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Can't understand the algebraic theory of spin.

  1. Dec 28, 2012 #1
    I've been reading Griffith's Introductory text on quantum mechanics, and I'm afraid I've hit a major roadblock. I don't understand his section on spin at all. I get the part about there being no classical analogy of quantum mechanical spin, but then he goes on to develop the algebraic theory of spin with no motivation apart from the fundamental angular momentum commutation relations. That suggests to me that spin has something, if not everything, to do with actual spinning. Where is the motivation behind there being an "intrinsic" angular momentum in addition to a particle's "extrinsic" angular momentum? Is it purely experimental ? And if you're developing the theory of "intrinsic" angular momentum entirely on the mathematical characteristics ( the commutation relations ) of the theory of "extrinsic" angular momentum, shouldn't they have a lot to do with each other? Further , in page 183 and 184 of the text, he does this
    S[itex]^{2}[/itex]{sm} = h[itex]^{2}[/itex]s(s+1){sm} ; S[itex]_{z}[/itex]{sm} = hm{sm} to
    S+-{sm} = h(s(s+1) - m(m+1))[itex]^{1/2}[/itex]{s(m+-1)}
    ( I couldn't find the appropriate mathematical notation in the menus, but the {} is a 'ket' , h is actually h/2[itex]\pi[/itex] and the +- should be a + on top of the - ) where S+- = S[itex]_{x}[/itex] +- iS[itex]_{y}[/itex]
    I don't get what he did here at all. And finally, he claims this - "But this time eigenvectors are not spherical harmonics ( not functions of phi and theta at all), and there is no "a priori" reason to exclude half-integer values of s and m."
    Why is the eigenvector not being a spherical harmonic a reason at all to include the half-integer values of s and m?
    Sorry for the long post, but this has been causing me considerable distress. I'd be so eternally grateful for some help figuring this out :)
  2. jcsd
  3. Dec 28, 2012 #2
    Ok, let’s clear out what spin is and how you can describe it. First of all, forget about intrinsic and extrinsic angular momentum; angular momentum, whatever it is, can be described with a single way.

    Now, consider Noether’s theorem of classical mechanics: continuous symmetries imply conservation laws and the conserved quantities are the generators of the symmetry transformations (translational symmetry implies conservation of momentum <-> momentum of the generator of translation, rotational symmetry implies conservation of angular momentum <-> angular momentum is the generator of rotations, etc.)
    This is a very useful statement, especially in QM, if you want to define physical quantities: just consider some symmetry and find the corresponding conserved quantity. You see that in this way, physical quantities are defined w.r.t. some symmetry transformation. This is extremely useful in QM, because of the ambiguity of notions as “position” and “momentum” (which in classical mechanics are used to define the dynamical variables). So, instead of using these notions to define your physical quantities, you can use the notion of symmetry transformations to define them. As stated before the conserved quantities are the generators of the transformation. So, in QM, the physical quantities (observables) will be defined as the generators of some symmetry.

    But how this is can be done exactly? The answer lies in group theory. Since the elements of some symmetry transformations can be composed and give a new transformation of the same kind, they can form a “group”. For continuous transformations, you can find the set of transformations that lie infinitesimal close to the identity transformation and expand them as linear combinations of the “generators”. If you demand this expansion to posses the group properties, then you get a set of commutation relations (Lie algebra).
    All you gave to do now, in order to transit to QM, is to find something to represent these generators as QM operators. This can be done by searching objects that satisfy the same commutation relations as the generators. Also, the properties of these operators will be hidden in the commutation relations.

    Now, about angular momentum. As stated before, angular momentum is the generator of rotational symmetry. So, to study angular momentum you have to find the group of rotations (SO(3)), then find it’s Lie algebra and something to represent it. When, you use the coordinate representation, you represent this algebra with the usual operator L = -i [itex]\hbar[/itex] x [itex]\times[/itex][itex]\nabla[/itex] and the eigen-values equation of this operator (actually of the Casimir operator L 2) are allowed (by only mathematical reasons) to have only integer values. This representation is suitable when one wants to describe kinematical notions, so this kind of angular momentum (that is represented by these operators) is the kinematical one (or extrinsic). When you choose matrices to represent the generators of SO(3)’s lie algebra then, as a mathematical consequence, you find that the angular momentum can take integer as well as half-integer values. This representation is suitable to define both intrinsic and extrinsic angular momentum.
  4. Dec 28, 2012 #3
    I understand that this is too much information and maybe it is quite ambiguous to you about how to use it to get results. When you clear these up, if you would like, I could guide you in some examples on how to use them.
  5. Dec 28, 2012 #4
    Thanks a ton for the reply. I'm sorry though, all this is a little too advanced for me (I've only recently picked up Griffith's QM, so I only have an introductory understanding )

    "When you choose matrices to represent the generators of SO(3)’s lie algebra then, as a mathematical consequence, you find that the angular momentum can take integer as well as half-integer values."

    I don't really understand all the symmetry and group theory bits, but here's what I get. When you solve for the eigenstates of angular momentum, you find that the allowed values of "l" , and hence of "m" are integers and half-integers. So, is it that you pick out the half-integer values and call these eigenstates "spin" eigenstates, and the integer values "angular momentum eigenstates"? If that were the case though, I really don't see the need for a separate algebraic theory of spin.
    If that's not the case, then my question is, why is it okay to allow half-integral values for 's' and 'm' , when these very half-integral values are excluded (I'm not entirely sure of this, but my textbook says they are ) when the eigenvectors are spherical harmonics( in the case of angular momentum eigenfunctions ) ? Secondly, the way my textbook does it, is to say that 'spin' is fundamentally different, yet related to angular momentum, and then go on to introduce the operators S etc. , and then work out the algebra the same way that it treated angular momentum. I don't understand how 'spin' arises naturally within the context of the theory ... I mean, it should right? Isn't the theory flawed otherwise? Could you give me an simpler (introductory course at an undergrad level ) explanation?
    Thanks again for all the help :)
  6. Dec 28, 2012 #5
    No, you don’t call the half-integer eigenstates “spin” and the others “angular momentum eigenstates” (<- the correct term is “orbital angular momentum”, but this is not the problem here). Since all of them are eigenstates of some angular momentum operator (which in algebraic theory of angular momentum is represented by a matrix), then they all are “angular momentum eigenstates”.
    The distinction between the two is not the integrity of the eigenvalue, but how this eigenvalue relates with some orbital quantity of the system.
    In particular, the “orbital angular momentum” operator is defined as: L = -i [itex]\hbar[/itex] x [itex]\times[/itex][itex]\nabla[/itex] ; this is because that operator does represent the angular momentum operator (i.e. has the same commutation relations as an.m.op. has) and is defined w.r.t. the operators x and p = -i [itex]\hbar[/itex][itex]\nabla[/itex] , which are orbital operators (i.e. they are related with the motion of the particle). So, when you solve the eigenvalue equation of Lz, then you find the eigenstate [itex]\exp \left( im\varphi \right)[/itex] and in order for it to be single-valued, then m has to be an integer.
    But the same requirement does not occur when solving the eigenvalue equation for angular momentum operators, when they are represented by matrices. So, in general, angular momentum can take integer as well half-integer values.
    Another distinction between orbital angular momentum and spin, is that the value of the latter does not go to zero in the rest frame of the particle, while orbital angular momentum value does.
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