Hi! I said in today’s class that it is just a random choice whether we use $\mathrm i$ or $-\mathrm i$. Here is the justification:

First, mathematically, conjugation is an automorphism of $\mathbb C$ (in the sense of being a field). This fact can be easily verified. It can be easily understood by considering $\mathbb C$ as the extension field $\mathbb R[X]/(X^2+1)$. Furthermore, due to this fact, all theorems in complex analysis are still valid if we replace every number by its conjugate.

Then, consider replacing $-\mathrm i$ with $\mathrm i$ in the SE, namely changing $\psi' = -\mathrm iH\psi$ into $\psi' = \mathrm iH\psi$. Due to the mathematical fact above, the new SE should lead to exactly the same theory as our familiar QM because all physically meaningful quantities are real (so that their conjugate are still themselves). The solution to the SE will be $\psi = \psi_0\exp(\mathrm iHt)$ instead of $\psi = \psi_0\exp(-\mathrm iHt)$, and they are exactly the same except an opposite phase (which does not matter) (given that $\psi_0$ also becomes its original counterpart’s conjugate in the new theory, where by saying “conjugate” here I mean taking the conjugate of all of its coordinates under the basis of eigenvectors of $H$).

What about time reversal? The time reversal is $t\to-t$ in the SE, which is actually slightly different from $\mathrm i\to-\mathrm i$ because when doing the latter I also assume that we make $\psi_0$ its conjugate, while $t\to-t$ leaves $\psi_0$ unchanged. However, the close connection between conjugate and time reversal does give us a hint about what the T-symmetry looks like in QM: QM does have T-symmetry, but $T$ cannot be a linear operator because it unavoidably involves conjugation. Actually, conjugation often does look like time reversal. For example, $[X,P]=\mathrm i$ becoming $[X,P]=-\mathrm i$ can be either due to conjugation (the $\mathrm i\to-\mathrm i$ here) or due to time reversal ($P\to-P$ while $X$ unchanged).

Other than saving some minus signs here or there, there is actually a benefit (though minor) about replacing our familiar QM with its conjugate: this makes equations in QM have the same convention as in electrical engineering. Specifically, QM uses $\exp(-\mathrm i E t)$ while EE uses $\exp(\mathrm i\omega t)$. I don’t know why, but conventions in EM seem to be the same as in QM because they also use $\exp(-\mathrm i\omega t)$. It seems strange that EE does not use the same conventions in EM.

Back to where this topic was brought up: why is infinitesimal translation identity minus $\mathrm i P \varepsilon$ instead of plus? The answer to this question is the choice we made when we wrote the SE, which is just a matter of convention. The question that can be genuinely asked is this: why is infinitesimal translation identity minus $\varepsilon \mathrm d/\mathrm dx$ instead of plus? The arguments made in class are then valid to answer this question.

After searching on the internet, I found something about what the mapping from the 3-dimensional Kepler problem to the 3-sphere is.

We choose the canonical coordinates as the 3 Cartesian coordinates and the 3 associated momenta. Then, use stereographic projection to map the 3-dimensional vector $\mathbf p/p_0$ (where $p_0\coloneqq\sqrt{-2mE}$) to a point on an 3-sphere (where we should already have built a coordinate system $\mathbf u$ for us to calculate with). The explicit expression for the projection is $$\mathbf u\coloneqq\frac{p^2-p_0^2}{p^2+p_0^2}\hat{\mathbf n}+\frac{2p_0}{p^2+p_0^2}\mathbf p,$$ where $\mathbf u$ is a unit 4-dimensional vector, and $\hat{\mathbf n}$ is a unit vector perpendicular to the hyperplane where $\mathbf p$ lies. I’m sorry for the abuse of notation, but the $\mathbf u$ in this email also stands for the (3-spherical) coordinates of it on the 3-sphere.

Now, we can use the coordinates on the 3-sphere to express the momenta $\mathbf p(\mathbf u)$. Then, write down the one-form $\mathbf p\cdot\mathrm d\mathbf x$ in terms of the 3-dimensional Cartesian coordinates $\mathbf x$ and the coordinates on the 3-sphere, and convert it into the form $\mathrm d(\mathbf p\cdot\mathbf x)-\mathbf P(\mathbf x,\mathbf u)\cdot\mathrm d\mathbf u$. Then, we get the canonical transform $(\mathbf x,\mathbf p)\mapsto(\mathbf u,\mathbf P)$. The hamiltonian will be correspondingly transformed into exactly the same as the hamiltonian of a free particle on the 3-sphere.

The explicit calculation for the 2-dimensional case can be found in this paper. It also covered the case where $E=0$ and $E>0$.

Figuring out this mapping is actually more straightforward if we consider it as a problem in quantum mechanics. Here is how Fock derived this historically. The Schrödinger equation of the Kepler problem, after a Fourier transformation, is $$\left(\frac{p^2}{2m}-E\right)\Phi(p)=\frac{k}{2\pi^2\hbar}\int\mathrm d^3q\frac{\Phi(q)}{|\mathbf p-\mathbf q|^2}.$$ Looking into it keenly enough, one may find that the factor $1/|\mathbf p-\mathbf q|^2$ is actually the Jacobian of the stereographic projection mentioned above. Some calculation should reduce this equation to the Schrödinger equation of a free particle on the 3-sphere.

Some online sources: several Wikipedia pages, Baez 2022, Egan 2013.