magnetic moment for a particular state with J, L and S can be calculated from the Lande g-factor:

[tex]g_J=g_L\frac{J(J+1)-S(S+1)+L(L+1)}{2J(J+1)} + g_S\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}[/tex]

This factor determines the scaling of the energy splitting of the m_J levels in a magnetic field with a given strength.

Note that the nuclear spin of 1/2 also interacts with the field, but the interaction is about 2000 times smaller due to the much larger mass of the proton, so that interaction can be safely ignored in the discussion above. However, it is no problem in principle to include the nuclear spin coupling .. that will give rise to what is called hyperfine splitting in the magnetic field.

Now the idea of spin hasn't changed in these years, so this isn't "out of date" -- though it is quite old.
The author also says in another place: "Since this is the same as equation. (52E) we see that while a spinning electron has a mechanical moment of 1/2h(bar), it has a magnetic moment of one Bohr magneton."

That is by definition ...

Thus, pointing back to what I said -- even though it is supposedly spin 1/2 -- it has the same magnetic field as an integer spin effect. (0,1,2) That is, the S orbital and the P orbital have the same magnetic strength multiplier minus some small deviation due to relativistic effects. Your questioning of my calling the particle as a whole "spin" 1 is fair -- I probably need to consider how to state what I meant in clearer terms.

That last is simply wrong ... see my explanation of Lande g-factors above to understand why.

All I need to show the attraction (which is clearly going to be weak) is that there is a coupling of "odd" spin with "odd" spin to make an even spin. eg: if it happens with two electrons of different spin -- it must also happen with the nucleus. The objects will align or anti-align.

This is correct, but it has to do with angular momentum only, not magnetic moments, for the reasons I have already explained.

The total angular momentum including nuclear angular momentum (F), is determined by the total electronic angular momentum J that I discussed above, and the nuclear spin angular momentum I, according to F= J + I. Again, angular momentum coupling rules, rather than simple arithmetic, are used to determine F. However, the relative sizes of the responses due to the electronic and nuclear contributions to the overall magnetic moment are still different by a factor of 2000. This is why the nuclear coupling just leads to a very small splitting of the electronic energy levels, called nuclear hyperfine splitting.

The proton is assigned a "spin" of 1/2h(bar) identical to that of the electron. So, I can't agree that they are different in the sense of the energy stored in them/HERTZ ..

No .. that is wrong ... the spin angular momentum is 1/2 hbar, but it is an angular momentum, not an energy .. and the units are joules times seconds not Hertz

... -- but I CAN agree that they are different in terms of mass and how quickly they will react to a magnetic field -- which appear to be what you are referring to. It's a puzzle to me as well -- so don't think I am proclaiming an defined answer.

There is no puzzle ... the energy splitting is not determined until the magnetic field is applied ... then the response is based on the appropriate magneton .. Bohr magneton for electrons, nuclear magneton for nuclei.

If you know of a free electron (1 electron) experiment that measures it's spin, then you would have my attention and I would happily reverse my remark. It may be possible, but I have never seen a true 1/2 spin experiment -- and that's why I don't believe with CERTAINTY that such a thing is possible.

Nothing with electrons as far as I know .. it's tricky, because the Lorentz force from the interaction of the charge with the magnetic field tends to swamp out the much smaller response due to the spin. However, neutrons also have spin 1/2, and I referenced that paper above on neutron deflection in a magnetic field ... it is just one such example .. there are plenty more in the literature.

The example you give of Rb-87 simply shows a far MORE shielded particle interacting with MANY more electrons being spin 1/2. That is, the single unpaired electron is not ISOLATED but attached quite nicely (if loosely?) to an atom. None of this is showing spin 1/2 in isolation which is all I was trying to say ...

You missed the point ... the Rb-87 atoms in the condensate are spin-0 bosons. By your arguments above, they should have no magnetic moment ... however it is the electronic magnetic moment that is used to trap them so that the Bose-Einstein condensation can be realized.

 

[andrewr]

Taylor and Phipps did the experiment in 1927. It was published in the Physical Review, volume 29, p. 309, (1927). It's hard to find layman's summaries of that experiment, however it showed similar behavior to the Stern-Gerlach experiment with silver atoms, namely, the beam of H-atoms split into two components, due to the interaction of the magnetic field gradient with the magnetic moment of the atoms arising from the unpaired electron (i.e. the Bohr magneton).
How then would you explain magnetic deflection experiments with neutrons? A beam of thermal neutrons is also deflected by magnetic fields. See for example, Phys. Rev. B, volume 7, p. 4142, (1973).

Exactly the same as I do with respect to hydrogen or any other pairing of spin 1/2 particles. A neutron is composite of an electron and a proton, both of which are spin 1/2.

Also, magnetic fields can be generated by the flow of current in conductors that are non paramagnetic or ferromagnetic, so I don't understand your point about spin-1/2 particles always being the generators of magnetic fields.

