Explaining Quantum Spin Without Math: Is It Possible?

[Hoof47]

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?

 

[cristo]

Spin is a type of "intrinsic" angular momentum-- i.e. a type of angular momentum that a particle has, regardless of its motion in space. marlon has posted in his blog an https://www.physicsforums.com/blogs/marlon-13790/what-is-spin-152/ to this, which you may wish to look at.

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.

 

[Cane_Toad]

Are all the terms associated with spin all analogous? E.g. "intrinsic angular momentum"; a particle has neither angular momentum or "spin" [EDIT: meaning rotation], right? Are all these just naming conventions?

I'm assuming the term "spin", etc., is due to the Stern-Gerlach experiment, where a particle's "spin" is measured in a way similar to what could be done for classical spinning object. True?

Since particles do deflect in a magnetic field as if they had angular momentum, but they don't actually have any, what is going on?

 

[jostpuur]

Spin is a type of "intrinsic" angular momentum-- i.e. a type of angular momentum that a particle has, regardless of its motion in space. marlon has posted in his blog an https://www.physicsforums.com/blogs/marlon-13790/what-is-spin-152/ to this, which you may wish to look at.

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've had some trouble with this "purely quantum mechanical" argument. Classical electromagnetic field has internal angular momentum, that is not cross product of location and momentum, and is pretty much spin angular momentum then.

When I asked what was spin a long time ago, one already graduated student responded to me that it is "angular momentum without rotation". That was quite good. But you can have internal angular momentum that is not result of any kind of rotation also in classical context.

 

[lightarrow]

Are all the terms associated with spin all analogous? E.g. "intrinsic angular momentum"; a particle has neither angular momentum or "spin", right? Are all these just naming conventions?
I'm assuming the term "spin", etc., is due to the Stern-Gerlach experiment, where a particle's "spin" is measured in a way similar to what could be done for classical spinning object. True?
Since particles do deflect in a magnetic field as if they had angular momentum, but they don't actually have any, what is going on?

Particles with spin *do* have angular momentum and magnetic momentum.
But they don't "spin".
If you wonder how could this be, you're not alone.

 

[Sojourner01]

Am I the only person who doesn't have a problem with writing it off as "a quantum mechanical property that a human being is not physiologically able to conceptualise"? In much the same way that I'm not able to comprehend what a dog dreams about - because I'm not a dog - I have no problem with the lack of realistic interpretation of spin, because I'm not a quantum mechanical object, I'm a macroscopic object, and I perceive macroscopic things.

So the answer to the question is, spin is spin. It doesn't need to go any deeper than that.

 

[Cane_Toad]

Particles with spin *do* have angular momentum and magnetic momentum.
But they don't "spin".
If you wonder how could this be, you're not alone.

But...it isn't *real* momemtum, is it (i.e. no rotation -> no velocity)? I'm wondering if it's just a model to account for deflection in a mag field?

(And you meant magnetic moment?)

 

[Cane_Toad]

...But you can have internal angular momentum that is not result of any kind of rotation also in classical context.

Please expound.

 

[Cane_Toad]

...
So the answer to the question is, spin is spin. It doesn't need to go any deeper than that.

If someone will tell me definitively that there is no explanation for how a particle is deflected in the Stern-Gerlach thing other than via a set of functions which have familiar sounds like "spin" and "angular momemtum", then I can be content, or at least as content as I am with duality (not!).

 

[jostpuur]

Please expound.

This is where I asked about the matter:

https://www.physicsforums.com/showthread.php?t=160778

My understanding of the Dirac's field has been rather "changing" lately, as result of discussions here in PF, and the OP of my thread probably isn't relevant for this matter anymore, but pay attention to the later comments on the classical electromagnetic field too.

 

[Tomsk]

My tutor explained it to me as a dynamical variable associated with the rotational symmetry of the particle. (Or of space, I can't remember!) Linear momentum is a variable associated with the translational symmetry of space, energy is linked to time in a similar way, and angular momentum comes from the rotational symmetry of space. So because a particle looks the same from every angle, there is a spin associated with it. But I think it is only noticable when there is something that breaks that symmetry, like a B field. This probably hasn't helped much, tbh I don't get it myself! :/

 

[Cane_Toad]

BTW, there is another thread with a lot of responses on this subject, in it I saw this:

https://www.physicsforums.com/showpost.php?p=1320028&postcount=29"

saying,

But it is angular momentum, and it contributes to the total macroscopic angular momentum of an object, as observed in the Einstein - de Haas effect.

Wikipedia has this to say:

the Einstein-de Haas effect demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics. This is remarkable, since electron spin, being quantized, cannot be described within the framework of classical mechanics.

