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Basis for an irreducible representation

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm having trouble understanding a concept in representation theory. I've been reading several texts on the application of rep. theory to quantum mechanics ("Group Theory and Quantum Mechanics" by Tinkham and "Group Theory and Its Application to Physical Problems" by Hamermesh). Their treatments are similar, and I feel like neither explains these one thing thoroughly enough:

    What does it mean when a function "belongs to a particular row of a particular irreducible representation"? More importantly, how is this property dependent on a particular choice of basis?

    2. Relevant equations

    Let [itex]P_R[/itex] be the operator associated with the group element [itex]R[/itex] and [itex]\{\psi_i^{(k)}\}[/itex] be a set of [itex]l_n[/itex] linearly independent functions such that:
    [itex]P_R\psi_v^{(n)} = \sum_{k=1}^{l_n}\psi_k^{(n)}\Gamma^{(n)}(R)_{kv}[/itex]
    Then the matrices [itex]\Gamma^{(n)}(R)[/itex] form an [itex]l_n[/itex] dimensional representation of the group.

    3. The attempt at a solution
    I understand the implications of the equation above. Tinkham and Hamermesh then both go on to refer to [itex]\psi_k^{(n)}[/itex] as "belonging to the k-th row of the n-th representation". Fine, but I just don't get the significance of the row part...it's obviously completely dependent on the choice of basis and the arbitrary ordering of the base functions within that basis. But then later, we come across this theorem:

    "Two functions which belong to different irreducible representations or to different rows of the same unitary representation are orthogonal."

    Huh? If a particular basis has been chosen, then a specific row of a specific unitary rep. indexes a single unique function...doesn't it? How can you have two different functions belonging to the same row of the same representation?
     
  2. jcsd
  3. Jul 22, 2012 #2
    Could you perhaps be more specific about the problem? I have neither of these books so it is difficult to discern context.

    Is there anything you can tell us about the group? About the vector space? What does it mean for a function to belong to a representation?
     
  4. Jul 23, 2012 #3
    The group is any group of symmetry operations.

    "What does it mean for a function to belong to a representation"
    That's what I'm asking. It seems to be standard terminology, not just in my textbooks. More specifically, I don't know if I understand what it means for a function to belong to a particular row[/row] of an irreducible representation.

    My question can be put this way: is it possible for two distinct functions to belong to the same row of the same irreducible representation? If so, how? If not, then how is the statement, "Two functions which belong to different irreducible representations or to different rows of the same unitary representation are orthogonal," supposed to be interpreted?
     
  5. Jul 23, 2012 #4
    Okay, I have spoken to a colleague in the physics department, since I have never seen this vernacular in mathematics. It seems to me that you are looking at a finite subspace of your vector space, since if it is infinite dimensional, how could you possibly choose a finite basis?

    This is how I understood our discussion:

    Consider the orbits of the induced G-action on V. There is a theorem that says that any finite dimensional subspace of V lives in a finite dimensional G-invariant subspace. For a fixed function, that function spans a one-dimensional subspace and so by theorem lives in a finite dimensional G-invariant subspace, say W. You have chosen a basis for this space, and the matrices given by decomposition are irreducible representations. Thus the direct sum of those matrices (namely, the space W) is also a representation. I believe this is what it means for a function to be long to a representation.

    If I am correctly seeing past the murky physical cloud that is obscuring the mathematics here, it seems to me that this essentially just says that the canonical representation of the symmetry group of a Hamiltonian system is completely reducible. Consequently, if one abusively thinks of elements of this canonical representation as infinite dimensional matrices, they look like block diagonal components of irreducible unitary representations. Given the discussion above, if two functions then belong to distinct unitary representations, they will belong to different "rows" of an element and hence be orthogonal.

    Edit: Though this is only my best interpretation. I am unfamiliar with this subject.
     
  6. Jul 23, 2012 #5
    Thank you for looking into this, but I'm afraid that this isn't quite what I mean. In the physicist's language, two functions may be said to belong to different rows of the same unitary representation. I'm afraid that the gap in formalism may be too great here; I assumed the physics version of representation theory used the same terminology as the mathematicians use. Apparently not.

    Nonetheless, I'll try again. I actually had a much more elaborate response prepared last time but the forum ate it when I tried to post it.

    Suppose the group we're looking at is the point group [itex]C_{3v}[/itex] (http://www.webqc.org/symmetrypointgroup-c3v.html). This is (approximately) the symmetry group of the ammonia molecule. Now imagine we've placed coordinate axes on the molecule, with the xy-plane aligned with the three hydrogen atoms. Then applying each of the group's operations to an arbitrary Cartesian vector (x,y,z) leaves the z-coordinate invariant. With each symmetry operator [itex]R[/itex] in the group (which operates on a Cartesian vector), we associate a function operator [itex]P_R[/itex] such that [itex]P_Rf(\vec{x}) = f(R^{-1}\vec{x})[/itex]. Now, from earlier, the function f(x,y,z)=z is invariant under each group operation, i.e. [itex]P_Rz = z[/itex] for each element in the group. For a general function, this process will yield a certain number (say, l) of linearly independent functions [itex]\{\psi_k\}[/itex] that mix together (as linear combinations) under each group operations:
    [itex]P_R\psi_v = \sum_{k=1}^{l}\psi_k\Gamma(R)_{kv}[/itex]
    It can then be shown that the matrices [itex]\Gamma(R)[/itex] form a (possibly reducible) representation of the group in the usual sense, and we say that the [itex]\{\psi_k\}[/itex] are a basis for the representation.

    If this an irreducible representation, we write (as I did in my original post) [itex]P_R\psi_v^{(n)} = \sum_{k=1}^{l_n}\psi_k^{(n)}\Gamma^{(n)}(R)_{kv}[/itex] and say that the function [itex]\psi_j^{(n)}[/itex] belongs to the the j-th row of the n-th irreducible representation (which has dimension [itex]l_n[/itex]). Thus, as shown above, f(x,y,z)=z is a basis for the totally symmetric representation [itex]A_1[/itex] of the group. We say that it belongs to the [itex]A_1[/itex] irreducible representation. The x and y coordinates of a Cartesian vector will clearly mix together under the group operations. Thus, they must form a basis for a 2-dimensional irreducible representation. Looking at the character table I linked to, we see that indeed they belong to the [itex]E[/itex] representation.

    My books then go on to say that two functions that belong to two different irreducible representations or to different rows of the same unitary representation are orthogonal. However, from the above definitions it seems clear that the property of belonging to a particular row within a representation is dependent on your choice of basis functions. Furthermore, if you've chosen a basis set, then exactly one function can be described as belonging to a particular row of it. However, the books sound like they are describing two distinct functions. Thus, my question: is there any unambiguous sense in which it can be said that two distinct functions to belong to the same row of the same irreducible representation? If so, then where in the above have I gone wrong?

    That's about as clear as I can make it without copying large sections of my books. I'm sorry if you find the notation perverse but unfortunately that's what I'm stuck with:frown:. If this still doesn't help, perhaps your physicist colleague would be willing to take a look?
     
  7. Jul 23, 2012 #6
    (Also, I see from your profile that you're a Canadian grad student. On the off chance you're at U of T, I probably know your physics colleague. If they wouldn't mind helping me figure this out, I can PM you my utoronto address.)
     
  8. Jul 23, 2012 #7
    By chance I am a UofT student. I'll just PM you my address and my physics friend's address.
     
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