# Quantum Mechanics in a non-inertial reference frame?

In non-relativistic Quantum Mechanics how does one write the momentum and position operator in a non-inertial reference frame? How is the Schrodinger Wave equation modified to account for non-zero acceleration? Does the commutation relation $px-xp= - i \hbar$ hold?

• I got interested and went to search for the answer. This is what I found, hope it helps. arxiv.org/abs/1303.6273 – matrp Aug 3 '16 at 18:24
• That's interesting he seems to claim the Equivalence principle is not respected by the Galelei transformations ... But my own calculations on flat Earth $U(y) = mgy$ one actually sees $\hat a = g$ from the reference frame of someone not moving – drewdles Aug 3 '16 at 18:33
• Thank you for posting an answer. Please would you make it more useful to the general user by summarising the relevant parts of the research in the link. Answers should be comprehensible on their own without the need to follow links. – sammy gerbil Aug 3 '16 at 18:40
• Anant Saxena, it seems like transformations using Galilei groups don't respect the Equivalence Principle in quantum mechanics, not on classical. sammy gerbil, I know, and I'm sorry, but as I said, I just got interested by the question, so I'm looking at the article right now. – matrp Aug 3 '16 at 19:20
• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – heather Aug 3 '16 at 22:27

## 1 Answer

Sometimes one can reason as follows:

We can write the action as $S=\int dS,$ where $dS=pdq-Hdt$, and the integral is along a curve in $P\times \mathbb{R}$, where $P$ is phase space and $\mathbb{R}$ is an extra time axis.

Now if one works in a non-inertial frame, one will be given non-inertial coordinates $\tilde q(q,t).$ If you manage to define additional coordinates $\tilde p(q,p,t)$ such that $dS=\tilde p d\tilde q-\tilde H dt$ for some other function $\tilde H$, then $\tilde p$ and $\tilde q$ will be a canonical pair at equal times and the Hamiltonian $\tilde H$ correctly takes into account any pseudo-forces arising due to the non-inertial frame.

The system can now be quantised in the normal manner by putting $\tilde q \mapsto x,$ $\tilde p \mapsto -id/dx$ and $\tilde H$ becomes the Hamiltonian operator.

Basically, a system where you can play the above trick with $dS$ is one where the pseudo forces allow for a Hamiltonian formulation. It should also be noted that the definition of $\tilde p$ can be quite complicated, and its physical interpretation can be quite complicated.