# Inequivalent quantizations of a classical system

Suppose to have a Hamiltonian $H(q,p)$ defined on a phase space with (q,p) and suppose that exist a canonical transformation $$(Q,P)=X(q,p)$$ such that the classical dynamics is equivalent using $H'(Q,P)$ and (Q,P). Now, what happens if I quantize the system using (q,p) or (Q,P)? I would say that I get two Hilbert spaces E and E' that are not isomorphic in general because a canonical transformations could be non linear. Is it true? If so, why we are used to quantize a system in a specific (q,p) phase space? I observed this ambiguity in the quantization of the hydrogen atom: the system is quantized in the phase space $(x,y,z,p_{x},p_{y},p_{z})$ and only in the quantum world we shift to polar coordinates: what would happen if this change is made at a classical level and then we quantize the phase space (r,$\theta$,$\phi$,p$_{r}$,p$_{\theta}$,p$_{\phi}$)?

• Yes, quantization of a classical system is not unique, cf. e.g. physics.stackexchange.com/q/22506/2451 – Qmechanic Apr 25 '17 at 9:46
• Does it mean that there is a preferred coordinate system (q,p) in which one has to quantize a system? From a physical point of view I mean – Yildiz Apr 25 '17 at 9:48
• No. In fact you don't need to make it so complicated. If you look at the $P$-, $Q$- and Wigner functions on the (same) canonical phase space obtained from coherent states, you will see they differ by the ordering of operator. Different quantizations will (presumably) provide different insights into the problem. – ZeroTheHero Apr 25 '17 at 12:32

Within your question lies an assumption that is important to state explicitly: you have a semiclassical picture in mind where you have a physically meaningful classical Hamilton function $H(q,p)$ on phase space (or some other classical observable) and you would like to promote it to a physically meaningful quantum Hamiltonian $\hat{H} = H(x,-\mathrm{i} \nabla - A)$. The fact that both need to be physically meaningful places restrictions on your quantization procedure.
However, choosing a non-symmetric operator ordering leads to mathematically well-defined operators that are usually not physically meaningful. On the other hand, the gauge freedom in choosing a vector potential is different in that respect: if $A' = A + \nabla \chi$ are two equivalent gauges, then $f \bigl ( x , - \mathrm{i} \hbar \nabla - A' \bigr )$ and $f \bigl ( x , -\mathrm{i} \hbar \nabla - A , x \bigr )$ defined via magnetic Weyl calculus are unitarily equivalent operators related by $\mathrm{e}^{+ \mathrm{i} \chi}$.