Explaining Quantum Spin Without Math: Is It Possible?

[jostpuur]

We can think that the integration domain has outer boundary that is a surface of some large ball B(0,R), and after integration we are taking the limit R\to\infty.

The asymptotic behavior of the integrand is

<br /> \Big|\frac{x-x_1}{|x-x_1|^3}\frac{1}{|x-x_2|}\Big|\approx \frac{1}{R^3}<br />

on the surface, when the ball is so big that we can assume the points x_1 and x_2 to be close to the origo. Since the area of the surface of the ball approaches infinity like R^2, the surface integral (that we get when the volume integral is made into a surface integral according to the Gauss's theorem) is going to give something like 1/R, and it will vanish on the limit R\to\infty.

 

[country boy]

I am interested in quantum mechanics, but my maths isn't up to understanding some of the concepts. I've tried to look at spin, but all explanations are mathematical. Is it possible for it to be explained in words?

In QM the angular momentum quantum number is a property of the symmetry of the wavefunction. The requirement of continuity around an orbit forces the angular momentum to be an integer number m of units of h/2pi. When you rotate the wavefunction through an angle 2pi it comes back to itself m times. We say that the wavefunction has m-fold symmetry.

The odd thing about 1/2-integer spin is that you have to rotate through 4pi, twice around, to get back to the original wavefunction. Classically, that's not intuitive. It's as if the continuity requirement is imposed not on the wavefunction but on the probability ("square" of the wavefunction).

 

[jostpuur]

Country boy's post reminded me of the OP. I have a question to you physicists. Do you know a physics book, that explains \pi_1 (SO(3))=\{0,1\} (or =\mathbb{Z}_2), when explaining spin-1/2 transformations?

 

[lightarrow]

We can think that the integration domain has outer boundary that is a surface of some large ball B(0,R), and after integration we are taking the limit R\to\infty.

The asymptotic behavior of the integrand is

<br /> \Big|\frac{x-x_1}{|x-x_1|^3}\frac{1}{|x-x_2|}\Big|\approx \frac{1}{R^3}<br />

on the surface, when the ball is so big that we can assume the points x_1 and x_2 to be close to the origo. Since the area of the surface of the ball approaches infinity like R^2, the surface integral (that we get when the volume integral is made into a surface integral according to the Gauss's theorem) is going to give something like 1/R, and it will vanish on the limit R\to\infty.

Suppose you don't have two point charges q1 and q2 but two spherical conductive charges of non-zero radius r.
The boundary is:
1.an infinitely far spherical surface
2. the suface of the first conductive sphere
3. the suface of the second conductive sphere

Now, make r go to zero...

 

[jostpuur]

Suppose you don't have two point charges q1 and q2 but two spherical conductive charges of non-zero radius r.
The boundary is:
1.an infinitely far spherical surface
2. the suface of the first conductive sphere
3. the suface of the second conductive sphere

Now, make r go to zero...

Ah, you were talking about those limits... I never paid much attention to those. The integral around x_1 seems nontrivial.

The integral around x_2 at least vanishes, because the area of the sphere approaches zero like r^2\to 0 and the integrand is approximately

<br /> \frac{1}{|x-x_2|}\approx \frac{1}{r} \to \infty<br />

So the surface integral gives something like r^2 \frac{1}{r} = r \to 0 which vanishes.

But

<br /> \frac{x-x_1}{|x-x_1|^3} \approx \frac{1}{r^2}<br />

seems tricky, I'll have to think about that more.

 

[jostpuur]

lightarrow, I just noticed I hadn't read your post 27 properly. The x_2 isn't problem, but x_1 seems to be. It could be that it has something to do with the fact that the delta function stuff was done carelessly in the calculation. I'll try to return to this later.

 

[jostpuur]

Okey, here's the calculation again!

Let us choose an integration domain A(R,r)=B(0,R)\backslash \big(B(x_1,r)\cup B(x_2,r)\big)

We want to calculate

<br /> \lim_{(R,r)\to(\infty,0)}\underset{A(R,r)}{\int} \frac{d^2x\; q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot \nabla\frac{-1}{|x-x_2|}<br />Since we have

<br /> \nabla\cdot \frac{x-x_1}{|x-x_1|^3} = 0<br />

everywhere in the integration domain, we can substitute

<br /> \frac{x-x_1}{|x-x_1|^3}\cdot\nabla\frac{-1}{|x-x_2|} = -\nabla\cdot\Big( \frac{x-x_1}{|x-x_1|^3}\frac{1}{|x-x_2|}\Big)<br />

