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Pyrus

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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?

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In summary, the electron has a spin of +- 1/2. This 1/2 signifies spin down or spin up. Other spins are also there like 1 and 2. What do they signify? There is no possibility of any particle with spin other than 1/2, 1 or 2.

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Pyrus

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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?

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- #2

mikeyork

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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

- #3

jtbell

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There are two separate but related quantum numbers associated with intrinsic angular momentum ("spin"). The first one, ##s##, is associated with thePyrus said:Like, electron has a spin of +- 1/2... What do this 1/2 signifies.

The second one, ##m_s##, is associated with the

##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

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Delta2

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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

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Nugatory

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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.Delta² said:

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 apoint can rotate around itself?. Classically this doesn't make sense though.

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.

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Nugatory

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If you will simply admit that maybe [Nature] does behave like this, you will find her a delightful, entrancing thing. Do not keep saying to yourself, if you can possibly avoid it, `But how can it be like that?’ because you will get ‘down the drain’, into a blind alley from which nobody has escaped. Nobody knows how it can be like that.

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muscaria

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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.

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muscaria

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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..Pyrus said:What do this 1/2 signifies

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weirdoguy

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muscaria said:So spin isn't a quantum mechanical property but a wave-like property associated with a certain type of energy flow in a field.

Can you show any source that claims that?

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jtbell

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Nugatory

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256bits

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weirdoguy said:Can you show any source that claims that?

http://cosmology.princeton.edu/~mcdonald/examples/EP/commins_arnps_62_133_12.pdfAt the present stage of knowledge, the electron is considered to be a particle with no spatial

extension, and the same is true of the other leptons as well as the quarks. Thus, the spin angular

momentum of a fundamental fermion, an observable with no classical analog, is considered by

many physicists to be describable only by abstract mathematical expressions, with no possibility of

intuitive visualization.However, a contrary view was provided by Ohanian (131) in 1984, based on

work by Belinfante (132) in 1939. Starting from a symmetric form of the energy-momentum tensor

associated with theDirac Lagrangian density, Ohanian showed that spin angular momentum arises

from a circular flow of energy in the electron wave field, and that the intrinsic spin magnetic

moment (gs= 2) is associated with a similar circular flow of charge. Furthermore, by way of

demonstration that these ideas are quite general, Ohanian used a symmetric form of the energymomentum tensor for a circularly polarized Maxwell field to derive separate expressions for the

orbital and spin angular momenta of that field, and thus he arrived at an intuitive picture of

photon spin. Finally, in 1952 a somewhat similar analysis for the electron, expressed in terms of

the zitterbewegung, was given by Huang (133).

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

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N88

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Many thanks. I like this, especially for its clarity. So now (hoping to be treated gently) I wonder about 3 things. Please:jtbell said:There are two separate but related quantum numbers associated with intrinsic angular momentum ("spin"). The first one, ##s##, is associated with themagnitudeof 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 theorientation(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 theEinstein-de Haas effect.

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

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edguy99

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N88 said:2. In Aspect's famous experiments with photons, his measurement outcomes are ±1. What happens to the 0 value in these cases? ...

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.

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N88 said: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?

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

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##).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

- #19

jtbell

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N88 said:1. The symbol σ\sigma is often associated with spin. How does it relate to your expressions?

Perhaps he's referring to the Pauli matrices:vanhees71 said:It's just a symbol. There's no deeper meaning

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

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N88

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jtbell said:

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##.

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$$\vec{S}=\frac{1}{2} \vec{\sigma}.$$

https://en.wikipedia.org/wiki/Pauli_matrices#SU.282.29

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N88

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Thanks for this and the link. This approach is new to me. Does it deliver the correct results for Aspect's experiments? Is it readily modified to deliver the correct EPRB results?edguy99 said: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.

- #23

edguy99

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Bit off topic here i suspect. Happy to discuss in new thread.

- #24

N88

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jtbell said:There are two separate but related quantum numbers associated with intrinsic angular momentum ("spin"). The first one, ##s##, is associated with themagnitudeof 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 theorientation(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}##.

For an electron (or any spin-half particle): Can both these quantum numbers be measured independently?*

For that would make it clear (at least to me) that the spin-vector precesses about ANY

So then, what are the correct symbols for representing that

* Then (hoping the above makes sense) to see how close the ratio comes to √3?

Thank you.

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jtbell

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Spin precession in a magnetic field oriented along the z-direction refers to the behavior ofN88 said:precessing spin-vector

http://farside.ph.utexas.edu/teaching/qmech/Quantum/node91.html

linked page said:Note, however, that ameasurementof ##S_x##, ##S_y## or ##S_z## will always yield either ##+\hbar/2## or ##-\hbar/2##. It is therelative probabilitiesof obtaining these two results which varies as the expectation value of a given component of the spin varies.

Introductory modern physics books etc. often show a diagram which gives the impression that a particle's spin vector literally precesses smoothly on the surface of an imaginary cone, like the axis of a spinning top, but you need to resist that impression. Because of the uncertainty relations among the three components of spin angular momentum, only one of them can have a well-defined value at any point in time.

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atyy

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So to understand spin, one first needs to understand what is meant by position is defined by its commutation relations. If one doesn't understand that, then one doesn't even understand position.

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ImStein

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σ for "quantum spin". Nice fit!

- #29

akvadrako

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Trevor Rempel and Laurent Freidel – Phys. Rev. D 95, 104014 – Published 11 May 2017

Spin in particle physics refers to the intrinsic angular momentum of a particle. It is a fundamental property of particles and is analogous to the spinning of a classical object around an axis.

Spin is measured in units of h-bar, which is a fundamental physical constant. It is typically measured using detectors such as particle accelerators, which can determine the spin of a particle by observing its interactions with other particles.

The concept of spin is crucial in understanding the properties and behavior of particles. It is a fundamental aspect of quantum mechanics and has implications in areas such as particle interactions, symmetry breaking, and the structure of matter.

Yes, particles can have either integer or half-integer spin values. This is a consequence of the spin-statistics theorem, which states that particles with integer spin are bosons, while those with half-integer spin are fermions.

Spin angular momentum is a physical quantity that describes the rotation of a particle around its axis. In quantum mechanics, spin is a form of intrinsic angular momentum and is related to the spin angular momentum through Planck's constant, h-bar.

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