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.


1 Answer 1


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

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

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