After this we can turn the volume integral into a surface integral according to Gauss's theorem. I already explained how the surface integrals vanish in the infinity, and around x_2, so the only thing we are left is

<br /> =\lim_{r\to 0} \underset{\partial B(x_1,r)}{\int} \frac{d^2x\; (-q_1q_2)}{16\pi^2\epsilon} \cdot \frac{x-x_1}{|x-x_1|^3} \frac{1}{|x-x_2|}<br />

In the limit r\to 0 the last factor becomes

<br /> \frac{1}{|x-x_2|}\to \frac{1}{|x_1-x_2|}<br />

and the integration can be carried out substituting d^2x = -\hat{r} r^2\sin\theta\; d\theta\;d\phi and

<br /> \frac{x-x_1}{|x-x_1|^3} = \frac{\hat{r}}{r^2}<br />

Then we are left with a simple integration of a constant and sine function.

<br /> =\lim_{r\to 0}\frac{q_1q_2}{16\pi^2\epsilon_0 |x_1-x_2|} \int\limits_{0}^{\pi} d\theta\; \int\limits_{0}^{2\pi} d\phi\; (\hat{r} r^2 \sin\theta)\cdot\frac{\hat{r}}{r^2}\; = \;\frac{q_1q_2}{4\pi \epsilon_0}\frac{1}{|x_1-x_2|}<br />

And there the Coulomb's potential is again. The delta function way, I suppose, is just a heuristic way of writing the same thing. This wasn't fully rigorous calculation still. We should have substituted a linear approximation

<br /> \frac{1}{|x-x_2|} = \frac{1}{|x_1-x_2|}\; -\; \frac{(x-x_1)\cdot(x_1-x_2)}{|x_1-x_2|^3} \;+\; |x-x_1|\varepsilon(x-x_1)<br />

and showed that the contribution from the correction terms really vanishes. Perhaps there's something else too.

 

[lightarrow]

It seems all correct. Good work.

 

[da_willem]

This is getting slightly off topic, but I could make a remark, that electric and magnetic fields can have rest mass too. The plane waves which travel at the speed of light are massless, but static electric fields around some charge distributions are not.

That's what I always thought! (see e.g. https://www.physicsforums.com/showthread.php?t=170928) Thanks for your calculations, they confirm what I always thought! Great work!

 

[da_willem]

A common misunderstanding is for one to interpret a paricle's spin as the "rotation about its own axis." However, spin is a purely quantum mechanical property, and it therefore makes no sense to interpret it in any classical sense.

I can't help to, again, point to the article of Ohanian (What is spin, Am j Ph '84) who continues calculations of Belinfante. His conclusion is that spin is associated to an intrinsic rotating energy flow around a particle. So, yes it is intrinsic, but it is associated to a rotation. Maybe not a rigid body rotation of the particle itself in a classical sense, but a rotating energy flow as described by the particles energy momentum tensor.

 

[jostpuur]

I must add, that they were not "my calculations" really I asked about this some years ago from some lecturer, and in the next lecture he had brought Jackson's book with him, and he showed some calculations where Coulomb's potential was derived. I didn't copy them back then, but when I later tried to go through the details, I was able to do it on my own (because I had seen it once before and I already new something about delta functions).

 

[jostpuur]

I can't help to, again, point to the article of Ohanian (What is spin, Am j Ph '84) who continues calculations of Belinfante. His conclusion is that spin is associated to an intrinsic rotating energy flow around a particle. So, yes it is intrinsic, but it is associated to a rotation. Maybe not a rigid body rotation of the particle itself in a classical sense, but a rotating energy flow as described by the particles energy momentum tensor.

This sounds confusing.

The quantum mechanical spin is confusing, I haven't understood it completely myself, but classical spins, that fields can have, are quite understandable. When a field (for example A^{\mu} of electromagnetism) has nontrivial transformation properties in rotations (that means it is not a scalar field), it usually has some kind of internal angular momentum. You can derive it starting with the Lagrangian formulation of the theory, and with the Noether's theorem. It's all about conserving quantities. Internal angular momentum doesn't necessarily have anything to do with physical rotation, but it is something that is part of the total angular momentum that is conserving.

In the earlier posts in this thread I already gave link https://www.physicsforums.com/showthread.php?t=160778 to another thread where internal angular momentum of the EM field got mentioned.

 

[jostpuur]

That's what I always thought! (see e.g. https://www.physicsforums.com/showthread.php?t=170928) Thanks for your calculations, they confirm what I always thought! Great work!

It seems I'll have to disagree with pmb_phy when he says the field wouldn't have invariant rest mass. Or perhaps there was a misunderstanding of some kind.

