Transfering spin onto a macroscopic object

  • #1
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Main Question or Discussion Point

Hi all.

I'm trying to understand (QM) spin. I have plenty of questions but this one is quite simple.
Lets say there is a tiny grain of coal (because it's black), far away from gravitational or EM fields.
I take a laser, and pick photons that have spin +1 in the z-direction, it doesn't really matter which way ##z## is.
(I assume that this is possible. If not, the experiment can be performed with electrons instead of photons, with some added difficulty).
Now I let the photons hit the grain.

Will the grain start to accumulate the spin?
Will it be radiated away using infrared photons?
Or does spin stay isolated in the electrons that absorbed the incoming photons? What happens after all electrons have switched to spin +1/2 ?

While I'm at it, 2 supplementary questions.
Is a photon's spin uniquely determined by its angular momentum? I tried searching but just can't figure out the right keywords.
Is there orbital spin, or only intrinsic spin?

I might be able to handle an I-level answer, but the simpler, the better.
 

Answers and Replies

  • #2
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When a spinning black grain radiates 0.01% of its mass-energy, it radiates 0.01% of its angular momentum. Spinning rate stays constant. That ratio can be adjusted by adjusting how much different parts of the grain radiate. Radiation from areas near the equator is relativistically beamed, that's why that radiation carries more angular momentum than radiation from near the poles.

If a spinning black grain absorbs 0.02% of its mass-energy and 0.01% of its angular momentum, then its spinning rate slows down.

Here's some info about how radiation affects atoms:
https://www.physicsforums.com/threads/heating-by-radiation-how-the-atoms-speed-is-increased.946827/#post-5993713
 
  • #3
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A note, I'm talking about quantum mechanical spin, not the normal rotation.
When a spinning black grain radiates 0.01% of its mass-energy, it radiates 0.01% of its angular momentum.
Does that mean that if I send a microgram of photons to a one-gram grain, this will transfer quite a lot of spin (depending on color), and when it gets radiated away as heat, only 0.1% of that spin gets lost? Then it should be possible to "pump" quite a lot of spin into the grain.
Thanks, that makes more sense.

I take it that neither option 2 or 3 are correct. But that means that the spin will be spread throughout the grain? Will this have no observable effect on the grain, just sitting there, "having" spin? I read in many places that spin is a purely quantum phenomenon that cannot be observed on macro scale.
 
  • #4
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But that means that the spin will be spread throughout the grain? Will this have no observable effect on the grain, just sitting there, "having" spin? I read in many places that spin is a purely quantum phenomenon that cannot be observed on macro scale.
It means that an atom on the grain's equator circles around the center of the grain. The spin did indeed spread through the grain. The spin was transferred from microscopic and quite quantum mechanical level to macroscopic and not obviously quantum mechanical level. That happens all the time. Kick an object with quantum spins pointing to the same direction so that the spins become disordered - the object starts spinning around.

https://en.wikipedia.org/wiki/Einstein–de_Haas_effect
 
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  • #5
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Does that mean that if I send a microgram of photons to a one-gram grain, this will transfer quite a lot of spin (depending on color), and when it gets radiated away as heat, only 0.1% of that spin gets lost? Then it should be possible to "pump" quite a lot of spin into the grain.
No, the pumping works much better than that: Grain lost one millionth of its mass - so it lost one millionth of its angular momentum.
 
  • #6
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It means that an atom on the grain's equator circles around the center of the grain.
Are you sure about all this? I don't think spin works like that.
Kick an object with quantum spins pointing to the same direction so that the spins become disordered - the object starts spinning around.

https://en.wikipedia.org/wiki/Einstein–de_Haas_effect
I came across this effect while reviewing other threads on spin, and while the Wikipedia article is not entirely clear, what I understand is that this effect only applies to orbital angular momentum, while the spin is unaffected. This is what allows measuring what part of magnetic field comes from spin, vs. from orbital angular momentum.
Orbital spin and intrinsic angular momentum aren't discussed very often, which is partly why I had the supplementary questions.
Is there orbital spin, or only intrinsic spin?
No, the pumping works much better than that: Grain lost one millionth of its mass - so it lost one millionth of its angular momentum.
Yes of course, somehow I thought that micro means 1/1000 o:)
 
  • #7
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Are you sure about all this? I don't think spin works like that.
Yes I'm sure.

