# I What is the basic concept of the spin of a particle

1. Jan 11, 2017

### Pyrus

What is the basic concept which decides the spin of a particle. Like, electron has a spin of +- 1/2... What do this 1/2 signifies.
Other spins are also there like 1 and 2. What do they signify?
Is there any possibility of any particle with spin other than 1/2, 1 or 2?

2. Jan 11, 2017

### mikeyork

For systems with non-zero mass, take the total angular momentum of the system and transform to the rest frame. The result is what we call spin. However, this does not mean that spin is angular momentum. It also doesn't tell us what the spin of a massless particle is.

The other way in which spin differs from the classical notion of angular momentum -- that is, in terms of something actually moving around in space with angular velocity -- is that orbital angular momentum, which more closely resembles classical angular momentum, can only be quantized as integer multiples of hbar, whereas spin can take on half-integer values and can be applied to particles that are assumed to be point-like (i.e. have no spatial extent). Nevertheless, one can add spin to orbital angular momentum to get total angular momentum.

So I would suggest it is more appropriate to think of spin as an intrinsic property that manifests as angular momentum in a space-time context.

3. Jan 11, 2017

### Staff: Mentor

There are two separate but related quantum numbers associated with intrinsic angular momentum ("spin"). The first one, $s$, is associated with the magnitude of a particle's spin: $S = \sqrt{s(s+1)}\hbar$ which for an electron equals $\sqrt{\frac 1 2 \cdot \frac 3 2}\hbar = \frac {\sqrt 3} 2 \hbar \approx 9.14 \times 10^{-35}~\rm{kg \cdot m^2 / s}$. This is a fixed, unchangeable value which (like the mass) is the same for all particles of a given type, e.g. 1/2 for electrons.

The second one, $m_s$, is associated with the orientation (direction) of the spin, specifically with the component of the spin along any given direction, which we customarily call the "z-component" although it doesn't have to be actually along the z-direction: $S_z = m_s \hbar$.

$m_s$ can have one of the values ranging from $-s$ to $+s$ in integer intervals. For $s=\frac 1 2$ (e.g. for an electron) these values are $m_s = - \frac 1 2$ ("spin down") and $m_s = + \frac 1 2$ ("spin up"). For $s = 1$ they are $m_s = -1, 0, +1$, for $s=\frac 3 2$ they are $m_s = -\frac 3 2, -\frac 1 2, +\frac 1 2, +\frac 3 2$, and similarly for other values of $s$.

For an electron, $S_z = \pm \frac 1 2 \hbar \approx \pm 5.28 \times 10^{-35}~\rm{kg \cdot m^2 / s}$.

Note that the spins of the electrons in a chunk of metal contribute to the chunk's total macroscopic angular momentum, as shown in an experiment described in chapter 37 of the Feynman Lectures on Physics, specifically, Fig. 37-3 and the preceding paragraph. This is known as the Einstein-de Haas effect.

4. Jan 11, 2017

### Delta²

I have the exact same question as the OP, what exactly is the basic concept behind the spin of a point like particle. A point particle cannot spin around itself (like say a rigid body rotates around an axis passing through it) cause it is a single point and therefore needs another point around which to rotate. But when it rotates around another point we call it orbital angular momentum.

So what is the spin (don't tell us what is the magnitude or the direction of the spin), it is an intrinsic property but what sort of intrinsic property? Do we actually accept that a point can rotate around itself?. Classically this doesn't make sense though.

5. Jan 11, 2017

### Staff: Mentor

You are asking for a way of visualizing quantum spin that works with your intuition for how classical objects work, so there may not be any answer that you will find satisfying. Quantum particles (and be aware that even that word "particle" is a misnomer - by the time it was clear that that wasn't a good name it had already stuck) just don't behave like classical objects.

What we do know:
1) An electron has an internal angular momentum that contributes to the overall conserved angular momentum of the system of which it is part in the same way that the internal angular momentum of a classical particle contributes to the overall conserved angular momentum of the system to which it belongs. "Internal angular momentum" is a mouthful, so we use the word "spin" instead.
2) The internal angular momentum of a classical object comes from it rotating about its axis (which is why we got in the habit of calling it "spin" in the first place). We know that the internal angular momentum of a quantum point particle doesn't come from rotating around its axis... But why does it have to "come from" anywhere? We don't insist on knowing where the electric charge of an electron "comes from"; we just accept that an electron has an intrinsic negative charge.

