# what does it mean for a particle with no size to have angular momenta?

I recently was reading about higgs boson and particle spin recently and I stubble upon an question that contains an answer to what a spin is.

It explains that electrons etc. have no size yet they have angular momenta. I dont understand what exactly is meant by that. Does it refer to the angular momenta of the magnetic field? I just dont imagine something with no size having any sort of angular momentum.

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You've check marked wrong answer. Spin is an attribute of individual electron.. not of a collection of electrons. –  Sachin Shekhar Jul 7 '12 at 10:10
Please do not accept wrong/vacuous answers, it dilutes the value of the site. Sachin Shekar's answer is not good. The spin angular momentum is a real honest to goodness angular momentum, not a mathematical analogy. It can be seen in the Einstein deHaas effect. –  Ron Maimon Jul 8 '12 at 7:33
Thank you for your comment I will look into it. I do not know enough about it to decide which answer is the best yet but i will read the Einstein deHaas effect and then try to judge. I dont know what to do in a situation like this If there is a way to start a discussion about which answer is better or have community resolve this problem in another way. Otherwise I will do my best. –  Xitcod13 Jul 8 '12 at 20:58

As a answer, first I'd like to ask why you're asking a quantum mechanics problem with classical mechanics mental model.

There are two types of angular momentum in quantum mechanics:

1. Orbital angular momentum, which is a generalization of angular momentum in classical mechanics (L=r×p). I think, you shouldn't have problem with this because Orbital has size.

2. Spin, which has no analogue in classical mechanics. You can understand it as a number appeared in quantum equation. It can be understood like charge(with physical dimension), which is a number to denote one of basic attributes of particles. Yes, Spin does have physical dimension of angular momentum. But, its because it is a type of angular momentum, mathematically.

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to answer your first question is because i would understand quantum mechanics i would not ask questions about it. Q.M. has a lot of confusing vocabulary which means something else in "regular" english. Also what exactly do you mean by "which is a number (with physical dimensions)" numbers cant have physical dimensions. You mean the charge with physical dimensions? –  Xitcod13 Jul 6 '12 at 13:08
@Xitcod13 I meant charge is ((just a number)) with physical dimension.. Ofcourse, numbers don't have physical dimensions. :) –  Sachin Shekhar Jul 6 '12 at 13:17
Spin is not a type of angular momentum, it is angular momentum, period. This is demonstrated by the Einstein deHaas experiment detailed in my answer. –  Ron Maimon Jul 8 '12 at 6:45
@Ron That experiment is out-dated. At that time, there wasn't any concept of Orbital. –  Sachin Shekhar Jul 8 '12 at 7:21

It means exactly what it says--- the point particle has an angular momentum. In quantum mechanics, angular momentum is dimensionless (since hbar has units of angular momentum), and saying the spinning electron has angular momentum means that if you have a large number of electrons with spin up sitting on a disk (like a disk magnetized with a B field going in one direction perpendicular to the disk), and you suddenly reverse the B, so that all the electrons flip their spin to the other direction, then the disk starts spinning to conserve the angular momentum of the flip. This is the famous Einstein deHaas experiment that established that magnetization is carried by electron spin.

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@SachinShekhar: This is incorrect--- the transfer is not due to field angular momentum, it isn't a macroscopic spinning up of the magnet because of field action. It's to do with the spin angular momentum of the electrons. When you flip the spin of all the electrons in a magnet, you start the magnet spinning macroscopically due to the angular momentum of the spinning electrons. The experimental paper might have assumed an electron radius (I didn't read it), but it doesn't need to, all it is using is that electrons have angular momentum. It's a reproduced classic experiment, why not use it? –  Ron Maimon Jul 7 '12 at 19:56
@SachinShekhar: yes, you are changing the spin of a sizable fraction of the electrons in the magnet by changing the direction of B. This is how it is done in a lab. "All" the electrons can have the same spin, they have different positions. By "all" I don't literally mean every single one, just the ones involved in making the magnet magnetized. When you flip B, you flip the magnetization, 2% of the electrons flip their spin, and the bar starts to rotate with the exact amount of angular momentum lost by the electrons. I prefer thinking to reading. –  Ron Maimon Jul 7 '12 at 20:40
@SachinShekhar: They are clear, you just didn't understand them. Two electrons in the "same quantum state" have the same position distribution as well as spin. Electrons on different atoms can have the same spin. The electrons on different atoms in a magnet are spinning in the same direction, this is why you have magnetization. When you flip the magnetization, you can detect the change in angular momentum of the electrons--- the magnet rotates. –  Ron Maimon Jul 7 '12 at 21:09
@SachinShekhar: The electrons with the same spin in a ferromagnetic are on different atoms. –  Ron Maimon Jul 8 '12 at 6:42
@SachinShekhar: I have not "accepted" anything! You have just misinterpreted my comments in absurd ways. Of course not all the electrons have the same spins, all the unpaired electrons have the same spins, the unpaired electrons in interior orbitals are responsible for magnetism. Orbital angular momentum of these electrons does not contribute to magnetism angular momentum change, because when you change the magnetism, the orbital is exactly the same, only the spin of the electron changes. This is the reason the Einstein deHaas experiment works to measure electron spin angular momentum. –  Ron Maimon Jul 8 '12 at 7:11