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Helicity and the Uncertainity Principle

  1. Feb 22, 2014 #1
    Hi all!

    So rumor has it, that the spin of a particle is preferably aligned opposite to the direction of the particle's momentum, whereas an antiparticle spins along the direction of motion. For the sake of simplicity let us assume that the (anti-)particle is massless and hence it is an eigenstate of both helicity and chirality operators. My question reads as follows: "How is the conception of such definite, predetermined spin direction compatible with the uncertainty princple, which forbids the simultaneous knowledge of all spin vector components?"
    I suspect that most textbooks refer only to the z-component of spin but they do not bother to elucidate that detail. If that's not the case, I would appreciate any kind of explanation, involving spectral theory concerning Dirac's solutions or relativistic redefinition of the spin operator.
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  3. Feb 22, 2014 #2


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    In fact it's not the z-component in general, but you align it to be the z-component.
    It's the projection of spin on the direction of momentum vector...once you do that projection, you can say that that's on the z-axis.
    The rest part of spin can still rotate around in an undetermined way.

    At least that's what I think, someone else could correct me if I'm wrong
  4. Feb 22, 2014 #3


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    Spin in a quantum object (a subatomic particle) is NOT "spin" in the classical mechanics sense. As I understand it, it does not HAVE an axis.
  5. Feb 22, 2014 #4


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    Spin and momentum components commute. So no particular limitation in this case.
  6. Feb 22, 2014 #5


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    What do you mean by that? in QM, spin is described by a vector operator. As a vector it has components. The main thing with spin, is that you can only know one component of it (you can only diagonalize one component's matrix element, for which we choose the 3rd,z-th, or whatever the notation can be).
  7. Feb 22, 2014 #6
    Yes of course: [pi,Sj]=0, but [Si,Sj]=i[itex]\hbar[/itex]εijkSk and that to my eye means, that [itex]\vec{S}[/itex] cant'be alinged to a certain direction [itex]\vec{p}[/itex], which in turn implies, that the famous picture of a particle with two arrows attached, [itex]\rightarrow[/itex] and [itex]\Rightarrow[/itex], representing momentum and spin, either in the same (right-handed) or opposite (left-handed) direction is somewhat false.

    Every one and single textbook I know speaks of spin direction. If @ChrisVer's view is not true, then sth more technical is going on here...
  8. Feb 22, 2014 #7


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    You must be careful at what you write... [itex]S_{i}[/itex] is just the i-th component of spin, nobody tells you which component it is. You would gain the same physics if you'd say that it's the x-component. The commutation relation just tells you that you cannot know simultaneously all 3 components of the spin vector (their matrix elements are not simultaneously diagonalizable). So if you choose your component to be
    [itex]\vec{S_{p}} = (\vec{S} \cdot \vec{p})\hat{p}[/itex]
    you would have the normal to it components undetermined (as when you know [itex]S_{3}[/itex] you have [itex]S_{1},S_{2}[/itex] undetermined).
  9. Feb 22, 2014 #8
    The axis convention is clearly not the problem here. The point is that textbooks and instructors after a specific point tend to neglect the nature of spin as an angular momentum and use phrases like "spin points to the direction of the external field" (see Zeeman Effect) or "direction of spin is the same as the direction of motion" (see Wu's experiment), which are conceptually false. We can never know the direction of spin. We can only prepare or know that nature prepares (e.g. outgoing neutrino after a weak decay) one and only one of the components of spin. At least that's the case in nonrelativistic QM.
  10. Feb 22, 2014 #9


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    they all consider the spin's component... (also knowing one component you know the overall spin's direction... the normal components to that component, will just give a circulation, but if z-component looks at positive z-axis, the spin vector will be in some cone opening towards the positive z's)...
  11. Feb 22, 2014 #10
    Yes that's exactly my understanding too. Since that particular spin component is of experimental significance, only theorists remember that distinction though.
  12. Feb 22, 2014 #11
    As far as the Zeeman effect and magnetism goes, I think people often tend to think of spin as a classical vector which points in some direction. It's not strictly right, but for many intents and purposes it works just fine as one tends to be more interested (at least in these applications) about the expectation values of the components than what the spin actually is.
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