# 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:

1. 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 $\mathcal{X}$?
2. 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?

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

4. What about an arbitrary diffeomorphism?

1. 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 $G\in\mathrm{Gal}(3)$.

2. 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.

3. 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.

4. 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.

• Fantastic answer. However: it does seem to me that there are situations in which general covariance might be desired in a description of localized, single-particle states. For example imagine performing the Bell experiment around a collapsing star. Is there really no simpler way to handle such situations? – AGML Nov 10 '15 at 13:06
• @AGML: Nope. You have to do at least QFT in curved spacetime to do that, and to use that in regions with a very strong gravitational field can only be a semiclassical approximation to the underlying quantum theory of gravity. – ACuriousMind Nov 10 '15 at 13:46

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.

• I was more wondering about how the operators change, but this is a useful perspective too. – AGML Nov 10 '15 at 13:07
• @AGML Almost none of the operators change at all, except for the ones sensitive to the kinetic energy of the center of mass. – Timaeus Nov 10 '15 at 14:55
• Wouldn't we have (for example) $\mathcal{X} -> \mathcal{X}\rho(G)$, per @ACuriousMind #1? Of course we could equivalently view the operators as remaining constant and apply $\rho$ directly to the state, which is perhaps what you're saying here. The equivalence in words: we can view the shift as a unitary rotation of psi or, equivalently, as a relabelling of which $\mathcal{X}$ eigenvectors correspond to which classical position. – AGML Nov 10 '15 at 19:25
• @AGML Most operators are really about the relative coordinates of the apparatus and the subject, in which case a Galilean transformation is irrelevant. It also depends on whether you want external potentials (possibly time dependent) and need to transform them, or whether you have internalized them. If you internalize everything and include the device and the subject then the Galilean frame only affects that overall free particle motion of the center of mass, which doesn't matter. – Timaeus Nov 10 '15 at 19:30
• @AGML You can do the same in classical mechanics, first write all the dynamics of the stuff relative to the center of mass and then have a simple equation for the evolution of the center of mass. Now when you change frames the only thing that changes is the simple motion of the center of mass. – Timaeus Nov 10 '15 at 19:31