Spinning basketball's angular momentum becomes angular momentum of the earth at some point. But electron's angular momentum doesn't. Why is that? Well, basketball has many possible angular momentums, while electron has one, that's probably one reason.

Have you seen this:

https://en.wikipedia.org/wiki/Angular_momentum_of_light#Exchange_of_spin_and_orbital_angular_momentum_with_matter

I'm not saying whether that is right or wrong.
 
  • #8
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It means that an atom on the grain's equator circles around the center of the grain.
Yes I'm sure.
So, are you saying that quantum spin can be converted to spatial rotation and back at will, i.e. they are essentially the same thing? That contradicts all I've heard about spin so far.
Until now I thought that ##S_{xt}##, ##S_{yt}##, ##S_{zt}##, ##L_{xy}##, ##L_{yz}##, ##L_{zx}## are 6 independently conserved quantities.
 
  • #9
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So, are you saying that quantum spin can be converted to spatial rotation and back at will, i.e. they are essentially the same thing? That contradicts all I've heard about spin so far.
Until now I thought that ##S_{xt}##, ##S_{yt}##, ##S_{zt}##, ##L_{xy}##, ##L_{yz}##, ##L_{zx}## are 6 independently conserved quantities.

Well, I guess at least one of us has some wrong idea then.:smile:


Let's go to a frame where a polarized laser pulse is a radio wave pulse. Now non-quantum physics can tell us what happens to the angular momentum of that wave when it is absorbed, right?

I mean the radio wave is not so special that it can't be absorbed by some antenna. Or is it?
 
  • #10
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Well, I guess at least one of us has some wrong idea then.:smile:


Let's go to a frame where a polarized laser pulse is a radio wave pulse. Now non-quantum physics can tell us what happens to the angular momentum of that wave when it is absorbed, right?

I mean the radio wave is not so special that it can't be absorbed by some antenna. Or is it?
Well I suppose it will scatter among the atoms that make up the antenna, and through molecular forces spread out to Earth. I have no problem imagining or understanding how rotation and angular momentum works.
I have trouble understanding spin.
 
  • #11
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Well I suppose it will scatter among the atoms that make up the antenna, and through molecular forces spread out to Earth. I have no problem imagining or understanding how rotation and angular momentum works.
I have trouble understanding spin.
Well I find it very instructive to think about a radio-wave photon and its spin. And I tried to make sure we have such radio-wave photon by saying it was created by a laser.

(Also thinking about black holes absorbing photons is quite instructive IMO)


Hmm ... let's say the argument that black holes convert all types of angular momentum into one type of angular momentum is my main argument.

And the argument that antennas maybe convert all types of angular momentum that radio-waves may have into one type of angular momentum is just an idea or something.
 
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  • #12
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Hmm ... let's say the argument that black holes convert all types of angular momentum into one type of angular momentum is my main argument.

And the argument that antennas maybe convert all types of angular momentum that radio-waves may have into one type of angular momentum is just an idea or something.
That doesn't seem quite right. I'd say spin is a part of the information that gets destroyed at singularity.
Would you say that ##S_{zt}## gets converted to ##L_{xy}## at 1:1 ratio, or does it go to all 3 projections? How do you even define whether it goes to ##L_{xy}## or ##-L_{xy}##?
 
  • #13
PeterDonis
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Spinning basketball's angular momentum becomes angular momentum of the earth at some point. But electron's angular momentum doesn't.
What kind of electron are you talking about?

Until now I thought that ##S_{xt}##, ##S_{yt}##, ##S_{zt}##, ##L_{xy}##, ##L_{yz}##, ##L_{zx}## are 6 independently conserved quantities.
Where are you getting these quantities from? In the usual formulations of non-relativistic QM, the spin and orbital angular momentum vectors ##S## and ##L## are each 3-vectors (more precisely, 3-vector operators), and whether they are conserved separately, or only their sum, the total angular momentum ##J##, is conserved, depends on the specific problem.
 
  • #14
PeterDonis
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Also thinking about black holes absorbing photons is quite instructive IMO
It's also way too much for this thread, particularly as it's a "B" level thread and any correct treatment of black holes is going to be at least "I" level and probably "A" level.

