How to show that particle spin includes angular momentum?

[snoopies622]

TL;DR

What kind of experiment demonstrates that a spinning subatomic particle has angular momentum?

I understand how a massive, electrically charged spinning ball would have both angular momentum and a magnetic dipole, and i can see how the
Stern–Gerlach experiment shows that the magnetic dipole of an electron is quantized.

What kind of experiment demonstrates
a connection between electron spin and angular momentum?

I imagine maybe a piece of iron hanging by a string, and changing a magnetic field that's parallel to the string causing the piece of iron to rotate?

 

[Hyperfine]

What kind of experiment demonstrates
a connection between electron spin and angular momentum?

Atomic spectroscopy. In particular the fine structure splittings.

 

[Hyperfine]

I neglected to point out that the magnetic moment of a particle such as the electron is a direct consequence of the electron's angular momentum: orbital angular momentum, spin angular momentum, or both.

For example, the spin magnetic moment is given by
μ_s=-g_sμ_B2πS/h

Sorry for the poor equation--seems LaTex is not working for some reason.

 

[snoopies622]

Thanks for the guidance, Hyperfine. I should clarify that what puzzles me is that introductions to the concept of particle spin seem to speak of angular momentum and magnetic dipole as if they were the same phenomenon, while for my classical brain the first is purely mechanical and the other includes electromagnetism, and of course they have different dimensions (kg m^2/sec versus coulomb m^2/sec).

 

[PeroK]

Thanks for the guidance, Hyperfine. I should clarify that what puzzles me is that introductions to the concept of particle spin seem to speak of angular momentum and magnetic dipole as if they were the same phenomenon, while for my classical brain the first is purely mechanical and the other includes electromagnetism, and of course they have different dimensions (kg m^2/sec versus coulomb m^2/sec).

You also have measurements of the electron mass and electric charge. In addition to its magnetic dipole moment.

 

[snoopies622]

Well, i suppose that dimensionally, angular momentum x charge density = magnetic moment, but this seems to leave out the question of the internal geometry of an electron, how far away from its center of rotation lies its mass and charge. But perhaps that's not necessary - will give this some thought.

 

[Nugatory]

Thanks for the guidance, Hyperfine. I should clarify that what puzzles me is that introductions to the concept of particle spin seem to speak of angular momentum and magnetic dipole as if they were the same phenomenon, while for my classical brain the first is purely mechanical and the other includes electromagnetism, and of course they have different dimensions (kg m^2/sec versus coulomb m^2/sec).

A charged particle with angular momentum has a non-zero magnetic dipole moment, so when we observe the dipole moment we infer the non-zero angular momentum.

 

[snoopies622]

A charged particle with angular momentum . .

How is it known that the particle has angular momentum in the first place? This is the essence of my question.

 

[PeroK]

How is it known that the particle has angular momentum in the first place? This is the essence of my question.

You mean it might have a magnetic dipole moment but no angular momentum? In a way that's what an electron has. Spin is sometimes called inherent angular momentum, as it is something the electron always has regardless of its state. Spin obeys the quantum theory of angular momentum. E.g. the spin of a system of two electrons obeys the rule for addition of angular momentum. That's the basis on which it is quantum angular momentum.

You can call this property what you like: spin, quantum spin, spin angular momentum, inherent angular momentum ... It all means the same thing. You can call it what you like. You could call it the snoop instead of spin and have a snoopie operator. It makes no difference to the experiments or the mathematics what you call it. In a way this is what has happened with the terms waves and particles in QM. You can tie yourself in knots by having preconceived notions of waves and particles and trying to reconcile QM with these preconceived concepts. In fact, I prefer to call electrons and photons quantum particles to emphasise that they are not classical objects. In the same way, by definition an electron has quantum angular momentum, because that is the name we have chosen for the observed property.

Note that in the relativistic theory only the sum of spin plus orbital angular momentum is conserved. These quantities are not conserved separately. That's another reason to call it angular momentum.

 

[WernerQH]

I imagine maybe a piece of iron hanging by a string, and changing a magnetic field that's parallel to the string causing the piece of iron to rotate?

It's called the Einstein - de Haas effect.

 

[snoopies622]

My impression now is that the Stern–Gerlach experiment demonstrates that an electron has an intrinsic magnetic moment, while the Einstein–de Haas effect shows a connection between electron magnetic moment and angular momentum (though complicated by the fact that both orbital and spin angular momentum are present).