There is no explanation how to make the leap from seeing a metal bar rotate to knowing it's because of the "spin angular momentum" of its [electrons/atoms?]. Heck, for all I know, spin/intrinsic angular momentum is defined by this effect.

 

[Cane_Toad]

This is where I asked about the matter:

https://www.physicsforums.com/showthread.php?t=160778

My understanding of the Dirac's field has been rather "changing" lately, as result of discussions here in PF, and the OP of my thread probably isn't relevant for this matter anymore, but pay attention to the later comments on the classical electromagnetic field too.

Ok, so you're talking about the internal angular momentum of an EM field, so at least there's not some wacky compound I didn't know about.

But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?

What the hell, I guess QM is re-defining all my CM terms according to its own fancy. Sigh, at least some of the QM fathers had the sense to use non-conflicting terms, like "flavor".

 

[Hootenanny]

But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?

Photons are massless and yet have linear momentum...

 

[jtbell]

But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?

Well, EM fields carry energy and linear momentum, so they pretty much have to have angular momentum too, based on the definition of angular momentum as $\vec r \times \vec p$.

 

[Cane_Toad]

But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?

Well, EM fields carry energy and linear momentum, so they pretty much have to have angular momentum too, based on the definition of angular momentum as $\vec r \times \vec p$.

Yeh, I remember now (I wasn't making the connection from photon momentum to fields...), but is that supposed to make it better? Anyway, that was a rhetorical question venting moment[um]ary outrage.

But now that you bring it up,
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photonMass.html"

Says,

Photons are traditionally said to be massless. This is a figure of speech that physicists use to describe something about how a photon's particle-like properties are described by the language of special relativity.
...
Because the energy of a particle just equals its relativistic mass times c2, physicists have learned to economise the language by only ever referring to a particle's energy. When they use the term "mass", they mean rest mass. This is purely a linguistic convention.

So, light does have "relativistic mass", but no rest/invariant mass.
It does go on to tell about a theoretical upper limit for the rest mass of light between 7 × 10-17 eV and 3 × 10-27 eV. I'm not sure why they bother to state the rest mass at zero, and then go on to give the estimate.

But anyway, now I feel better.

 

[Hootenanny]

It does go on to tell about a theoretical upper limit for the rest mass of light between 7 × 10-17 eV and 3 × 10-27 eV. I'm not sure why they bother to state the rest mass at zero, and then go on to give the estimate.

That is an experimentally derived upper limit, from observations. The theoretical mass of the photon is defined as zero. Period.

 

[jostpuur]

Ok, so you're talking about the internal angular momentum of an EM field, so at least there's not some wacky compound I didn't know about.

But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?

What the hell, I guess QM is re-defining all my CM terms according to its own fancy. Sigh, at least some of the QM fathers had the sense to use non-conflicting terms, like "flavor".

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.

 

[lightarrow]

But...it isn't *real* momemtum, is it (i.e. no rotation -> no velocity)? I'm wondering if it's just a model to account for deflection in a mag field?
(And you meant magnetic moment?)

You can define angular momentum without the need to have a rotating mass: think about light, as jtbell wrote. What is striking in that case is that *statics* electric and magnetic fields can have angular momentum: you only need a circulation of the Poyinting vector.
Yes I meant magnetic moment.

 

[lightarrow]

Ok, so you're talking about the internal angular momentum of an EM field, so at least there's not some wacky compound I didn't know about.
But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?
What the hell, I guess QM is re-defining all my CM terms according to its own fancy. Sigh, at least some of the QM fathers had the sense to use non-conflicting terms, like "flavor".

Example: if you shoot a very strong light pulse in a specific direction, you receive a recoil in the opposite direction. If you want to hold true the law of momentum conservation (higly advisable!) you are forced to conclude that light too carries momentum. That is a more general definition than p = m*v (for non zero rest mass and not relativistically) or p = m*v*gamma (for non zero rest mass and relativistically).
The more general formula which comes out is:
p = Sqrt[(E/c)^2 - (mc)^2]
where E is the total energy and m is the rest mass. For a photon:
p = Sqrt[(E/c)^2 - (0)^2] = E/c
To have angular momentum, you only have to compute $$\vec r \times \vec p$$

 

[lightarrow]

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.

Please explain this (that seems as a nonsense to me).

 

[jostpuur]

Please explain this (that seems as a nonsense to me).

Just $$E^2=p^2c^2+(mc^2)^2$$. For a plane wave we have $$E=pc$$ and $$m=0$$, but for a static field instead $$p=0$$ and $$E=mc^2$$. Suppose you have electron and proton close to each other, and you then drag them far away from each other. You need energy to do this, and after the procedure the mass of the system has increased by the energy you inserted there. The energy is in the electric field, and that is also where the mass increase is.

