0
$\begingroup$

For the infinitesimal time-evolution operator, [Sakurai] has the following equation [2.1.21]:

$$ \mathcal U\left( t_0 + dt, t_0 \right) = 1 - \frac{iHdt}{\hbar},$$

where $H$ is the Hamiltonian. Now they derive the Schrödinger equation for the time-evolution operator [2.1.25] as follows:

enter image description here The composition property they're referring to is [2.1.12] $$\mathcal U\left( t_2, t_0\right) = \mathcal U\left( t_2, t_1 \right)\mathcal U\left( t_1, t_0\right).$$

Does anybody see why $\mathcal U\left( t+dt, t_0 \right) = 1-\frac{iHdt}{\hbar},$ as it indirectly says in their Eq. [2.1.23]? According to my very first formula, this this would hold if the first argument were $t_0 + dt$ and not $t+dt$?

[Sakurai] J.J. Sakurai, Jim Napolitano, "Modern Quantum Mechanics", 2nd Edition, Pearson Education

$\endgroup$
2
$\begingroup$

Does anybody see why $\mathcal U\left( t+dt, t_0 \right) = 1-\frac{iHdt}{\hbar}$, as it indirectly says in their Eq. [2.1.23]?

That's not what the equation says. The second argument of the first time-evolution operator in the middle part of the equation is $t$, not $t_0$. They are using the the formula for the infinitesimal time-evolution operator given in (2.1.21) to say $$\mathcal U\left( t + dt, t \right) = 1 - \frac{iHdt}{\hbar}$$

Note that there is still a $\mathcal U(t,t_0)$ on the right in (2,1,23).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy