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I Does the electron really spin 720 degrees?

  1. Oct 4, 2016 #1
    A photon has a spin of 1, while an electron is 1/2. I've heard it said that the electron needs to spin 720 degrees to complete a full cycle. Is there any proof that the electron is spinning 720 degrees and not just spinning 360 degrees, but at a lower harmonic of the photon-- 1/2 the photon's spin frequency?

    As an analogy, 2 pendulums of proportionately different mass, suspended on the same string will "feed" off each-other. If you start the lighter pendulum swinging, the heavier one will start swinging too, but at a lower frequency (and vice versa). Depending on the mass proportions, you can get the heavier one at frequency 1/2, 1/3, 1/4 or whatever you like along that line. Let's say you have it at 1/2... if you measure a whole cycle in the lighter pendulum, you'll only measure 1/2 cycle in the heavier one, but both of them are still covering the same amount of degrees.
     
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  3. Oct 5, 2016 #2

    Drakkith

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    Quantum mechanical spin should not be interpreted as the electron physically spinning around an axis. It's more like the electron simply has a certain amount of intrinsic angular momentum that cannot be gotten rid of any more than it can get rid of its electric charge.

    I'm sure others can explain things in much more detail, but you can check out the following article for a brief overview: https://en.wikipedia.org/wiki/Spin_(physics)
     
  4. Oct 5, 2016 #3
    Yes there isn't any real meaning to a point particle "spinning", that's an intrinsically macroscopic concept. A better way to think of it is that if you were to rotate the electron (or any fermion having spin 1/2) then it is symmetrical under 4π (720deg) rotation not 2π (360deg) as would be the case with the integer spin particles (e.g. photon). There *is* a physical meaning to rotating fermions and bosons, since they transform in representations of the Lorentz group, which includes the subgroup of SO(3) i.e. rotations in 3D space.
     
  5. Oct 5, 2016 #4

    Demystifier

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    When a physical object is rotated by 360 degrees, one obtains the same object. The wave function ##\psi## of the electron does not satisfy that property, but ##\psi## is not a physical object. The product such as ##\psi^{\dagger}\psi## is more closely related to an actual physical quantity (e.g. it can be interpreted as probability density, at least in the non-relativistic limit), and ##\psi^{\dagger}\psi## does not change when you rotate it by 360 degrees.
     
  6. Oct 5, 2016 #5

    rubi

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    The space of physical states is actually the space of density matrices, rather than the Hilbert space of the quantum system. Mathematically, a trace preserving representation of the rotation group (##SO(3)##) on the space of density matrices lifts to a unitary representation of its centrally extended universal cover, which happens to be ##SU(2)##. The unitary operators corresponding to ##2\pi## and ##4\pi## rotations in ##SU(2)## both map to a ##2\pi## rotation on the space of density matrices. Physical states don't feel the difference between ##2\pi## and ##4\pi## rotations. If they did, we would not be considering unitary representations of ##SU(2)## at all.
     
  7. Oct 5, 2016 #6
    Thank you for the response. I understand that a particle conceived as a point-particle couldn't rotate all by itself. But then the very idea of a point only works when you have other points to relate it to. We know that when a particle poses spin, it can translate that into actual rotation in a mass of particles-- actual angular momentum in that mass. Definition for angular momentum: the quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity.

    So... if we are going to say a particle has spin, we can consider that is has rotation, or at least rotation potential. This universe, on the macroscopic scale, is considered to have 3 spacial dimensions, and 360 degrees in a circle.

    Back to my original question:
    This question really has to do with understanding the derivation of an electron needing to spin "720 degrees to return to its original state" as I've heard it described. Perhaps the electron is merely spinning 360 degrees, but at half the rate of the EM energy we use to measure it, as I analogized with the pendulums. Or perhaps the fact that the electron has more mass/density, that density will sort of neatly expand out under spin/rotation, thus acting like it has 720 degrees...?

    Note: Anyone who wants to contribute to this thread, please keep the terminology in your answers comparable to the terminology in my question. I have no interest in taking an advanced college course to simply understand a single response in this thread ;-)
     
  8. Oct 5, 2016 #7

    vanhees71

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    Well, if you don't want to learn the adequate terminology, why do you ask then in the first place?

