# Tensor product and representations

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1. Feb 17, 2015

Hi, I that $<I|M|J>=M_{I}^{J}$ is just a way to define the elements of a matrix. But what is $|I>M_{I}^{J}<J|=M$ ? 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:
$\hat{G_{i}}=q^{I}(G_{i})_{I}^{J}p_{j}$ where q and p are the coordinates and the conjugate momenta.

Can someone give me an easy example? :)

2. Feb 17, 2015

### Staff: Mentor

Which book?

3. Feb 17, 2015

The bible....no jk I'm reading "Fields" from Warren Siegel. It's free on arxiv.

Last edited: Feb 17, 2015
4. Feb 18, 2015

### samalkhaiat

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

Sam

5. Feb 19, 2015

Ok thanks Sam. Could you also help me with coordinate representations? How do I apply this formula
$\hat{G_{i}}=q^{I}(G_{i})_{I}^{J}p_{j}$
Is this like an inner product of the coordinates and the conjugate momentum?
Like this
$\hat{G_{1}}=q^{1}(G_{1})_{1}^{1}p_{1}+q^{2}(G_{1})_{2}^{2}p_{2}+....$

6. Feb 19, 2015

### Demystifier

7. Feb 19, 2015

8. Feb 19, 2015

### Demystifier

9. Feb 20, 2015

### samalkhaiat

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

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

Sam

Last edited: Feb 20, 2015
10. Feb 20, 2015

Ok I see. Now I understood this:
$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}=$$\begin{pmatrix} A_{11} & A_{12} \\ A_{12} & A_{22} \end{pmatrix}$
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 $|^{I}>$ in the context of a vector $V$ it just means $|^{I}>=|e_{i}>$. Then in the context of matrices or generators he uses it as $|^{I}>=|e_{i}><e_{j}|$? 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:
$[D(g)]_{ij}=<e_{i}|D(g)|e_{j}>$?
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
11. Feb 21, 2015

Ok I think I'm confused about what $q^{I}$ and $p_{J}$ are. Coordinates and conjugate momentum ok. But are they the basis of the space or what? If not how does this make sense $\hat{G_{i}}= q^{1}(G_{1})_{1}^{1}p_{1}+....$? So it's just a number or what?

Last edited: Feb 21, 2015
12. Feb 24, 2015

### samalkhaiat

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