Spin - magnitude & direction

In summary, the angular momentum of a photon does not have an analogy in classical mechanics, however, it can be thought of as a particle "having" spin.
  • #1
San K
911
1
The below might need lots of corrections/modifications, since spin is a hard concept to grasp and I don't understand it much.

We know that:

The angular momentum (spin) of a photon does not have an analogy in classical mechanics.

however let's use a loose/inaccurate analogy of something that is rotating very fast along the axis in the direction of motion...and that generates angular momentum (?)

1. Is the spin generated from the EM waves (in a sense "rotating" very fast) across the direction of travel of the photon?

2. The angular momentum remains same for same type of elementary particle. Why? Can we not have faster or slower rotation? (if we can have different frequencies for a same type of elementary particle, why not "rotation"?

3. Why is the vector different from classical mechanics?

4. If the angular momentum is quanitzed, can we assume the "fields/forces" producing it must have some sort of quantization as well?
 
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  • #2
Spin is angular momentum a particle just "has." Just like a particle "has" a certain mass or a certain charge, it has some inherent angular momentum that can be measured and is quantized.

Photons aren't the only particles with spin. Electrons, quarks, mesons, etc also have spin. The spin of particles can change, but it can only take on certain values depending on the system. Spin isn't litteral rotation in the classical sense. We just call it angular momentum because when you do experiments with particles that have no classical angular momentum, they still produce results you would expect to see as if it were "spinning" about an axis.

The vector is different because it doesn't point in any specific direction.

Yeah, fields are quantized. For example, the electromagnetic filed quanta is something familiar to you: the photon.
 
  • #3
I would say spin is not angular momentum.

The photon has l = 1, s = 0, this can be pictured from the resultant vector of the E x B field.

Experimentally you can measure spin, see stern gerlach experiment.
 
  • #4
RGauld said:
I would say spin is not angular momentum.
Spin IS angular momentum.

RGauld said:
The photon has l = 1, s = 0, this can be pictured from the resultant vector of the E x B field.
The photon has s = 1, and any value ℓ ≥ 0. E x B is the momentum density, whereas the angular momentum density is x x (E x B).

RGauld said:
Experimentally you can measure spin, see stern gerlach experiment.
Good luck doing the Stern-Gerlach experiment on a photon. (It only works for charged particles.)
 
  • #5
The spin of the photon corresponds to the polarization of the EM wave. Since the photon has spin 1, one would expect 3 different possibilities to align the spin (+1,0,-1) which would correspond to 3 different polarizations (x,y,z). However, the polarization in the direction of propagation is ruled out due to the transversal nature of EM waves. So we are left with only two polarizations (x,y) which is the reason why photons behave like spin-1/2 particles (possible alignments +1/2, -1/2) in experiments like Bell tests.
 
  • #6
In the Bell tests the inequality, n[y+z-] + n[x-y+] ≥ n [x-z-] with spin 1/2 photons is derived
from perfect anti correlations when the detector settings are parallel. Where P(α|λ| is
considered to be deterministic based on conservation laws - a total spin of zero.
Since the inequality is compared to experiments where detector settings are not parallel.
Can P(α|λ| be considered to be non deterministic with respect to magnetic field in detector when
the detector settings are not parallel, and how this may affect the spin of particle at time of measurement ? Also what happens to the conservation law that is in effect when detector
settings are parallel but then do not seem to be in effect when inequality is disproven ?
 
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  • #7
Bill, I think you are confusing quantum numbers.

When I say spin I do not mean total angular momentum j.

Therefore the stern gerlach is for testing spin, presuming I do not mean fermions without em charge.
 
  • #8
From pdg;

I(J PC) = 0,1 ( 1 - - )
 
  • #9
RGauld said:
Bill, I think you are confusing quantum numbers. When I say spin I do not mean total angular momentum j.
Particle physicists often use J to denote a particle's spin, as you see in the PDG tables. But in ordinary parlance, J = L + S, where S is the spin and L the orbital angular momentum. The photon has spin S = 1, and may have any value of orbital angular momentum. Consequently given the value of L there are only three possibilities for J: either J = L or J = L ± 1. The various cases are known as multipole radiation, and are important whenever a photon is emitted, such as X-rays from an atom or gamma rays from a nucleus.
 
  • #10
wotanub said:
We just call it angular momentum because when you do experiments with particles that have no classical angular momentum, they still produce results you would expect to see as if it were "spinning" about an axis.

The vector is different because it doesn't point in any specific direction.

.

Thanks wotanub, bill k and others. Great answers.

What is "spinning"? The excitation of the em field? What is the speed of the (hypothetical) spin? ...say relative to c.
 
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  • #11
San K said:
Thanks wotanub, bill k and others. Great answers.

What is "spinning"? The excitation of the em field? What is the speed of the (hypothetical) spin? ...say relative to c.

That's the thing... Nothing is actually spinning, but it still has this angular momentum. Ferimions spin in integer multiples of hbar/2 and Bosons have integer multiples of hbar.
 
  • #12
morrobay said:
In the Bell tests the inequality, n[y+z-] + n[x-y+] ≥ n [x-z-] with spin 1/2 photons is derived
from perfect anti correlations when the detector settings are parallel. Where P(α|λ| is
considered to be deterministic based on conservation laws - a total spin of zero.
Since the inequality is compared to experiments where detector settings are not parallel.
Can P(α|λ| be considered to be non deterministic with respect to magnetic field in detector when
the detector settings are not parallel, and how this may affect the spin of particle at time of measurement ? Also what happens to the conservation law that is in effect when detector
settings are parallel but then do not seem to be in effect when inequality is disproven ?
Let me sort this out for above : The question about the conservation law and spin applies to entangled photons.The question on detector magnetic field affecting spin applies to electrons.
Maybe this should be a new topic, but I am looking for why Bells' inequalities are violated from the
physics of how spin changes.
 

1. What is spin in relation to magnitude and direction?

Spin refers to the intrinsic angular momentum of a particle, which is a fundamental property of subatomic particles. It is a vector quantity with both magnitude and direction, and it plays a crucial role in determining the behavior and properties of particles.

2. How is spin measured?

Spin is measured in units of h-bar, which is a combination of Planck's constant and 2π. It is typically represented by the symbol "s" and can have values of 0, 1/2, 1, 3/2, etc. These values correspond to the different possible orientations of the spin vector.

3. What is the difference between spin magnitude and spin direction?

The magnitude of spin refers to the strength of the spin, while the direction refers to the orientation of the spin vector. For example, a spin of 1/2 can have two possible directions - up or down. The direction of spin can also be changed through interactions with other particles.

4. How does spin affect the behavior of particles?

Spin plays a crucial role in determining the properties and behavior of particles. For example, particles with half-integer spin are fermions and are subject to the Pauli exclusion principle, while particles with integer spin are bosons and can occupy the same quantum state. Spin also affects how particles interact with magnetic fields, and it is a key factor in determining the stability of atoms and molecules.

5. Can spin change over time?

Spin is a fundamental property of particles and is considered to be conserved, meaning it does not change over time. However, it can change direction through interactions with other particles, such as when a particle absorbs or emits a photon. In these cases, the spin of the particle will change to conserve the total angular momentum of the system.

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