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Some confusion about Ballentine Sec 3.3 Generators of Gallilei Groups

  1. Jul 4, 2009 #1
    I am trying to do some self-study and plow through Ballentine's book on Quantum Mechanics. I thought I was following the majority of it until I got to Sec. 3.3, in particular the derivation of the multiples of identity for the commutators of these generators.

    For example, the commutator of two rotations is given as

    [tex][J_\alpha,J_\beta]=i\epsilon_{\alpha\beta\gamma}J_\gamma + i\epsilon_{\alpha\beta\gamma}b_\gamma I[/tex]

    He then states that one can redefine the phase of certain vectors to transform the phase term to 0 and does so with the substitutions

    [tex]J_\alpha + b_\alpha I \rightarrow J_\alpha[/tex]

    leading to

    [tex][J_\alpha,J_\beta]=i\epsilon_{\alpha\beta\gamma}J_\gamma[/tex]

    I feel I am missing something key in what is going on here because I am not understanding why one can't just arbitrarily say the extra phase piece in the first equation is just 0 by setting [tex]b_\gamma = 0[/tex]. Where are these phase differences coming from that one needs to change your generator as Ballentine does?
     
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  3. Jul 4, 2009 #2
    I agree that Ballentine is not very clear about the logic of what's going on. The idea is this: In quantum mechanics we want to assign an unitary operator [itex]U_g[/itex] to each space-time transformation [itex]g[/itex]. However, we can notice that two unitary operators differing by a unimodular constant [itex]U_g[/itex] and [itex]\alpha U_g[/itex] are physically equivalent, so there is a great freedom in choosing unitary representatives of space-time transformations. In terms of Hermitian representatives of generators of space-time transformations, this means that one can add arbitrary constants to such representatives without modifying the theory.

    This arbitrariness is not a big problem. We can simply choose these arbitrary constants as we like (e.g., zeros). The only inconvenience is that commutators of these chosen Hermitrian generators may not necessarily repeat the commutation relations of the Galilei Lie algebra. It is possible that right hand sides of commutators contain additional constants as shown in eqs. (3.14) - (3.24) in Ballentine's book.

    Then the idea is to choose better representatives of Galilei generators to get rid of these inconvenient additional constants on the right hand sides of commutation relations. These better representatives are obtained by adding constants to our originally chosen operators.
     
  4. Jul 4, 2009 #3
    Thanks for this clarification. After reading your post and rereading the section its becoming clearer.
     
  5. Jul 5, 2009 #4

    dextercioby

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    There's a thorough and rather no too technical discussion in the second chapter of Weinberg's QFT book on central charges in a Lie algebra and how they can be made to disappear for the Poincare Lie algebra.
     
  6. Jul 5, 2009 #5

    Fredrik

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    I think the key is to understand that when you rewrite the left-hand side of (3.11) as a single exponential, (3.7) says that you have to include an ω. This ω is what shows up as the b's in the first equation you posted. But you have probably realized this already.

    Wigner's theorem guarantees that (3.7) must hold, but it doesn't say that ω is non-zero for all symmetry groups, or anything like that. You can think of what we're doing here as an investigation of whether we can take an arbitrary ω and eliminate it by suitable redefinitions of the generators. We are only partially successful in this case. We can eliminate the b's from the commutation relations, and this constrains ω to only take values that make exp(iω)±1, but we're not getting rid of the -1 so easily. That can be traced back to the fact that the rotation group SO(3) isn't simply connected.

    I haven't actually studied this section of Ballentine yet, so I don't know if this is what he does, but you can actually get rid of the -1, by throwing out SO(3) and choosing to consider SU(2) instead. It makes sense to do this, because the quantum theory with SU(2) as the symmetry group is essentially the same as the quantum theory with SO(3) as the symmetry group. There's a good discussion about this in Weinberg's book. ("The quantum theory of fields", vol. 1, chapter 2).

    SU(2) is homeomorphic* to a 3-sphere, and is therefore simply connected. SO(3) is homeomorphic to a 3-sphere with opposite points identified, so there's a 2-1 correspondence between members of SU(2) and members of SO(3).

    *) X is said to be homeomorphic to Y if there exists a continuous bijection f:X→Y that has a continuous inverse.
     
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