I think I heard somewhere, that this has not been verified experimentally with electric interactions yet, but only with nuclear interactions. You know, in nuclear physics the mass excess stuff is plain clear. Anyway, it should be true also for electric interactions.

 

[jtbell]

It does go on to tell about a theoretical upper limit for the rest mass of light between 7 × 10-17 eV and 3 × 10-27 eV. I'm not sure why they bother to state the rest mass at zero, and then go on to give the estimate.

No, that's an experimental upper limit on the mass of the photon. It means that if the photon has a nonzero mass, it must be smaller than this value. This is derived from the limitations of the experimental apparatus. If the mass were smaller than this value, we wouldn't be able to detect it, because the apparatus that we've used so far isn't good enough.

 

[Sojourner01]

If someone will tell me definitively that there is no explanation for how a particle is deflected in the Stern-Gerlach thing other than via a set of functions which have familiar sounds like "spin" and "angular momemtum", then I can be content, or at least as content as I am with duality (not!).

Oh dear, not what I meant at all. I didn't mean to imply that this was the truth, only tht that line of reasoning is my rationalisation for not minding that I don't have a better explanation. I merely don't feel it reasonable to expect a more common-sense explanation when there's little justification (in my mind) for expecting one to be forthcoming.

Wow, that was a mouthful. In all seriousness, I imagine we'd all have to study quantum field theory to gain a satisfactory understanding. Even then, it's probably not cut and dried.

 

[lightarrow]

Just $$E^2=p^2c^2+(mc^2)^2$$. For a plane wave we have $$E=pc$$ and $$m=0$$, but for a static field instead $$p=0$$ and $$E=mc^2$$. Suppose you have electron and proton close to each other, and you then drag them far away from each other. You need energy to do this, and after the procedure the mass of the system has increased by the energy you inserted there. The energy is in the electric field, and that is also where the mass increase is.
I think I heard somewhere, that this has not been verified experimentally with electric interactions yet, but only with nuclear interactions. You know, in nuclear physics the mass excess stuff is plain clear. Anyway, it should be true also for electric interactions.

But to evaluate the electric field's energy and compare it with the energy you give to the system to establish if the mass is in the field or in the particles, you should (also) know the electron's shape and dimensions. Do you know them?

 

[jostpuur]

But to evaluate the electric field's energy and compare it with the energy you give to the system to establish if the mass is in the field or in the particles, you should (also) know the electron's shape and dimensions. Do you know them?

Hello lightarrow, I didn't have energy to answer this back then when the thread was active, and then I forgot this. But anyway, yes, the energy of the interaction is precisely the energy of the field. Point charges bring a little problem of infinite rest energies, but they can be ignored and the interaction energy can be solved. You don't need to know the shape of the charges really. The Coulomb's potential is valid, if the distance between two charges is much larger than the size of the charges. Here's the calculation for point charges (in the physicist way).

If we have a charge $q$ at location $x'$, then the electric field is

$$E(x) = \frac{q}{4\pi \epsilon_0}\frac{x-x'}{|x-x'|^3}$$

The energy density of the electric field is

$$\frac{1}{2}\epsilon_0 |E|^2$$

So the total energy of the field around the point charge is

$$\int d^3x\; \frac{q^2}{32\pi^2 \epsilon_0}\frac{1}{|x-x'|^4} = \infty$$

This is the already mentioned infinite rest energy. Now suppose we have two charges, $q_1$ and $q_2$, at locations $x_1$ and $x_2$. The the electric field is

$$E(x) = \frac{1}{4\pi\epsilon_0}\Big(\frac{q_1(x-x_1)}{|x-x_1|^3} + \frac{q_2(x-x_2)}{|x-x_2|^3}\Big)$$

The energy of the electric field is now

$$\int\frac{d^3x}{32\pi^2\epsilon_0} \Big(\frac{q_1^2}{|x-x_1|^4} + \frac{q_2^2}{|x-x_2|^4} + \frac{2q_1q_2(x-x_1)\cdot(x-x_2)}{|x-x_1|^3 |x-x_2|^3}\Big)$$

The first two terms give the same infinities that are the rest energies of the both particles, so we can ignore them. Only the last term contributes to the interaction. Substituting

$$\nabla\frac{1}{|x-x_2|} = -\frac{x-x_2}{|x-x_2|^3}$$

makes the interaction energy to be

$$\int\frac{d^3x\; q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot \nabla\frac{-1}{|x-x_2|}$$

The nabla can be moved onto the first factor using integration by parts, and this gives

$$\int \frac{d^3x\;q_1q_2}{16\pi^2\epsilon_0}\Big( \nabla\cdot\frac{x-x_1}{|x-x_1|^3}\Big)\frac{1}{|x-x_2|}$$

Now using an delta function representation

$$\nabla\cdot\frac{x-x_1}{|x-x_1|^3} = 4\pi\delta^3(x-x_1)$$

the interaction energy becomes

$$\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{|x_1-x_2|}$$

which is just the Coulomb's potential. So when you speak about the potential energy given by the Coulomb's potential formula, you are speaking about the energy in the electric field. And when the mass of the system is changing as result of changing potential energy, it is the field that is carrying the changing mass.

