What exactly is electron spin?

[jobsism]

What exactly is electron spin?The thing is, I had read in some general books on QM that spin of an electron is a non-classical degree of freedom,i.e.,an intrinsic angular momentum it possesses, and that it is NOT to be confused with a physically spinning electron. But in my physics textbook, it says that an electron physically spins about it's axis,due to which it has a magnetic dipole moment. So which is it?

 

[G01]

No. It is not correct to think of the electrons intrinsic angular momentum as resulting from the electron physically rotating about an axis.

In fact this problem is usually solved during a standard junior level quantum mechanics course. You can treat the electron as a spinning spherical shell of radius r_c=\frac{e^2}{4\pi\epsilon_omc^2}, the "classical electron radius." Now, calculate how fast the electron must be spinning in order to have angular momentum equal to \hbar/2.

You'll see that the tangential speed of a point on the electrons equator would have to spinning faster than the speed of light for everything to work out and give the correct angular momentum. Thus, the intrinsic angular momentum of an electron cannot be due to the electron rotating about an axis.

 

[edguy99]

No. It is not correct to think of the electrons intrinsic angular momentum as resulting from the electron physically rotating about an axis.

In fact this problem is usually solved during a standard junior level quantum mechanics course. You can treat the electron as a spinning spherical shell of radius r_c=\frac{e^2}{4\pi\epsilon_omc^2}, the "classical electron radius." Now, calculate how fast the electron must be spinning in order to have angular momentum equal to \hbar/2.

You'll see that the tangential speed of a point on the electrons equator would have to spinning faster than the speed of light for everything to work out and give the correct angular momentum. Thus, the intrinsic angular momentum of an electron cannot be due to the electron rotating about an axis.

You often see this arguement, but consider http://en.wikipedia.org/wiki/List_of_moments_of_inertia" how different shapes have different moments of inertia. Clearly you have proven that an electron is not a spinning shape that you suggest - "spinning spherical shell", but I don't think I have ever seen it proven that ANY type of spinning object could not have the properties of an electron.

 

[Drakkith]

Hasn't the electron been shown to have spherical symmetry or whatever in its magnetic and electric properties all the way down to within something like 1 part in 1 billion? Does that not mean that the electron is spherical in regards to the spinning electric field if it were in fact spinning?

 

[G01]

You often see this arguement, but consider http://en.wikipedia.org/wiki/List_of_moments_of_inertia" how different shapes have different moments of inertia. Clearly you have proven that an electron is not a spinning shape that you suggest - "spinning spherical shell", but I don't think I have ever seen it proven that ANY type of spinning object could not have the properties of an electron.

Try working it out for some other shape on that list. All the shapes have the same dependence on the length dimension in their moment of inertia. The only thing that changes is a multiplicative factor of order 1. The order of magnitude of the resulting speed of the furthest point from the center will remain the same.

It will be around 100 times greater than c.

Also, remember that the actual experimental radius is much smaller than r_c. By using r_c we are being "generous," and the result is still impossible.

 

[jobsism]

I see...thanks everyone for the help! Just one more thing: What would be the physical significance of this intrinsic angular momentum? Wouldn't it mean that the electron possesses some kind of motion,in addition to orbital motion around the nucleus?

 

[Drakkith]

In a classical object, physical spinning around an axis is required for it to have angular momentum. However this is not the case for an Electron. That is the physical significance.

 

[granpa]

spin is 2 things

the first can be THOUGHT OF AS the spinning of the electron.
(even though the electron doesn't really spin)

the other is how it behaves in an external magnetic field
either it aligns with the magnetic field (paramagnetic) or it aligns against it (diamagnetic)

 

[jobsism]

So, am I to understand that (please correct me if I'm wrong) electron spin is this non-classical property than an electron posseses, due to which it has an angular momentum(which does not mean a spinning electron), which determines its alignment with the magnetic field, in the form of the spin quantum number, which can be either +1/2 or -1/2?

 

[granpa]

the angular momentum does not determine how it aligns.

the spin quantum number does determine how it aligns.

 

[jobsism]

Then what exactly is the use of introducing this angular momentum? I mean, is it brought about just so that "everything fits into place"?

 

[granpa]

spin is 2 things

the first can be THOUGHT OF AS the spinning of the electron.
(even though the electron doesn't really spin)[/color]

the other is how it behaves in an external magnetic field
either it aligns with the magnetic field (paramagnetic) or it aligns against it (diamagnetic)

see above

 

[jobsism]

I'm sorry, but I still don't completely get it...where does the spin quantum number come into all of this? And doesn't spin refer to the angular momentum itself? Then according to what you have said, it HAS something to do with the alignment in magnetic field,right?

 

[edguy99]

Hasn't the electron been shown to have spherical symmetry or whatever in its magnetic and electric properties all the way down to within something like 1 part in 1 billion? Does that not mean that the electron is spherical in regards to the spinning electric field if it were in fact spinning?

