# How can a point particle spin?

1. Feb 24, 2009

### Sumo

Hello. Obviously I have only an amateur understanding of physics, but I am trying to gain a bit more knowledge about the concept of quantum spin.

I understand that 'spin' does not refer exactly to angular momentum, but I believe more to a kind of mathematical degree of freedom which has similar properties to intrinsic angular momentum. But I was wondering if there is more to concept of spin than this in our understanding. As wikipedia says:
"As the name indicates, the spin has originally been thought of as a rotation of particles around their own axis. This picture is correct insofar as spins obey the same mathematical laws as do quantized angular momenta."

How can an indivisible point rotate?

Also on wikipedia it claims Richard Feynman made this statement about spin when asked for a layman explanation:
"What he said was that groups of particles with spin 1/2 "repel", whereas groups with integer spin "clump""

I don't really understand this, as don't groups of fermions, like electrons, repel one another because of their like charge. What approximately did he mean by this?

Thank you.

2. Feb 24, 2009

### Staff: Mentor

"Spin" is shorthand for "intrinsic angular momentum." The macroscopic angular momentum of an object includes the intrinsic angular momentum of the particles that comprise it. This can be demonstrated by e.g. the Einstein - de Haas effect, in which "flipping" the spins of electrons in a magnetized object causes the object to start rotating. This is similar to the common demonstration of conservation of angular momentum by a person sitting on a rotating stool, initially stationary, while holding a spinning wheel. When he flips the wheel over, he and the stool start to rotate in the opposite direction.

3. Feb 24, 2009

### clem

The QM "spin" of electrons and other elementary particles is angular momentum without rotation of the particle. In elementary physics class you need rotation to have angular momentum, but even a static classical electromagnetic field can have angular momentum without rotation.

4. Feb 24, 2009

### atyy

He's probably thinking of the Pauli exclusion principle. It makes Neutron stars possible. Apparently responsible for the stability of matter too.

5. Feb 24, 2009

### muppet

He almost certainly is. I'll try and explain this for the OP assuming as little technical knowledge as I can:

The pauli exclusion principle states that no two fermions (particles with spin 1/2, 3/2, etc) can be in the same "quantum state". In this simple form, you can account for the stability of atoms, neutron stars etc. In QM it can be the case that physical quantities such as energy can only assume discrete values, rather than a continuum. The most obvious example is that of the energy levels of electrons in an atom. Each of these energy levels corresponds to a collection of states which have the same energy, but specify other observables (spin, angular momentum, etc) differently. Electrons are fermions, so you can fit exactly one of them into each state. The result is that you have to "stack" the electrons into the "slots", "one on top of the other" (the inverted commas being intended to emphasise that this is nothing to do with physical space or slots, but is a metaphor for the arrangements of the energy states). If you look at the periodic table, the different rows of the table correspond to the highest occupied energy level in those atoms; the values of other quanties change as you move across the table. Protons and neutrons are also fermions, so a similar thing happens in the nucleus, but the details are more complicated.

To explain Feynmann's assertion that fermions "repel each other" whilst bosons (particles with integer values of spin- 1,2,3 etc) "clump" together is slightly harder to do non-technically. It's not some physical force as the phrase "repel" might suggest. Rather, it's a statistical effect; a consequence of the fact the QM only predicts probabilities, that emerges when you're discussing systems containing multiple particles.
The general idea is as follows. You've probably heard that the state of some particle is described by a "wavefunction", which (amongst other things) tells you the probability of finding that particle in some region of space. Here, you have to construct the wavefunction for the whole system, which you do by piecing together individual wavefunctions in a particular way. When you do this, you have to bear in mind that no two individual fermions can be in the same state, so their spatial distributions differ, which leads to them being separated on average when you measure their relative positions. Bosons, by contrast, are quite happy to be in the same state, so their statistical spread of positions leaves them clumped together.

A bit technical, but I hope that helped...

6. Feb 24, 2009

### v2kkim

When I worked on electron scattering in semiconductors we treat electron as a wave not a point particle mostly, meaning that we can think electrons as a small wave or cloud roughly
so it has a volume. I think this idea is not quite wrong, and helps on accepting spin.

7. Feb 24, 2009

### ice109

waves don't have anymore "volume" than a point does.

8. Feb 25, 2009

### StatusX

It might be helpful to think of a particle with spin, say spin 1, not as a little point, but as a little arrow. Then it's less mysterious that it can spin. Other spins correspond to slightly more complicated geometric objects, tensors and spinors, but the idea is the same. Still, it is true they have no spatial extent: these must be thought of as infinitesimally small arrows.