Yes, I see that. What particle is flowing in the conductor? is it "light" or an electron or a proton ?
What is the spin of the object flowing?

You have misinterpreted that quotation.

Considering you didn't understand my previous comment -- I don't think it appropriate to rebut something that I don't find has to do with my argument in a clear way.

This is correct, but it has to do with angular momentum only, not magnetic moments, for the reasons I have already explained.

And the reason I mentioned the old experiment on spin was because of how the magnetic moment is affected by changes in spin of the electron. It is integerial; Spin always increases by h(bar) between levels of increasing spin. Yet -- empirically -- the value of the bohr magneton does not start at 1/2 of something, it starts at zero -- which is the same as I would expect of a boson. It is the NET magnetic field I am speaking of -- that is all. I am not even bothering to take into account small deviations that are clearly caused by the nucleus -- they don't matter to my argument.

However, the relative sizes of the responses due to the electronic and nuclear contributions to the overall magnetic moment are still different by a factor of 2000. This is why the nuclear coupling just leads to a very small splitting of the electronic energy levels, called nuclear hyperfine splitting.

Sure, and I mentioned those to begin with and said that I wouldn't say a "spin 0" composite particle is NOT affected by a magnetic field. You are merely reinforcing my point -- not rebutting it.

No .. that is wrong ... the spin angular momentum is 1/2 hbar, but it is an angular momentum, not an energy .. and the units are joules times seconds not Hertz

BY DEFINITION JOULES times TIME is ENERGY/HERTZ. Look at the units of h(bar). It is energy in ratio to *something*, and is therefore energy -- unless TIME is energy in your world.

Nothing with electrons as far as I know .. it's tricky, because the Lorentz force from the interaction of the charge with the magnetic field tends to swamp out the much smaller response due to the spin. However, neutrons also have spin 1/2, and I referenced that paper above on neutron deflection in a magnetic field ... it is just one such example .. there are plenty more in the literature.

I'm sure there are -- and I'll get to some of those as I learn more.

You missed the point ... the Rb-87 atoms in the condensate are spin-0 bosons. By your arguments above, they should have no magnetic moment ... however it is the electronic magnetic moment that is used to trap them so that the Bose-Einstein condensation can be realized.

No, I didn't miss your point. I didn't argue they should have no magnetic moment -- you have added words to my mouth, and I don't much like that. I argued from the beginning that a spin 0 particle CAN be affected by a magnetic field -- see: fine and hyperfine and spin orbit coupling. I also said things that you might have made erroneous inferences from, and I apologize; I didn't know how to say it better at the time -- and for all these replies, I still don't.

 

[SpectraCat]

@andrewr You seem to be confused about this stuff. I encourage you to re-read the primary sources and reconsider your arguments ... some of them may be correct, but are written in such a way as to seem very wrong to someone familiar with the field because you seem to be using the wrong terms and units to describe physical phenomena (More on this below). I think I understand part of why your posts seem a bit muddled ... but first let's deal with some more erroneous statements you made in your last post ... to wit:

A neutron is composite of an electron and a proton, both of which are spin 1/2.

That is not correct ... a neutron is a spin 1/2 particle .. it can decay ([itex]\beta^-[/itex]- decay) into an electron and a proton, but must emit another particle called an electron anti-neutrino (which also has spin 1/2) in order to conserve total angular momentum. Prior to the decay event, the neutron is a single particle (according to the Standard Model anyway).

Just in case you want to still stick to your argument that a neutron is a "composite of an electron and a proton", consider the case of [itex]\beta^+[/itex]- decay, where a proton absorbs some energy and decays into 3 spin 1/2 particles: a neutron plus a positron (anti-electron) and an electron neutrino. So are you going to say that a proton is a "composite of a neutron and a positron"?

BY DEFINITION JOULES times TIME is ENERGY/HERTZ. Look at the units of h(bar). It is energy in ratio to *something*, and is therefore energy -- unless TIME is energy in your world.

It is most certainly not energy .. it is angular momentum. Force has units of energy per unit length, is force energy? Acceleration has units of energy per unit mass per unit length, is acceleration energy?

Now, what may be confusing you is that multiplying an angular momentum by an angular frequency DOES give energy as a result. So once you know the angular frequency, for example by calculating the response of the magnetic dipole to an applied magnetic field, then you can multiply that result by the angular momentum (in units of hbar) to get the energy. However I don't think it is correct to associate an energy with an electron spin in isolation ... you need a magnetic field to get a non-zero result.

Yet -- empirically -- the value of the bohr magneton does not start at 1/2 of something, it starts at zero -- which is the same as I would expect of a boson.

No .. a Bohr magneton is an arbitrary composition of physical constants (see http://en.wikipedia.org/wiki/Bohr_magneton). It happens to be handy because it is dimensioned properly to easily express the magnetic dipole moment of electrons. It certainly doesn't "start at zero" ... it has precisely one value for a given system of units. As I mentioned, there is also a nuclear magneton constant, which is dimensioned appropriately to express the magnetic dipole moments of nuclei.

