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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)

Thanks in advance

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    $\begingroup$ Isn't this essentially the point of Noether's theorem? $\endgroup$ – Ben Webster May 9 '13 at 12:05
  • $\begingroup$ Noether’s theorem does not talk about quantized quantities. Well, in this simple case it is not that difficult—the conserved quantities are just the canonical momenta which are (we have a free particle) just derivatives (times mass) and when quantizing these derivatives of course match the result from the representation (you do the derivative when you look at the representation Lie algebra). But I am wondering whether there is a more general way to look at it. $\endgroup$ – The User May 9 '13 at 17:08
  • $\begingroup$ I have also seen this assumption in physicist’s depictions of quantum field theory and they do assume the same thing, but they use conserved charges given by Noether’s theorem for classical field theories, quantize it and assume that these operators generate the symmetries. Hm, maybe the connection is quite obvious and not due to something “more abstract”, and I am currently too confused—that might be the case. Maybe somebody has a clever description which makes it clearer—but it is not that important. $\endgroup$ – The User May 9 '13 at 17:25
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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.

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  • $\begingroup$ Yep, I think it is just that and nothing else. Thanks for your concise answer. $\endgroup$ – The User May 10 '13 at 1:38

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