9. Feb 25, 2009

### per.sundqvist

You could think about it rather as an intrinsic magnetic dipole m. Classically this could be calculated for a circular loop with current I, enclosing an area A, as: m=I*A, where A=pi*r^2. But here (for the electron) we don't know either the current I or the radius r. In the limit r->0 we could expect m to be expressed in terms of the natural constants so that we get the right unit: [m]=A*m^2. The Bohr magnetron does have this unit $$\mu_B=e\hbar/2m$$.

The spin is a purely relativistic property, which could be derived from Dirac's equation. It is similar to the discovery of E=m0*c^2, that mass, even at rest, contains energy. In the same way relativity theory predict that there is a magnetic dipole falling out "intrinsic", below roughly expressed as:

$$E\sim E_{cl}+m_0c^2$$
$$\vec{m}\rightarrow \vec{m}_{cl}+\vec{m}_{rel}$$

The spin also obeys angular momentum properties like
$$\hat{s}^2\Psi=s(s+1)\Psi, \;\hat{s}_z\Psi=s_z\Psi$$, (sz=1/2)
which is like angular momentum L.

Finally you could think of the spin as a an operator of the magnetic dipole, like you do for the momentum in quantum mechanics. In classical physics a magnetic dipole is a vector, and with this vector you gain an energy in a magnetic field $$E_m=\vec{m}\cdot\vec{B}$$. Now the operator is of matrix form. Lets summarize it as follows:

$$\vec{p}\rightarrow -i\hbar\nabla$$
$$\vec{m}\rightarrow \mu_B\hat{\vec{s}}=\mu_B\frac{1}{2}\hat{\vec{\sigma}}$$

where $$\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$$ are the Pauli 2x2 matrices. This implies that the wave function is a 2 element column vector.

10. Feb 25, 2009

### Sumo

Thank you everyone, that's helped a lot. If you'll indulge me just a bit longer, so I'm sure I understand this correctly.
Is this because the wavefunction for a single particle 'expands' over time? Meaning, the probability of finding a particle in one state becomes less likely over time, finding it in other states (other positions say) becomes more likely. You would have to calculate every possible state for the system from every possible state for each particle, and the number would be much greater if you were restricted by the pauli exclusion principle (fermions) than if you were not (bosons)? Therefore a system of fermions would seem to 'repel' one another over time, the wavefunction of the system would expand more over time.

As far as the comments on how spin works with a 'point' particle, I will have to look at these replies more closely. I don't understand how something can have spin and not be rotating, but I'm guessing this is from my attempt to visualize the thing.

11. Feb 25, 2009

### Naty1

A "point" particle likely can't actually have spin....it's a convenient mathematical idealization. So we represent some characteristics as a point particle, assign some physical features, like spin or maybe mass, and voila, we have a model we can utilize. I prefer to think of particles as a bit extended in space, like strings...

Identical Force particles, like photons, tend to clump, stay together, where as identical matter particles (fermions) cannot occupy the same quantum state simultaneously via the Pauli exclusion principle.....they sort of "repel" ...All known fermions are particles with half-integer spin. In the framework of nonrelativistic quantum mechanics, this is a purely empirical observation. However, in relativistic quantum field theory, a "spin-statistics theorem" shows that half-integer spin particles cannot be bosons and integer spin particles cannot be fermions.

Yes, electrically charged particles, like electrons or protons, do repel like particles; Neutrons and other uncharged particles would tend to follow Feynmann's summary. Don't take it TOO literally.

12. Feb 25, 2009

### atyy

Most phenomena associated with the Pauli exclusion principle occur even for stationary states that don't evolve in time except for a global phase. The principle is equivalent to saying that the states are anti-symmetric.

In everyday life, we are used to seeing bar magnets that interact with magnetic fields even though they are not spinning. So you can think of "spin" as the particle being like a little bar magnet, causing it to interact with magnetic fields. It could be called "magnetic property of a particle", but it's called "spin" because the mathematics describing it is algebraically analogous to circular movement.

13. Feb 25, 2009

### alxm

Seems you're taking the old "spin is a relativistic effect" to the next level by claiming it's also a purely empirical property.

AFAIK that's an old misconception that's still very often repeated. Spin was thought for a while to be a relativistic effect, due to the Dirac equation. It's actually neither a purely relativistic effect nor a purely quantum one without classical analogues. In fact it can be, and has been, derived without using relativity.

