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More spin questions

  1. Jun 7, 2010 #1
    1. Can spin be measured without an external magnetic field applied? If not, how do we know it exists without the external field and what makes it intrinsic to the particle in that case.

    2. The states of a Hydrogen atom without any angular momentum would not be subject to spectral splitting under the influence of an external magnetic field. Would we be able to measure the spin of a particle in a state with no orbital angular momentum?
  2. jcsd
  3. Jun 7, 2010 #2


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    The intrinsic angular momentum of an object's electrons contributes to its total macroscopic angular momentum. Look up the Einstein - de Haas effect.
  4. Jun 7, 2010 #3


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    1. No. A field is required for the most direct observation, but you couldn't explain any magnetic phenomena (para-, ferromagnetism), or antisymmetry of fermions, or why they obey Fermi-Dirac statistics, or even the structure of the periodic table without spin.

    2. Well if you're talking 4He+, in a magnetic field you'd see a splitting of the ground state (l=0) into two levels, because of the spin. You still get a spllitting. With hydrogen (spin 1/2 nucleus), you also have nuclear-spin coupling so it'll split into more than one line. (Finally a mention of EPR in this forum that isn't about that Einstein paper)
  5. Jun 7, 2010 #4
    Thanks alxm,

    That clears it up somewhat.

    It seems to me like there is no proof that the extra angular momentum is from the particle itself.

    Could it be an uncertainty in the particle's path?
  6. Jun 8, 2010 #5


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    QM particles don't have paths. (Or the Feynman perspective: They take all paths at once)

    In any case, there's ample evidence that spin is a property of the particle itself, starting with the Stern-Gerlach experiment. If it was simply a case of "extra" angular momentum, then you would expect free electrons to have continuous values and not just two. Theoretically, spin is a quite deep property, but it takes relativistic QM and quite a bit of group theory. You could summarize it as simply being a necessary property in order to satisfy Lorentz-invariance (hence, a relativistic effect). But there's already been quite a lot of "What is spin?" threads around here.

    I should underline that spin doesn't just work as 'extra angular momentum' but that it also leads to antisymmetry, and hence, the Pauli principle. So it's got quite a huge impact even on things that don't have anything at all to do with angular momentum.
  7. Jun 8, 2010 #6


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    Sorry, but they have path. You can observe electron path in bubble chamber.

    Maybe it is better to say that it could be related to uncertainty in position?
    At least I think there are experiments that utilize spin dependent tunneling.
  8. Jun 8, 2010 #7
    I have heard this statement before, never understood what it means though. I am familiar with Lorentz transformations and invariance under transformation of coordinates, also familiar with invariance of Hamiltonian and conservation consequences.

    I can't understand how coordinate transformations and invariance leads to angular momentum though.
  9. Jun 8, 2010 #8


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    A bubble chamber is a macroscopic device that measures macroscopic-scaled paths at a precision far, FAR lower than quantum indeterminancy.
    All that says is that particles have paths in the classical domain.
  10. Jun 8, 2010 #9
    Are you familiar with Noether's theorem? It states that any continuous symmetry of the action gives rise to a conserved current. Lorentz transformations are a symmetry of nature (as far as we know) and are composed of rotations (in 3d space) and boosts ((hyperbolic) rotations between space and time), the rotations give rise to angular momentum as a conserved current, invariance under boosts preserves the velocity of the centre of mass (like inertia maybe...?) The Noether current of Lorentz transformations is called the relativistic angular momentum. It is given by [tex]M^\rho_{\mu\nu} = x_\mu T^\rho_\nu - x_\nu T^\rho_\mu[/tex] where T is the stress energy tensor. However, that isn't really where spin comes from - actually I'm shaky on the details myself.

    I don't think spin is a purely relativistic phenomenon. For starters, it's detectable at v<<c. Historically, Dirac found that it emerges in a nice way when you demand the probability density in relativistic QM is positive, and Pauli kind of had to add it on to non relativistic QM by hand, but really it arises in both limits from the way the rotation group works. The rotation group in non-relativistic physics is SO(3) (3-d orthogonal rotations) and relativistically you have SO(3,1) - the Lorentz transformations. Then you use the dark magicks of representation theory, and you demand that your fields transform under a certain representation of the rotation group in question. It's a mystery to me but there are representations that act on spinors, so then you end up with spinor fields sometimes. The point is it can arise both relativistically and non-relativistically just from the way the rotation group works.

    The only geometrical way to understand spinors I know is through orientation entanglement (nothing to do with quantum entanglement) - twist the end of a belt through 360 degrees, it's twisted. But twist it through 720 degrees, pass one end under the rest of the belt without rotating it - and the twist is gone! Really crazy. When you rotate a spinor by 360 degrees you change its sign, but 720 gets you back to normal... maybe someone else can explain this...
  11. Jun 8, 2010 #10
    I am familiar with the idea that if the Hamiltonian is invariant under rotations then angular momentum is conserved. This seems logical though. It is just saying that if the potential is symmetric the momentum should be conserved throughout the system. This theorem seems to be a more general explanation of that from what I see on wikipedia, I am not familiar with it though.

    :bugeye: :confused: :uhh: over my head.

    Seems similar to derivatives of a sinusoid
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