Last edited: Jan 11, 2017
6. Jan 11, 2017

### Staff: Mentor

Quite coincidentally, while following a different thread I came across a quote from Richard Feynman:

7. Jan 12, 2017

### muscaria

Spin is essentially a wave property which describes a circulating flow of energy in the physical field in question. This isn't like an energy "lump" flowing along a circle in the field (that would be orbital angular momentum) but a circulating energy flow so that it closes in on itself so to speak. Of course this circular flow isn't constrained to a circle but felt throughout the whole field, as if there was some kind of torsion/twist/vorticity within the field. This could happen in all physical fields mediating energy flow, which contain vector fields, tensors, spinors..but not simply scalars, e.g. the E-M field displays spin angular momentum, Dirac field etc..

So spin isn't a quantum mechanical property but a wave-like property associated with a certain type of energy flow in a field. Whether it is classical or quantum mechanical is secondary and simply determines whether the circulating flow can take on a continuous range of values or whether this is discrete.

You can generally figure out the spin of a field through the symmetrised stress-energy-momentum tensor of the field. If you do this for the Dirac field of an electron, you find that the momentum density splits off into a translational part and an additional part which describes circulating energy flow in the rest frame of the electron which is numerically identical to electron spin. So if you imagine a localised packet in the Dirac field, you have circulating energy flow throughout the whole field which is centered about the packet and follows the packet around so to speak. It's intrinsic in the sense that it doesn't depend on what other external fields are doing etc, but it manifests throughout the field and not really within the internal structure of a particle in the usual sense.

8. Jan 12, 2017

### muscaria

I'm not certain about this and it's been a while since I've looked into this but that comes from the spinor algebra. A spinor is a 2-d complex vector which is specified by 4 real numbers and can be visualised in 3d space in terms of a flagpole i.e. a vector with a flag instead of an arrow head. 3 numbers specify the orientation and magnitude of the vector and the 4th number specifies the angle of the flag about the pole. For a standard 3-d vector, if you rotate the vector about the z-axis so that it traces out a cone, you return to the same state after a 360 rotation. The situation is different for the spinor. When you rotate the flagpole about the z-axis, the spinor algebra is such that after a 360 rotation the flag on the end of the pole comes out in the opposite direction and you need to rotate around the z-axis twice, there's an inherent twist similar to a mobius strip. Maybe someone else may be able to give more details on this but I remember getting the impression that the denominator in the fraction for the spin is related to the number of windings you have to carry out before returning to the same state e.g. spin $1/2$ needs 2 windings, spin $1$ needs only 1, spin $1/4$ would need 4 etc.. so if you exchange 2 bosons by rotating through by $2\pi$, you only needed $2\pi$ given that they have spin 1, in order to come back to same state, meaning the state is symmetric under exchange. For spin $1/2$ you need to exchange them twice due to $4/pi$ rotation and the state is antisymmetric. Hopefully someone can comment on the validity of this statement..

9. Jan 12, 2017

### weirdoguy

Can you show any source that claims that?

10. Jan 12, 2017

### Staff: Mentor

Yes, please post a reference to a textbook or published journal article, or if you don't have those handy, at least a journal or author name to search for.

11. Jan 12, 2017

### vanhees71

Well, that's written in a bit unusual way, and I'm not sure if it's entirely correct, but indeed there's angular momentum carried by fields, and for the electromagnetic field it's even there in the classical sense. However, it's not clear, how to split it into spin and orbital angular momentum, particularly not for massless fields as the electromagnetic field. You can find the correct treatment in any good textbook on electrodynamics. It's formally described by the multipole expansion of the electromagnetic field (see, e.g., Jackson).

12. Jan 12, 2017

### Staff: Mentor

It is, unfortunately, very difficult (perhaps impossible) to produce a math-free natural language explanation that doesn't mislead by omission.