I have often wondered about this. Neither wiki or hyperphysics refer to the electron as spherical symmetrical (although maybe I missed it). Do you have a reference for this?

TIA

 

[granpa]

the electron can be thought of as spinning and producing a magnetic field even though it isn't really spinning.

it also has a second property[/color] which determines how it responds to a magnetic field.

diamagnetic materials align against an external magnetic field
and paramagnetic materials align with an external magnetic field.

I can't make it any clearer than that.

 

[Dickfore]

Angular momentum in quantum mechanics classifies the irreducible representations of the Lie group associated with rotations, which is called SO(3). The generators of this group (3 of them) are actually the components of angular momentum J_{i} (measured in units of \hbar) and they satisfy the fundamental commutation relations:
<br /> [J_{i}, J_{k}] = i \, \varepsilon_{i k l} \, J_{l}<br />
Another Lie group, called SU(2) (the group of all unitary 2 \times 2 matrices, satisfying U \cdot U^{\dagger} = U^{\dagger} \cdot U = 1, with determinant one \mathrm{det}U = 1) has 3 generators that satisfy the same Lie algebra. The half-integer eigenvalues for J are actually representations of SU(2) and not of SO(3), which only allows the integer eigenvalues. The reason for this is that a 360^o does not bring back the system in its original state, as it should be for proper rotations.

One may wonder how can angular momentum attain the half-integer eigenvalues if it is a generator of rotations. My answer to this, and I do not know if it is true or not, is that the group of proper rotations is a subgroup of a larger group of Lorentz transformation, involving Lorentz boosts in addition to proper rotations. This group has 6 generators (3 more for the boosts) and one can form two su(2) algebras, or:
<br /> \mathcal{so}(3, 1) \sim \mathcal{su}(2) \times \mathcal{su}(2)<br />
and the two subalgebras are connected by hermitian conjugation. If we call the generators of the first group N_{j}, then the other one are N^{\dagger}_{j} and the generators corresponding to rotations from before are expressed as:
<br /> J_{j} = N_{j} + N^{\dagger}_{j}<br />
So, we have to perform `angular momentum addition' to find the possible values for angular momentum in a particular irreducible representation of the Lorentz group, labeled as (j_{1}, j_{2}).

So, I would say 'spin' is a remnant of what we used to call 'angular momentum' when we consider the extended group associated with relativistic covariance.

 

[strangerep]

One may wonder how can angular momentum attain the half-integer eigenvalues if it is a generator of rotations. My answer to this, and I do not know if it is true or not, is that the group of proper rotations is a subgroup of a larger group of Lorentz transformation, involving Lorentz boosts in addition to proper rotations
[...]
So, I would say 'spin' is a remnant of what we used to call 'angular momentum' when we consider the extended group associated with relativistic covariance.

We don't need SR to get half-integer eigenvalues for angular momentum.
All we need are the so(3) commutation relations, and the requirement that the generators
of the SO(3) group be represented as Hermitian operators on a Hilbert space.
Then it falls out by turning the mathematical handle.

See, e.g., Ballentine, or Greiner & Muller, or probably lots of other QM textbooks. :-)

 

[Dickfore]

We don't need SR to get half-integer eigenvalues for angular momentum.
All we need are the so(3) commutation relations, and the requirement that the generators
of the SO(3) group be represented as Hermitian operators on a Hilbert space.
Then it falls out by turning the mathematical handle.

You didn't read the upper part of my post carefully where I said that the Lie algebra of so(3) is isomorphic to su(2), thus making the corresponding Lie groups locally isomporphic. However, globally SU(2) is a double cover of SO(3) and SU(2) is the only simply connected group.

According to the theorems for the relations of Lie groups and Lie algebras, the correspondence between the representations of the Lie algebra and the Lie group from it is generated via the exponential map is only true for the simply connected groups. Thus, the representations of the lie algebra given above is a representation for SU(2).

 

[Dickfore]

From http://en.wikipedia.org/wiki/Spin_(physics)#Spin_and_rotations":

Spin is a type of angular momentum, where angular momentum is defined in the modern way as the "generator of rotations" (see Noether's theorem).[1][2] This modern definition of angular momentum is not the same as the historical classical mechanics definition, L = r × p. (The historical definition, which does not include spin, is more specifically called "orbital angular momentum".)

If you go to the section Spin and rotations, it says:

Mathematically speaking, these matrices (referring to n dimensional unitary transformation matrices) furnish a unitary projective representation of the rotation group SO(3). Each such representation corresponds to a representation of the covering group of SO(3), which is SU(2). There is one n-dimensional irreducible representation of SU(2) for each dimension, though this representation is n-dimensional real for odd n and n-dimensional complex for even n (hence of real dimension 2n).

 

[strangerep]

You didn't read the upper part of my post carefully

I think you did not read the first sentence of mine carefully. :-)
I was more interested in whether SR is necessary to account for
half-integer spin. Afaict, it's not -- but you seemed to be saying
the contrary in your earlier post.

