# Sakurai: Time evolution operator

How does Sakurai derive the infinitesimal time-evolution operator from scratch without Hamiltonian? $$\mathcal{U}(t_0+dt,t_0) = 1 - i\Omega dt.$$ It is definitely from Taylor's expansion. But complex $$i$$'s emergence and the sign aren't quite clear here.

From the axioms of quantum mechanics, we know that time evolution from time $$t_0$$ up to time $$t_1$$ is given by a unitary operator $$U(t_1,t_0)$$. The family of such operators satisfies : \begin{align} U(t_2,t_1)U(t_1,t_0) &= U(t_2,t_0)\\ U(t_0,t_0) &= 1 \end{align}

Now, if $$t_1\mapsto U(t_1,t_0)$$ is differentiable, since it is unitary, we have : \begin{align} U^\dagger(t_1,t_0)U(t_1,t_0) &= 1 \\ \partial_{t_1}U(t_1,t_0)^\dagger|_{t_1=t_0} + \partial_{t_1}U(t_1,t_0)|_{t_1=t_0}& = 0 \end{align}

ie the operator $$\partial_{t_1}U(t_1,t_0)|_{t_1=t_0}$$ is anti-hermitian. We can therefore write it : $$\partial_{t_1}U(t_1,t_0)|_{t_1=t_0} = -i\Omega(t_0)$$

with $$\Omega(t_0)$$ hermitian (which turns out to be the Hamiltonian.)

Then, the Taylor expansion gives : $$U(t_0 + \text{d}t,t_0) = 1 - i\Omega(t_0)\text dt$$

• Great! Thank you Commented May 14, 2021 at 14:06
• Is your 2nd last eqn the only possible way of making the LHS anti hermit Ian Commented May 14, 2021 at 14:25
• @Shashaank if $A$ is antihermitian then $B:=iA$ is Hermitian, hence $A=-iB$ where $B$ is Hermitian. Commented May 14, 2021 at 16:21