# Stone’s Theorem and Time Ordered Exponentials

The time evolution operator of quantum mechanics seems (at least to me) to form a strongly continuous, one parameter group of unitary operators. Hence, by Stone’s theorem, we should have that $$U(t) = \exp(it\hat{h})$$ for some self-adjoint operator $$\hat{h}$$. But, outside of the Heisenberg picture, we have in general that $$U(t) = T\exp\left(\int_{t_0}^{t}\hat{H}(t’)\,dt’\right)$$ which is not of the simple form guaranteed by Stone’s theorem. Does this mean that there is a simpler (though possibly less physical) realization of $$U(t)$$ lying around, or that one of the hypotheses of Stone’s theorem was violated (or something else)?

• There should be a treatment of the general time evolution somewhere in the four volumes of Reed & Simon. Commented Nov 25, 2019 at 9:03

The conditions needed for Stone's theorem require that $$U(t,t') = U(t-t',0)$$ for any $$t,t'$$. This is certainly not satisfied by a general time-dependent Hamiltonian $$H(t)$$. (Consider e.g. turning on a potential at time $$t = 0$$, then $$U(s+\delta s,s)$$ will depend on whether $$s < 0$$ or $$s>0$$.)
Here's a nitpick (maybe the crucial one?): $$U(t)$$ is not a $$C_0$$ semigroup because it fails the second condition:
1. $$\forall t,r\ U(t+r)=U(t)\,U(r)$$
since the choice of $$t_0$$ from which we evolve the system affects the second product differently than the first: The first operator evolves the system to time $$t_0+t+r$$ whereas the second (in the best case scenario, time invariant Hamiltonian) evolves the system to $$t_0+r+(t_0-t)$$, and in the worst case scenario the operation is not physically meaningful since we are trying to evolve the system at time $$t_0+r$$ using a Hamiltonian that is retarded (starts from time $$t_0$$).