# Quantized conserved quantities appearing from the Lie-algebra

Hi,

consider a simple situation in quantum mechanics: Your Hilbert space is $\mathcal{H}=L^2(\mathbb{R}^3)$ and you use the obvious unitary representation $\pi\colon G=O(3)\times\mathbb{R}^3\to U(\mathcal{H})$ given by acting on the underlying space. From the group representation you get a representation of the Lie algebra of $G$ which maps elements of the algebra generating translations to momentum operators and elements of the algebra generating rotations (use Euler angles) to angular momentum operators. You notice: The elements of the Lie algebra are mapped to the operators which are the quantized conserved quantities corresponding the the element of the Lie algebra (using the Lagrangian of a free particle and Noether’s theorem). Is there any deep mathematical reason why that works? (why you can get the quantized conserved quantity in two ways: 1. use the representation 2. use the free Lagrangian, apply Noether’s theorem, quantize the quantity)

Suppose we have a $G$-group action on a configuration manifold $Q$ (in your case $G=O(3)\ltimes\mathbb{R}^3$ and $Q=\mathbb{R}^3$). Then for each element $\xi\in\mathfrak{g}$ in the Lie algebra of $G$, there is a corresponding vector field $\xi_Q$ on $Q$. Following approach 2: the conserved quantity (i.e. momentum map) associated with $\xi$ is the phase space function $J^\xi(q,p)=\sum_{i}p_i\;\xi_Q^i(q)$ (on the Lagrangian side it looks like $\sum_i\frac{\partial L}{\partial \dot{q}^i}(q,\dot{q})\;\xi^i_Q(q)$). Whatever quantization procedure you choose (e.g. canonical, geometric etc.), it will quantize $p_i$ to the operator $-i\hbar\frac{\partial}{\partial q^i}$ and $\xi_Q^i(q)$ to the operator $\psi(q)\mapsto \xi_Q^i(q)\;\psi(q)$. Also most quantization procedures allow quantization of classical observables linear in momentum $p_i$. So with an appropriate choice of operator ordering, $J^\xi(q,p)$ will quantize to the operator $$\psi(q)\mapsto -i\hbar\; \xi^i_Q(q)\frac{\partial \psi}{\partial q^i}(q) = -i\hbar \;\xi\_Q(q)\psi.$$ This is the operator $\widehat{J^\xi}$ obtained by approach 1. In case this isn't obvious, it equals $$i\hbar \frac{d}{dt}\Bigg|\_{t=0} \psi(\phi^\xi_{-t}(q)) = i\hbar\frac{d}{dt}\Bigg|\_{t=0} \left(U^\xi_t\psi\right)(q) = (\widehat{J^\xi}\psi)(q)$$ where $\phi^\xi_t:Q\rightarrow Q$ is the flow on $Q$ generated by $\xi\in\mathfrak{g}$, $U^\xi_t$ the induced flow on $L^2(Q)$, and $\widehat{J^\xi}$ the generator of $U^\xi_t$ (i.e. $U^\xi_t = e^{-\frac{i}{\hbar}\widehat{J^\xi}\;t}$). So the two approaches agree.