13. Jan 12, 2017

### 256bits

http://cosmology.princeton.edu/~mcdonald/examples/EP/commins_arnps_62_133_12.pdf
Eugene D. Commins
Section 11
Also includes a discussion on the Electric dipole Moment
Section 10

https://www.physics.mcmaster.ca/phys3mm3/notes/whatisspin.pdf
Hans C. Ohanian

Last edited: Jan 12, 2017
14. Jan 13, 2017

### muscaria

In addition to Ohanian's paper, attached is an extract on Field angular momentum and the symmetrised energy-momentum-stress tensor from Davison's "Classical Field Theory".

#### Attached Files:

• ###### Field_angular_Davison.pdf
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15. Jan 14, 2017

### vanhees71

Well, the there proposed "spin component" is physically meaningless since it is gauge dependent since it contains explicity components of the four-potential. Only the total angular momentum is a well-defined physical quantity and of course also the helicity $\vec{J} \cdot \vec{p}/|\vec{p}|$.

16. Mar 19, 2017

### N88

Many thanks. I like this, especially for its clarity. So now (hoping to be treated gently) I wonder about 3 things. Please:

1. The symbol $\sigma$ is often associated with spin. How does it relate to your expressions?

2. In Aspect's famous experiments with photons, his measurement outcomes are ±1. What happens to the 0 value in these cases?

3. Given: we can only ever determine the spin of a spin-half particle in one direction at a time. So, in a Cartesian coordinate system, we might regard the related outcomes (of the related (but independent) tests along each axis) as projections. Then we can the say that the intrinsic spin is a vector, such that:

$S^2= S^2_x+S^2_y+S^2_z=3$ (in appropriate units; from looking at the √3 in your $S$ equation above). (1)

Now a vector bound by eqn (1) is certainly "special" (as is QM) and workable.

But what if they were not projections-onto an axis but "rotations-onto" because (say) the spin-vector was a special type of vector (say a "spinctor") that (under measurement) always rotated to be parallel or antiparallel to the testing field. Then we'd have:

$S^2= S^2_x=S^2_y=S^2_z=1$ (in appropriate units; from looking at the ±1 in your $S_z$ equation above). (2)

So, it seems to me (a QM beginner), that a 'spinctor" bound by (2) is "less special" than the vector in (1). But in questioning whether it is workable, is its failure obvious?

With my thanks again, N88

Last edited: Mar 20, 2017
17. Mar 20, 2017

### edguy99

I would respectfully disagree with this interpretation. Consider the normalized Jones Vector as the axis of spin (ie. polarization angle). The 3 states of the spin axis are either some fixed direction (spin0 will not impart angular momentum on anything it hits), the spin axis is spinning right (spin+1 and will impart angular momentum of +hbar on anything it hits) and the spin axis spinning left (spin-1 and will impart angular momentum of -hbar on anything it hits).

As I understand Aspect type experiments, they are determining the "instantaneous" axis of spin as being either "+1" or "-1" based on whether the photon can get through a polarizer set at a specific angle.

Last edited: Mar 20, 2017
18. Mar 20, 2017

### vanhees71

It's just a symbol. There's no deeper meaning

An electromagnetic wave (photons are photon-number-one Fock states of the electromagnetic field) is a massless spin-1 field, and thus has only two polarization degrees of freedom. Usually one uses the helicity eigenstates as a basis, where helicity is the projection of the total angular momentum to the direction of the momentum of the photon. For a photon (spin 1) the possible eigenvalues of helicity are $h \in \{-1,1\}$ (in units, where $\hbar=1$).

I don't understand your logic here. The spin eigenstates are determined as the common eigenstates of $\vec{s}^2$ and $s_z$. The possible eigenvalues are $s(s+1)$ with $s \in \{0,1/2,1,\ldots \}$ and $s_z \in \{-s,-s+1,\ldots,s-1,s \}$ (in units, where $\hbar=1$).

19. Mar 20, 2017

### Staff: Mentor

Perhaps he's referring to the Pauli matrices:

https://en.wikipedia.org/wiki/Pauli_matrices

20. Mar 20, 2017

### N88

Yes, thanks; from the link it's called the Pauli vector; which I understand to be a "vector of matrices". I was interested in how the Pauli vector should be linked to the equations that you gave us; ie, with your use of $S$ and $S_z$.