 

[lightarrow]

...
makes the interaction energy to be

$$\int\frac{d^3x\; q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot \nabla\frac{-1}{|x-x_2|}$$

The nabla can be moved onto the first factor using integration by parts, and this gives

$$\int \frac{d^3x\;q_1q_2}{16\pi^2\epsilon_0}\Big( \nabla\cdot\frac{x-x_1}{|x-x_1|^3}\Big)\frac{1}{|x-x_2|}$$

Now using an delta function representation

$$\nabla\cdot\frac{x-x_1}{|x-x_1|^3} = 4\pi\delta^3(x-x_1)$$

the interaction energy becomes

$$\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{|x_1-x_2|}$$

which is just the Coulomb's potential. So when you speak about the potential energy given by the Coulomb's potential formula, you are speaking about the energy in the electric field. And when the mass of the system is changing as result of changing potential energy, it is the field that is carrying the changing mass.

Ok, nice work; however I don't understand why the other term of the integration by parts vanishes. Apart the constant terms, you have to evaluate

$$\frac{x-x_1}{|x-x_1|^3}\cdot\frac{1}{|x-x_2|}$$

In the limiting values of x:

for x = oo there is no problem, the term vanishes, but not for

x = x_1 and x = x_2.

 

[jostpuur]

I was afraid you were going to ask something about the delta function, because I don't know it very well. After substituting the delta function representation the energy is

$$\int \frac{d^3x\; q_1q_2}{4\pi \epsilon_0}\delta^3(x-x_1) \frac{1}{|x-x_2|}$$

and the delta function works so that the integration can be carried out by substituting $x_1$ to there where $x$ is elsewhere. You know this stuff:

$$\int\limits_{-\infty}^{\infty} dx\; \delta(x-x') f(x) = f(x')?$$

I have seen other calculations, where similar techniques were used rigorously, and there's a change that I could maybe understand what happens here too, but you should first be able to understand the hand waving calculation so that you see what is supposed to happen.

 

[lightarrow]

I was afraid you were going to ask something about the delta function, because I don't know it very well. After substituting the delta function representation the energy is

$$\int \frac{d^3x\; q_1q_2}{4\pi \epsilon_0}\delta^3(x-x_1) \frac{1}{|x-x_2|}$$

and the delta function works so that the integration can be carried out by substituting $x_1$ to there where $x$ is elsewhere. You know this stuff:

$$\int\limits_{-\infty}^{\infty} dx\; \delta(x-x') f(x) = f(x')?$$

I have seen other calculations, where similar techniques were used rigorously, and there's a change that I could maybe understand what happens here too, but you should first be able to understand the hand waving calculation so that you see what is supposed to happen.

Yes, this is straightforward, but what I was asking, is before that substitution, in the evaluation of the integral by parts:

$$\int\frac{d^3x\; q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot \nabla\frac{-1}{|x-x_2|} = \int \frac{d^3x\;q_1q_2}{16\pi^2\epsilon_0}\Big( \nabla\cdot\frac{x-x_1}{|x-x_1|^3}\Big)\frac{1}{|x-x_2|}+$$

$$-(\frac{q_1q_2}{16\pi^2\epsilon_0} \frac{x-x_1}{|x-x_1|^3}\cdot\frac{1}{|x-x_2|})(x=x_1,x_2,\infty)$$

The last term doesn't seem to vanish.

 

[country boy]

This is getting slightly off topic, but I could make a remark, that electric and magnetic fields can have rest mass too. ...

Here is an example of an EM field with rest mass. If two EM waves with equal frequency are traveling in opposite directions, their momenta cancel and their energies adds. The system is at rest but it has energy. That's rest mass.

If you think that's cheating, remember that anything that appears to be at rest is actually composed of a collection of moving parts.

 

[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

$$\Big|\frac{x-x_1}{|x-x_1|^3}\frac{1}{|x-x_2|}\Big|\approx \frac{1}{R^3}$$

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

$$\Big|\frac{x-x_1}{|x-x_1|^3}\frac{1}{|x-x_2|}\Big|\approx \frac{1}{R^3}$$

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

$$\frac{1}{|x-x_2|}\approx \frac{1}{r} \to \infty$$

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

But

$$\frac{x-x_1}{|x-x_1|^3} \approx \frac{1}{r^2}$$

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