JonnyMaddox said:
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}+...
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