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Greiner Mueller's QM II Symmetry

  1. Oct 11, 2011 #1
    Is anyone in PF reading the titled book? For me, it is slow work because I have not done physics for a while - decades! I am retired and read some old physics books just for challenge. Mueller gives me that, but also reward. From time to time, I find little stumbling blocks, sometimes a misprint. If I had a co-reader, I might make better progress. On the other hand, I might slow his or her progress!

    I am in Chapter 3, and reading past Cartan's Theorem.

    My homework is self assigned :smile:
  2. jcsd
  3. Oct 12, 2011 #2


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    I am moving this to "Quantum Physics".
  4. Oct 12, 2011 #3


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    The book is a useful reading, indeed, even though there are other (perhaps better, or richer in information) sources on this matter (Fonda & Ghirardi comes to my mind). You can post your questions here, if there are some staments you need help with.
  5. Oct 12, 2011 #4
    I have read some reviews of the Greiner book. It has its good and bad poiints. Most of my difficulties have come from those bad points, I realize. Thanks for the referral to Fonda & Ghirardi.
  6. Oct 12, 2011 #5
    To dextercioby: If you have a copy of G-M and the time, could you look at eq (13) on top of page 100 (2nd Ed.)? It shows the generator of a Lie group L as a linear combination of others with coefficients that are solutions of a system of linear equations in the metric g.
    My belief is there is one and only one such set of coefficients that is an independent set. Yet, eq (13) shows a subspace of generators which are linear combinations of others from the group. The coefficients, a, which have only one index in eq (12) (a superscript) have a subscript as well in eq (13). But the a's are just a single set of numbers that solve eq (12), as I see it. Where did the subscript in eq (13) come from?
  7. Oct 13, 2011 #6
    Mr Lewin: I have searched my local library and Worldcat for Fonda & Ghirardi. No luck! Amazon sell used copies (2) for $199 and for $89. Too expensive for my need! I have posted what is holding me up in my message to dextercioby. If you have a chance, please see if you can shed any light on it.
  8. Oct 13, 2011 #7


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    To me the i and λ share the same values, as per the chosen regular representation (eq. on page 98), so that the set of a's which solve the equation 12 is not a 1-dimensional vector space, but an i/λ one, so that's why the a's can be further indexed by i, not only by λ.
  9. Oct 13, 2011 #8
    Thanks for your reply, but I am apparently not following. The a's are defined by eq 12. They are its unique solution, a set of numbers equal to the order of the group, G, (the number of L's in the group). Turning to eq 13, we see L' = aL , where the summation index (sigma) is, I thought, also the order of the L group, G. The L' is subscripted, as if one had several sets of a's. Are you writing that a different subset (of order I) of the Ls is chosen for each L' and the index i labels the particular subset chosen? Are you also writing that the number of subsets G/I and that is the range of the index i? I don't think you mean all this, because you wrote that the index i and lambda share the same values. The values for lambda come from the structure constants. That number of values is G, the order of the group.

    To add to my confusion, eq 14 seems to be missing a prime marking on the rhs for one of the Ls.

    Could you be more explicit? Thanks, again.
  10. Oct 13, 2011 #9
    PS You stated, too, that the i and the lambda share the same values. If so, then their quotient is one. No?
  11. Oct 13, 2011 #10
    dextercioby: I suggest that what Greiner meant was that not only is the determinant of the metric g zero, but that its matrix be reducible to a number of square matrices along the diagonal AND that the determinant of each such matrix be zero. In other words, he means to divide the group into a number of subgroups along the diagonal of the L,L matrix. With each sub-determinant of the g matrix equal to zero, the L' in that sub-group commute. Maybe that's what you have in mind. The hyp that det g = 0 must be then be augmented for each sub matrix of g.

    So it's not an "if and only if" theorem. Moving on! Thanks for your thoughts.
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