    Anyway, again you should forget about quasi-classical pictures of a particle when it comes to specific quantum-mechanical phenomena. Spin is one of them. No matter, whether it's integer or half-integer spin, it's not something spinning in the sense of an extended body. What comes closest to this intrinsic quantum number in classical physics is a classical point particle with some dipole (e.g., a magnetic dipole, and even in the quantum theoretical generalization you have magnetic dipole moments associated with the spin of elementary particles).

    Now to you question concerning the 720 degrees to return to its original state. Already the question imposes a wrong assumption, and you have to relearn the concepts, but for this you don't need an advanced college course, but perhaps the first one or two lectures on quantum theory (provided you have a modern teacher who starts right away with the modern version of QT ;-)). What's correct is that if ##|\psi \rangle## is a vector in the Hilbert space of an half-integer-spin particle, then the rotion by ##2 \pi## gives ##-|\psi \rangle##. Now pure states are not represented by the Hilbert-space vectors but by rays in Hilbert space, i.e., Hilbert-space vectors up to a (phase) factor. The phase factor is unobservable and thus ##|\psi \rangle## and ##-|\psi \rangle## represents the same state.

    A more convenient and equivalent definition of a state representative is the statistical operator, which is a positive semidefinite self-adjoint operator with trace 1. A pure state is represented by the special class projection operators, i.e., given a normalized Hilbert-space vector ##|\psi \rangle## then the corresponding state is given by the statistical operator ##\hat{\rho}=|\psi \rangle \langle \psi|##, and this doesn't change under rotations by ##2 \pi ## as it must be.

    As we are at it, we go to the end of the college QT course and mention that the above arguments imply a socalled superselection rule, i.e., a rule forbidding some superpositions: You cannot have a superposition of an integer-spin particle's with and half-integer-spin particle's (or any other systems with spin), because then a rotation by ##2 \pi## would indeed lead to a different state, since an integer-spin vector is unchanged while an half-integer-spin vector flips its sign. In a superposition the relative phase between the two vectors would change, and this has observable consequences. This, however, contradicts the fact that a rotation by ##2 \pi## must not change the state, because it's the identity operator of the SO(3) rotation group.
     
  9. Oct 5, 2016 #8
    Oh, pardon me for wanting to understand.
    I already addressed this.
    ???
     
  10. Oct 5, 2016 #9

    rubi

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    The answer to your question is: There is nothing physical that requires a 720° degree rotation to go back to its original state. If you rotate 360°, you are already back in the initial physical state. Your premise isn't correct.

    Every explanation beyond this requires some math. Since you have marked this thread as intermediate level, people naturally assume that you are familiar with the content of an introductory course on quantum mechanics. I think vanhees' explanation in his 3nd paragraph is as simple as it possibly gets.
     
  11. Oct 5, 2016 #10

    vanhees71

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    As I said, studying physics always includes the danger that you have to change your worldview, and if you are not willing to do so, it's the wrong subject for you. Then study philosophy and write long books full of footnotes defending your prejudices ;-)). SCNR.
     
  12. Oct 5, 2016 #11
    You seem to have a fixed notion/delusion that I'm attacking physics. That's actually not the case; I am simply looking for answers that are tangible.
     
  13. Oct 5, 2016 #12

    vanhees71

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    One should make physics as simple as possible but not simpler (Einstein)!
     
  14. Oct 5, 2016 #13
    That may be true, but it's hard to wade through the abstract math.
     
  15. Oct 5, 2016 #14

    vanhees71

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    Well, it's hard but a lot of fun too. Without math there's no way to even talk about physics in an adequate way. So if you look for tangible answers in regions where all of us have no direct intuition, because we don't observe it directly, which is true for most quantum effects, you have to learn the math and do many applications of it in QT to get an intuition about the quantum world. There's no shortcut! On the other hand, as I said, it's very fascinating and a lot of fund to learn this stuff too.

