2

I have read that the electron spin is represented by a vector of 2 complex numbers. And a frequently asked question is how can it be that an electron must be rotated 720 degrees to return to its original state. I have read various posts and I am not satisfied with the answers given.

From what I gather in the mathematics there is a spin operator and when applied with 360 degrees the vector of 2 complex numbers is not the original.

But how does one "rotate" an electron as such? Can you apply a magnetic field that does that and only that operator?

And why should the operator by $n$ degrees count as rotating by $n$ degrees. I feel like all the explanations I read failed to justify the "rotation" or "spin" terminology. In comparison with quarks we speak of their color but we don't literally expect the property to behave like color and don't try to justify it. What I mean is if electron spin was called electron goomy then I wouldn't have a problem with the 720 degrees because goomy has no preconceived notion of what it should be. Why should we call the vector of 2 complex numbers spin instead of inventing a completely new term?

Qmechanic
  • 201,751
user782220
  • 165
  • An electron is described by a Dirac bi-spinor, which transforms under $\text{Spin}^\mathbb{C}(1,3)$, where $\text{Spin}(1,3) = \text{SL}(2,\mathbb{C})$, the double cover of the Lorentz group. In order to rotate a spinor, one uses the representation of $\text{Spin}^\mathbb{C}(1,3)$ acting on $\mathbb{C}^4$, and selects the spatial rotation among all the possible transformations. More on Wikipedia – Jeanbaptiste Roux Dec 08 '22 at 11:11
  • I think Sashwat's answer here is good: https://physics.stackexchange.com/a/469024/65074 . But in the end it is not necessarily a real rotation of the electron.. we don't even know if an electron has an orientation. It just considers a situation where a classical electron would be rotated, and the remark is that $\psi$ gets a negative sign after what would have been a classical $2\pi$ rotation. But normally this $(-1)$ cannot be observed, because it doesn't affect $|\psi|^2$. It is only observable if it is a term in a superposition, which is I think what was apparently experimentally confirmed. – doublefelix Dec 08 '22 at 11:24
  • 1
    Does this answer your question? Rotation of a Spinor – John Rennie Dec 08 '22 at 11:32
  • @JohnRonnie I think as far as intuition goes, rotation of a "spinor" is not quite the same as rotation of a point-like creature called electron. There may be no such intuitive answer but I for one am not satisfied with any offered so far being intuitive. – hyportnex Dec 08 '22 at 17:26
  • 1
    @JohnRennie Watching and learning from you over the years, I have noticed that you have an uncanny ability to explain obscure issues in an unusually simple and intuitive manner but I have also noticed that you have not even attempted to answer this particular question that way unlike the others, so... – hyportnex Dec 08 '22 at 17:33
  • I've been reading a great paper "How Electrons Spin" by Charles Sebens. https://arxiv.org/abs/1806.01121 In it he describes an idea that it is not the electron itself that is spinning, because that would violate the speed of light. Rather it is the Dirac field surrounding the electron that spins. – foolishmuse Dec 08 '22 at 18:36
  • 1
    @hyportnex a spinor is not a physical object in the same sense that a wavefunction is not a physical object. A transformation of a spinor will affect physical observables, but rotation of the physically observable properties is not the same as rotation of the spinor. For example if you rotate the magnetic moment of the electron (which is done routinely in NMR spectrometers) you will find that rotating it by 2 does indeed return it to its original state. – John Rennie Dec 10 '22 at 07:10
  • @JohnRennie yes, so please write this into a full length answer with examples and the world of physics students and similar enthusiasts will be eternally grateful to you. – hyportnex Dec 10 '22 at 12:43
  • @hyportnex ACuriousMind wrote the definitive answer for this here. – John Rennie Dec 12 '22 at 08:32
  • @JohnRennie this is indeed the definitive answer and it being the "definitive" is exactly the problem... let me rephrase then the question: how would you explain this "rotation thing" to a 12-year old curious student who does not know Hilbert from Schmilbert, or, even better, to his grandmother? – hyportnex Dec 12 '22 at 12:07
  • @hyportnex You cannot. The best you can say is what ACM says right at the start of his answer. Rotating a spinor is not the same as rotating the particle that the spinor represents. To understand why it's not the same you need to know about Schmilbert spaces. There is simply no shortcut. – John Rennie Dec 12 '22 at 12:38
  • @JohnRennie OK, in that case then I will just keep paying my rent in Hilbert spaces then; thanks. – hyportnex Dec 12 '22 at 12:41
  • Anyone remember the Feynnman quote "If you can't explain something to a first year student, then you haven't really understood." – user782220 Dec 12 '22 at 17:21

0 Answers0