Since this thread started with a Feynman quote, I'll refer to another one (from "Quantum Electrodynamics"):

14. Feb 25, 2009

### Tac-Tics

Wouldn't a photon's spin as its polarization also be evidence of this? It seems like light, holding its speed constant, wouldn't even have a non-relativistic interpretation.

15. Feb 25, 2009

### muppet

Not quite, and unfortunately the answer is again a technical one. The number of states is not the relevant thing. In QM a "state" really means "the most complete possible description of a system". A system only has one state, and the phrase "superposition of states" which you may have come across is really a shorthand for "superposition of eigenstates", an eigenstate being a state possesing a definite value of some observable quantity. Typically, an eigenstate of one quantity can be expressed equivalently by a sum of eigenstates of some other quantity. To try and make things clearer with an example, something possessing a definite energy could be measured in one of many positions.
You could keep a lid on the range of possible positions of the components of your system by trapping it in a box. The consequence of Pauli exclusion is that when you lift the lid of the box to see where the particles are, a group of fermions will tend to have spread themselves out in comparison to a group of bosons, which you would be likely to find near each other.

The reason you can't see this is because the word spin means rotation in every context outside of this one. Here, it doesn't. The term is used because a small (but finite) spinning charge would interact with magnetic fields in the same way as an electron does. People knew there was something wrong with the picture from the outset (they calculated that a point on the surface of a rotating electron would have to be moving many times faster than the speed of light) but the behaviour of spin so closely parallels the quantum mechanical description of angular momentum that the term persists, not completely unhelpfully once you understand that the particle is not a small rotating sphere.

16. Feb 26, 2009

### alxm

Yes, a circularly polarized light wave is a macroscopic and classically derivable property (the angular momentum) that's the result of the addition of a large number of quantum spins. (which are quantized, unlike a classical analogue.

Well, it's true of course that light cannot be described properly non-relativistically. And spin cannot be described correctly classically or non-relativistically. I'm just pointing out that spin isn't, as is often assumed, a property that's intrinsically quantum, intrinsically relativistic, or without a classical analogue.

I was always told it was "a relativistic effect" but then I noticed that this was denied as routinely by the relativists I encountered as it was repeated by people working with non-relativistic theory. After looking into it, I wasn't surprised to find the relativists were correct.

A decent article on the classical analogy, btw: http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf" [Broken]

Last edited by a moderator: May 4, 2017
17. Feb 27, 2009

### Naty1

What is spin? mentions angular momentum/spin as a consequence of the "wave field" of the electron rather than a feauture of its "internal" structure...which seems an odd statement since I thought an electron was a wave/field....anyway, as others have noted or implied, the idea of "spin" has evolved over a long period of time; people did not agree on it's interpretation when it was first discovered and many don't agree on what it means today....

I think it was Lee Smolin who mentions in one of his books doing quantum calculations where a number of physicsts working together thought they were making progress...by comparing calculational results as a check...and then made the mistake of trying to interpret that mathematics...they could not agree on thye meaning!!!!!
....
hence Feynmanns "shut up and calculate" which I posted about already, seems to remain a feature of quantum discussions today.

18. Feb 27, 2009

### per.sundqvist

Even though the spin was investigated non-relativistic before Dirac, it doesn't mean that it is not relativistic. First came the experiment, later a model by Pauli, then Dirac explains its origin, from relativity theory.

As I pointed out in #9, the spin is more like a magnetic dipole. It requires the particle to have a charge, otherwise there cannot be any "circular electric current", which is how magnetic dipoles appear. Klein-Gordan is thus valid relativity theory for particles without charge. Something with charge has spin, as a relativistic effect.

19. Feb 27, 2009

### Tac-Tics

I think this quote is misattributed to Feynman.

I think it's important to keep in mind what the symbols on the page mean (or could mean). Your intuition guides your math. Your math tells you if you're making any sense at all. And experiment tells you if you're right or wrong, regardless of how much sense you make.

Something that bothers me about every book and text I've read on QM is that it doesn't provide a firm grounding in the intuition. At least, not in a clear, concise way. The history of the subject gives you the intuition from a historical perspective, including all the false assumptions and clunky results. The mathematical approach to the subject make assertions without justification. (Eg: why do we model measurements during an experiment modeled as a Hermitian operator?) It's maybe a bit tangent to the topic, but I'd really like a book in the style of Feynman's QED that doesn't hand-wave through the math. Giving a "physicist's history" of QM, high-lighting a stumble-free path from its early history to its modern form.

20. Feb 27, 2009

### atyy

Because the eigenvalues of the operator are what we measure, which is presumably a real number?