http://math.stackexchange.com/a/38963
The above link contains a simple, understandable answer to the frustrating question: What are irreducible representations ?
If you do Quantum Espresso Phonon calculations with ph.x you have heard about them as irreps.
What the code does is that, once you define which phonon wavenumber (q) you want the dynamical matrix to be calculated for, the code looks for the small group of q.
Small group of q is found by selecting , among the point group symmetries of the crystal, the operations that leave the vector q unchanged, or carries it to -q+G , where G is a reciprocal space vector.
And once you know the small group of q, you can write this group in terms of its irreducible representations.
Then the code carries on the understand, for each phonon mode, which irreducible representation they belong to:
The outcome looks like this:
The above link contains a simple, understandable answer to the frustrating question: What are irreducible representations ?
If you do Quantum Espresso Phonon calculations with ph.x you have heard about them as irreps.
What the code does is that, once you define which phonon wavenumber (q) you want the dynamical matrix to be calculated for, the code looks for the small group of q.
Small group of q is found by selecting , among the point group symmetries of the crystal, the operations that leave the vector q unchanged, or carries it to -q+G , where G is a reciprocal space vector.
And once you know the small group of q, you can write this group in terms of its irreducible representations.
Then the code carries on the understand, for each phonon mode, which irreducible representation they belong to:
The outcome looks like this:
There are 3 irreducible representations
Representation 1 2 modes - To be done
Representation 2 2 modes - To be done
Representation 3 2 modes - To be done
I like the explanation by Arturo Magidin in the mentioned link so much that I decided to mirror it here so that perhaps it will be better protected from the perils of online existence :)
All credits go to original author
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A representation of the group G means a homomorphism from G into the group of automorphisms of a vector space V . Essentially, you are trying to interpret each element of G as an invertible linear transformation V→V , in order to try to understand the group G by how it "acts on V ."
If you have an actionρ1 of G on a vector space W (that is, one representation), and you have some other action ρ2 of G on another vector space Z (another representation), then you can use these two actions to construct an action of G on the vector space W⊕Z : just let G act on the first coordinate using the old action on W , and let it act on the second coordinate using the old action on Z .
The point to observe, however, is that the action ofG on W⊕Z defined this way does not give you any new insights into the structure of G : anything you can glean about G from this action, you can learn about G by considering the original actions ρ1 and ρ2 . So this new action does not give us anything new.
Conversely, suppose you have one representationρ , with G acting on V , and that there are proper subspaces W and Z of V that satisfy the following properties:
G on W to get a representation, and the restriction on Z
to get another representation; and these two representations will give
you all the information from the original representation, the same way
we had before. The advantage being that since W and Z are proper subspaces of V , they have smaller dimension and, presumably, it's easier to understand a subgroup of linear automorphisms for them than for V .
So the moral is that we want to find representations that cannot be "broken up" into smaller ones, because there's no point in trying to understand ones that do break up, we can focus our attention on those that don't, because all the other representations can be built up in terms of the ones that cannot be broken up.
The irreducible representations are precisely the ones that cannot be broken up into smaller pieces. There is a theorem that says that if you have a representationρ of G acting on V , and W is a subspace of V such that for all g∈G , the image of W under the action of g is W itself, then you can find a subspace Z of V such that V=W⊕Z and every g∈G maps Z to itself (that is, in order to break up ρ into two smaller pieces, it is enough to find a single proper piece on which ρ acts; then you can find a complement for it). With this in mind, we say:
SO(3) will be a representation of SO(3) that is irreducible. SO(3) acts naturally on the vector space R3 : it consists of all automorphisms of R3 that respect the inner product, so this is itself a representation of SO(3) (which is irreducible, because no proper subspace of R3 is sent to itself by all elements of SO(3) ).
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If you have an action
The point to observe, however, is that the action of
Conversely, suppose you have one representation
V=W⊕Z ; and- The action of every
g∈G onV mapsW to itself; and - The action of every
g∈G onV mapsZ to itself.
So the moral is that we want to find representations that cannot be "broken up" into smaller ones, because there's no point in trying to understand ones that do break up, we can focus our attention on those that don't, because all the other representations can be built up in terms of the ones that cannot be broken up.
The irreducible representations are precisely the ones that cannot be broken up into smaller pieces. There is a theorem that says that if you have a representation
LetAn irreducible representation ofρ:G→Aut(V) be a representation ofG . We say thatρ is irreducible if and only ifV is not the zero vector space, and the only subspaces ofV that are mapped to themselves under the action of everyg∈G are{0} andV itself.
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1 comment:
http://users.ictp.it/~pub_off/lectures/lns024/10-giannozzi/10-giannozzi.pdf
Please go through this link page 169, you get a physical reasoning as well as idea of this 'irrep'.
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