# Generality of the Schrödinger equation

According to the Schrödinger equation $$i \hbar \frac{d}{d t}\Psi(t) = H\Psi(t) \tag 1,$$ the transformation $$U_t:\mathcal H\to \mathcal H: \Psi\mapsto \Psi(t)=e^{-itH}\Psi$$ for every $$t\in\mathbb R$$ is a unitary transformation of the Hilbert space $$\mathcal H$$, and since from this equation,

$$U_tU_s=U_{t+s}\tag 2,$$ the mapping, $$t\mapsto U_t$$ it is a one-parameter subgroup of the unitary group.

Starting from the other direction, if we suppose that the evolution of the states is a one-parameter subgroup of the unitary group, Stone's theorem yields (1).

But, since quantum mechanical states are not elements of $$\mathcal H$$, but they are elements of the projective Hilbert space $$\mathcal P(\mathcal H)$$, time evolution isn't a one-parameter subgroup of the unitary group, but it is a one-parameter subgroup of the projective unitary group, that is, by Wigner's theorem, instead of (2), only $$U_tU_s=\omega(t,s)U_{t+s}\tag 3$$ must hold, where $$\omega(t,s)$$ is a complex number of modulus $$1$$ depending on $$t$$ and $$s$$.

From (2), Stone's theorem yields (1), but from (3), what theorem yields what?

• $U(1)$ has the second Chevalley-Eilenberg cohomology group trivial, hence by Bargmann's theorem, you can choose the phase as $+1$. Nov 18, 2022 at 7:33
• @DanielC Could you expand this to a complete Answer?
– mma
Nov 18, 2022 at 20:58
• @DanielC Are you sure that the cohomology of $U(1)$ is relevant, not the cohomology of $\mathbb R$?
– mma
Nov 20, 2022 at 7:20
• Yes, the cohomology group of the Lie algebra of U(1). Nov 21, 2022 at 21:28
• This is a cocycle in the cohomology of the (additive) group $\mathbb R$, not one in the lie algebra cohomology of anything. Nov 23, 2022 at 2:39

There are two theorems relevant here. The former proves that actually, under natural hypotheses, the multipliers $$\omega$$ can be removed and one can safely apply the Stone theorem. This theorem is an immediate corollary of the Bargmann theorem as the Lie group $$\mathbb{R}$$ is (simply connected and) Abelian and its Lie algebra cohomology is trivial, or also can be directly proved (it was established by Wigner independently).

1. (Wigner-Bargmann) If $$\mathbb{R}\ni t \mapsto U_t$$ is a projective-unitary (so there are your $$\omega$$) representation such that $$|\langle \psi|U_t\phi\rangle|$$ is a continuous function of $$t$$ for every choice of $$\psi,\phi$$ in the Hilbert space, then it is possible to re-arrange the multipliers $$\omega$$ (by multiplying the $$U_t$$ with suitable phases $$U’_t:= e^{if(t)}U_t$$) in order to obtain a strongly-continuos properly unitary representation.

The latter shows that, if the representation is unitary, continuity is almost automatic, since non Borel-measurable functions are really difficult to find.

1. (von Neumann) if $$\mathbb{R} \ni t \mapsto \langle \psi| U_t\phi\rangle$$ is Borel measurable for every choice of $$\psi,\phi$$ of a separable Hilbert space where each $$U_t$$ acts as a unitary operator and the $$U_t$$ form a group representation of $$\mathbb{R}$$, then $$t\to U_t$$ is strongly continuous. So we can apply the Stone theorem.
• It is nothing but the celebrated Bargmann theorem applied to the Lie group $\mathbb{R}$. Nov 20, 2022 at 9:37
• The Bargmann theorem states that, for a simply connected finite-dimensional LIe group $G$, $G \ni g \mapsto U_g$ is a projective-unitary representation which is also continuous in the sense I wrote above, then the representation can be transformed in a strongly-continuous properly unirtari rep of $G$ if a cohomological condition is true. This condition states that, if $\alpha: T_eG \times T_eG \to T_eG$ is antisymmetric map satisfying a sort of Jacobi condition then it has the form $\alpha(g,g')= \beta([g,g'])$, for some linear map $\beta$. Nov 20, 2022 at 10:04
• However, Wigner established an independent proof of the case $G=\mathbb{R}$. You may find all these things for instance in my books Spectral Theory and Quantum Mechanics (2018?) 2nd edition Springer and Foundamental Mathematical Structures of Quantum Theory (2019) Springer. However those results are quite famous and you can find them (and generalisations) in several modern textbooks on related subgects. Nov 20, 2022 at 10:15