    Also the math of quantum theory is not that difficult. You need linear algebra of complex Hilbert spaces and some calculus about partial differential equations. A very nice book is Susskind's "The theoretical minimum" Quantum Mechanics.
     
  16. Oct 5, 2016 #15

    Strilanc

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    I can't really speak to the actual physics of rotating electrons, but I do know that on a quantum computer the answer is: it depends.

    Specifically, what I mean is that just saying for example "a 90 degree rotation around the X axis" is not enough to unambiguously describe what you are doing to a qubit. Both the operation ##e^{i \pi X / 2} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1&i\\i&1 \end{bmatrix}## and the operation ##X^{1/2} = \frac{1}{2} \begin{bmatrix} 1-i&1+i\\1-i&1+i \end{bmatrix}## are consistent with that description. But ##\big( e^{i \pi X / 2} \big)^4 = -I## (four quarter turns negates the global phase) while ##\big( X^{1/2} \big)^4 = I## (four quarter turns is equivalent to doing nothing).

    Normally the difference between ##e^{i \pi X / 2}## and ##X^{1/2}## doesn't matter. They only differ in how they affect the global phase of the qubit. But once you start applying operations conditionally, controlled by the state of other qubits, that global phase becomes a relative phase.

    I don't know enough about physics to say which is more common. I'd bet it's ##e^{i \pi X / 2}##, since it's ##e^\text{bla}## often shows up in the solutions of differential equations and also because, you know, people say things like "720 degrees to get back to the starting state".
     
  17. Oct 5, 2016 #16

    rubi

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    You don't need to understand all the math to extract the basic reasoning: By requiring that all physical objects are invariant under rotation symmetry, mathematics naturally leads us to consider spin (Wigner's theorem + Bargmann's theorem). It just so happens that some unphysical objects only return to their initial state after a 720° rotation, but that's not problematic, since they are unphysical. All physical answers that can be extracted don't depend on this ambiguity. The same thing also happens in electrodynamics: You can add an arbitrary constant to the electric potential, but all physical answers stay the same.
     
  18. Oct 5, 2016 #17
    Fair enough. That's helpful. Still though, I believe that a complete understanding of these problems would allow for an answer that tends towards tangability with basic English.
     
  19. Oct 5, 2016 #18
    Ok. How do you define "unphysical" here?
     
  20. Oct 5, 2016 #19

    stevendaryl

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    If you rotate an electron through 360o, the electron's state picks up a minus sign. That's considered to have no physical importance, since the only thing of physical significance is the square of the state, and the sign disappears when you square it.

    However, it seems to me that the minus sign would come into play in interference effects. I don't actually know how one would accomplish this, but conceptually, you can imagine an experiment similar to the two-slit experiment, where an electron make take one of two paths to get to a destination. Along one of the paths, the electron is rotated 360o by its journey, while along the other path, no rotation takes place. Then (assuming an electron is equally likely to take either path), there would be total destructive interference at the destination, so the electron would have no probability of arriving there.

    I'm not sure what would reliably rotate an electron, though. Maybe one path has a magnetic field, and the other doesn't?
     
  21. Oct 5, 2016 #20

    strangerep

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    You seem to be saying that fermions are unphysical, which is clearly not correct since one can observe effects in, e.g., neutron interferometers (by using a magnetic field to rotate the neutrons on one side of the interferometer and then recombining).

    Perhaps you meant something else?

    (And let's not forget ye olde Plate trick, aka "Belt trick" or "Dirac Scissors", which shows that there are macroscopic situations where rotation by 360deg is not equivalent to rotation by 720deg.) :oldbiggrin:

    @Chris Frisella : as someone else noted, your thread has become a bit of a mess because you marked it as "I", but then rejected any basic math. I'd suggest you mark such threads as "B" in future. Also, it takes a lot of effort to descramble the gobbledegook in your opening post since you're conflating a rotational transformation of a single-electron system with spinning motion. Those 2 things first need to be dissected apart. But I'm not sure how far the other answers in this thread have gone towards achieving this, so I'll wait to see what you say before I try to descramble it.
     
    Last edited: Oct 5, 2016
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