So I'll tidy up my sloppiness a little and restate...

All we need are the so(3) (or su(2)) commutation relations, and the requirement
that the usual basis elements of the so(3) (or su(2)) Lie algebra are represented
as Hermitian operators on a Hilbert space (and also quadratic products of
elements thereof, such as J^2, etc).

That's why I referred to derivations in those textbooks on non-relativistic QM.

 

[Dickfore]

what is a basis element of a lie algebra?

 

[dextercioby]

I'm surprised you ask. A Lie algebra is a vector space with one additional internal mapping, the Lie product. As a vector space, it has at least a basis with at least one vector as its element.

 

[Dickfore]

I'm surprised you ask. A Lie algebra is a vector space with one additional internal mapping, the Lie product. As a vector space, it has at least a basis with at least one vector as its element.

Which one is it?

 

[dextercioby]

Depends on the algebra, of course. For u(1), it's simply 'x', a real number.

 

[akhmeteli]

No. It is not correct to think of the electrons intrinsic angular momentum as resulting from the electron physically rotating about an axis.

In fact this problem is usually solved during a standard junior level quantum mechanics course. You can treat the electron as a spinning spherical shell of radius r_c=\frac{e^2}{4\pi\epsilon_omc^2}, the "classical electron radius." Now, calculate how fast the electron must be spinning in order to have angular momentum equal to \hbar/2.

You'll see that the tangential speed of a point on the electrons equator would have to spinning faster than the speed of light for everything to work out and give the correct angular momentum. Thus, the intrinsic angular momentum of an electron cannot be due to the electron rotating about an axis.

I have read this argument many times in several books, but somehow failed to understand it. Maybe you could explain it? My problem is I consider a point particle with a mass equal to that of electron and having a circular orbit with a radius of the order of the Compton wavelength (or even classical electron radius). If the (tangential) velocity of this particle is very close to the speed of light, its momentum can be arbitrarily high, so the angular momentum can also be arbitrarily high. I don't quite see how this reasoning is faulty. Do you?

 

[Goldstone1]

What exactly is electron spin?The thing is, I had read in some general books on QM that spin of an electron is a non-classical degree of freedom,i.e.,an intrinsic angular momentum it possesses, and that it is NOT to be confused with a physically spinning electron. But in my physics textbook, it says that an electron physically spins about it's axis,due to which it has a magnetic dipole moment. So which is it?

According to modern physics, it does not spin round an axis. It would need to spin twice the angle it began with just to return to it's original restarting point - this is because we say particles have no structure - they are dimensionless.

Spin is simply an angular momentum. It is an intrinsic property of all matter, and there have been no psuedoscalar fields which predict zero-spin particles ever been detected.

 

[vanhees71]

Pions are (pseudo-)scalar particles with spin 0. However, they are not elementary but the bound state of a quark-antiquark pair (coming in three flavors, the charged pions and the neutral pion).

By definition, the spin of an elementary massive particle s \in \{0,1/2,1,3/2,\ldots \} denotes the representation of the rotation group of particle's rest state (momentum 0). I.e., the momentum eigenspace for momentum eigenvalue 0 is (2s+1)-dimensional.

For massless particles the situation is slightly more complicated, and for each s there are only two helicity states with h=\pm s for each momentum state (of course for s=0 there's only one helicity state with helicity 0).

 

[ytuab]

I have read this argument many times in several books, but somehow failed to understand it. Maybe you could explain it? My problem is I consider a point particle with a mass equal to that of electron and having a circular orbit with a radius of the order of the Compton wavelength (or even classical electron radius). If the (tangential) velocity of this particle is very close to the speed of light, its momentum can be arbitrarily high, so the angular momentum can also be arbitrarily high. I don't quite see how this reasoning is faulty. Do you?

In your spin state, the spnning speed is almost light speed, and the relativistic mass is too heavy to move it.<br>
But the electron has orbital motion (kinetic energy) and orbital angular momentum, too.

If you consider the orbital kinetic energy, its speed exceeds the speed of light. ( v = c + a )
So the spinning electron must stop. (This is the reason why its mass is very big.)
Of course, you can not imagine the free spinning electron, which is flying in the air.
And you can not accelerate the free electron any more by V, because it exceeds the light speed.
(Electron spin 1/2 always exists.)

Originally the correct energy of Schrödinger equation can be gotten from the electron's rest mass (or reduced mass).
If you use the heavy mass , the electron is always stopping near the nucleus, which gives different energy result.

 

[akhmeteli]

In your spin state, the spnning speed is almost light speed, and the relativistic mass is too heavy to move it.<br>
But the electron has orbital motion (kinetic energy) and orbital angular momentum, too.

If you consider the orbital kinetic energy, its speed exceeds the speed of light. ( v = c + a )

I would think you can only add speeds like you do when the speeds are not relativistic, which is not the case.

 

[granpa]

the speed of the electron in ground state of the bohr model of hydrogen is ac