 

[vanhees71]

The relation between the magnetic moment and angular momentum can be understood already classically. To that end consider magnetostatics in Coulomb gauge,
$$\Delta \vec{A}=-\mu_0 \vec{j}, \quad \vec{\nabla} \cdot \vec{A}=0.$$
Now consider a current distribution, which is entirely contained within a sphere of radius, $a$, i.e., $\vec{j}(\vec{r})=0$ for $|\vec{r}| \geq a$. Then for $|\vec{r}| \gg a$,
$$\vec{A}(\vec{r})=\frac{\mu_0}{4 \pi} \int_{K_a} \frac{\vec{j}(\vec{r}')}{|\vec{r}-\vec{r}'|}=\frac{\mu_0}{4 \pi r} \int_{K_a} \vec{j}(\vec{r}') \left (1 + \frac{\vec{r} \cdot \vec{r}'}{r^2} \right).$$
The first term vanishes, because from $\vec{\nabla} \cdot \vec{j}=\partial_j j_j=0$
$$j_k=\partial_j (x_k j_j)$$
and Gauß's theorem
$$\int_{K_a} \mathrm{d}^3 r' j_k (\vec{r}') = \int_{\partial K_a} \mathrm{d}^2 f_j x_k j_j=0,$$
because by assumption $\vec{j}(\vec{r'})=0$ for $|\vec{r}'|=r' \geq a$. So we are left with
$$A_k(\vec{r})=\frac{\mu_0 4 \pi r^3} \int_{K_a} \mathrm{d}^3 r' r_j r_j' j_k.$$
Now we can write
$$r_j r_j' j_k=\frac{1}{2} r_j (r_j' j_k-r_k' j_j) + \frac{1}{2} r_j (r_j' j_k+r_k' j_j).$$
The symmetric part can be written as
$$r_j' j_k + r_k' j_j=\partial_l' (r_j' r_k' j_l),$$
and thus the corresponding integral vanishes.

The antisymmetric piece can be written as
$$\vec{e}_k r_j (r_j' j_k-r_k' j_j)=\vec{r} \cdot \vec{r}' \vec{j}-\vec{r}' \vec{r} \cdot \vec{j} = -\vec{r} \times (\vec{r}' \times \vec{j}).$$
So finally we have
$$\vec{A}(\vec{r}) \simeq \frac{\mu_0}{4 \pi r^3} \vec{m} \times \vec{r}, \quad \vec{m}=\frac{1}{2} \int_{K_a} \mathrm{d}^3 r' \vec{r}' \times \vec{j}.$$
Now let the particle-number density of the charges making up the current be $n(\vec{r}')$. Then we have $\vec{r}' \time \vec{j}=q n \vec{r}' \times \vec{v}=\frac{q}{m} n m \vec{r}' \times \vec{v}$ and thus
$$\vec{m}=\frac{q}{2m} \vec{L},$$
where $\vec{L}$ is the total (orbital) angular momentum of the charges.

That's why in 1915 Einstein insistent on this relation between angular momentum and magnetic moment, and that's why de Haas missed the discovery of the non-trivial gyromagnetic factors. It was pretty soon clear that the gyrofactor for usual ferromagnetics is close to 2 rather than 1, and indeed for an elementary spin-1/2 fermion the Dirac equation together with the gauge principle and minimal coupling leads to $g=2$ (modulo corrections due to higher-order QED contributions).

 

[snoopies622]

Thanks vanhees71, in a way that addresses my question about the internal geometry of an electron, if that has any scientific meaning.

 

[Hyperfine]

The following book might be of interest:

Sin-itiro Tomonaga, The Story of Spin, The University of Chicago Press, 1998.

 

[vanhees71]

Thanks vanhees71, in a way that addresses my question about the internal geometry of an electron, if that has any scientific meaning.

In a sense yes, but you must not think that this classical picture is literally describing an electron, which is (according to what we know today) a generic quantum object (aka an "elementary particle"). The magnetic moment due to the spin formally leads to a contribution to a current, but this cannot be understood as "moving electric charge". At this point, of course, our "classical intuition" fails and has to be substituted by a much more abstract, mathematical "quantum intuition".