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Tensor product and representations

  1. Feb 17, 2015 #1
    Hi, I that [itex]<I|M|J>=M_{I}^{J}[/itex] is just a way to define the elements of a matrix. But what is [itex]|I>M_{I}^{J}<J|=M[/itex] ? I don't know how to calculate that because the normal multiplication for matrices don't seem to work. I'm reading a book where I think this is used to get a coordinate representation of a group with a matrix representation as:
    [itex]\hat{G_{i}}=q^{I}(G_{i})_{I}^{J}p_{j}[/itex] where q and p are the coordinates and the conjugate momenta.

    Can someone give me an easy example? :)
     
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  3. Feb 17, 2015 #2

    Nugatory

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    Which book?
     
  4. Feb 17, 2015 #3
    The bible....no jk I'm reading "Fields" from Warren Siegel. It's free on arxiv.
     
    Last edited: Feb 17, 2015
  5. Feb 18, 2015 #4

    samalkhaiat

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    Consider an abstract, n-dimensional linear vector space spanned by complete and ortho-normal basis vectors, i.e. [tex]\langle I | J \rangle = \delta_{I}^{J} , \ \ \ \ \ (1)[/tex] [tex]\sum_{K}^{n} | K \rangle \langle K | = \mathbb{I}_{n} . \ \ \ \ (2)[/tex] In this n-dimensional (index) space, a [itex]n \times n[/itex] matrix [itex]M[/itex] acts as operator: [tex]M | J \rangle = | M J \rangle = \sum_{K}^{n} M^{J}_{K} \ | K \rangle , \ \ \ (3)[/tex] where the numbers [itex]M^{J}_{I}[/itex] , i.e. the expansion coefficients, are determined from the orthonomality condition (1): [tex]\langle I | M | J \rangle = \sum_{K} M^{J}_{K} \ \delta^{K}_{I} = M^{J}_{I} .[/tex] To construct a matrix representation of operator [itex]M[/itex], we use the completeness relation (2) and the expansion (3): [tex]M \sum_{J} | J \rangle \langle J | = M_{n \times n} = \sum_{I , J} M^{J}_{I} \ | I \rangle \langle J | . \ \ \ (4)[/tex] As an example, consider 2-dimentional index space with basis vectors [tex]| 1 \rangle = \langle 1 |^{ \dagger } = ( 1 \ , \ 0 )^{T} ,[/tex] [tex]| 2 \rangle = \langle 2 |^{ \dagger } = ( 0 \ , \ 1 )^{T} .[/tex] In this case, equation (4) should give a [itex]2 \times 2[/itex] matrix representation for the operator [itex]M[/itex]: [tex]M_{ 2 \times 2 } = \sum_{I , J}^{2} M^{J}_{I} \ | I \rangle \langle J | . \ \ \ (5)[/tex] Indeed, since [tex]| 1 \rangle \langle 1 | = \left( \begin {array} {c c} 1 & 0 \\ 0 & 0 \end {array} \right) , \ \ | 1 \rangle \langle 2 | = \left( \begin{array} {c c} 0 & 1 \\ 0 & 0 \end{array} \right) , [/tex] [tex]| 2 \rangle \langle 1 | = \left( \begin{array} {c c} 0 & 0 \\ 1 & 0 \end{array} \right) , \ \ | 2 \rangle \langle 2 | = \left( \begin{array} {c c} 0 & 0 \\ 0 & 1 \end{array} \right) ,[/tex] equation (5) gives [tex]M_{2 \times 2} = \left( \begin{array} {c c} M^{1}_{1} & M^{2}_{1} \\ M^{1}_{2} & M^{2}_{2} \end{array} \right) .[/tex]

    Sam
     
  6. Feb 19, 2015 #5
    Ok thanks Sam. Could you also help me with coordinate representations? How do I apply this formula
    [itex]\hat{G_{i}}=q^{I}(G_{i})_{I}^{J}p_{j}[/itex]
    Is this like an inner product of the coordinates and the conjugate momentum?
    Like this
    [itex]\hat{G_{1}}=q^{1}(G_{1})_{1}^{1}p_{1}+q^{2}(G_{1})_{2}^{2}p_{2}+....[/itex]
     
  7. Feb 19, 2015 #6

    Demystifier

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  8. Feb 19, 2015 #7
  9. Feb 19, 2015 #8

