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Group multiplication laws

  1. Sep 15, 2005 #1
    Greetings--I was wondering if someone could recommend some background reading on group theory in physics that explains group multiplication laws (I guess "tensor multiplication" is more appropriate).

    I.e. in SO(3), 3 x 3 = 1 + 3 + 5

    So that is to say if I had two vectors, [tex]\bar{x}=(x_1, x_2, x+3), \bar{y} = (y_1, y_2, y_3)[/tex], then the tensor product of the two vectors is equal to a direct sum of something that transforms as a singlet, triplet, and 5-let in SO(3). I guess the singlet ends up being the trace ([tex]\sum x_iy_i[/tex]), and the triplet ends up being the cross product, while the 5-let is some linear combination of the left over pairs. How do I construct these? (i.e. in components, what *are* the 5-let components?)
     
  2. jcsd
  3. Sep 15, 2005 #2
    I read S. Hassani's Mathematical Physics. I am not competant to judge, but I heard other people saying good things about it.

    I don't understand. You have two vectors, you can take their tensor products?
     
    Last edited: Sep 15, 2005
  4. Sep 15, 2005 #3
    I have the book. The only good thing is that you have bibliographies of all possible mathematicians and theoretical physicsts in one place, but other than that I really didn't like it. Few of my friends recommended Greiner: "Quantum mechanics: symmetries". Also there are some free lectures here:

    http://www.sst.ph.ic.ac.uk/people/d.vvedensky/courses.html

    but I didn't have time to look at it. Hope this helps.
     
  5. Sep 15, 2005 #4

    Physics Monkey

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    Wu Ti Kung's "Group Theory in Physics"
    Harry Lipkin's "Lie Groups for Pedestrians"

    The first is a good comprehensive reference, one of the best I've seen. The second is different kind of book, and you can get an amazing knowledge just by reading through it in a few days.
     
  6. Sep 17, 2005 #5

    CarlB

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    The physicists do this sort of thing by using raising and lowering operators. The lowering operator, when applied to a product of states gives a sum with one entry for each element in the product. The term corresponding to the nth product element will have the nth product element lowered, and all the other elements left alone. An example will suffice to explain:

    [tex]L (x_3 y_2 z_7)
    = x_2 y_2 z_7 + x_3 y_1 z_7 + x_3 y_2 z_6[/tex]

    If the nth product element can't be lowered (because it's already the lowest possible, then that product term will be zero.

    I will use state notation. [tex]|n;m>[/tex] means the element of a spin-m irrep that has value of n. Define the x1 and y1 states as the "-1", the x2 and y2 as "0" and the x3 and y3 as "+1". Then to get the highest state in the 5, you have:

    [tex]|+2;2> = x_3 y_3[/tex]

    Applying a lowering operator to this (which drops the level of one index), we get the following states:

    [tex]|+2;2> = x_3 y_3[/tex]
    [tex]|+1;2> = x_2 y_3 + x_3 y_2[/tex]
    [tex]|+0;2> = x_1 y_3 + 2x_2 y_2 + x_3 y_1[/tex]
    [tex]|-1;2> = x_1y_2 + x_2 y_1[/tex]
    [tex]|-2;2> = x_1 y_1[/tex]

    The next highest state will be the +1 of the triplet. Applying the lowering operator to that gives:

    [tex]|+1;1> = x_2 y_3 - x_3 y_2[/tex]
    [tex]|+0;1> = x_1 y_3 - x_3 y_1[/tex]
    [tex]|-1;1> = x_1 y_2 - x_2 y_1[/tex]

    The remaining state is:

    [tex]|+0;0> = x_1y_1 + x_2 y_2 +x_3y_3[/tex]

    The lesson here is that one begins by quite arbitrarily defining what "high" means, and then one works out the irreducible representations by first finding the highest possible state, and then applying the lowering operator to it to work out the remaining elements of the irrep. Then one repeats, beginning with the highest remaining state. Thus we find the 5, the 3 and finally the 1.

    All this applies pretty much the same way when you go to a more complicated group like SU(3), except that you need to have more lowering operators. SU(3), for example, requires two, one operating in each of the two directions that one requires quantum numbers for SU(3) type objects.

    I think Georgi's book is the best. And thanks for mentioning that 3x3 includes the inner product and the cross product. I didn't know this.

    Carl
     
    Last edited: Sep 17, 2005
  7. Sep 17, 2005 #6

    vanesch

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    Eh, aren't we forgetting the coefficients in the way you present things ? I mean: J_ |j,m> = Sqrt[(j+m)(j-m+1)] |j,m-1>
     
  8. Sep 18, 2005 #7

    CarlB

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    Yeah, well, sort of...

    The problem was to find a way of defining the 5-plet of 3x3 = 5+3+1, if one defines the 3 and the 1 to be the cross product and dot product, respectively.

