Commutation relation under an arbitrary Lie algebra representation

Question

This is an exercise in Woit's book, B9, Problem 2:

For the case of the Euclidean group $E(2)$, show that in any representation $\pi'$ of its Lie algebra, there is a Casimir operator $$ |\vec{P}|^2 = \pi'(p_1)\pi'(p_1) +\pi'(p_2)\pi'(p_2) $$ that commutes with all the Lie algebra operators $\pi'(p_1), \pi'(p_2), \pi'(l)$.

I have a couple of doubts regarding this computation. Suppose we want to prove that it commutes with $\pi'(p_1)$. One has to compute $$ [\pi'(p_1)\pi'(p_1) +\pi'(p_2)\pi'(p_2),\pi'(p_1)] = 0, $$

but since $\pi'(\cdot)$ is arbitrary, I would like to "use" the definition of the Lie algebra representation

A Lie algebra representation $(\phi,V)$ of a Lie algebra $\mathbb{g}$ on an $n$-dimensional complex vector space $V$ is given by a real-linear map $$\phi: X\in \mathbb{g}\rightarrow \phi(X) \in \mathbb{gl}(n,\mathbb{C})$$ satisfying $$\phi([X,Y]) = [\phi(X),\phi(Y)].$$

Of course this definition is for finite vector spaces (it is the one in my book), but still I can use it in my case (e.g. the Schrödinger representation is still unitary and fulfills the commutator relation above), right?

If this holds then $$ \pi'([p_1^2+p_2^2,p_1]), $$ where the Lie bracket is now the Poisson bracket and thus $\pi'(0)$.

Is the process above valid? Is it $\pi'(0)=0$?

Answer 1

Is the process above valid? Is it $\pi'(0)=0$?

Yes, it is. The phase-space origin in your ill-met infinite-dimensional representations maps to the trivial zero operator in Hilbert space, just like the zero matrix for the 3×3 matrices provided.

Specifically, you have the E(2) brackets, $$ [\pi'(p_1),\pi'(p_2)]=0, \qquad [\pi'(l),\pi'(p_1)]=\pi'(p_2), \qquad [\pi'(l),\pi'(p_2)]=-\pi'(p_1), $$ where your Casimir works; but it is not in the Lie algebra.

In point of fact, your target bracket can also be computed by inspection as a plain commutator of the Schroedinger realization, $$ \pi'(l)= -q_1\partial/\partial q_2 + q_2\partial/\partial q_1 ,\qquad \pi'(p_1)=-\partial/\partial q_1, \qquad \pi'(p_2)=-\partial/\partial q_2, \\ \pi^{'~ 2}(p_1)+ \pi^ {'~ 2}(p_2)= \frac{\partial^2 }{\partial q_1^2}+ \frac{\partial^2 }{\partial q_2^2}~~. $$ Note the last line is not in the Lie algebra, but it is still "represented" faithfully by this "crypto-quantization" realization.