    Demystifier

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  10. Feb 20, 2015 #9

    samalkhaiat

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    First, you need to understand the connection between this and equation (4) in my previous post. In here, [itex]( q^{I} , p_{I} )[/itex] are continuous coordinate numbers, i.e. they don’t (directly) span the index space of matrices. However, if we write [tex]|Q \rangle = \sum_{I}^{n} q_{I} \ | I \rangle , \ \ \Rightarrow \ q^{I} = ( q_{I} )^{T} = \langle Q | I \rangle ,[/tex] [tex]| P \rangle = \sum_{J}^{n} p_{J} \ | J \rangle , \ \ \Rightarrow \ p_{J} = \langle J | P \rangle ,[/tex] then we can transform equation (4) into equation similar to the one you wrote: [tex]M ( p , q ) \equiv \langle Q | M | P \rangle = \sum_{I , J} M^{J}_{I} \ \langle Q | I \rangle \langle J | P \rangle = \sum_{I , J} M^{J}_{I} \ q^{I} \ p_{J} .[/tex] In fact (see the exercise below) [itex]G^{I}_{J} \equiv i | I \rangle \langle J |[/itex] and [itex]J^{I}_{J} \equiv q^{I} \ p_{J}[/itex] generate “isomorphic” Lie algebras.
    Now, let us talk about coordinate representation. Let [itex]G_{a}[/itex] , [itex]a = 1 , 2 , \cdots , m[/itex] be basis in an m-dimensional Lie algebra [itex]\mathcal{L}^{m}[/itex] with the following Lie bracket relations [tex][ G_{a} , G_{b} ] = C_{a b}{}^{c} \ G_{c} . \ \ \ \ \ (1)[/tex] Since every Lie algebra has a faithful matrix representation, we may the [itex]G_{a}[/itex]’s to be a set of [itex](m)[/itex] matrices ([itex]n \times n[/itex]) and, therefore, realizing the Lie brackets by commutation relations [tex][ G_{a} , G_{b} ]^{J}_{I} = C_{a b}{}^{c} \ ( G_{c} )^{J}_{I} , \ \ I , J = 1 , 2 , \cdots , n . \ \ \ (2)[/tex] Now, we take n-pairs of real munbers [itex]( q^{I} , p_{I} )[/itex] and define m numbers (functionals) [itex]J_{a} ( q , p )[/itex] by [tex]J_{a} = ( G_{a} )^{J}_{I} \ q^{I} \ p_{J} , \ \ \ a = 1 , 2 , \cdots , m . \ \ \ \ (3)[/tex] Clearly, the set of numbers [itex]J_{a}[/itex] forms a representation of [itex]\mathcal{L}^{m}[/itex] , i.e. they satisfy a Lie bracket relation. To see this, take [itex]( q^{I} , p_{I} )[/itex] to be local coordinates on Poisson manifold [itex]\mathcal{P}^{2 n}[/itex] and evaluate the Poisson bracket [tex]\{ J_{a} , J_{b} \} = \sum_{K}^{n} \left( \frac{ \delta J_{a} }{ \delta q^{K} } \frac{ \delta J_{b} }{ \delta p_{K} } - \frac{ \delta J_{a} }{ \delta p_{K} } \frac{ \delta J_{b} }{ \delta q^{K} } \right) .[/tex] Using (2) and (3), we find [tex]\{ J_{a} , J_{b} \} = C_{a b}{}^{c} \ J_{c} . \ \ \ \ \ (4)[/tex] Since the structure constants of [itex]\mathcal{L}^{m}[/itex] appear on the RHS of (4), then [itex]\{ J_{a} , J_{b} \}[/itex] is a Lie bracket on [itex]\mathcal{L}^{m}[/itex]. Mathematically speaking, to every (associative) Lie algebra there corresponds a Poisson structure, i.e., the universal enveloping algebra of [itex]\mathcal{L}^{m}[/itex] is a Poisson-Lie algebra.

    Okay, now I leave you with the following exercise. Define [tex]M^{I}_{J} = i \ | I \rangle \langle J | , \ \ \mbox{and} \ \ G^{I}_{J} = q^{I} \ p_{J} ,[/tex] then prove the following Lie brackets [tex][ M^{I}_{J} , M^{L}_{K} ] = \delta^{I}_{K} \ M^{L}_{J} - \delta^{L}_{J} \ M^{I}_{K} ,[/tex] [tex]\{ G^{I}_{J} , G^{L}_{K} \} = \delta^{I}_{K} \ G^{L}_{J} - \delta^{L}_{J} \ G^{I}_{K} .[/tex] What is the corresponding Lie group?


    Sam
     
    Last edited: Feb 20, 2015
  11. Feb 20, 2015 #10
    Ok I see. Now I understood this:
    [itex]A_{ij}|e_{i}><e_{j}|= A_{11}|e_{1}><e_{1}|+A_{21}|e_{2}><e_{1}|+A_{12}|e_{1}><e_{2}|+A_{22}|e_{2}><e_{2}|=A_{11}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}+A_{21}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}+A_{12}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}+A_{22}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}=[/itex][itex]\begin{pmatrix} A_{11} & A_{12} \\ A_{12} & A_{22} \end{pmatrix}[/itex]
    I think I have to wrap my head a little around how he (in his book) uses this in all the different ways it can be used. So when he uses [itex]|^{I}>[/itex] in the context of a vector [itex]V[/itex] it just means [itex]|^{I}>=|e_{i}>[/itex]. Then in the context of matrices or generators he uses it as [itex]|^{I}>=|e_{i}><e_{j}|[/itex]? Same with the conjugate representations and so on.
    And I'm right that this equation I asked you about is analogous to this equation with which you can define a regular representation of a finite group:
    [itex][D(g)]_{ij}=<e_{i}|D(g)|e_{j}>[/itex]?
    Ok I think I have to play a little around with all this and then I will try to solve your exercise. Thank you Sam !
     
    Last edited: Feb 20, 2015
  12. Feb 21, 2015 #11
    Ok I think I'm confused about what [itex]q^{I}[/itex] and [itex]p_{J}[/itex] are. Coordinates and conjugate momentum ok. But are they the basis of the space or what? If not how does this make sense [itex]\hat{G_{i}}= q^{1}(G_{1})_{1}^{1}p_{1}+....[/itex]? So it's just a number or what?
     
    Last edited: Feb 21, 2015
  13. Feb 24, 2015 #12

    samalkhaiat

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    Did you read the first two lines in post #9 ?
    Did you read the line just before equation (3) in post #9 ?
     
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