That doesn't seem quite right. I'd say spin is a part of the information that gets destroyed at singularity.
A spinning black hole's singularity is not spacelike and an object falling into the hole doesn't have to hit it. (Arguably, the whole spacetime inside the inner horizon of a spinning black hole, including the singularity, is unphysical anyway.) In any case, as above, dragging in black holes is way too much for this thread. Please keep the discussion focused on the simplest cases.
 
  • #15
Charles Link
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I still need to read the previous inputs more carefully, so my apologies if I am stating something that has already been mentioned, but from what the OP is asking, this question really gets answered by the de Haas-Einstein experiment: https://en.wikipedia.org/wiki/Einstein–de_Haas_effect
 
  • #16
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Until now I thought that ##S_{xt}##, ##S_{yt}##, ##S_{zt}##, ##L_{xy}##, ##L_{yz}##, ##L_{zx}## are 6 independently conserved quantities.
Where are you getting these quantities from? In the usual formulations of non-relativistic QM, the spin and orbital angular momentum vectors ##S## and ##L## are each 3-vectors (more precisely, 3-vector operators)
It was my understanding from some online lectures that they essentially come from Noether's theorem. And, since the space is invariant after rotating along all 3 axes, then all 3 momenta should be conserved separately.
At this point everyone usually says something like "and it's the same for spin", which is where I get lost.
whether they are conserved separately, or only their sum, the total angular momentum ##J##, is conserved, depends on the specific problem.
Is it a vector sum ##L+S##, or more like ##L_x^2+L_y^2+L_z^2+S_x^2+S_y^2+S_z^2## ?
 
  • #17
PeterDonis
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It was my understanding from some online lectures that they essentially come from Noether's theorem.
Then you're going to need to give a specific reference for those online lectures, because it looks to me like you've misunderstood or misread something. This is why we always want references.

In particular, while it is possible, in 3 spatial dimensions (but only in 3 spatial dimensions), to view angular momentum operators as tied to planes instead of vector directions (for example, the component ##L_z##, pointing along the ##z## direction, can also be viewed as the component ##L_{xy}##, in the ##x y## plane), that holds for both ##L## and ##S##. There is no formulation in which either of them has double indexes including the time ##t##.

In relativity, where we work in 4 dimensions (3 space, 1 time, in a standard inertial frame), it is possible to have an antisymmetric 4-index tensor that generalizes angular momentum, with six independent components, whose purely spatial components (##xy##, ##xz##, ##yz##) correspond to the total ordinary angular momentum ##J##. But the "space-time" components of this tensor (##xt##, ##yt##, ##zt##) correspond to boosts; they have nothing to do with spin. How to split ##J## into "orbital" and "spin" parts in a relativistic context is a complicated subject that is probably beyond the scope of this thread; I would advise sticking with non-relativistic QM for this discussion.

Is it a vector sum ##L+S##, or more like ##L_x^2+L_y^2+L_z^2+S_x^2+S_y^2+S_z^2## ?
The vector sum gives the 3-vector operator ##J##; the sum of the squares gives the scalar operator ##J^2## (which is also the sum of the two scalar operators ##L^2## and ##S^2##).
 
  • #18
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Then you're going to need to give a specific reference for those online lectures, because it looks to me like you've misunderstood or misread something. This is why we always want references.
Unfortunately it's not in English:
You may be able to hear the word "spin" as he writes ##s_x## at time 1:35:19. The lecture's title (shown at time 0) is Conservation Laws and the Noether's Theorem.

He's a respected teacher so I had no reason to doubt this claim. However, I tried to find an English reference and indeed it seems that either he oversimplified or misunderstood something. He explicitly says that spin is the quantity conserved under boosts but that doesn't seem to be the case. I'll try to find some text about this that I can understand.

So, let me refine my original question.
Can a macroscopic object have spin? Or does it always stay at the particle level?

The vector sum gives the 3-vector operator ##J##; the sum of the squares gives the scalar operator ##J^2## (which is also the sum of the two scalar operators ##L^2## and ##S^2##).
So, can an isolated apparatus convert its total spin to its total angular momentum, and vice versa? (total meaning sum over all parts)
 
  • #19
Charles Link
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So, can an isolated apparatus convert its total spin to its total angular momentum, and vice versa? (total meaning sum over all parts)
Please see my post 15. I think Einstein and de Haas successfully answered this question. ## \\ ## A careful reading of the "link" in post 15 shows that even though the effect and the experiment were named after Einstein and de Haas, there were subsequent experiments by others that obtained more accurate results.
 