However, the relative sizes of the responses due to the electronic and nuclear contributions to the overall magnetic moment are still different by a factor of 2000. This is why the nuclear coupling just leads to a very small splitting of the electronic energy levels, called nuclear hyperfine splitting.

Sure, and I mentioned those to begin with and said that I wouldn't say a "spin 0" composite particle is NOT affected by a magnetic field. You are merely reinforcing my point -- not rebutting it.

No .. I am making a different point. Your post implied that ONLY the hyperfine and spin-orbit interactions would cause non-zero interaction with the magnetic field. I am saying that if you neglect nuclear hyperfine coupling, an H-atom behaves EXACTLY like a spin-1/2 particle in a magnetic field. It does not behave like a spin-0 particle with a nuclear hyperfine interaction ... furthermore spin-orbit coupling for a spin-0 particle is non-sensical. The spin must be non-zero to have any spin-orbit coupling.

No, I didn't miss your point. I didn't argue they should have no magnetic moment -- you have added words to my mouth, and I don't much like that. I argued from the beginning that a spin 0 particle CAN be affected by a magnetic field -- see: fine and hyperfine and spin orbit coupling. I also said things that you might have made erroneous inferences from, and I apologize; I didn't know how to say it better at the time -- and for all these replies, I still don't.

I am not putting words in your mouth .. I am addressing your arguments as I understand them .. I cannot help it if you are mis-using physical terms in a way that makes your arguments hard to understand.

Anyway, I think the problem here is that all the stuff you are saying basically boils down to angular momentum conservation. When you say, "the Bohr magneton increases in integer steps" (or something similar), what you really mean is that angular momentum is quantized, and comes in units of hbar. As I mentioned above, the Bohr magneton is the unit we use to quantify the response of the magnetic dipole moment associated with that angular momentum to an applied magnetic field. As I pointed out in my last post (the part you didn't address), it is the angular momentum that is increasing in integer steps as you increase the orbital angular momentum quantum number, and that is what gives rise to the experimentally observed phenomena.

Remember that the original point I objected to was your statement from your first post that:

For example, an electron in Hydrogen is coupled to the spin of the proton nucleus -- and the total magnetic effect is spin 1 or 0. (usually/eventually 0...) Thus experiments don't measure 1/2 bohr magneton -- they always measure in integer units as far as I know.

That is incorrect as stated. It is true (as I said earlier) that the total spin of a hydrogen atom (considering both nuclear and electronic effects) is 0 or 1, depending on the relative orientations of the spins of the electron and proton. However, its "total magnetic effect" does NOT behave as a spin 1 or spin 0 particle ... a spin 0 particle would have NO response to an applied electric field, and the energy states of a (theoretical) spin 1 particle would be split into 3 levels. As I pointed out earlier, in a magnetic field, an H-atom behaves essentially as a spin 1/2 particle, and nominal electronic ground state energy level is split into two sub-levels ... (you actually agreed with this above when you said you were neglecting the nuclear spin terms). Now, I think what you were trying to say is that the *splitting* between the two states is always an integral multiple of the field strength times the Bohr magneton (you need to put the g-factor in there as well, but that's just a dimensionless constant). That is certainly true, and arises from the fact that angular momentum is quantized ... the two sub-levels correspond to m_s=+1/2 and m_s=-1/2, and the difference between them is obviously one unit (i.e. hbar).

So in one sense, the splitting between the levels is always an integer multiple of the Bohr magneton (for a given field strength and g-factor). However you can also talk about the shift of the ground state from the field-free case to the case with the field ... that shift will be 1/2 Bohr magneton (again times the field strength and g-factor).

 

[Drakkith]

does spin +1 mean a particle is 'spinning' in a different direction to a -1 particle or is it a magnitude of spin?
does antimatter spin in the opposite direction to normal matter?

The + or - sign merely indicates which way the particle is spinning around its axis. This gives rise to electrons and other 1/2 spin particles having "up" or "down" spin. The fact that there are only two spin states available to an electron is one of the primary reasons that the Pauli Exclusion Principle works the way it does, as the spin projection quantum number is one of the 4 quantum states that determines where electrons are located in their orbitals. The direction can be changed, but not the magnitude or value of the spin. IE you can never make a 1/2 spin particle a 3/2 spin or anything else.

Antimatter does NOT spin differently than normal matter. A positron is still a spin 1/2 particle just like an electron is.

where would the 1/2 particle go half north south and thank you for your explanations but why 'If the magnetic direction of the particle and antiparticle were the same, the direction of spin would be opposite'

The difference is that the electric charge is opposite and will result in an opposite magnetic moment from it's antiparticle. (Meaning that if you think of the magnetic moment like a little bar magnet, spin 1/2 for an electron would be with the south pole up and for a positron it would be the north pole up)