    If we're going to use a lowering operator to define the states it clearly is not going to be the usual physics spin-z lowering operator because, while that will divide the degrees of freedom correctly into the multiplets, it will not give the cross product. So I used a modified lowering operator that was designed to reproduce the cross product.

    Written out in differential language, what I used was the operator:

    [tex]x_2\partial_{x3} + x_1\partial_{x2} + y_2\partial_{y3} + y_1\partial_{y2}[/tex]

    It gave close to the right answer, (as far as the degrees of freedom of the 5, 3 and 1), but it fails because it mixes the 5 and the 1. This is not so obvious with the 5, but it's obvious with the 1 because when you apply it to the singlet you get a triplet sequence:

    [tex]x_1y_1 + x_2 y_2 + x_3y_3[/tex]
    [tex]x_1y_1 + x_2 y_2[/tex]
    [tex]x_1y_1[/tex]

    The basic problem is that the lowering operator I defined does not commute with total angular momentum. On the other hand, it does commute with symmetry (that is, symmetric or antisymmetric sums of products remain that way after application of that lowering operator). Since it preserves symmetry, it works on the 3.

    As an aside, the usual physics lowering operator has the disadvantage that it is written with imaginary numbers. So it produces a sequence of states in a multiplet that typically are complex. The lowering operator I wrote uses no imaginary numbers so it preserves real states.

    While physicists prefer their multiplets split into the complex states associated with particular values of spin-z, the chemists prefer those same multiplets split into real states oriented in different directions. It was the chemist version of the multiplet splitting that fliptomato was looking for. Here's a link for drawings of the orbital wave functions as used by the chemists:
    http://swarna.ncsa.uiuc.edu/chemvizngi.html

    In the above link, note that the "3" series orbitals form the 1+3+5 that we've been discussing, but from the chemical standpoint. The 3 orbitals, and their products in the terminology of fliptomato are then:

    3S [tex]\;\;\; \equiv x_1y_1+x_2y_2+x_3y_3[/tex]

    3Px [tex]\;\;\; \equiv x_2y_3-x_3y_2[/tex]
    3Py [tex]\;\;\; \equiv x_3y_1-x_1y_3[/tex]
    3Pz [tex]\;\;\; \equiv x_1y_2-x_2y_1[/tex]

    3Dxy [tex]\;\;\; \equiv x_1y_2+x_2y_1 [/tex]
    3Dxz [tex]\;\;\; \equiv x_1y_3+x_3y_1 [/tex]
    3Dyz [tex]\;\;\; \equiv x_2y_3+x_3y_2 [/tex]
    3Dz2 [tex]\;\;\; \equiv x_3y_3 [/tex]
    3Dx2-y2 [tex] \equiv x_1y_1 - x_2y_2[/tex]

    I'm tempted to see if I can find a (real) lowering operator that commutes with total angular momentum and that gives the cross product for the 3, but it will have to wait as I'm having too much fun with density matrices. Such a lowering operator should give an interesting sequence for the 3D orbitals.

    Carl
     
    Last edited: Sep 18, 2005
  9. Sep 19, 2005 #8

    dextercioby

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    Simply put

    [tex] 2 a_{i} b_{j}=\left( a_{i}b_{j}-a_{j}b_{i}\right) +\left(a_{i}b_{j}+a_{j}b_{i}-\frac{2}{3}\delta_{ij}a_{k}b_{k}\right) + \left(\frac{2}{3}\delta_{ij}a_{k}b_{k}\right) [/tex]

    and search Sakurai's text or some other one under "spherical tensor operators". And of course, some group theory from Hammermesh.

    Daniel.
     
  10. Sep 24, 2005 #9
    Sorry, how do you go from the third line to the fourth line? That is, what happens to the coefficient?

    Wouldn't we get:

    [tex]|-1;2> =3 x_1y_2 +3 x_2 y_1[/tex]

    Since we get a [tex] x_1y_2[/tex] from the first term of the [tex]|+0;2>[/tex], a 2x_1y_2 from the second term, and similarly for the [tex]x_2 y_1[/tex] term.

    Is there something about normalization that I'm missing?

    Thanks!
    Flip
     
  11. Sep 25, 2005 #10
    Apologies, but this is also unclear to me. In the previous cases, each term satisfied the constraint on [tex]n[/tex]. Here, however, the [tex]x_1y_1[/tex] and [tex]x_3y_3[/tex] have [tex]n[/tex] equal to -1 and 1 respectively, not zero.

    I suppose another way of phrasing my question is "why is this an [tex]n=0[/tex] state?" If we apply an operator [tex]N_z[/tex] that tells us the "angular momentum in the z-direction" (as measured by [tex]n[/tex]) of the state, wouldn't we would get:

    [tex]N(x_1y_1 + x_2 y_2 +x_3y_3) = -x_1y_1 + 0 +x_3y_3[/tex]

    ?

    Thanks once again,
    Flip
     
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