How do classical reference frames manifest in quantum mechanics? Consider a particle in flat spacetime. We wish to model its spatial position. We do this in the standard way: by mapping each point $x_i$ in space to a basis vector $|x_i>$ in a Hilbert space, then drawing a state vector $|\psi>$ such that the squared projections of $|\psi>$ onto $|x_i>$ give our certainty of detecting the particle in the neighborhood of $x_i$. These bases vectors form the eigenvectors of some "position operator" $\mathcal{X}$.
Now the question: how does the spatial metric - the choice of reference frame - come into all this? Since I've used up all other available structures, it must manifest in the details of the eigenspectrum: the specific eigenvalue associated with each $|x_i>$ by $\mathcal{X}$ must encode the spatial metric somehow. But how?
In particular:


*

*Suppose I perform some Galilean frame shift from $x_i$ to $x_i'$. For example I want to describe the same state in terms of two measurement set-ups, one slightly translated from the other. How can I express this referring only to $\mathcal{X}$?

*What if I don't change frames, but just labels, for example from Cartesian to polar coordinates? In particular what happens to the parts of the spectrum which just got eaten by coordinate singularities?

*What about a Lorentz transformation? Take $\mathcal{X}$ here to represent a 4-position.

*What about an arbitrary diffeomorphism? 

 A: *

*A Galilean frame shift is encoded by a representation $\rho:\mathrm{Gal}(3)\to\mathrm{U}(\mathcal{H})$ of the Galilean group $\mathrm{Gal}(3) = \mathbb{R}^4 \rtimes (\mathbb{R}^3\rtimes \mathrm{SO}(3))$ upon the Hilbert space $\mathcal{H}$. This representation is such that $\rho(G)\lvert \vec x \rangle = \lvert G\vec x\rangle$ for any Galilean transformation $G\in\mathrm{Gal}(3)$.

*A coordinate change to non-Cartesian coordinates is subtle in quantum mechanics, and the naive quantization procedure leads to the wrong operators and/or the wrong Hamiltonian, see for example this question and links therein.

*To do a Lorentz transformation in the sense you imagine we would have to have a four-position operator. This cannot be, because time is not an operator. The closest you get to a relativistic position operator are Newton-Wigner operators, but the proper way to do relativity and quantum mechanics is to go to quantum field theory, where the Lorentz transformations are implemented as transformations of the fields.

*For the same reasons as in 2. and more, the general action of a diffeomorphism is complicated. The usual quantum theories we look at are covariant under Galilean or Lorentz transformations, but not under arbitrary diffeomorphisms. A full quantum theory of gravity should possess general covariance, but we do not yet possess a full quantum theory of gravity, and, in any case, it would be a quantum field theory (there are effective field theories of gravity), not ordinary quantum mechanics.
A: A wave function is not a function of space, it is a function of configuration space.
And since fundamentally the Hamiltonian is determined solely by relative positions and relative velocities you can express the wave as a linear combination of states like $\Psi(\vec r_1, ..., \vec r_N)$=$$\Psi(\frac{m_1\vec r_1+...+m_N\vec r_N}{m_1+...m_N})\Psi(\vec r_2-\vec r_1)\cdot\cdot\cdot\Psi(\vec r_N-\vec r_{N-1}),$$ where the $\Psi(\frac{m_1\vec r_1+...+m_N\vec r_N}{m_1+...m_N})$ is a free particle solution.
And then when you perform a frame transformation the only part that changes is the free particle center of mass part.
So nothing really changes except a part you usually ignored anyway. So you can talk about the spectrum of a center of mass observable changing, but don't get all excited now if before now you never even noticed it.