In fact in a more axiomatic approach to physics I would consider m=\sqrt{E^2-p^2} as the definition of the mass. Energy and momentum must be first solved with the Noether's theorem, and mass comes next.

I remember struggling with this stuff earlier. People seem to have difficulty in taking the fields seriously. Lecturers often don't want to talk about mass or internal angular momentum of the electromagnetic field.

 

[premagg]

Spin is an intrinsic property.just like mass and charge.e.g. you cannot define what is amount of matter ??It is there ,we know this much and based on this we calculate other things.llly you cannot tell what a coloumb is?(You might say charge of 1 electron,but i would ask what the coloumb is made of?)So you just cannot the three intrinsic proprties namely mass,charge and spin in quantum mechanics.Though they play many important roles in the structure of materials.

 

[Hans de Vries]

I can't help to, again, point to the article of Ohanian (What is spin, Am j Ph '84) who continues calculations of Belinfante. His conclusion is that spin is associated to an intrinsic rotating energy flow around a particle. So, yes it is intrinsic, but it is associated to a rotation. Maybe not a rigid body rotation of the particle itself in a classical sense, but a rotating energy flow as described by the particles energy momentum tensor.

This follows straightforwardly from the Gordon decomposition for the
vector current of the Dirac equation:

j_\mu\ =\ iec\ {\bar \phi}\gamma_\mu\phi

Which is decomposed in:

j^{(1)}_\mu\ =\ \frac{ie\hbar}{2m}\left( \frac{\partial {\bar \phi}}{\partial x_\mu}\phi-{\bar \phi}\frac{\partial \phi}{\partial x_\mu}\ \right)

for the charge, and:

j^{(2)}_\mu\ =\ -\frac{e\hbar}{2m}\frac{\partial }{\partial x_\mu}<br /> ({\bar \phi}\sigma_{\nu\mu}\phi)

For the spin magnetic moment. The latter term has the correct factor 2
and represents the curl of the spin-density. This effect is very similar
to that of a classical magnetic material which exhibits an effective
current equal to the curl of the magnetic dipole density:

j_{eff}\ =\ c\nabla \times {\cal M}

Now, recall Stokes law: Inside the magnetic material the little circular
currents cancel each other if the spin density is constant. However they
don't cancel at the edge or where there is a gradient. This then gives
rise to a large effective current surrounding the magnetic material
at the edge.

This is the same which happens with the electron's wave function, both
electrically in the form of an effective current surrounding the wave
function, as well as inertially in the form of an effective momentum flow
around the wave function.Regards, Hans.

P.S. See for instance Sakurai 3-5: "Gordon decomposition of the vector current.

 

[jostpuur]

what is \sigma_{\nu\mu}?

 

[dextercioby]

The spin matrix for a massive Dirac field.

\sigma_{\mu\nu}=\frac{i}{2}\left[\gamma_{\mu},\gamma_{\nu}\right]_{-}

 

[jostpuur]

Okey...

btw. Now when I looked that more closely, it seems that Hans's third equation has mistake, because \mu is being summed on the right hand side, but it still appears on the left. You can guess what it is supposed to be, except that it is impossible to guess the sign.

 

[Hans de Vries]

Okey...

btw. Now when I looked that more closely, it seems that Hans's third equation has mistake, because \mu is being summed on the right hand side, but it still appears on the left. You can guess what it is supposed to be, except that it is impossible to guess the sign.

Yep, typo indeed. The first mu must be a nu at the RHS, like this:

j^{(2)}_\mu\ =\ -\frac{e\hbar}{2m}\frac{\partial }{\partial x_\nu}<br /> ({\bar \phi}\sigma_{\nu\mu}\phi)Regards, Hans

 

[JDługosz]

I found a particularly enlightening explanation in The Force of Symmetry by Vincent Icke.

He illustrates that under Special Relativity a series of translations cannot be combined to a single translation. Rather, the general case is a translation and a rotation. That is, a series of Lorentz Boosts does not form a group, but makes you drag in rotation as well to form a closure.

Meanwhile, you have quantum uncertainty caused by the existence of a "smallest scale".

This means that given a system of two objects rotating around each other, you cannot separate the angular momentum into definite orbital angular momentum and definite rotational components.

To represent the system, and preserve Lorentz Invariance, you must add a mathematical quantity to the expression that becomes "spin".

Spin is still hard to fathom, but it is comforting to know that it is not some arbitrary thing that just happens to be. It is a necessary feature, which falls out of the symmetry of spacetime and the existence of a "small" scale.