Register to reply 
Representations, states and tensors 
Share this thread: 
#1
Jul1114, 11:53 AM

P: 53

Hi. I am currently studying about representations of Lie algebras. I have two questions:
1. As I understand, when we say a "representation" in the context of Lie algebras, we don't mean the matrices (with the appropriate Lie algebra) but rather the states on which they act. But then, the states contain less information than matrices, since there could be several representations (for a given dimension) acting on the same state. I am trying to understand why these states contain the same information as matrices. Is it because by "states" we actually mean weight vectors, which  in addition to containing vectors  contain the eigenvalues, so we could build up the matrices from them, and therefore they are equivalent to finding the generator matrices? 2. When we speak about "tensor representations" (for example, SU(n) tensors, which can also be described by Young Tableaux), what is their relation to the representations I mentioned in question 1 above? Thanks. 


#2
Jul1114, 02:16 PM

Sci Advisor
P: 908

[tex][2] \otimes [2] = [3] \oplus [1][/tex] Do you know the meaning of those numbers and their connection to "representation spaces", "states", "matrices" and "tensor representation" of [itex]SU(2)[/itex] and its algebra? 


#3
Jul1114, 03:25 PM

P: 53

I know the connection to states: the LHS is the direct product of 2D states and the RHS is the 3D and 1D states corresponding to the irreps.



#4
Jul1114, 05:33 PM

Sci Advisor
P: 908

Representations, states and tensors
Ok, [itex][2][/itex] is the smallest dimensional (defining) representation space, i.e., a 2dimensional linear vector space [itex]V^{ (2) }[/itex]. The elements of such space are 2component vectors [itex]v_{ i }[/itex]. The [itex]2 \times 2[/itex] matrices of [itex]SU(2)[/itex] act naturally on [itex]v_{ i }[/itex] mixing its components (states).
So, [itex][2] \otimes [2][/itex] is the 4dimentional (reducible) representation space [itex]V^{ (4) }[/itex] which contains 2 invariant (irreducible) subspaces [itex]V^{ (3) }[/itex] and [itex]V^{ (1) }[/itex]. Again, you can think of the elements of [itex][3][/itex] (3 states) as 3component vectors or rank2 symmetric tensor. And the elements of the space [itex][1][/itex] (one state) as 1component scalar or antisymmetric rank2 tensor. So, we can translate [itex]V^{ (4) } = V^{ (3) } \oplus V^{ (1) }[/itex] into a statement about reducing the tensor product [itex]u_{ i } v_{ j } \equiv T_{ i j } \in V^{ (4) }[/itex] into irreducible tensors. Simply, decompose the tensor product into symmetric and antisymmetric tensors: [tex]u_{ i } v_{ j } = \frac{ 1 }{ 2 } ( u_{ i } v_{ j } + u_{ j }v_{ i } ) + \frac{ 1 }{ 2 } ( u_{ i } v_{ j }  u_{ j } v_{ i } ) ,[/tex] or [tex]T_{ i j } = T_{ ( i j ) } + T_{ [ i j ] }[/tex] For SU(2), the (irreducible) symmetric tensor [itex]T_{ ( i j )}[/itex] has 3 independent components (3 states) and the antisymmetric tensor [itex]T_{ [ i j] }[/itex] has only one component (one state). The (3 + 1) states, vectors or tensors do not mix. Sam 


#5
Jul1114, 10:36 PM

P: 53

Thanks. That answers my second question, i.e. connect the tensors to the state representations. So the irreducible representations are represented by symmetric and antisymmetric tensors. But I still don't know why this is true. I read that in general the irreps of SU(N) are tensors with definite symmetry and antisymmetry, but I don't know why this is true.
Also, I would like to get an answer to my question 1: why the states are the representations in the first place? I thought representations were supposed to be matrices (generators) satisfying the Lie algebra. 


#6
Jul1114, 10:53 PM

P: 864




#7
Jul1214, 04:16 AM

Sci Advisor
P: 1,924

Cf. this Wiki page. People often say that the matrices "represent" the abstract group generators. Similarly, the linear space is sometimes said to be the "carrier space" for the representation, or that it "carries the representation". 


#8
Jul1214, 04:57 AM

P: 53




#9
Jul1214, 01:44 PM

Sci Advisor
P: 908

In general, a Lie algebra admits matrix as well as operator representations. Sam 


#10
Jul1214, 03:32 PM

P: 53




#11
Jul1214, 05:47 PM

Sci Advisor
P: 908

No, as well as the generators (matrices or operators) you also need the eigenvalues and eigenvectors of the invariant Casmir operators of the group. Pauli matrices alone do not let you formulate theory for spin 1/2 particle. You also need the simultaneous eigenvectors of [itex]S^{ 2 }[/itex] and [itex]S_{ z }[/itex].



#12
Jul1214, 11:00 PM

P: 53




Register to reply 
Related Discussions  
QM I  Decomposition of countable basic states into coherent states  Advanced Physics Homework  1  
Stationary States and timeindependent states (aren't they the same?)  Quantum Physics  4  
Help clarify the probability of pure states and mixture of states  Quantum Physics  5  
Quantum Mechanics, Glauber Coherent States of QHO, and superpositions of states  Advanced Physics Homework  0  
Vector Representations of Quantum States  Quantum Physics  15 