1
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.