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  • #20
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Please see my post 15. I think Einstein and de Haas successfully answered this question. ## \\ ## A careful reading of the "link" in post 15 shows that even though the effect and the experiment were named after Einstein and de Haas, there were subsequent experiments by others that obtained more accurate results.
Thies requires external magnetic field. While its lines are parallel to the change of momentum, it's still an external field. Can it be done with a field generated inside the device itself?
while the Wikipedia article is not entirely clear, what I understand is that this effect only applies to orbital angular momentum, while the spin is unaffected. This is what allows measuring what part of magnetic field comes from spin, vs. from orbital angular momentum.
Is that wrong?
 
  • #21
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Thies requires external magnetic field. While its lines are parallel to the change of momentum, it's still an external field. Can it be done with a field generated inside the device itself?

Is that wrong?
The article along with the discussion farther down treats the subject in detail, and yes, it does apply to spin angular momentum as well as orbital angular momentum. As for making an apparatus where everything is inside the system, that would be difficult because the angular momentum is quite small. As the article discusses, although simple in principle, it is a very difficult experiment to perform and get meaningful and accurate results. See the section at the bottom "Later measurements and Applications" for experiments that show how much of the magnetization in iron is due to spin. These are the results of very precise measurements of ## g' ## with a de Haas-Einstein type apparatus. ## \\ ## For some additional detail, magnetic moment ## \vec{\mu}=g \, \mu_B \frac{\vec{J}}{\hbar} ## where ## \mu_B ## is the Bohr magneton. In cgs units ## \mu_B=\frac{e \hbar}{2mc} ##. ## g_L=1.00 ## and ## g_s=2.0023 ##. If magnetization is fraction ## f_s ## of spin, and ##1-f_s ## of orbital, we get that ## M=\mu_B(g_s f_s+g_L (1-f_s)) J ##. If measured ## \frac{M}{\mu_B \, J}=g' ##, this gives ##g'=g f_s+(1-f_s ) ## where we used ## g_s=g ## and ## g_L=1 ##. Solving, we get ## f_s=\frac{g'-1}{g-1} ##. I didn't quite duplicate their formula=yet= I need one more step: ##\frac{M_s}{M}=\frac{f_s \, g \, J}{g' \, J}=\frac{(g'-1)g}{(g-1)g'} ##, and yes, now my result is in agreement with theirs.
 
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  • #22
PeterDonis
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Can a macroscopic object have spin?
If you choose an appropriate way to model it, yes. For example, consider the Earth-Moon system. You can model it as a system containing two objects, each of which has orbital angular momentum and spin angular momentum. Then you can view, for example, the tides from the Moon gradually making the Earth's day longer as transferring some angular momentum from the Earth's spin angular momentum to the Moon's orbital angular momentum (because as the Earth's day lengthens, the Moon's mean orbital radius increases). Note, though, that this is a classical model, not a quantum model.

can an isolated apparatus convert its total spin to its total angular momentum
Total spin is already part of total angular momentum.

If you meant to ask whether an isolated system can transfer angular momentum between spin and orbital, the answer is yes; I just gave an example above. However, note that that's a macroscopic example. I'm not sure there are microscopic examples, because in quantum mechanics ##L##, ##S##, and ##J## are operators, and the possible eigenvalues of those operators will depend on the specific system, and can be counterintuitive if you're use to the treatment of angular momentum in classical mechanics (such as the example I gave above).
 
  • #23
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I didn't quite duplicate their formula=yet=
I picked ##J_o=1## and solved for ##J_s## then plugged it back to get
$$J_s = \frac{q'-1}{q-q'}$$
and
$$\frac{M_S}{M} = \frac{q(q'-1)}{q'(q-1)}$$
but that's not so important here.
 
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  • #24
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Earth's spin angular momentum
So there's really nothing special about spin, except that a massless and/or size-less particle can have it?
In that case I think I understand the previous posts.
 
  • #25
Charles Link
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I'm getting
$$\frac{M_S}{M} = \frac{q'-1}{q-1}$$
as well but that's not so important here.
It may not be clear that I got the same result that they did, so let me edit the post.
 

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