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I found this image for the classical description of the electron spin at hyperphysics

enter image description here

Can you explain why the axis of rotation makes an angle of 60° with the z-axis and how this particular alignment is detected and justified?

I am aware that this view of the spin is rejected by QM in which spin (and its value) is an intrinsic property not dependent on actual rotation. What made classical physicists come to that conclusion?

Clarification after the edit in the answer:

I am aware of of the history of the discovery of the spin. The point is that classical spin violated the known laws of physics at any tilt. I read also elsewhere that the spin-axis was always described at a tilt and a well defined tilt, at that. I am just wondering why did they not imagine (more naturally) an electron spinning normally to the plane of the orbit? There surely must be a physical reason. Anybody knows?

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  • $\begingroup$ $\cos 60^\circ = 1/2$ - the red line is not $60^\circ$ with $z$-axis. $\endgroup$ – innisfree Mar 14 '15 at 14:32
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    $\begingroup$ I don't know why they've drawn it in that way. I don't think the angles in that diagram are significant. $\endgroup$ – innisfree Mar 14 '15 at 14:37
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First of all you should not think of spin of an electron as a ball spinning on is own axis. At least this comparison should not go beyond your intuitive understanding of an electron spin. The quantity $s=\sqrt{s(s+1)}= \frac{\sqrt3}{2} \hbar$ is an intrinsic property of an electron with spin one-half and has nothing to do with the axis of rotation. I don't know why they have drawn the red arrows in the drawing. In my opinion they try to show that the electron spin can take one of the possible configurations of spin by drawing those arrows but as I've said I am not 100% sure.

Although it depends on what you mean by this view, the view shown in the website seems to be perfectly fine although I didn't read everything too carefully.

Edit after the change in the question:

I think quoting Wikipedia would suffice to answer your new question.

The physical interpretation of Pauli's "degree of freedom" was initially unknown. Ralph Kronig, one of Landé's assistants, suggested in early 1925 that it was produced by the self-rotation of the electron. When Pauli heard about the idea, he criticized it severely, noting that the electron's hypothetical surface would have to be moving faster than the speed of light in order for it to rotate quickly enough to produce the necessary angular momentum. This would violate the theory of relativity.

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  • $\begingroup$ On the website you mentioned it says that: In the pattern of other quantized angular momenta, this gives total angular momentum I think that is what you mean by “classical approach”. However the derivation of this formula also bases on Schrödinger Eqn, which is far from classical approach. Frankly I don't see any classical approach other than that on this website. $\endgroup$ – Gonenc Mogol Mar 14 '15 at 14:28
  • $\begingroup$ With the new version of your question the answer is completely different. I see now what you mean by classical approach. $\endgroup$ – Gonenc Mogol Mar 14 '15 at 14:30
  • $\begingroup$ The picture is just a way to visualize something that we cannot see and think of other than making analogies. But it has a description at least in a mathematical sense. You can describe a spin of an electron beautifully with pauli matrices and the justification of the spin of the electrons is the stern gerlach experiment. $\endgroup$ – Gonenc Mogol Mar 14 '15 at 14:50
  • $\begingroup$ @basil because that wouldn't fit the data and require too many assumptions, which wouldn't hold under other circumstances. $\endgroup$ – Gonenc Mogol Mar 14 '15 at 14:55
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    $\begingroup$ Zeeman effect and the result of stern gerlach experiment these are the two I know. $\endgroup$ – Gonenc Mogol Mar 14 '15 at 15:04
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The diagram is simply a way to remember two variables at the same time, one called the total spin $\hbar\sqrt{3}/2$, the other called the z-component which has two possibilities $\pm\hbar/2$. If you draw a picture of a vector of length $\hbar\sqrt{3}/2$ that is at an angle so that the z-component of the vector is $\pm\hbar/2$ (so it lies on two cones, one with $+\hbar/2$ and one with $-\hbar/2$) the picture can help you remember those two numbers (and match them to the right name).

The names are very misleading however, so a picture to help you remember does not help much, and probably hurts more than it could ever help. That's because if you "measure" the component in any direction whatsoever you always get $\pm\hbar/2$, so clearly you aren't measuring a pre-existing component of some preexisting vector. If there were a preexisting vector what would happen if you measured the component orthogonal to that vector? You can't get zero since you always get $\pm\hbar/2$. What you are really doing is polarizing the spin so that later it will give the same result on future measurements as what you just got. So the spin direction should be telling you how it is polarized, specifically a direction that gives $+\hbar/2$ (and where if you measure in the 100% opposite direction for your z-axis you get $-\hbar/2$) and where if you measure in any other directions besides those two you get a statistical spread of answers, sometimes $+\hbar/2$ and sometimes $-\hbar/2$. The real physics is learning how to get that statistical spread, and that's what a quantum theory course would cover.

And since you always get $\pm\hbar/2$ for any component, that is exactly why the total spin is $\hbar\sqrt{3}/2$: $S=\sqrt{S^2}=\sqrt{S_x^2+S_y^2+S_z^2}=\sqrt{(\pm\hbar/2)^2+(\pm\hbar/2)^2+(\pm\hbar/2)^2}=\sqrt{3\hbar^2/4}=\hbar\sqrt{3}/2.$ There is nothing but harm coming from thinking there is a vector that points in a certain direction and you measure an already existing component of that vector.

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It is true that for the electron the eigenvalue of the operator $ \ \hat {S^2} \ $ is $ \ \frac {\hbar^2}{2} \left(1 + \frac {1}{2} \right)$ s.t. the square root is as you say $\frac {\sqrt {3}}{2}$, however, about the angle of 60°. it is not clear where it comes from. In the spin-up state $|\uparrow\rangle$ the spin projection is parallel with the axis $z$, while in the spin-down state $|\downarrow\rangle$ it is oriented in the direction $-z$.

The electron spin has nothing to do with classical physics, it is at all a relativistic effect explained by the Dirac equation. The model by which the electron is an electric charge that rotates around an axis, does not fit the magnetic dipole of the electron.

Now, I just guess that 60° comes from $\text {sin} 30°= \frac {1}{2}$ i.e. some idea to represent on the $z$ axis the spin-up eigenvalue as $\frac {1}{2}$, and the spin-down eigenvalue as $- \frac {1}{2}$. But is it not correct from QM point of view.

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